Datasets:

Hack90

/

libre_chem_textbooks

like 14

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.06%3A_The_Arrhenius_Law_-_Pre-exponential_Factors

The pre-exponential factor (\(A\)) is an important component of the Arrhenius equation, which was formulated by the Swedish chemist Svante Arrhenius in 1889. The pre-exponential factor is also known as the and represents the frequency of collisions between reactant molecules at a standard concentration. Although often described as temperature independent, it is actually dependent on temperature because it is related to molecular collision, which is a function of temperature. The units of the pre-exponential factor vary depending on the order of the reaction. In first order reactions, the units of the pre-exponential factor are reciprocal time (e.g., 1/s). Because the pre-exponential factor depends on frequency of collisions, it is related to and . \[ k = A e^{E_a/RT} \label{eq1} \] The Arrhenius equation introduces the relationships between rate and \(A\), \(E_a\), and \(T\), where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, and \(T\) is the temperature. The pre-exponential factor, \(A\), is a constant that can be derived experimentally or numerically. It is also called the and describes how often two molecules collide. To first approximation, the pre-exponential factor is considered constant. When dealing with the collision theory, the pre-exponential factor is defined as \(Z\) and can be derived by considering the factors that affect the frequency of collision for a given molecule. Consider the most elementary bimolecular reaction: \[A + A \rightarrow Product\nonumber \] An underlying factor to the frequency of collisions is the space or volume in which this reaction is allowed to occur. Intuitively, it makes sense for the frequency of collisions between two molecules to be dependent upon the dimensions of their respective containers. By this logic, \(Z\) is defined the following way: \[Z = \dfrac{(\text{Volume of the cylinder}) (\text{Density of the particles})}{\text{time}}\nonumber \] Using this relationship, an equation for the collision frequency, \(Z\), of molecule \(A\) with \(A\) can be derived: \[Z_{AA} = 2N^2_Ad^2 \sqrt{\dfrac{\pi{k_{b}T}}{m_a}}\nonumber \] A similar reasoning is used for bimolecular reactions that involve the collisions of \(A\) and \(B\) \[A + B \rightarrow Product\nonumber \] for deriving the collision frequency, \(Z\) between \(A\) and \(B\). \(Z_{AB} = N_AN_Bd^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\) Substituting the collision factor back into the original Arrhenius equation yields: \[\begin{align*} k &= Z_{AB}e^{\frac{-E_a}{RT}} \\[4pt] &= N_A\, N_B\, d^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\,e^{\frac{-E_a}{RT}}\end{align*}\nonumber \] This equation produces a rate constant with the standard units of (M s ); however, on a molecular level, a rate constant with molecular units would be more useful. To obtain this constant, the rate is divided by \(N_A\,N_B\). This produces a rate constant with units (m molecule s ) and provides the following equation: \[k = Z_{AB} e^{\frac{-E_a}{RT}}\nonumber \] Divide both sides by \(N_AN_B\) \[\dfrac{k}{N_AN_B} = d^2_{AB} \sqrt{\dfrac{8 k_b T}{\mu}}e^{\frac{-E_a}{RT}}\nonumber \] \(Z_{AB}\) becomes \(z_{AB}\): \[\dfrac{Z_{AB}}{N{_A}N{_B }} = z_{AB}\nonumber \] Substituting back into the Arrhenius equation (Equation \ref{eq1}): \[k = z_{AB}e^{\frac{-E_a}{RT}}\nonumber \] The pre-exponential factor is now defined within the collision theory as the following: \[d^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\nonumber \] \(A\) and \(Z\) are practically interchangeable terms for collision frequency. The derivations for \(Z\) often ignores the steric effect of molecules. For a reaction to occur, two molecules must collide in the correct orientation. Not every collision results in the proper orientation, and thus some do not yield a corresponding product. To account for this steric effect, the variable \(P\), which represents the probability of two atoms colliding with the proper orientation, is introduced. The Arrhenius equation is as follows: \[k = Pze^{\frac{-E_a}{RT}}\nonumber \] The probability factor, \(P\), is very difficult to assess and still leaves the Arrhenius equation imperfect. The collision theory deals with gases and neglects to account for structural complexities in atoms and molecules. Therefore, the collision theory estimation for probability is not accurate for species other than gases. The transition state theory attempts to resolve this discrepancy. It uses the foundations of thermodynamics to give a representation of the most accurate pre-exponential factor that yields the corresponding rate. The equation is derived through laws concerning Gibbs free energy, enthalpy and entropy: \[k = \dfrac{k_bT}{h} e^{\frac{\Delta S^o}{R}} e^{\frac{-\Delta H^o}{RT}}(M^{1-m})\nonumber \] \(d^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\) The pre-exponential factor is a function of temperature. As indicated in Table 1, the factor for the collision theory and the transition state theory are both responsive to temperature changes. The collision theory factor is proportional to the square root of \(T\), whereas that of the transition state theory is proportional to \(T\). The empirical factor is also sensitive to temperature. As temperature increases, molecules move faster; as molecules move faster, they are more likely to collide and therefore affect the collision frequency, \(A\).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/The_Solubility_of_the_Hydroxides_Sulfates_and_Carbonates

This page discusses the solubility of the hydroxides, sulfates and carbonates of the Group 2 elements—beryllium, magnesium, calcium, strontium and barium—in water Group II metal oxide basicity and hydroxide solubility in water increase as you go down the column. BeO and Be(OH)2 are amphoteric and react with acids and strong bases such as NaOH. MgO is basic and Mg(OH)2 is weakly basic and do not dissolve in NaOH solution. The oxides of calcium, strontium, and barium are basic and the hydroxides are strongly basic. The solubilities of the hydroxides in water follow the order: Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2. Group II metal oxides become more basic as you go down the column. This trend is easily seen if you compare the electronegativity of the group II metal to the electronegativity of oxygen. As you can see the electronegativities of the metals decrease down the column making the change in electronegativities increases down the group. The greater the difference in electronegativity the more ionic the metal-oxygen bond becomes. The more ionic the metal-oxygen bond the more basic the oxide is Group II metal hydroxides become more soluble in water as you go down the column. This trend can be explained by the decrease in the lattice energy of the hydroxide salt and by the increase in the coordination number of the metal ion as you go down the column. The larger the lattice energy the more energy it takes to break the lattice apart into metal and hydroxide ions. Since the atomic radii increase down the group it makes sense that the coordination numbers also increases because the larger the metal ion the more room there is for water molecules to coordinate to it. The following examples illustrate this trend: This simple trend is true provided hydrated beryllium sulfate is considered, but not anhydrous beryllium sulfate. The Nuffield Data Book quotes anyhydrous beryllium sulfate, BeSO , as insoluble, whereas the hydrated form, BeSO .4H O is soluble, with a solubility of about 39 g of BeSO per 100 g of water at room temperature. Solubility figures for magnesium sulfate and calcium sulfate also vary depending on whether the salt is hydrated or not, but the variations are less dramatic. Two common examples illustrate this trend: The carbonates become less soluble down the group. All the Group 2 carbonates are very sparingly soluble. Magnesium carbonate, for example, has a solubility of about 0.02 g per 100 g of water at room temperature. There is little data for beryllium carbonate, but as it reacts with water, the trend is obscured. The trend to lower solubility is, however, broken at the bottom of the group: barium carbonate is slightly more soluble than strontium sulfate. There are no simple examples of this trend. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/Reactions_of_Group_2_Elements_with_Oxygen

Group 2 elements (beryllium, magnesium, calcium, strontium and barium) react oxygen. to generate metal oxides. This Module addressed why it is difficult to observe a tidy pattern of this reactivity. On the whole, the metals burn in oxygen to form a simple metal oxide. Beryllium is reluctant to burn unless it is in the form of dust or powder. Beryllium has a very strong (but very thin) layer of beryllium oxide on its surface, and this prevents any new oxygen getting at the underlying beryllium to react with it. \[ 2X_{(s)} + O_{2(g)} \rightarrow 2XO_{(s)}\] with \(X\) representing any group 2 metal. It is almost impossible to find any trend in the way the metals react with oxygen. It would be quite untrue to say that they burn more vigorously as you go down the Group. To be able to make any sensible comparison, you would have to have pieces of metal which were all equally free of oxide coating, with exactly the same surface area and shape, exactly the same flow of oxygen around them, and heated to exactly the same extent to get them started. What the metals look like when they burn is a bit problematical! Strontium and barium will also react with oxygen to form strontium or barium peroxide. Strontium forms this if it is heated in oxygen under high pressures, but barium forms barium peroxide just on normal heating in oxygen. Mixtures of barium oxide and barium peroxide will be produced. \[ Ba_{(s)} + O_{2(s)} \rightarrow BaO_{2(s)}\] The strontium equation would look just the same. The reactions of the Group 2 metals with air rather than oxygen is complicated by the fact that they all react with nitrogen to produce nitrides. In each case, you will get a mixture of the metal oxide and the metal nitride. The general equation for the Group is: \[ 3X_{(s)} + N_{2(g)} \rightarrow X_3N_{2(s)}\] For example, the familiar white ash you get when you burn magnesium ribbon in air is a mixture of magnesium oxide and magnesium nitride. \[ 2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)}\] \[ 3Mg_{(s)} + N_{2(g)} \rightarrow Mg_3N_{2(s)}\] There are no simple patterns in the way the metals burn. While it would be tempting to say that the reactions get more vigorous as you go down the Group, but it is not true. The overall amount of heat evolved when one mole of oxide is produced from the metal and oxygen also shows no simple pattern: If anything, there is a slight tendency for the amount of heat evolved to decrease as you go down the Group. But how reactive a metal seems to be depends on how fast the reaction happens (i.e., Kinetics) - not the overall amount of heat evolved (i.e., Thermodynamics). The speed is controlled by factors like the presence of surface coatings on the metal and the size of the activation energy. You could argue that the activation energy will fall as you go down the Group and that will make the reaction go faster. The activation energy will fall because the ionization energies of the metals fall. In this case, though, the effect of the fall in the activation energy is masked by other factors - for example, the presence of existing oxide layers on the metals, and the impossibility of controlling precisely how much heat you are supplying to the metal in order to get it to start burning. Beryllium, magnesium and calcium don't form peroxides when heated in oxygen, but strontium and barium do. There is an increase in the tendency to form the peroxide as you go down the Group. The peroxide ion, O looks like this: The covalent bond between the two oxygen atoms is relatively weak. Now imagine bringing a small 2+ ion close to the peroxide ion. Electrons in the peroxide ion will be strongly attracted towards the positive ion. This is then well on the way to forming a simple oxide ion if the right-hand oxygen atom (as drawn below) breaks off. We say that the positive ion the negative ion. This works best if the positive ion is small and highly charged - if it has a high charge density. A high charge density simply means that you have a lot of charge packed into a small volume. Ions of the metals at the top of the Group have such a high charge density (because they are so small) that any peroxide ion near them falls to pieces to give an oxide and oxygen. As you go down the Group and the positive ions get bigger, they don't have so much effect on the peroxide ion. For example, Barium peroxide can form because the barium ion is so large that it doesn't have such a devastating effect on the peroxide ions as the metals further up the Group. Nitrogen is often thought of as being fairly unreactive, and yet all these metals combine with it to produce nitrides, X N , containing X and N ions. Nitrogen is fairly unreactive because of the very large amount of energy is required to break the triple bond joining the two atoms in the nitrogen molecule, N . When something like magnesium nitride forms, you have to supply all the energy needed to form the magnesium ions as well as breaking the nitrogen-nitrogen bonds and then forming N ions. All of these processes absorb energy. This energy has to be recovered from somewhere to give an overall exothermic reaction - if the energy can't be recovered, the overall change will be endothermic and will not happen. Energy is evolved when the ions come together to produce the crystal lattice ( or enthalpy). The size of the lattice energy depends on the attractions between the ions. The lattice energy is greatest if the ions are small and highly charged - the ions will be close together with very strong attractions. In the whole of Group 2, the attractions between the 2+ metal ions and the 3- nitride ions are big enough to produce very high lattice energies. When the crystal lattices form, so much energy is released that it more than compensates for the energy needed to produce the various ions in the first place. The excess energy evolved makes the overall process exothermic. This is in contrast to what happens in of the Periodic Table (lithium, sodium, potassium, rubidium and cesium). Their ions only carry one positive charge, and so the lattice energies of their nitrides will be much less. Lithium is the only metal in Group 1 to form a nitride. Lithium has by far the smallest ion in the Group, and so lithium nitride has the largest lattice energy of any possible Group 1 nitride. Only in lithium's case is enough energy released to compensate for the energy needed to ionize the metal and the nitrogen - and so produce an exothermic reaction overall. In all the other Group 1 elements, the overall reaction would be endothermic. Those reactions don't happen, and the nitrides of sodium and the rest are not formed. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/The_Thermal_Stability_of_the_Nitrates_and_Carbonates

This page examines at the effect of heat on the carbonates and nitrates of the Group 2 elements (beryllium, magnesium, calcium, strontium and barium). It explains how the thermal stability of the compounds changes down the group. All the carbonates in this group undergo thermal decomposition to the metal oxide and carbon dioxide gas. The term "thermal decomposition" describes splitting up a compound by heating it. All the Group 2 carbonates and their resulting oxides exist as white solids. If "X" represents any one of the elements, the following describes this decomposition: \[XCO_3(s) \rightarrow XO(s) + CO_2(g)\] Down the group, the carbonates require more heating to decompose. The Group 2 nitrates undergo thermal decomposition to the metal oxide, nitrogen dioxide and oxygen gas. These compounds are white solids and brown nitrogen dioxide and oxygen gases are also given off when heated. Magnesium and calcium nitrates normally crystallize with water, and the solid may dissolve in its own water of crystallization to make a colorless solution before it starts to decompose. Again, if "X" represents any one of the elements: \[ 2X(NO_3)_2(s) \rightarrow 2XO(s) + 4NO_2(g) + O_2 (g)\] Down the group, the nitrates must also be heated more strongly before they will decompose. Group 2 nitrates also become more thermally stable down the group. Both carbonates and nitrates of Group 2 elements become more thermally stable down the group. The larger compounds further down require more heat than the lighter compounds in order to decompose. This page offers two different explanations for these properties: polarizability and energetics. Detailed explanations are given for the carbonates because the diagrams are easier to draw, and their equations are also easier. Exactly the same arguments apply to the nitrates. A small 2+ ion has a lot of charge packed into a small volume of space. In other words, it has a high charge density and has a marked distorting effect on any negative ions which happen to be near it. A bigger 2+ ion has the same charge spread over a larger volume of space, so its charge density is lower; it causes less distortion to nearby negative ions. A shorthand structure for the carbonate ion is given below: This structure two single carbon-oxygen bonds and one double bond, with two of the oxygen atoms each carrying a negative charge. In real carbonate ions all the bonds are identical, and the charges are distributed over the whole ion, with greater density concentrated on the oxygen atoms.In other words, the charges are delocalized. The next diagram shows the delocalized electrons. The shading is intended to show that there is a greater electron density around the oxygen atoms than near the carbon. If this ion is placed next to a cation, such as a Group 2 ion, the cation attracts the delocalized electrons in the carbonate ion, drawing electron density toward itself. The carbonate ion becomes polarized. If the carbonate is heated the carbon dioxide breaks free, leaving the metal oxide. The amount of heating required depends on the degree to which the ion is polarized. More polarization requires less heat. The smaller the positive ion is, the higher the charge density, and the greater effect it will have on the carbonate ion. As the positive ions get larger down the group, they affect on the carbonate ions near them less. More heat must be supplied for the carbon dioxide to leave the metal oxide. In other words, the carbonates become more thermally stable down the group. The argument is exactly the same for the Group 2 nitrates. The small cations at the top of the group polarize the nitrate ions more than the larger cations at the bottom do. This process is much more difficult to visualize due to interactions involving multiple nitrate ions. The enthalpy changes for the decomposition of the various carbonates indicate that the reactions are strongly endothermic, implying that the reactions likely require constant heating to proceed. Remember that the reaction in question is the following: \[XCO_{3(s)} \rightarrow XO_{(s)} + CO_{2(g)}\] The calculated enthalpy changes (in kJ mol ) are given in the table below (there is no available data for beryllium carbonate). The reactions are more endothermic down the group, as expected, because the carbonates become more thermally stable, as discussed above. Here's where things start to get difficult! If you aren't familiar with Hess's Law cycles (or with Born-Haber cycles) and with lattice enthalpies (lattice energies), you aren't going to understand the next bit. Don't waste your time looking at it. You can dig around to find the underlying causes of the increasingly endothermic changes as you go down the Group by drawing an enthalpy cycle involving the lattice enthalpies of the metal carbonates and the metal oxides. is the heat needed to split one mole of crystal in its standard state into its separate gaseous ions. For example, for magnesium oxide, it is the heat needed to carry out 1 mole of this change: \[ MgO_{(s)} \rightarrow Mg^{2+}_{(g)} + O^{2-}_{(g)}\] Lattice Energy (LE): + 3889 kJ/mol Here's where things start to get difficult! If you aren't familiar with Hess's Law cycles (or with Born-Haber cycles) and with lattice enthalpies (lattice energies), you aren't going to understand the next bit. Don't waste your time looking at it. Lattice enthalpy is more usually defined as the heat evolved when 1 mole of crystal is formed from its gaseous ions. In that case, the lattice enthalpy for magnesium oxide would be -3889 kJ mol . The term we are using here should more accurately be called the "lattice dissociation enthalpy". The cycle we are interested in looks like this: You can apply Hess's Law to this, and find two routes which will have an equal enthalpy change because they start and end in the same places. For reasons we will look at shortly, the lattice enthalpies of both the oxides and carbonates fall as you go down the Group. But they don't fall at the same rate. The oxide lattice enthalpy falls faster than the carbonate one. If you think carefully about what happens to the value of the overall enthalpy change of the decomposition reaction, you will see that it gradually becomes more positive as you go down the Group. The size of the lattice enthalpy is governed by several factors, one of which is the distance between the centres of the positive and negative ions in the lattice. Forces of attraction are greatest if the distances between the ions are small. If the attractions are large, then a lot of energy will have to be used to separate the ions - the lattice enthalpy will be large. The lattice enthalpies of both carbonates and oxides fall as you go down the Group because the positive ions are getting bigger. The inter-ionic distances are increasing and so the attractions become weaker. The lattice enthalpies fall at different rates because of the different sizes of the two negative ions - oxide and carbonate. The oxide ion is relatively small for a negative ion (0.140 nm), whereas the carbonate ion is large (no figure available). In the oxides, when you go from magnesium oxide to calcium oxide, for example, the inter-ionic distance increases from 0.205 nm (0.140 + 0.065) to 0.239 nm (0.140 + 0.099) - an increase of about 17%. In the carbonates, the inter-ionic distance is dominated by the much larger carbonate ion. Although the inter-ionic distance will increase by the same amount as you go from magnesium carbonate to calcium carbonate, as a percentage of the total distance the increase will be much less. Some made-up figures show this clearly. I can't find a value for the radius of a carbonate ion, and so can't use real figures. For the sake of argument, suppose that the carbonate ion radius was 0.3 nm. The inter-ionic distances in the two cases we are talking about would increase from 0.365 nm to 0.399 nm - an increase of only about 9%. The rates at which the two lattice energies fall as you go down the Group depends on the percentage change as you go from one compound to the next. On that basis, the oxide lattice enthalpies are bound to fall faster than those of the carbonates. The nitrate ion is bigger than an oxide ion, and so its radius tends to dominate the inter-ionic distance. The lattice enthalpy of the oxide will again fall faster than the nitrate. if you constructed a cycle like that further up the page, the same arguments would apply. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/Reactions_of_Group_2_Elements_with_Water

This page discusses the reactions of the Group 2 elements ( , , , and barium) with water, using these reactions to describe the trend in reactivity in . Beryllium reacts with steam at high temperatures (typically around 700°C or more) to give white beryllium oxide and hydrogen. \[ Be_{(s)} + H_2O_{(g)} \rightarrow BeO_{(s)} + H_{2(g)} \label{0}\] There is an additional reason for the lack of reactivity of beryllium compared with the rest of the Group. Beryllium has a strong resistant layer of oxide on its surface which lowers its reactivity at ordinary temperatures. However, the oxide layer breaks up above 750°C and exposes the beryllium metal surface below it, and so the protection then fails. Magnesium burns in steam to produce white magnesium oxide and hydrogen gas. \[ Mg_{(s)} + H_2O_{(g)} \rightarrow MgO_{(s)} + H_{2(g)} \label{1}\] Very clean magnesium ribbon has a mild reaction with cold water, given below. After several minutes, hydrogen gas bubbles form on its surface, and the coil of magnesium ribbon usually floats to the surface. However, the reaction is short-lived because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction. \[ Mg_{(s)} + 2H_2O_{(l)} \rightarrow Mg(OH)_{2(s)} + H_{2(g)} \label{2}\] As a general rule, if a metal reacts with cold water, the metal hydroxide is produced. If it reacts with steam, the metal oxide is formed. This is because the metal hydroxides thermally decompose to the oxide and water. These metals react with cold water with increasing vigor to give the metal hydroxide and hydrogen. Strontium and barium have reactivities similar to that of lithium. Calcium, for example, reacts fairly vigorously and exothermically with cold water. Bubbles of hydrogen gas are given off, and a white precipitate (of calcium hydroxide) is formed, together with an alkaline solution (also of calcium hydroxide, which is slightly water-soluble). The equation for the reactions of any of these metals would is as follows: \[ X_{(s)} + 2H_2O_{(l)} + X(OH) \rightarrow X(OH)_{2 (aq\, or\, s)} + H_{2(g)} \label{3}\] The hydroxide solubilities increase down the group. Calcium hydroxide is mainly formed as a white precipitate (although some does dissolve). Less precipitate is formed down the group with increasing solubility. The enthalpy change of a reaction is a measure of the amount of heat absorbed or evolved when the reaction takes place. An enthalpy change is negative if heat is evolved, and positive if it is absorbed. Calculate the enthalpy change for the possible reactions between beryllium or magnesium and steam gives the following values: \[Be_{(s)} + H_2O_{(g)} \rightarrow BeO_{(s)} + H_{2(g)} \;\;\; \Delta H = -369\; kJ/mol\] \[Mg_{(s)} + H_2O_{(g)} \rightarrow MgO_{(s)} + H_{2(g)} \;\;\; \Delta H = -360\; kJ/mol \] Notice that both possible reactions are strongly exothermic, giving out almost identical amounts of heat. However, only the magnesium reaction actually happens. The explanation for the different reactivities must lie somewhere else. Similarly, calculating the enthalpy changes for the reactions between calcium, strontium or barium and cold water reveals that the amount of heat evolved in each case is almost exactly the same—about -430 kJ mol . The reason for the increase in reactivity must again lie elsewhere. The activation energy for a reaction is the minimum amount of energy which is needed in order for the reaction to take place. It does not matter how exothermic the reaction would be once it got started - if there is a high activation energy barrier, the reaction will take place very slowly, if at all. When Group 2 metals react to form oxides or hydroxides, metal ions are formed. The formation of the ions from the original metal involves various stages all of which require the input of energy - contributing to the activation energy of the reaction. These stages involve the input of: After this, there will be a number of steps which give out heat again - leading to the formation of the products, and overall exothermic reactions. The graph shows the effect of these important energy-absorbing stages as you go down Group 2. Notice that the ionization energies dominate this - particularly the second ionization energies. Ionization energies fall down the group. Because it gets easier to form the ions, the reactions will happen more quickly. The reactions of the Group 2 elements proceed more readily as the energy needed to form positive ions falls. This is mainly due to a decrease in ionization energy down the group. This leads to lower activation energies, and therefore faster reactions. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.03%3A_The_Arrhenius_Law-_Activation_Energies

All molecules possess a certain minimum amount of energy. The energy can be in the form of kinetic energy or potential energy. When molecules collide, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. If molecules move too slowly with little kinetic energy, or collide with improper orientation, they do not react and simply bounce off each other. However, if the molecules are moving fast enough with a proper collision orientation, such that the kinetic energy upon collision is greater than the minimum energy barrier, then a reaction occurs. The minimum energy requirement that must be met for a chemical reaction to occur is called the activation energy, \(E_a\). The reaction pathway is similar to what happens in Figure 1. To get to the other end of the road, an object must roll with enough speed to completely roll over the hill of a certain height. The faster the object moves, the more kinetic energy it has. If the object moves too slowly, it does not have enough kinetic energy necessary to overcome the barrier; as a result, it eventually rolls back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming occur, forming a new product (provided that the molecules collide with the proper orientation). The activation energy (\(E_a\)), labeled \(\Delta{G^{\ddagger}}\) in Figure 2, is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa. However, if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed, as shown in Figure 3. Enzymes can be thought of as biological catalysts that lower activation energy. Enzymes are proteins or RNA molecules that provide alternate reaction pathways with lower activation energies than the original pathways. Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the (k) increases. As indicated by Figure 3 above, a catalyst helps lower the activation energy barrier, increasing the reaction rate. In the case of a biological reaction, when an enzyme (a form of catalyst) binds to a substrate, the activation energy necessary to overcome the barrier is lowered, increasing the rate of the reaction for both the forward and reverse reaction. See below for the effects of an enzyme on activation energy. Catalysts do not just reduce the energy barrier, but induced a completely different reaction pathways typically with multiple energy barriers that must be overcome. For example: In thermodynamics, the change in , ΔG, is defined as: where \( \Delta G^o \) is the change in Gibbs energy when the reaction happens at Standard State (1 atm, 298 K, pH 7). To calculate a reaction's change in Gibbs free energy that did not happen in standard state, the Gibbs free energy equation can be written as: \[ \Delta G = \Delta G^o + RT\ \ln K \label{2} \] where When the reaction is at , \( \Delta G = 0\). The equation above becomes: \[ 0 = \Delta G^o + RT\ln K \nonumber \] Solve for ΔG : \[ \Delta G^o = -RT \ln K \nonumber \] Similarly, in transition state theory, the Gibbs energy of activation, \( \Delta G ^{\ddagger} \), is defined by: \[ \Delta G ^{\ddagger} = -RT \ln K^{\ddagger} \label{3} \] and \[ \Delta G ^{\ddagger} = \Delta H^{\ddagger} - T\Delta S^{\ddagger}\label{4} \] where Combining equations 3 and 4 and then solve for \(\ln K^{\ddagger}\) we have the : \[ \ln K^{\ddagger} = -\dfrac{\Delta H^{\ddagger}}{RT} + \dfrac{\Delta S^{\ddagger}}{R} \nonumber \] As shown in the figure above, activation enthalpy, \(\Delta{H}^{\ddagger} \), represents the difference in energy between the ground state and the transition state in a chemical reaction. The higher the activation enthalpy, the more energy is required for the products to form. Note that this activation enthalpy quantity, \( \Delta{H}^{\ddagger} \), is analogous to the activation energy quantity, E , when comparing the Arrhenius equation (described below) with the Eyring equation: \[E_a = \Delta{H}^{\ddagger} + RT \nonumber \] In general, a reaction proceeds faster if E and \(\Delta{H}^{\ddagger} \) are small. Conversely, if E and \( \Delta{H}^{\ddagger} \) are large, the reaction rate is slower. As temperature increases, gas molecule velocity also increases (according to the ). This is also true for liquid and solid substances. The (translational) kinetic energy of a molecule is proportional to the velocity of the molecules (KE = 1/2 mv ). Therefore, when temperature increases, KE also increases; as temperature increases, more molecules have higher KE, and thus the fraction of molecules that have high enough KE to overcome the energy barrier also increases. The fraction of molecules with energy equal to or greater than E is given by the exponential term \(e^{\frac{-E_a}{RT}}\) in the Arrhenius equation: \[k = Ae^{\frac{-E_a}{RT}} \label{5} \] Taking the natural log of both sides of Equation \(\ref{5}\) yields the following: \[\ln k = \ln A - \frac{E_a}{RT} \label{6} \] Equation \(\ref{4}\) has the linear form y = mx + b. Graphing ln k vs 1/T yields a straight line with a slope of -E /R and a y-intercept of ln A., as shown in Figure 4. As indicated in Figure 5, the reaction with a higher E has a steeper slope; the reaction rate is thus very sensitive to temperature change. In contrast, the reaction with a lower E is less sensitive to a temperature change. Because radicals are extremely reactive, E for a radical reaction is 0; an arrhenius plot of a radical reaction has no slope and is independent of temperature. The activation energy can also be calculated directly given two known temperatures and a rate constant at each temperature. Using Equation (2), suppose that at two different temperatures T and T , reaction rate constants k and k : \[\ln\; k_1 = - \frac{E_a}{RT_1} + \ln A \label{7} \] and \[\ln\; k_2 = - \frac{E_a}{RT_2} + \ln A \label{8} \] Subtract \(ln\; k_2\) from \(ln\; k_1\): \[ \ln\; k_1 - \ln\; k_2 = \left (- \dfrac{E_a}{RT_1} + \ln A \right ) - \left(- \dfrac{E_a}{RT_2} + \ln A \right) \label{9} \] After rearrangement: \[ \ln \left (\dfrac{k_1}{k_2} \right ) = \left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)\dfrac{E_a}{R} \label{10} \] Use the Arrhenius Equation: \(k = Ae^{-E_a/RT}\) Use the equation: \( \ln \left (\dfrac{k_1}{k_2} \right ) = \dfrac{-E_a}{R} \left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)\) Use the equation \(\Delta{G} = \Delta{H} - T \Delta{S}\) Use the equation \(\ln k = \ln A - \dfrac{E_a}{RT}\) to calculate the activation energy of the forward reaction No. Most enzymes denature at high temperatures. At some point, the rate of the reaction and rate constant will decrease significantly and eventually drop to zero. Once the enzyme is denatured, the alternate pathway is lost, and the original pathway will take more time to complete.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/Alkaline_Earth_(Group_II)_Trends

1. Beryllium is the least reactive and does not react with water even at red heat and does not react with N . Magnesium only reacts at reasonable rate with steam, calcium and strontium readily tarnish in moist air and barium tarnishes readily. 2. The ionic character of the compounds increases down the Group. Beryllium forms highly covalent compounds generally with tetrahedral geometries, [BF ] , [BeCl ] (infinite linear polymer). Magnesium forms more polar compounds with 6-coordination. Calcium, strontium, and barium form increasingly ionic compounds with higher coordination numbers (8 is particularly common). 3. The organometallic compounds of beryllium are covalent and rather unreactive. Magnesium forms two important series of organometallic compounds, RMgX and R Mg, which provide convenient sources of carbanions for organic synthesis. The organometallic compounds of calcium, strontium, and barium are generally more reactive and are insoluble in organic solvents. 4. The oxides become progressively more basic down the Group. BeO and Be(OH) are amphoteric and react with acids and strong bases such as NaOH. MgO is basic and Mg(OH) is weakly basic and do not dissolve in NaOH solution. The oxides of calcium, strontium, and barium are basic and the hydroxides are strongly basic. The solubilities of the hydroxides in water follow the order: Be(OH) < Mg(OH) < Ca(OH) < Sr(OH) < Ba(OH) 5. BeX (X = F, Cl, Br, or I) are covalent polymers, which are readily hydrolyzed and are Lewis acids forming adducts BeX L (L = Lewis base). Magnesium, calcium, strontium, and barium halides are essentially ionic and are soluble in water. 6. BeH is a covalent polymer, magnesium hydride is partially ionic and the hydrides of calcium, strontium, and barium are very ionic and hydridic in their properties. 7. Mg and Ca have the greatest tendency to form complexes especially with ligands which have oxygen donor atoms. For small highly charged anions the order of stability is generally: Mg > Ca > Sr > Ba but for the anions, NO , SO , and IO the stability order is: Mg < Ca < Sr < Ba The most important complexes of these metals are with EDTA . The order of stability for this and related polydentate ligands is: Mg < Ca > Sr > Ba The calcium complex, [Ca(EDTA)] , is particularly important because it is water soluble and allows EDTA to solubilize calcium carbonate. Polyphosphates, P O and P O are able to function similarly to solubilize hard water deposits of CaCO . The crown polyethers and cryptate ligands also form stable complexes with Ca and Mg . 8. Both Mg and Ca have important roles in biology due to their fast to moderate ligand exchange rates; slower than K and Na but faster than most transition metal cations. 9. The thermal stabilities of the nitrates, carbonates, and peroxides increase down the column. 10. The solubilities of the sulfates, nitrates, and chlorides increase down the group. 11. The solubilities of the halides in alcohols increase down the group. 1. The high enthalpy of atomization of beryllium causes it to be mechanically harder, higher melting, less dense, and less reactive than the heavier elements of the group. 2. The high charge/size ratio of Be leads to compounds that are more covalent and complexes that are more stable than those of the remaining Group II cations. Many of the compounds have anomalously low melting points, enthalpies of formation, and are more soluble in organic solvents. The compounds are stronger Lewis acids. The halides are hygroscopic and fume when exposed to moist air. 3. Unlike magnesium and the heavier metals of the group, beryllium oxide and hydroxide are amphoteric. 4. Beryllium salts are much less thermally stable because the high lattice energy of the oxide lowers the Gibbs energy change for the decomposition reaction. Similarly it odes not form a peroxide or superoxide. With ethyne, beryllium forms the carbide, Be C, rather than the ethnide presumably because the lattice energy of the carbide is very favorable.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Prediction_of_Aromatic_Anti_Aromatic_and_Non_Aromatic_Character_of_Heterocyclic_Compounds_along_with_their_Omission_Behavior-_Innovative_Mnemonics

In this article, formulae based mnemonics by classifying lone pair of electrons (localized or delocalized) have been highlighted in an innovative and time economic way to enhance interest of students’ on heterocyclic chemistry for determination of planarity by calculating Hybridization state of hetero atom and by prediction of Aromatic, Anti aromatic, non aromatic behavior of different heterocyclic compounds . Here, I have tried to hub three (03) time economic mnemonics by including two (02) formulae for the prediction of hybridization of hetero atom, aromatic and anti aromatic behavior of heterocyclic compounds. This article encourages students to solve multiple choice type questions (MCQs) on ‘Aromaticity of Heterocyclic compounds’ at different competitive examinations in a time economic way. The conventional methods1-7 for determination of hybridization state of hetero atom (planarity of molecule), prediction of aromatic and anti aromatic nature of heterocyclic compound is time consuming. Keeping this in mind, in this article, I have introduced three time economic innovative mnemonics by using two formulae for the prediction of hybridization state of hetero atom of heterocyclic compounds to determine its planarity and aromatic / anti aromatic / non aromatic nature of heterocyclic compounds containing one, two or more number of hetero atoms to make heterocyclic chemistry metabolic and interesting for students. This study also shows omission behavior of some heterocyclic compounds with respect to their aromatic/anti aromatic/non aromatic nature due to presence or absence of vacant d orbitals in DLP based hetero atoms and how lone pair electron discriminates prediction of hybridization state of hetero atom in heterocyclic compound with prediction of its Aromatic and Anti Aromatic nature. Time Economic Innovative Mnemonics in Heterocyclic Chemistry Lone Pair of electrons can be generally classified into two types as Delocalized lone pair of electron (DLP) and Localized lone pair of electron (LLP) as follows: i)Delocalized lone pair of electron (DLP): When lone pair of electron of hetero atom undergo delocalization through conjugation then it is to be treated as delocalized lone pair of electron (DLP). Hetero atom (atom containing lone pair of electron) which is directly attached with single bonds only from all ends is to be considered as DLP containing hetero atom and its lone pair is to be treated as (DLP). Eg. In Pyrrole lone pair of N atom is to be treated as DLP because it is directly attached with three single bonds only. ii)Localized lone pair of electron (LLP): When lone pair of electron of hetero atom does not undergo delocalization through conjugation then it is to be treated as Localized lone pair of electron (LLP). Hetero atom (atom containing lone pair of electron) which is directly attached with single and double bonds with the ring system is to be considered as LLP containing hetero atom and its lone pair is to be treated as localized lone pair of electron (LLP). Eg. In Pyridine lone pair of N atom is to be treated as LLP because it is directly attached with double and single bonds with the ring system. It was first devised by Hückel in 1931. Conventional method for prediction of Aromatic nature of organic compound: Conventional method for Anti Aromatic nature of organic Compound: Cyclic molecule, Planer molecule in which all bonded atoms lie in same plane (having sp2 hybridized) Conjugated molecule with conjugated π-electron system, 4nπ electrons, where, n is a positive integer (n = 0,1,2,3 etc.) Conventional method for identification of Non Aromatic Nature of organic Compound: If a compound violates any one of the above three conditions (1 or 2 or 3) then it is non aromatic in nature. Planarity of Heterocyclic Compounds with the prediction of Hybridization State Planarity of heterocyclic compounds depends on the nature of the hybridization state of carbon and hetero atoms present in it. When all atoms (carbon and hetero) in the heterocyclic compounds having sp2 hybridized then it is planar but when there is a mixing of sp2 and sp3 hybridization state then it is treated as non planar. Hybridization state theory Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. Conventional method for prediction of hybridization state: Hybridization state for a molecule can be calculated by the formula 0.5 (V+H−C+A), Where, V = Number of valance electrons in central atom, H = Number of surrounding monovalent atoms, C = Cationic charge, A = Anionic charge Formula 1: Prediction of hybridization state of hetero atom Power on the Hybridization state of the hetero atom = (Total no of σ bonds around each hetero atom - 1) This formula should be applicable up to 4 σ bonds. If the power of the hybridization state will be 03, 02 and 01 then the hybridization state will be sp3, sp2 and sp respectively. All single (-) bonds are σ bond, in double bond (=) there is one σ and one π. In addition to these each localized lone pair of electron (LLP) can be treated as one σ bond. Hybridization State of Hetero atom with the help of LLP to find out the planarity in Heterocyclic Compounds are shown in Table-1 below. Heterocyclic Compounds (Planar/non planar) Number of σ bonds around hetero atom (from single and double bonds) (A) LLP (localized Lone Pair of e-s) (B) Total Number of σ bonds around hetero atom (A+B) Power of the Hybridization state of the hetero atom (Corresponding Hybridization state) = (A+B)-1 Pyrrole (Planar) 03 0 (lone pair of electron undergo delocalization,DLP with the ring system) 03 02 (sp2 N) Furan (Planar) 02 01 (out of two lone pair of electrons, one undergo delocalization,DLP and other remain as LLP) 03 02 (sp2 O) Thiophene (Planar) 02 01 (out of two lone pair of electrons, one undergo delocalization, DLP and other remain as LLP) 03 02 (sp2 S) Pyridine (Planar) 02 01 03 02 (sp2 N) Indole (Planar) 03 0 03 02 (sp2 N) Quinoline (Planar) 02 01 03 02 (sp2 N) (Planar) 02 01 03 02 (sp2 N) (Planar) 03 (N1) 02 (N3) 0 (N1) 01 (N3) 03 03 02 (sp2 N1) 02 (sp2 N3) (Planar) 02 (N1) 02 (N3) 01 (N1) 01 (N3) 03 03 02 (sp2 N1) 02 (sp2 N3) (Planar) 02 (N1) 02 (N3) 02 (N7) 03 (N9) 01 (N1) 01 (N3) 01 (N7) 0 (N9) 03 03 03 03 02 (sp2 N1) 02 (sp2 N3) 02 (sp2 N7) 02 (sp2 N9) (Planar) 02 (N) 02 (S) 01 (N) 01 (S) (out of two lone pair of electrons on S, one undergo delocalization,DLP and other remain as LLP) 03 03 02 (sp2 N) 02 (sp2 S) (Planar) 02 (N) 02 (S) 01 (N) 01 (S) (out of two lone pair of electrons on S, one undergo delocalization,DLP and other remain as LLP) 03 03 02 (sp2 N) 02 (sp2 S) (Planar) 02 (N1) 02 (N1) 01 (N1) 01 (N1) 03 03 02 (sp2 N1) 02 (sp2 N4) (Planar) 02 (N1,N3 and N5) 01 (N1,N3 and N5) 03 02 (sp2 N1,N3,N5) (Planar) 03 (N) 02 (S) 0 (N) 01 (S) (out of two lone pair of electrons on S, one undergo delocalization, DLP and other is LLP) 03 03 02 (sp2 N) 02 (sp2 S) (Planar) 02 (both N) 01 (both N) 03 02 (sp2 both N) (Planar) 02 (N1,N2,N3,N4) 01 (N1,N2,N3,N4) 03 02 (sp2 All N) (Planar) 02 01 03 02 (sp2 N) (Planar) 02 01 03 02 (sp2 N) (Non Planar) 03 01 04 03 (sp3 N) (Non Planar) 02 02 04 03 (sp3 O) The present study will be an innovative mnemonic involving calculation of ‘A’ value by just manipulating the no of π bonds within the ring system and delocalized lone pair of electron (DLP) with one (01). The heterocyclic compound having cyclic, planar, conjugated (i.e. all the carbon atoms having same state of hybridization, sp2) with even number of ‘A’ value will be treated as aromatic in nature and with odd number of ‘A’ value will be treated as anti aromatic in nature. Formula 2: Evaluation of A Value to predict Aromatic and Anti Aromatic Nature A = πb+DLP+1(constant) = even no = Aromatic A = πb+DLP+1(constant) = odd no = Anti Aromatic where, πb = number of π bonds with in the ring system; DLP = Delocalized lone pair of electron. In case of a multi hetero atom based heterocyclic compound, containing both DLP and LLP hetero atoms, Aromatic and Anti Aromatic behaviour should be predicted with respect to DLP based hetero atom only. Benzothiazole (Figure 1), is a multi hetero atom based heterocyclic compound, containing both DLP and LLP hetero atoms. Here, for N, DLP = 0 , LLP = 1 and for S, DLP = 1, LLP =1, so, in this case ‘A’ value should be calculated with respect to S only not N. Here, A = 4 + 1 + 1 = 6 (even no) = Aromatic. But when heterocyclic compounds contain both LLP based hetero atoms then Aromaticity should be predicted with respect to that hetero atom which contains lowest possible position number as per IUPAC nomenclature or any one of the hetero atom. Imidazole (Figure 2) is a multi hetero atom based hetero cyclic compound in which, N1 is DLP based hetero atom and N3 is LLP based hetero atom. In this case Aromaticity should be predicted with respect to the DLP based hetero atom N1. For N1, A = πb+DLP+1(constant) = 2+1+1 = 4 (even No) - Aromatic Eg. Pyrimidine (Figure 3) is a multi hetero atom based hetero cyclic compound in which, both N1 & N3 are in same environment based hetero atoms (LLP based hetero atoms). In this case Aromaticity should be predicted with respect to N1 (lowest possible position number as per IUPAC nemenclature). For N1, A = πb+DLP+1(constant) = 3+0+1 = 4 (even no) - Aromatic Aromaticity of heterocyclic compounds have been illustrated in Table-2 Hetero Cyclic Compound (Cyclic, Planar, Conjugated) πb value [πb =number of π bonds with in the ring system] DLP A value [A = πb + DLP + 1(constant)] (even No /odd No) Remark on Nature of compound (Aromatic/Anti Aromatic) Pyrrole 2 1 2 + 1 + 1 = 4 (even No) Aromatic Furan 2 1 ( Here out of two lone pairs on O only one LP take part in delocalization) 2 + 1 + 1 = 4 (even No) Aromatic Thiophene 2 1 (Here out of two lone pairs on O only one LP take part in delocalization) 2 + 1 + 1 = 4 (even No) Aromatic Pyridine 3 0 3 + 0 + 1 = 4 (even No) Aromatic Indole 4 1 4 + 1 + 1 = 6 (even No) Aromatic Quinoline 5 0 5 + 0 + 1 = 6 (even No) Aromatic 05 0 5 + 0 + 1 = 6 (even No) Aromatic 02 01 (N1) 2 + 1 + 1 = 4 (even No) Aromatic (m-diazine) 03 0 (N1) 3 + 0 + 1 = 4 (even No) Aromatic 04 01 (N9) 4 + 1 + 1 = 6 (even No) Aromatic 02 01 (S) 2 + 1 + 1 = 4 (even No) Aromatic 04 01 (S) 4 + 1 + 1 = 6 (even No) Aromatic 03 0 3 + 0 + 1 = 4 (even No) Aromatic 03 0 3 + 0 + 1 = 4 (even No) Aromatic 07 0 7 + 0 + 1 = 8 (even No) Aromatic 03 0 3 + 0 + 1 = 4 (even No) Aromatic 04 0 4 + 0 + 1 = 5 (odd No) Anti aromatic 02 0 2 + 0 + 1 = 3 (odd No) Anti aromatic Hetero Cyclic Compound (Cyclic, non-planar) πb value [πb =number of π bonds with in the ring system] DLP A value [A = πb + DLP + 1(constant)] (even No/odd No) Remark on Nature of compound - - - Non Aromatic (non planar – sp3) - - - Non Aromatic (non planar – sp3) Omission behavior of some heterocyclic compounds with respect to their Aromatic / Anti Aromatic and Non Aromatic nature : Aromatic Behavior of some heterocyclic compounds containing different DLP based hetero atoms (one contains vacant d orbitals) : In Phenothiazine (Figure 4), there is two DLP based hetero atoms N and S. In between N and S, since S having vacant d orbitals, so, in this case ‘A’ value will be predicted with respect to DLP based S hetero atom which contains vacant d orbitals only. Here, A = πb + DLP + 1(constant) = 6 +1+1 = 8 (even no) = Aromatic. Non Aromatic Behavior of some heterocyclic compounds containing same DLP based heteroatom having no d orbitals: Omission behavior of some heterocyclic compounds will be observed (Figure 5 and 6),when there, is at least two hetero atoms (same or different) but both the hetero atoms do not have any d orbitals (such as O,N etc.) and they are in DLP based environment in the ring system. These molecules have been studied with advanced molecular orbital techniques known as ‘ab initio calculations’. ‘Ab initio quantum chemistry methods’ are computational chemistry methods based on quantum chemistry8. In the case of 1,2-dioxin, 1,4-dioxin and dibenzo-1,4-dioxin there is DLP based O atoms in all the molecules but still they will be non aromatic due to prevention of significant free electron delocalization (makes non conjugated). The π electrons from the carbon bonds and the lone pair electrons on the oxygen atoms do not overlap to a significant degree due to absence of vacant d orbitals in both O atoms in each case (pπ-dπ overlap is not possible here in conjugation). It makes these molecules non conjugated and thus allows the molecules to become non aromatic instead of aromatic (A value = even No). In the heterocyclic compounds, where, there is two DLP based N atoms instead of two DLP based O atoms or there is one DLP N atom along with one DLP O atom, the same phenomena of non aromatic behavior will be observed. Because, both N and O atoms do not have any vacant d orbitals, and hence pπ-dπ overlap is not possible here in conjugation. Anti Aromatic Behavior of some heterocyclic compounds containing same DLP based hetero atoms having vacant d orbitals: These compounds (Figure 7) are anti aromatic, here both S atoms, having vacant d orbitals, contain one DLP and one LLP and here both DLP of both S atoms participate in the delocalization. Hence, for the prediction of ‘A’ value, consider both DLP (DLP = 2). Here, A = πb + DLP + 1 (Constant) = 2 + 2 + 1 = 5 (odd No) = Anti Aromatic. It may be expected that these three time economic innovative mnemonics of heterocyclic chemistry will help the students of Undergraduate, Senior Undergraduate and Post-Graduate level to predict aromatic, anti aromatic and non aromatic character of heterocyclic compounds along with their omission behaviour. Experiment in vitro on 100 students showed that by using these two formulae students can save up to 5-10 minutes time in the examination hall to predict the aromatic, anti aromatic and non aromatic character of any heterocyclic compounds and their comparative study including omission behaviour with respect to the DLP and LLP based hetero atoms present on them. On the basis of this, I can strongly recommend to use these three time economic innovative mnemonics in the field of heterocyclic chemistry. External Links:

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/Group_2%3A_General_Properties

The elements in the group include beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). Group 2 contains soft, silver metals that are less metallic in character than the Group 1 elements. Although many characteristics are common throughout the group, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the . All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2. This enables the metals to easily lose electrons, which increases their stability and allows them to form compounds via ionic bonds. The following diagram shows the location of these metals in the Periodic Table: The table below gives a detailed account of the descriptive chemistry of each of the individual elements. Notice an increase down the group in atomic number, mass, and atomic radius, and a decrease down the group for ionization energy. These common periodic trends are consistent across the whole periodic table. 1560 K 973 K The reactions of the alkaline earth metals differ from those of the Group 1 metals. Radium is radioactive and is not considered in this section. \[ Ca_{(s)} + H_{2\, (g)} \rightarrow CaH_{2\, (s)} \] \[ Sr_{(s)} + O_{2 (g)} \rightarrow SrO_{2\, (s)}\] \[ 3Mg_{(s)} + N_{(g)} \rightarrow Mg_3N_{2 (s)} \] \[ Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)} \] \[ Ba_{(s)} + 2H_2O_{(l)} \rightarrow Ba(OH)_{2(aq)} + H_{2(g)} \] Beryllium was first identified in 1798 by Louis-Nicolas Vauquelin, while performing a chemical analysis on aluminum silicates. The element was originally named glucinum, and it was first isolated in 1828 by Antoine Bussy and Friedrich Wohler. In 1898, Paul Lebeau was able to produce the first pure samples of Beryllium by electrolyzing molten beryllium fluoride and sodium fluoride. It was later renamed beryllium ( from the Greek word meaning "to become pale") after a pale, beryllium-containing gemstone called a beryl. Beryllium is the very first element in Group 2, and has the highest melting point (1560 K) of any element in the group. It is very rare on Earth as well as in the universe and is not considered important for plant or animal life. In nature, it can only be found in compounds with other elements. In solutions, it remains in elemental form only for pH values below 5.5. Beryllium is extremely light with a high ionization energy, and it is used primarily to strengthen alloys. Beryllium has a strong affinity for oxygen at high temperatures, and thus it is extremely difficult to extract from ores. This beryllium is not commercially available as it cannot be economically mass produced. Since 1957, the majority of industrial beryllium is produced by reducing BeF with magnesium, making it more readily available. : Because beryllium is relatively light and has a wide temperature range, it has many mechanical uses. It can be used in aircraft production in nozzles of liquid-fueled spacecrafts, and mirrors in meteorological satellites. The famous Spitzer Space Telescope's optics are composed entirely of beryllium. One of the most important applications of beryllium is the production of radiation windows. As beryllium is almost transparent to x-rays, it can be used in windows for x-ray tubes. The minimal absorption by Beryllium greatly reduces heating effects due to intense radiation. : Beryllium is a monoisotopic element—it has only one stable isotope, Be. Another notable isotope is cosmogenic Be, which is produced by cosmic ray spallation of oxygen and nitrogen. This isotope has a relatively long half-life of 1.51 million years, and is useful in examining soil erosion and formation, as well as the age of ice cores. : Beryllium forms compounds with most non-metals. The most common compound is beryllium oxide (\(BeO\)) which does not react with water and dissolves in strongly basic solutions. Because of its high melting point, \(BeO\) is a good heat conductor in electrical insulators. It is also an amphoteric oxide, meaning it can react with both strong acids and bases. Cation: \(H_2O_{(l)} + BeO_{(s)} + 2H_3O^+_{(aq)} \rightarrow [Be(H_2O)_4]^{2+}_{(aq)}\) Anion: \(H_2O_{(l)} + BeO_{(s)} + 2OH^-_{(aq)} \rightarrow [Be(OH)_4]^{2-}_{(aq)}\) Magnesium was first discovered in 1808 by Sir Humphry Davy in England by the electrolysis of magnesia and mercury oxide. Antoine Bussy was the first to produce it in consistent form in 1831. It is the 8th most abundant element in the Earth's crust, constituting 2% by mass. It is also the 11th most common element in the human body: fifty percent of magnesium ions are found in bones, and it is a required catalyst for over three hundred different enzymes. Magnesium has a melting point of 923 K and reacts with water at room temperature, although extremely slowly. It is also highly flammable and extremely difficult to extinguish once ignited. As a precaution, when burning or ligh magnesium, UV-protected goggles should be worn, as the bright white light produced can permanently damage the retina. Magnesium can be found in over 100 different minerals, but most commercial magnesium is extracted from dolomite and olivine. The Mg ion is extremely common in seawater and can be filtered and then electrolyzed to produce pure magnesium. : In its elemental form, magnesium is used for structural purposes in car engines, pencil sharpeners, and many electronic devices such as laptops and cell phones. Due to it's bright white flame color, magnesium is also often used in fireworks. In a biological sense, magnesium is vital to the body's health: the Mg ion is a component of every cell type. Magnesium can be obtained by eating foods rich in magnesium, such as nuts and certain vegetables, or by eating supplementary diet pills. Chlorophyll, the pigment that absorbs light in plants, interacts heavily with magnesium and is necessary for photosynthesis. : The three stable isotopes of magnesium are Mg, Mg, and Mg. The lightest isotope, Mg, composes about 74% of Mg, whereas Mg is associated with meteorites in the solar system. Mg is the only known radioactive isotope of Magnesium and it has a half-life of around 21 hours. : Magnesium ions are essential for all life on Earth, and can be found mainly in seawater and the mineral carnallite. Some examples of magnesium compounds include: magnesium carbonate (MgCO ), a white powder used by athletes and gymnasts to dry their hands for a firm grip; and magnesium hydroxide (Mg(OH) ), or milk of magnesia, used as a common component of laxatives. Calcium was isolated in 1808 by Sir Humphry Davy by the electrolysis of lime and mercuric oxide. In nature, it is only found in combination with other elements. It is the 5th most abundant element in the Earth's crust, and is essential for living organisms. Calcium, in the presence of Vitamin D, is well known for its role in building stronger, denser bones early in the lives of humans and other animals. Calcium can be found in leafy green vegetables as well as in milk, cheese, and other dairy products. Calcium has a melting point of 1115 K and gives off a red flame when ignited. Calcium was not readily available until the early 20th Century. : Calcium is an important component in cement and mortars, and thus is necessary for construction. It is also used to aid cheese production. : The four stable isotopes of calcium are \(\ce{^{40}Ca}\), \(\ce{^{42}Ca}\), \(\ce{^{43}Ca}\), \(\ce{^{44}Ca}\). The most abundant isotope, Ca, composes about 97% of naturally occurring calcium. \(\ce{^{41}Ca}\) is the only radioactive isotope of calcium with a half life of 103,000 years. : The most common calcium compound is calcium carbonate (\(CaCO_3\)). Calcium carbonate is a component of shells in living organisms, and is used as a commercial antacid. It is also the main component of limestone. As shown below, three steps are required to to obtain pure \(CaCO_3\) from limestone: calcination, slaking, and carbonation. CaCO (s) → CaO(s) + CO (g) CaO(s) + H O(l) → Ca(OH) (s) Ca(OH) + CO (g) → CaCO (s) + H O(l) Another important calcium compound is calcium hydroxide (\(Ca(OH)_2\)). Often referred to as 'slack lime', it can be refined to form cement. The formation of calcium hydroxide is given below: \[CaO_{(s)}+H_2O_{(l)} \rightarrow Ca(OH)_2\] Strontium was first discovered in 1790 by Adair Crawford in Scotland and is named after the village it was discovered in, Strontian. In nature, it is only found in combination with other elements as it is extremely reactive. It is the 15th most abundant element on Earth and is commonly found in the form of the mineral celestite. Strontium metal is a slightly softer than calcium and has a melting point of K. : In it's pure form, Strontium is used in alloys. It can also be used in fireworks as it produces a scarlet flame color. strontium ranelate (\(C_{12}H_6N_2O_8SSr_2\)) is used to treat sufferers of osteoporosis and strontium chloride (\(SrCl_2\)) is used to make toothpaste for sensitive teeth. : Strontium has four stable isotopes: Sr, Sr, Sr, and Sr. About 82% of naturally occurring strontium comes in the form of Sr. : Some applications of strontium compounds include strontium carbonate (\(SrCO_3\)), strontium sulfate (\(SrSO_4\)), and strontium nitrate (\(Sr(NO_3)_2\)), which can be used as a red flame in fireworks. Radium was first discovered in 1898 by Marie Sklodowska-Curie, and her husband, Pierre Curie, in a pitchblende uranium ore in North Bohemia in the Czech Republic; however, it was not isolated as a pure metal until 1902. Radium is the heaviest and most radioactive of the alkaline earth metals and it reacts explosively with water. Radium appears pure white but when exposed to air it immediately oxidizes and turns black. Because radium is a decay product of uranium, it can be found in trace amounts in all uranium ores. The exposure or inhalation of radium can cause great harm in the form of cancer and other disorders. : There are 25 isotopes of radium are that known to exist, but only 4 are found in nature. However, none are stable. The isotope which has the longest half life is Ra, which is produced by the decay of Uranium. The four most stable isotopes are Ra, Ra, Ra, and Ra. The three most abundant: Ra, Ra and Ra decay by emitting alpha particles, whereas Ra decays emitting a beta particle. Most radium isotopes have relatively short half-lives. : Radium compounds are extremely rare in nature because of its short half-life and intense radioactivity. As such, radium compounds are found almost entirely in uranium and thorium ores. All known radium compounds have a crimson colored flame. The most important compound of radium is radium chloride (\(RaCl_2\)). Previously, it had only been found in a mixture with barium chloride, but as \(RaCl_2\) appeared to be less soluble than barium chloride, the mixture could continually be treated to form a precipitate. This procedure was repeated several times until the radioactivity of the precipitate no longer increased, as radium chloride could be electrolyzed using a mercury cathode to produce pure radium. Currently, \(RaCl_2\) is still used to separate radium from barium and it is also used to produce Radon gas, which can be used to treat cancer. \[Be_{(s)}+F_2 \rightarrow BeF\]

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.04%3A_The_Arrhenius_Law_-_Arrhenius_Plots

In 1889, Svante Arrhenius proposed the from his direct observations of the plots of rate constants vs. temperatures: \[k = Ae^{-\frac{E_a}{RT}} \label{eq1} \] The , E , is the minimum energy molecules must possess in order to react to form a product. The slope of the Arrhenius plot can be used to find the activation energy. The Arrhenius plot can also be used by extrapolating the line back to the y-intercept to obtain the pre-exponential factor, A. This factor is significant because A=p×Z, where p is a steric factor and Z is the collision frequency. The pre-exponential, or frequency, factor is related to the amount of times molecules will hit in the orientation necessary to cause a reaction. It is important to note that the Arrhenius equation is based on the . It states that particles must collide with proper orientation and with enough energy. Now that we have obtained the activation energy and pre-exponential factor from the Arrhenius plot, we can solve for the rate constant at any temperature using the Arrhenius equation. The Arrhenius plot is obtained by plotting the logarithm of the rate constant, k, versus the inverse temperature, 1/T. The resulting negatively-sloped line is useful in finding the missing components of the Arrhenius equation. Extrapolation of the line back to the y-intercept yields the value for ln A. The slope of the line is equal to the negative activation energy divided by the gas constant, R. As a rule of thumb in most biological and chemical reactions, the reaction rate doubles when the temperature increases every 10 degrees Celsius. Looking at the Arrhenius equation, the denominator of the exponential function contains the gas constant, R, and the temperature, T. This is only the case when dealing with moles of a substance, because R has the units of J/molK. When dealing with molecules of a substance, the gas constant in the dominator of the exponential function of the Arrhenius equation is replaced by the Boltzmann constant, k . The Boltzmann constant has the units J/K. At room temperature, k T, is the available energy for a molecule at 25 C or 273K, and is equal to approximately 200 wave numbers. It is important to note that the decision to use the gas constant or the Boltzmann constant in the Arrhenius equation depends primarily on the canceling of the units. To take the inverse log of a number, the number must be unitless. Therefore all the units in the exponential factor must cancel out. If the activation energy is in terms of joules per moles, then the gas constant should be used in the dominator. However, if the activation energy is in unit of joules per molecule, then the constant, K, should be used. The Arrhenius equation (Equation \ref{eq1}) can be rearranged to deal with specific situations. For example, taking the logarithm of both sides yields the equation above in the form y=-mx+b. \[\ln k = \dfrac{-E_a}{RT}+\ln A \label{eq2} \] Then, a plot of \(\ln k\) vs. \(1/T\) and all variables can be found. This form of the Arrhenius equation makes it easy to determine the slope and y-intercept from an Arrhenius plot. It is also convenient to note that the above equation shows the connection between temperature and rate constant. As the temperature increases, the rate constant decreases according to the plot. From this connection we can infer that the rate constant is inversely proportional to temperature. The integrated form of the Arrhenius equation is also useful (Equation \ref{eq3}). This variation of the Arrhenius equation involves the use of two Arrhenius plots constructed on the same graph to determine the activation energy. The above equation, shows temperature's effect on multiple rate constants. This allows easy inference of the rate constants' sensitivity to activation energy and temperature changes. If the activation energy is high for a given temperature range, then the rate constant is highly sensitive; changes in temperature have a significant effect on the rate constant. If the activation energy is low for a given temperature range, then the rate constant is less sensitive, and changes in temperature have little effect on the rate constant. This phenomenon is graphically illustrated in the example below: The graph above shows that the plot with the steeper slope has a higher activation energy and the plot with the flatter slope has a smaller activation energy. This means that over the same temperature range, a reaction with a higher activation energy changes more rapidly than a reaction with a lower activation energy. The Arrhenius plot may become non-linear if steps become rate-limiting at different temperatures. Such an example can be found with Fox and co-workers in 1972 with beta-glycoside transport in . . The differences in the transition temperatures are due to fatty acid composition in cell membranes. The transition state difference is a result of the sharp change of fluidity of the membrane. Another example includes a sudden drop at low 1/T (high temperatures), a result of protein denaturation. 1. T/F The E calculated from the Arrhenius equation gives an exact value. 2. Describe the relationship between temperature and E and give examples. 3. Using the following information: A= 1×10 sec E = 75×10 J/mol R= 8.314 J mol/K Calculate k at 27° C with proper units. 4. Using information from problem 3, calculate k at 37° C with proper units. 5. Using the integrated equation solve for E using: k =7.78×10 at T =273 K k =3.46×10 at T =298 K

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al)/09%3A_The_Electronic_States_of_the_Multielectron_Atoms/9.06%3A_Electron_Configurations%2C_The_Pauli_Exclusion_Principle%2C_The_Aufbau_Principle%2C_and_Slater_Determinants

To discuss the electronic states of atoms we need a system of notation for multi-electron wavefunctions. As we saw in Chapter 8, the assignment of electrons to orbitals is called the electron configuration of the atom. One creates an electronic configuration representing the electronic structure of a multi-electron atom or ion in its ground or lowest-energy state as follows. First, obey the , which requires that each electron in an atom or molecule must be described by a different spin-orbital. Second, assign the electrons to the lowest energy spin-orbitals, then to those at higher energy. This procedure is called the (which translates from German as build-up principle). The mathematical analog of this process is the construction of the approximate multi-electron wavefunction as a product of the single-electron atomic orbitals. For example, the configuration of the boron atom, shown schematically in the energy level diagram in Figure \(\Page {1}\), is written in shorthand form as 1s 2s 2p . As we saw in , the degeneracy of the 2s and 2p orbitals is broken by the electron-electron interactions in multi-electron systems. Rather than showing the individual spin-orbitals in the diagram or in the shorthand notation, we commonly say that up to two electrons can be described by each spatial orbital, one with spin function \(\alpha\) (electron denoted by an arrow pointing up) and the other with spin function \(\beta\) (arrow pointing down). This restriction is a manifestation of the Pauli Exclusion Principle mentioned above. An equivalent statement of the Pauli Exclusion Principle is that each electron in an atom has a unique set of quantum numbers (n,\(l , m_l , m_s\)). Since the two spin functions are degenerate in the absence of a magnetic field, the energy of the two electrons with different spin functions in a given spatial orbital is the same, and they are shown on the same line in the energy diagram. Write the electronic configuration of the carbon atom and draw the corresponding energy level diagram. Write the values for the quantum numbers (n, \(l , m_l , m_s\)) for each of the six electrons in carbon. We can deepen our understanding of the quantum mechanical description of multi-electron atoms by examining the concepts of electron indistinguishability and the Pauli Exclusion Principle in detail. We will use the following statement as a guide to keep our explorations focused on the development of a clear picture of the multi-electron atom: “When a multi-electron wavefunction is built as a product of single-electron wavefunctions, the corresponding concept is that exactly one electron’s worth of charge density is described by each atomic spin-orbital.” A subtle, but important part of the conceptual picture is that the electrons in a multi-electron system are not distinguishable from one another by any experimental means. Since the electrons are indistinguishable, the probability density we calculate by squaring the modulus of our multi-electron wavefunction also cannot change when the electrons are interchanged (permuted) between different orbitals. In general, if we interchange two identical particles, the world does not change. As we will see below, this requirement leads to the idea that the world can be divided into two types of particles based on their behavior with respect to permutation or interchange. For the probability density to remain unchanged when two particles are permuted, the wavefunction itself can change only by a factor of \(e^{i\varphi}\), which represents a complex number, when the particles described by that wavefunction are permuted. As we will show below, the \(e^{i\varphi}\) factor is possible because the probability density depends on the absolute square of the function and all expectation values involve \(\psi \psi ^*\). Consequently \(e^{i\varphi}\) disappears in any calculation that relates to the real world because \(e^{i\varphi} e^{-i\varphi} = 1\). We could symbolically write an approximate two-particle wavefunction as \(\psi (r_1, r_2)\). This could be, for example, a two-electron wavefunction for helium. To exchange the two particles, we simply substitute the coordinates of particle 1 (\(r_l\)) for the coordinates of particle 2 (\(r_2\)) and vice versa, to get the new wavefunction \(\psi (r_1, r_2)\). This new wavefunction must have the property that \[|\psi (r_1, r_2)|^2 = \psi (r_2, r_1)^*\psi (r_2, r_1) = \psi (r_1, r_2)^* \psi (r_1, r_2) \label {9-38}\] since the probability density of the electrons in the atom does not change upon permutation of the electrons. Permute the electrons in Equation \(\ref{9-13}\) (the product function for He wavefunction.) Equation \(\ref{9-38}\) will be true only if the wavefunctions before and after permutation are related by a factor of \(e^{i\varphi}\), \[\psi (r_1, r_2) = e^{i\varphi} \psi (r_1, r_2) \label {9-39}\] so that \[ \left ( e^{-i\varphi} \psi (r_1, r_2) ^*\right ) \left ( e^{i\varphi} \psi (r_1, r_2) ^*\right ) = \psi (r_1 , r_2 ) ^* \psi (r_1 , r_2) \label {9-40}\] If we exchange or permute two identical particles twice, we are (by definition) back to the original situation. If each permutation changes the wavefunction by \(e^{i \varphi}\), the double permutation must change the wavefunction by \(e^{i\varphi} e^{i\varphi}\). Since we then are back to the original state, the effect of the double permutation must equal 1; i.e., \[e^{i\varphi} e^{i\varphi} = e^{i 2\varphi} = 1 \label {9-41}\] which is true only if \(\varphi = 0 \) or an integer multiple of π. The requirement that a double permutation reproduce the original situation limits the acceptable values for \(e^{i\varphi}\) to either +1 (when \(\varphi = 0\)) or -1 (when \(\varphi = \pi\)). Both possibilities are found in nature. Use Euler’s Equality to show that \(e^{12\varphi} = 1\) when \(\varphi = 0\) or \(n \pi\) and consequently \(e^{i \varphi} = \pm 1\). Wavefunctions for which \(e^{i \varphi} = +1\) are defined as symmetric with respect to permutation, because the wavefunction is identical before and after a single permutation. Wavefunctions that are symmetric with respect to interchange of the particles obey the following mathematical relationship: \[e^{i\varphi} e^{i\varphi} = e^{i 2\varphi} = 1 \label {9-42}\] The behavior of some particles requires that the wavefunction be symmetric with respect to permutation. These particles are called bosons and have integer spin such as deuterium nuclei, photons, and gluons. The behavior of other particles requires that the wavefunction be antisymmetric with respect to permutation \((e^{i\varphi} = -1)\). A wavefunction that is antisymmetric with respect to electron interchange is one whose output changes sign when the electron coordinates are interchanged, as shown below: \[ \psi (r_2 , r_1) = e^{i\varphi} \psi (r_1, r_2) = - \psi (r_1, r_2) \label {9-43}\] These particles, called fermions, have half-integer spin and include electrons, protons, and neutrinos. Explain without any equations why there are only two kinds of particles in the world: bosons and fermions. In fact, an elegant statement of the Pauli Exclusion Principle is simply “electrons are fermions.” This statement means that any wavefunction used to describe multiple electrons must be antisymmetric with respect to permutation of the electrons, providing yet another statement of the Pauli Exclusion Principle. The requirement that the wavefunction be antisymmetric applies to all multi-electron functions \(\psi (r_1, r_2, \cdots r_i)\), including those written as products of single electron functions \(\varphi _1 (r_1) \varphi _2 (r_2) \cdots \varphi _i (r_i)\). Another way to simply restate the Pauli Exclusion Principle is that “electrons are fermions.” The first statement of the Pauli Exclusion Principle was that two electrons could not be described by the same spin orbital. To see the relationship between this statement and the requirement that the wavefunction be antisymmetric for electrons, try to construct an for two electrons that are described by the same spin-orbital. We can try to do so for helium. Write the He approximate two-electron wavefunction as a product of identical 1s spin-orbitals for each electron,\(\varphi _{1s_{\alpha}} (r_1) \) and \(\varphi _{1s_{\alpha}} (r_2) \): \[ \psi (r_1, r_2 ) = \varphi _{1s\alpha} (r_1) \varphi _{1s\alpha} (r_2) \label {9-44}\] To permute the electrons in this two-electron wavefunction, we simply substitute the coordinates of electron 1 (\(r_l\)) for the coordinates of electron 2 (\(r_2\)) and vice versa, to get \[ \psi (r_2, r_1 ) = \varphi _{1s\alpha} (r_2) \varphi _{1s\alpha} (r_1) \label {9-45}\] This is identical to the original function (Equatin \(\ref{9-44}\)) since the two single-electron component functions . The two-electron function has not changed sign, as it must for fermions. We can construct a wavefunction that is antisymmetric with respect to permutation symmetry only if each electron is described by a different function. What is meant by the term permutation symmetry? Explain why the product function \(\varphi (r_1) \varphi (r_2)\) could describe two bosons (deuterium nuclei) but can not describe two fermions (e.g. electrons). Let’s try to construct an antisymmetric function that describes the two electrons in the ground state of helium. Blindly following the first statement of the Pauli Exclusion Principle, that each electron in a multi-electron atom must be described by a different spin-orbital, we try constructing a simple product wavefunction for helium using two different spin-orbitals. Both have the 1s spatial component but one has spin function \(\alpha\) and the other has spin function \(\beta\) so the product wavefunction matches the form of the ground state electron configuration for He, \(1s^2\). \[ \psi (r_1, r_2 ) = \varphi _{1s\alpha} (r_1) \varphi _{1s\beta} (r_2) \label {9-46}\] After permutation of the electrons, this becomes \[ \psi (r_2, r_1 ) = \varphi _{1s\alpha} (r_2) \varphi _{1s\beta} (r_1) \label {9-47}\] which is different from the starting function (Equation \(\ref{9-46}\)) since \(\varphi _{1s\alpha}\) and \(\varphi _{1s\beta}\) are functions. However, an antisymmetric function must produce the same function multiplied by (–1) after permutation, and that is not the case here. We must try something else. To avoid getting a totally different function when we permute the electrons, we can make a linear combination of functions. A very simple way of taking a linear combination involves making a new function by simply adding or subtracting functions. The function that is created by subtracting the right-hand side of Equation \(\ref{9-47}\) from the right-hand side of Equation \(\ref{9-46}\) has the desired antisymmetric behavior. \[\psi (r_1, r_2) = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(r_1) \varphi _{1s\beta}(r_2) - \varphi _{1s\alpha}(r_2) \varphi _{1s\beta}(r_1)] \label {9-48}\] The constant on the right-hand side accounts for the fact that the total wavefunction must be normalized. Show that the linear combination in Equation \(\ref{9-48}\) is antisymmetric with respect to permutation of the two electrons. Replace the minus sign with a plus sign (i.e. take the positive linear combination of the same two functions) and show that the resultant linear combination is symmetric. Write a similar linear combination to describe the \(1s^12s^1\) excited configuration of helium. A linear combination that describes an appropriately antisymmetrized multi-electron wavefunction for any desired orbital configuration is easy to construct for a two-electron system. However, interesting chemical systems usually contain more than two electrons. For these multi-electron systems a relatively simple scheme for constructing an antisymmetric wavefunction from a product of one-electron functions is to write the wavefunction in the form of a determinant. John Slater introduced this idea so the determinant is called a . The Slater determinant for the two-electron wavefunction of helium is \[ \psi (r_1, r_2) = \frac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) \end {vmatrix} \label {9-49}\] and a shorthand notation for this determinant is \[ \psi (r_1 , r_2) = 2^{-\frac {1}{2}} Det | \varphi _{1s} (r_1) \varphi _{1s} (r_2) | \label {9-50}\] The determinant is written so the electron coordinate changes in going from one row to the next, and the spin orbital changes in going from one column to the next. The advantage of having this recipe is clear if you try to construct an antisymmetric wavefunction that describes the orbital configuration for uranium! Note that the normalization constant is \((N!)^{-\frac {1}{2}}\) for N electrons. Show that the determinant form is the same as the form for the helium wavefunction that is given in Equation \(\ref{9-48}\). Expand the Slater determinant in Equation \(\ref{9-49}\) for the He atom. Write and expand the Slater determinant for the electronic wavefunction of the Li atom. Write the Slater determinant for the carbon atom. If you expanded this determinant, how many terms would be in the linear combination of functions? Write the Slater determinant for the \(1s^12s^1\) excited state orbital configuration of the helium atom. Now that we have seen how acceptable multi-electron wavefunctions can be constructed, it is time to revisit the “guide” statement of conceptual understanding with which we began our deeper consideration of electron indistinguishability and the Pauli Exclusion Principle. What does a multi-electron wavefunction constructed by taking specific linear combinations of product wavefunctions mean for our physical picture of the electrons in multi-electron atoms? Overall, the antisymmetrized product function describes the configuration (the orbitals, regions of electron density) for the multi-electron atom. Because of the requirement that electrons be indistinguishable, we can’t visualize specific electrons assigned to specific spin-orbitals. Instead, we construct functions that allow each electron’s probability distribution to be dispersed across each spin-orbital. The total charge density described by any one spin-orbital cannot exceed one electron’s worth of charge, and each electron in the system is contributing a portion of that charge density. Critique the energy level diagram and shorthand electron configuration notation from the perspective of the indistinguishability criterion. Can you imagine a way to represent the wavefunction expressed as a Slater determinant in a schematic or shorthand notation that more accurately represents the electrons? (This is not a solved problem!)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/Reactions_of_Group_2_Elements_with_Acids

This page discusses the reactions of the Group 2 elements (beryllium, magnesium, calcium, strontium and barium) with common acids. Each metal reacts with dilute hydrochloric acid, producing bubbles of hydrogen gas and a colorless solution of the metal chloride: \[ X + 2HCl \rightarrow XCl_2 + H_2\] These reactions become more vigorous down the group. These are more complicated, because of the formation of insoluble sulfates. These metals react with with dilute sulfuric acid just as they did with dilute hydrochloric acid; the reaction between magnesium and dilute sulfuric is familiar to many beginning chemists. Hydrogen gas is formed, along with colorless solutions of beryllium or magnesium sulfate. For example: \[ Mg + H_2SO_4 \rightarrow MgSO_4 + H_2\] Calcium sulfate is slightly soluble, and strontium and barium sulfates are essentially insoluble. When exposed to sulfuric acid, a layer of insoluble sulfate is formed on each of these metals, slowing or stopping the reaction entirely. In the calcium case, some hydrogen is produced, along with a white precipitate of calcium sulfate. These reactions are more complicated. When a metal reacts with an acid, the metal usually reduces hydrogen ions to hydrogen gas. The elemental metal is oxidized to metal cations in the process. However, nitrate ions are easily reduced to nitrogen monoxide and nitrogen dioxide. Metals reacting with nitric acid, therefore, tend to produce oxides of nitrogen rather than hydrogen gas. If the acid is relatively dilute, the reaction produces nitrogen monoxide, although this immediately reacts with atmospheric oxygen, forming nitrogen dioxide. If c Various sources disagree on whether beryllium reacts with nitric acid. Beryllium forms a strong oxide layer (similar to that of aluminum) which slows reactions down until it has been removed. Some sources say that beryllium does not react with nitric acid. However, procedures for making beryllium nitrate by reacting beryllium powder with nitric acid are readily available. One source uses semi-concentrated nitric acid, claiming that the gas evolved is nitrogen monoxide. This is a reasonable conclusion. The reactivity of beryllium seems to depend on its source, and how it was manufactured. It is possible that small amounts of impurities in the metal can affect its reactivity. The rest of the Group 2 metals produce hydrogen gas from very dilute nitric acid, but this gas is contaminated with nitrogen oxides. Colorless solutions of the metal nitrates are also formed. Taking magnesium as an example, if the solution is very dilute: \[ Mg + 2HNO_3 \rightarrow Mg(NO_3)_2 + H_2\] At moderate concentrations (even with very dilute acid, this occurs to some extent): \[ 3Mg + 8HNO_3 \rightarrow 3Mg(NO_3)_2 + 2NO + 4H_2O\] And with concentrated acid: \[ Mg + 4HNO_3 \rightarrow Mg(NO_3)_2 + 2NO_2 + 2H_2O\] Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.05%3A_Orientation_in_Addition_to_Alkenes

Addition of an unsymmetrical substance such as \(\ce{HX}\) to an unsymmetrical alkene theoretically can give two products: and Both products are seldom formed in equal amounts; in fact, one isomer usually is formed to the exclusion of the other. For example, the hydration of propene gives 2-propanol (not 1-propanol), and hydrogen chloride adds to 2-methylpropene to give -butyl chloride (not isobutyl chloride): To understand the reason for the pronounced selectivity in the orientation of addition of electrophiles, it will help to consider one example, hydrogen bromide addition to 2-methylpropene. Two different carbocation intermediates could be formed by attachment of a proton to one or the other of the double bond carbons: Subsequent reactions of the cations with bromide ion give -butyl bromide and isobutyl bromide. In the usual way of running these additions, the product is very pure -butyl bromide. How could be have predicted which product would be favored? The first step is to decide whether the prediction is to be based on (1) which of the two products is the , or (2) which of the two products if formed . If we make a decision on the basis of product stabilities, we take into account \(\Delta H^0\) values, entropy effects, and so on, to estimate the equilibrium constants \(K_\text{eq}\) for the reactants and each product. When the ratio of the products is determined by the ratio of their equilibrium constants, we say the overall reaction is subject to (or ) . Equilibrium control requires that the reaction be . When a reaction is carried out under conditions in which it is , the ratio of the products is determined by the relative rates of formation of the various products. Such reactions are said to be under . The products obtained in a reaction subject to kinetic control are not necessarily the same as those obtained under equilibrium control. Indeed, the equilibrium constant for interconversion of -butyl bromide and isobutyl bromide at \(25^\text{o}\) is 4.5, and if the addition of hydrogen bromide to 2-methylpropene were under equilibrium control, the products would be formed in this ratio: \[K_\text{eq} = \dfrac{\left[ \text{tert-butyl bromide} \right]}{\left[ \text{isobutyl bromide} \right]} = 4.5\] But the addition product is \(99+\%\) -butyl bromide so the reaction clearly is kinetically controlled, -butyl being formed considerably faster than isobutyl bromide. . So to account for the formation of -butyl bromide we have to consider why the -butyl cation is formed more rapidly than the isobutyl cation: As we have seen in Section 8-7B, alkyl groups are more electron donating than hydrogen. This means that the more alkyl groups there are on the positive carbon of the cation, the more stable and the more easily formed the cation will be. The reason is that electron-donating groups can partially compensate for the electron deficiency of the positive carbon. As a result, we can predict that the -butyl cation with three alkyl groups attached to the positive center will be formed more readily than the primary isobutyl cation with one alkyl group attached to the positive center. Thus the problem of predicting which of the two possible products will be favored in the addition of unsymmetrical reagents to alkenes under kinetic control reduces to predicting which of two possible carbocation intermediates will be formed most readily. With simple alkenes, we shall expect the preference of formation of the carbocations to be in the order: > > . The reaction scheme can be represented conveniently in the form of an energy diagram (Figure 10-10). The activation energy, \(\Delta H^1_\text{tert}\) for the formation of the -butyl cation is less than \(\Delta H^1_\text{prim}\) for the formation of the isobutyl cation because the tertiary ion is much more stable (relative to the reactants) than the primary ion, and therefore is formed at the faster rate. The second step, to form the product from the intermediate cation, is very rapid and requires little activation energy. Provided that the reaction is , it will take the lowest-energy path and form exclusively -butyl bromide. However, if the reaction mixture is allowed to stand for a long time, isobutyl bromide begins to form. Over a long period, the products equilibrate and, at equilibrium, the product distribution reflects the relative stabilities of the rather than the stability of the transition states for formation of the intermediates. A rather simple rule, formulated in 1870 and known as , correlates the direction of additions of \(\ce{HX}\) to unsymmetrical alkenes. This rule, an important early generalization of organic reactions, may be stated as follows: . It should be clear that Markownikoff's rule predicts that addition of hydrogen bromide to 2-methylpropene will give -butyl bromide. We can extend Markownikoff's rule to cover additions of substances of the general type \(\ce{X-Y}\) to unsymmetrically substituted alkenes when a clear-cut decision is possible as to whether \(\ce{X}\) or \(\ce{Y}\) is the more electrophilic atom of \(\ce{X-Y}\). If the polarization of the \(\ce{X-Y}\) bond is such that \(\ce{X}\) is positive, \(^{\delta \oplus} \ce{X-Y} ^{\delta \ominus}\), then \(\ce{X}\) will be expected to add as \(\ce{X}^\oplus\) to the alkene to form the more stable carbocation. This step will determine the direction of addition. For example, if we know that the \(\ce{O-Br}\) bond of \(\ce{HOBr}\) is polarized as \(\overset{\delta \ominus}{\ce{HO}} - \overset{\delta \oplus}{\ce{Br}}\), then we can predict that addition of \(\ce{HOBr}\) to 2-methylpropene will give 1-bromo-2-methyl-2-propanol: Pauling's value for the electronegativity of carbon makes it slightly more electron-attracting than hydrogen. However, we expect that the electron-attracting power of a carbon atom (or of other elements) will depend also on the electronegativities of the groups to which it is attached. In fact, many experimental observations indicate that carbon in methyl or other alkyl groups is significantly electron-attracting than hydrogen. Conversely, the \(\ce{CF_3}-\) group is, as expected, far electron-attracting than hydrogen. The direction of polarization of bonds between various elements may be predicted from Figure 10-11. For example, an \(\ce{O-Cl}\) bond should be polarized so the oxygen is negative; a \(\ce{C-N}\) bond should be polarized so the nitrogen is negative: \[\overset{\delta \ominus}{\ce{O}}---\overset{\delta \oplus}{\ce{Cl}} \: \: \: \: \: \overset{\delta \oplus}{\ce{C}}---\overset{\delta \ominus}{\ce{N}}\] We then can predict that, in the addition of \(\ce{HOCl}\) to an alkene, the chlorine will add preferentially to form the more stable of two possible carbon cations. Generally, this means that chlorine will bond to the carbon carrying the greater number of hydrogens: A number of reagents that are useful sources of electrophilic halogen are included in Table 10-2. Some of these reagents, notably those with \(\ce{O}-\)halogen or \(\ce{N}-\)halogen bonds, actually are sources of hypohalous acids, \(\ce{HOX}\), and function to introduce halogen and hydroxyl groups at carbon. There are very few good fluorinating agents whereby the fluorine is added as \(\ce{F}^\oplus\). For alkenes that have halogen or similar substituents at the doubly bonded carbons, the same principles apply as with the simple alkenes. That is, under kinetic control the preferred product will be the one derived from the more stable of the two possible intermediate carbon cations. Consider a compound of the type \(\ce{Y-CH=CH_2}\). If \(\ce{Y}\) is electron-attracting than hydrogen, then hydrogen halide should add in such a way as to put the proton of \(\ce{HX}\) on the \(\ce{YCH=}\) end and \(\ce{X}\) on the \(\ce{=CH_2}\) end. The reason is that the positive carbon is expected to be more favorably located if it is not attached directly to an electron-attracting substituent: The addition goes as predicted, . For example, \[\ce{CF_3-CH=CH_2} + \ce{HCl} \rightarrow \ce{CF_3-CH_2-H_2-Cl}\] Such substituents are relatively uncommon, and most of the reported \(\ce{H-X}\) additions have been carried out with \(\ce{Y}\) groups having unshared electron pairs on an atom connected directly to a carbon of the double bond: These substituents usually are strongly electronegative relative to hydrogen, and this often causes diminished reactivity of the double bond toward electrophiles. Nonetheless, : The electron-attracting power of the substituent is more than counterbalanced by stabilization of the intermediate cation by the ability of the substituents to delocalize their electrons to the adjacent positive carbon (see ). and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.05%3A_The_Arrhenius_Law_-_Direction_Matters

Complicated molecular structures increase the likelihood that the rate constant depends on the trajectories with which the reactants approach each other. This kind of is well-known to all students of organic chemistry. Consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane: Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown in below: The reason for this is apparent: HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The double bond of ethene consists of two clouds of negative charge corresponding to the σ ( ) and π ( ) molecular orbitals. The latter, which extends above and below the plane of the C H molecule, interacts with and attracts the HCl molecule. If, instead, the HCl approaches with its chlorine end leading as in , electrostatic repulsion between the like charges causes the two molecules to repel each other before any reaction can take place. The same thing happens in : the electronegativity difference between carbon and hydrogen is too small to make the C–H bond sufficiently polar to attract the incoming chlorine atom. )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.01%3A_Arrhenius_Equation

It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator; butter goes rancid more quickly in the summer than in the winter; and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be more lethargic on cold days. The reason for this is not hard to understand. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelihood of bond cleavages and rearrangements. Whether it is through the , transition state theory, or just common sense, chemical reactions are typically expected to proceed faster at higher temperatures and slower at lower temperatures. By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry: Take a moment to focus on the meaning of this equation, neglecting the factor for the time being. First, note that this is another form of the exponential decay law discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent . And what is the significance of this quantity? Recalling that is the , it becomes apparent that the exponent is just the ratio of the activation energy to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign). This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. Because these terms occur in an exponent, their effects on the rate are quite substantial. The two plots below show the effects of the activation energy (denoted here by ) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 10 . Looking at the role of temperature, a similar effect is observed. (If the -axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.) The Arrhenius equation, \[k = A e^{-E_a/RT} \label{1} \] can be written in a non-exponential form that is often more convenient to use and to interpret graphically. Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields \[\begin{align} \ln k &= \ln \left(Ae^{-E_a/RT} \right) \\[4pt] &= \ln A + \ln \left(e^{-E_a/RT}\right) \label{2} \\[4pt] &= \left(\dfrac{-E_a}{R}\right) \left(\dfrac{1}{T}\right) + \ln A \label{3} \end{align} \] Equation \ref{3} is in the form of \(y = mx + b\) - the equation of a straight line. \[ \ln k=\ln A - \dfrac{E_{a}}{RT} \nonumber \] So if one were given a data set of various values of \(k\), the rate constant of a certain chemical reaction at varying temperature \(T\), one could graph \(\ln (k)\) versus \(1/T\). From the graph, one can then determine the slope of the line and realize that this value is equal to \(-E_a/R\). One can then solve for the activation energy by multiplying through by -R, where R is the gas constant. This affords a simple way of determining the activation energy from values of observed at different temperatures, by plotting \(\ln k\) as a function of \(1/T\). For the isomerization of cyclopropane to propene, the following data were obtained (calculated values shaded in pink): From the calculated slope, we have \[\begin{align*} –\left(\dfrac{E_a}{R}\right) &= –3.27 \times 10^4 K \\ E_a &=– (8.314\, J\, mol^{–1} K^{–1}) (–3.27 \times 10^4\, K) \\[4pt] &= 273\, kJ\, mol^{–1} \end{align*} \] This activation energy is high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature. Because the ln -vs.-1/ plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. To see how this is done, consider that \[\begin{align*} \ln k_2 -\ln k_1 &= \left(\ln A - \frac{E_a}{RT_2} \right)\left(\ln A - \frac{E_a}{RT_1} \right) \\[4pt] &= \color{red}{\boxed{\color{black}{ \frac{E_a}{R}\left( \frac{1}{T_1}-\frac{1}{T_2} \right) }}} \end{align*} \] The ln- term is eliminated by subtracting the expressions for the two ln- terms.) Solving the expression on the right for the activation energy yields \[ E_a = \dfrac{R \ln \dfrac{k_2}{k_1}}{\dfrac{1}{T_1}-\dfrac{1}{T_2}} \nonumber \] A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be? Center the ten degree interval at 300 K. Substituting into the above expression yields \[\begin{align*} E_a &= \dfrac{(8.314)(\ln 2/1)}{\dfrac{1}{295} – \dfrac{1}{305}} \\[4pt] &= \dfrac{(8.314\text{ J mol}^{-1}\text{ K}^{-1})(0.693)}{0.00339\,\text{K}^{-1} – 0.00328 \, \text{K}^{-1}} \\[4pt] &= \dfrac{5.76\, J\, mol^{–1} K^{–1}}{(0.00011\, K^{–1}} \\[4pt] &= 52,400\, J\, mol^{–1} = 52.4 \,kJ \,mol^{–1} \end{align*} \] It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein. The ratio of the rate constants at the elevations of Los Angeles and Denver is 4.5/3.0 = 1.5, and the respective temperatures are \(373 \; \rm{K }\) and \(365\; \rm{K}\). With the subscripts 2 and 1 referring to Los Angeles and Denver respectively: \[\begin{align*} E_a &= \dfrac{(8.314)(\ln 1.5)}{\dfrac{1}{365\; \rm{K}} – \dfrac{1}{373 \; \rm{K}}} \\[4pt] &= \dfrac{(8.314)(0.405)}{0.00274 \; \rm{K^{-1}} – 0.00268 \; \rm{K^{-1}}} \\ &= \dfrac{(3.37\; \rm{J\; mol^{–1} K^{–1}})}{5.87 \times 10^{-5}\; \rm{K^{–1}}} \\[4pt] &= 57,400\; \rm{ J\; mol^{–1}} \\[4pt] &= 57.4 \; \rm{kJ \;mol^{–1}} \end{align*} \] : This low value seems reasonable because thermal denaturation of proteins primarily involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken (although can interfere with this interpretation). Up to this point, the pre-exponential term, \(A\) in the Arrhenius equation (Equation \ref{1}), has been ignored because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. However, because \(A\) multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate. Recall that the exponential part of the Arrhenius equation expresses the fraction of reactant molecules that possess enough kinetic energy to react, as governed by the . This fraction can run from zero to nearly unity, depending on the magnitudes of \(E_a\) and of the temperature. If this fraction were 0, the Arrhenius law would reduce to \[k = A \nonumber \] In other words, \(A\) is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded \(E_a\) — admittedly, an uncommon scenario (although barrierless reactions have been characterized). What would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the or , \(Z\). In some reactions, the relative orientation of the molecules at the point of collision is important, so a geometrical or (commonly denoted by \(\rho\)) can be defined. In general, we can express \(A\) as the product of these two factors: \[A = Z\rho \nonumber \] Values of \(ρ\) are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which \(A\) is assumed to be the same as \(Z\). The exponential term in the Arrhenius equation implies that the rate constant of a reaction increases exponentially when the decreases. Because the rate of a reaction is directly proportional to the rate constant of a reaction, the rate increases exponentially as well. Because a reaction with a small activation energy does not require much energy to reach the transition state, it should proceed faster than a reaction with a larger activation energy. In addition, the Arrhenius equation implies that the rate of an uncatalyzed reaction is more affected by temperature than the rate of a catalyzed reaction. This is because the activation energy of an uncatalyzed reaction is greater than the activation energy of the corresponding catalyzed reaction. Since the exponential term includes the activation energy as the numerator and the temperature as the denominator, a smaller activation energy will have less of an impact on the rate constant compared to a larger activation energy. Hence, the rate of an uncatalyzed reaction is more affected by temperature changes than a catalyzed reaction. \[ \ln k_{1}=\ln A - \dfrac{E_{a}}{k_{B}T_1} \label{a1} \] at \(T_1\) and \[ \ln k_{2}=\ln A - \dfrac{E_{a}}{k_{B}T_2} \label{a2} \] at \(T_2\). By \[ \ln A = \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} \label{a3} \] and substitute for \(\ln A\) into Equation \ref{a1}: \[ \ln k_{1}= \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} - \dfrac{E_{a}}{k_{B}T_1} \label{a4} \] This simplifies to: \[\begin{align*} \ln k_{1} - \ln k_{2} &= -\dfrac{E_{a}}{k_{B}T_1} + \dfrac{E_{a}}{k_{B}T_2} \\[4pt] \ln \dfrac{k_{1}}{k_{2}} &= -\dfrac{E_{a}}{k_{B}} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right ) \end{align*} \] \[ k=Ae^{^{\frac{-E_{a}}{RT}}} \nonumber \] or \[\ln k=\ln A - \frac{E_{a}}{RT} \nonumber \] with the following terms: \(k\): \(A\): The or frequency factor \(E_a\): The is the threshold energy that the reactant(s) must acquire before reaching the transition state. \(R\): The gas constant. \(T\): The absolute temperature at which the reaction takes place. E is the factor the question asks to be solved. Therefore it is much simpler to use \(\large \ln k = -\frac{E_a}{RT} + \ln A\) To find E , subtract ln A from both sides and multiply by -RT. This will give us: \( E_a=\ln A -\ln k)RT\) Substitute the numbers into the equation: \(\ ln k = \frac{-(200 \times 1000\text{ J}) }{ (8.314\text{ J mol}^{-1}\text{K}^{-1})(289\text{ K})} + \ln 9\) Use the equation ln(7/k )=-[(900 X 1000)/8.314](1/370-1/310) Use the equation 12 = 15e Use the equatioin ln(15/7)=-[(600 X 1000)/8.314](1/T - 1/389)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.06%3A_Addition_of_Boron_Hydrides_to_Alkenes._Organoboranes

An especially valuable group of intermediates can be prepared by addition of an compound to carbon-carbon double or triple bonds: The reaction is called and is a versatile synthesis of organoboron compounds. One example is the addition of diborane, \(\ce{B_2H_6}\), to ethene. Diborane behaves as though it is in equilibrium with \(\ce{BH_3}\) \(\left( \ce{B_2H_6} \rightleftharpoons 2 \ce{BH_3} \right)\), and addition proceeds in three stages: The monoalkylborane, \(\ce{RBH_2}\), and the dialkylborane, \(\ce{R_2BH}\), seldom are isolated because they rapidly add to the alkene. These additions amount to reduction of both carbons of the double bond: Organoboranes can be considered to be organometallic compounds. Elemental boron does not have the properties of a metal, and boron-carbon bonds are more covalent than ionic. However, boron is more electropositive than either carbon or hydrogen and when bonded to carbon behaves like most metals in the sense that bonds are polarized with \(\ce{R}\) negative and boron positive: Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes. The simplest borane, \(\ce{BH_3}\), exists as the dimer, \(\ce{B_2H_6}\), or in complexed form with certain ethers or sulfides: Any of these \(\ce{BH_3}\) compounds adds readily to most alkenes at room temperature or lower temperatures. The reactions usually are carried out in ether solvents, although hydrocarbon solvents can be used with the borane-dimethyl sulfide complex. When diborane is the reagent, it can be generated either or externally through the reaction of boron trifluoride with sodium borohydride: \[3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{H_4} + 4 \ce{BF_3} \rightarrow 2 \ce{B_2H_6} + 3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{F_4}\] Hydroborations have to be carried out with some care, because diborane and many alkylboranes are highly reactive and toxic substances; many are spontaneously flammable in air. With unsymmetrical alkenes, hydroboration occurs so that : These additions are : Furthermore, when there is a choice, addition occurs preferentially from the less crowded side of the double bond: If the alkene is a bulky molecule, borane may add only one or two alkene molecules to give either mono- or dialkylborane, \(\ce{RBH_2}\) or \(\ce{R_2BH}\), respectively, as the following reactions show: These bulky boranes still possess \(\ce{B-H}\) bonds and can add further to a multiple bond, but they are highly selective reagents and add only if the alkene or alkyne is unhindered. This selectivity can be useful, particularly to 1-alkynes, which are difficult to stop at the alkenylborane stage when using diborane: With a bulky dialkylborane, such as di-(1,2-dimethylpropyl)borane, further addition to the alkenylborane does not occur. An especially selective hydroborating reagent is prepared from 1,5-cyclooctadiene and borane. The product is a bicyclic compound of structure \(1\) (often abbreviated as 9-BBN), in which the residual \(\ce{B-H}\) bond adds to unhindered alkenes with much greater selectivity than is observed with other hydroborating reagents. It is also one of the few boranes that reacts sufficiently slowly with oxygen that it can be manipulated in air. An example of the difference in selectivity in the hydroboration of -4-methyl-2-pentene with \(\ce{B_2H_6}\) and \(1\) follows: According to the electronegativity chart (Figure 10-11), the boron-hydrogen bond is polarized in the sense \(\overset{\delta \oplus}{\ce{B}} --- \overset{\delta \ominus}{\ce{H}}\). Therefore the direction of addition of \(\ce{B_2H_6}\) to propene is that expected of a mechanism whereby the electrophilic boron atom becomes bonded to the less-substituted carbon of the double bond. However, there is no firm evidence to suggest that a carbocation intermediate is formed through a stepwise electrophilic addition reaction. For this reason, the reaction often is considered to be a . The stepwise formulation explains why boron becomes attached to the less-substituted carbon, but does not account for the fact that the reactions show no other characteristics of carbocation reactions. This could be because of an expected, extraordinarily fast rate of hydride-ion transfer to the carbocation. A more serious objection to the stepwise mechanism is that alkynes react more rapidly than alkenes, something which normally is not observed for stepwise electrophilic additions (cf. ). Some alkylboranes rearrange at elevated temperatures \(\left( 160^\text{o} \right)\) to form more stable isomers. For example, the alkylborane \(2\), produced by hydroboration of 3-ethyl-2-pentene, rearranges to \(3\) on heating: In general, the boron in alkylboranes prefers to be at the of a hydrocarbon chain so it is bonded to a carbon where steric crowding around boron is least severe. Thus rearrangement tends to proceed in the direction Rearrangement is associated with the fact that hydroboration is reversible at elevated temperatures. This makes possible a sequence of elimination-addition reactions in which boron becomes attached to different carbons and ultimately leads to the most stable product that has boron bonded to the carbon at the end of the chain: Rearrangement of alkylboranes can be used to transform alkenes with double bonds in the middle of the chain into 1-alkenes; for example, \(\ce{RCH=CHCH_3} \rightarrow \ce{RCH_2-CH=CH_2}\). The procedure involves hydroboration of the starting alkene in the usual manner; the borane then is isomerized by heating. An excess of 1-decene (bp \(170^\text{o}\)) then is added to the rearranged borane and the mixture is reheated. Heating causes the alkylborane to dissociate into 1-alkene and \(\ce{HBR_2}\); the 1-decene "scavenges" the \(\ce{HBR_2}\) as it forms, thereby allowing a more volatile 1-alkene (bp \(<170^\text{o}\)) to be removed by simple distillation. Thus, for the rearrangement of 3-ethyl-2-pentene to 3-ethyl-1-pentene, Alkylboranes formed in the hydroboration of alkenes and alkynes seldom are isolated; for the most part they are used as reactive intermediates for the synthesis of other substances. In the reactions of alkylboranes, the \(\ce{B-C}\) bond is cleaved in the sense \(\ce{B}^\oplus - \ce{C}^\ominus\) so that carbon is transferred to other atoms, such as \(\ce{H}\), \(\ce{O}\), \(\ce{N}\), and \(\ce{C}\), its bonding electron pair: In the first of these reactions (Equation 11-2), a hydrocarbon is produced by the cleavage of a borane, \(\ce{R_3B}\), with aqueous acid, or better, with anhydrous propanoic acid, \(\ce{CH_3CH_2CO_2H}\). The overall sequence of hydroboration-acid hydrolysis achieves the reduction of a carbon-carbon multiple bond without using hydrogen and a metal catalyst or diimide (Table 11-3): The second reaction (Equation 11-3) achieves the synthesis of a alcohol by the oxidation of the alkylborane with hydrogen peroxide in basic solution. Starting with a 1-alkene, one can prepare a primary alcohol in two steps: This sequence complements the direct hydration of 1-alkenes, which gives alcohols: Hydroboration of an alkene and subsequent reactions of the product trialkylborane, either with hydrogen peroxide or with acid, appear to be highly stereospecific. For example, 1-methylcyclopentene gives exclusively -2-methylcyclopentanol on hydroboration followed by reaction with alkaline hydrogen peroxide. This indicates that, overall, : Hydroboration of an alkyne followed by treatment of the alkenylborane with basic peroxide provides a method of synthesis of aldehydes and ketones. Thus hydroboration of 1-hexyne and oxidation of the 1-hexenylborane, \(4\), with hydrogen peroxide gives hexanal by way of the enol: If \(4\) is treated with deuteriopropanoic acid, replacement of \(\ce{-BR_2}\) by deuterium occurs with of configuration, forming -hexene-1-\(\ce{D_1}\): The stereospecific oxidation of alkylboranes occurs with hydrogen peroxide by an interesting and important general type of rearrangement which, for these reactions, involves migration of an organic group from boron to oxygen. The first step in the oxidation depends on the fact that tricoordinate boron has only six electrons in its valence shell and therefore behaves as if it were electron-deficient. The first step is bond formation at boron by the strongly nucleophilic peroxide anion (from \(\ce{H_2O_2} + \ce{OH}^\ominus \rightleftharpoons ^\ominus \ce{OOH} + \ce{H_2O}\)) to give a tetracovalent boron intermediate: In the second step, an alkyl group moves from boron to the neighboring oxygen and, in so doing, displaces hydroxide ion. Reaction is completed by hydrolysis of the \(\ce{B-O}\) bond: All three groups on boron are replaced in this manner. The rearrangement step (Equation 11-5) is an example of many related rearrangements in which a group, \(\ce{R}\), migrates with its bonding electrons from one atom to an adjacent atom. We already have encountered an example in the rearrangement of carbocations ( ): The difference between the carbocation rearrangement and the rearrangement of Equation 11-5 is that \(\ce{R}\) migrates from boron to oxygen as \(\ce{HO}^\ominus\) departs in what might be considered an internal \(S_\text{N}2\) reaction. We can generalize this kind of reaction of boron with a substance, \(\ce{X-Y}\), as in Equation 11-6: An example of the use of an \(\ce{X-Y}\) reagent is conversion of alkylboranes to primary amines with hydroxylaminesulfonic acid, \(\ce{H_2NOSO_3H}\) (Equation 11-4). The key steps are attack of the nucleophilic nitrogen at boron, followed by rearrangement, and hydrolysis, and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/17%3A_Carbonyl_Compounds_II-_Enols_and_Enolate_Anions._Unsaturated_and_Polycarbonyl_Compounds/17.04%3A_Nucleophilic_Addition_Reactions_of_Enolate_Anions

A most important property of enolate anions, at least as far as synthesis is concerned, is their excellent nucleophilicity, which enables them to to double bonds and to participate in nucleophilic substitution. When the addition is to a carbonyl double bond, it is called an (Equation 17-4). Additions of enolate anions to carbon-carbon double bonds usually are classified as (Equation 17-5), and these are discussed in and . The principles of \(S_\text{N}\) nucleophilic reactions of enolate anions (Equation 17-6) will be considered in , and their synthetic applications in detail in Chapter 18. The products of aldol addition are \(\beta\)-hydroxy aldehydes ( ) or \(\beta\)-hydroxyl ketones ( ). A typical example is the reaction of ethanal with base and, if the conditions are reasonably mild, the product is 3-hydroxybutanal: The overall reaction corresponds to a dimerization of ethanal, that is, an addition of one ethanal molecule to another with formation of a new carbon-carbon bond. The synthetic value of the reaction lies in the fact that it can be used to build large molecules from smaller molecules (see ). Formation of the enolate anion, \(7\), by removal of an \(\alpha\) hydrogen by base is the first step in the aldol addition: The anion then adds to the carbonyl group of a second molecule of ethanal in a manner analogous to the addition of other nucleophiles to carbonyl groups (e.g., cyanide ion, ). The adduct so formed, \(8\), rapidly adds a proton to the alkoxide oxygen to form the aldol, 3-hydroxybutanal. This last step regenerates the basic catalyst, \(\ce{OH}^\ominus\): The two possible valence-bond structures of the enolate anion, \(7a\) and \(7b\), show that the anion should act as an - a nucleophile with nucleophilic properties associated with carbon and oxygen. The addition step in the aldol reaction therefore may be expected to take place in either of two ways: The anion could attack as a nucleophile to form a carbon-carbon bond, \(8\), leading ultimately to the aldol, \(9\), or it might attack as an nucleophile to form a carbon-oxygen bond, thereby leading to the hemiacetal, \(10\). By this reasoning, we should obtain a mixture of products \(9\) and \(10\). However, the aldol \(9\) is the only one of these two possible products that can be isolated: Why is only one of these products formed? To understand this, you must recognize that aldol reactions are reversible and therefore are subject to rather than control ( ). Although the formation of \(10\) is mechanistically reasonable, it is not reasonable on thermodynamic grounds. Indeed, while the overall \(\Delta H^0\) (for the vapor) calculated from bond energies is \(-4 \: \text{kcal mol}^{-1}\) for the formation of the aldol, it is \(+20.4 \: \text{kcal mol}^{-1}\) for the formation of \(10\).\(^2\) Therefore, the reaction is overwhelmingly in favor of the aldol as the more stable of the two possible products. The equilibrium constant is favorable for the aldol addition of ethanal, as in fact it is for most aldehydes. For ketones, however, the reaction is much less favorable. With 2-propanone (acetone) only a few percent of the addition product "diacetone alcohol", \(11\) is present at equilibrium: The 2-propanone is boiled and the hot condensate from the reflux condenser flows back over solid barium hydroxide contained in the porous thimble and comes to equilibrium with the addition product \(11\). The barium hydroxide is retained by the porous thimble and the liquid phase returns to the boiler where the 2-propanone, which boils \(110^\text{o}\) below the temperature at which \(11\) boils, is selectively vaporized and returns to the reaction zone to furnish more adduct. The key step in aldol addition requires an electron-pair donor (nucleophile) and an electron-pair acceptor (electrophile). In the formation of 3-hydroxybutanal or \(11\), both roles are played by one kind of molecule, but there is no reason why this should be a necessary condition for reaction. Many kinds of mixed aldol additions are possible. Consider the combination of methanal and 2-propanone. Methanal cannot form an enolate anion because it has no \(\alpha\) hydrogens. However, it is expected to be a particularly good electron-pair acceptor because of freedom from steric hindrance and the fact that it has an unusually weak carbonyl bond (\(166 \: \text{kcal}\) compared to \(179 \: \text{kcal}\) for 2-propanone). In contrast, 2-propanone forms an enolate anion easily but is relatively poor as the electrophile. Consequently the addition of 2-propanone to methanal should and does occur readily: The problem is not to get addition, but rather to keep it from going too far. Indeed, all six \(\alpha\) hydrogens of 2-propanone can be replaced easily by \(\ce{-CH_2OH}\) groups: A commercially important mixed addition involves ethanal and an excess of methanal in the presence of calcium hydroxide. Addition occurs three times and the resulting trihydroxymethylethanal (which has no \(\alpha\) hydrogens) undergoes a "crossed Cannizzaro" reaction with more methanal to give a tetrahydroxy alcohol known as "pentaerythritol:: Pentaerythritol is used widely in the preparation of surface coatings and in the formation of its tetranitrate ester, pentaerythrityl tetranitrate [PETN, \(\ce{C(CH_2ONO_2)_4}\)], which is an important high explosive. An important property of aldol addition products is the ease with which they eliminate water in the presence of either acids or bases. For example, when 3-hydroxybutanal is heated in the basic solution in which it is formed (by aldol addition of ethanal), 2-butenol results: The ease of dehydration compared with simple alcohols is related to the fact that the product is a alkenone. The stabilization energy of the conjugated system makes the equilibrium constant for dehydration especially favorable. In many cases the aldol adduct is only an intermediate in aldol reactions because it dehydrates more rapidly than it can be isolated. Such is most often the case when the dehydration product is a polyunsaturated conjugated aldehyde or ketone. 2-Propanone and bezenecarbaldehyde (benzaldehyde), for instance, give the unsaturated ketone \(12\) in cold aqueous sodium hydroxide solution: Although the equilibrium for aldol addition may be unfavorable, when dehydration of the aldol product is rapid, \(\ce{C-C}\) bond formation may be pushed to completion by conversion of the aldol to the \(\alpha\),\(\beta\)-unsaturated ketone. The mechanism of base-catalyzed dehydration of aldols involves formation of an enolate anion by removal of a proton from the \(\ce{C2}\) or carbon and subsequent elimination of the hydroxyl group as hydroxide ion: This last step is one of the rare examples in which the leaving group is \(\ce{OH}^\ominus\). Generally, hydroxide is a poor leaving group in substitution (\(S_\text{N}1\) or \(S_\text{N}2\)) or elimination (\(E1\) or \(E2\)) reactions (see Section 8-7C). Dehydration of aldols to \(\alpha\),\(\beta\)-unsaturated carbonyl compounds usually is achieved best with acidic catalysts. An example is the dehydration of the aldol from 2-propanone to give 4-methyl-3-penten-2-one: If this reaction were attempted under basic conditions, extensive reversion of the aldol to 2-propanone would occur (see ). Under acidic conditions, however, the process is a straightforward proton transfer to oxygen followed by elimination of water and proton transfer from carbon: Aldol reactions provide a valuable synthetic method for forming carbon-carbon bonds. They can be adapted to extend the length of a carbon chain, to form cyclic compounds, and to provide intermediates that can be transformed into more useful materials. An important feature of these intermediates is that functional groups useful for later reactions are located close to or on the carbons of the newly formed \(\ce{C-C}\) bond. There is an almost bewildering number of variations on the aldol reaction and we shall not mention all of them. The main thing to recognize in all of these reactions is that the acceptor molecule always is a carbonyl compound, best an aldehyde, sometimes a ketone, even an ester (see ). The donor molecule is some type of carbanion; usually, but not always, an enolate anion. However, any substance that has a \(\ce{C-H}\) acidity in the p\(K_\text{a}\) range of 25 or less can be converted easily to a carbanion, which in principle may serve as the donor in aldol additions. Examples are listed in Table 17-1 and include not only aldehydes and ketones but esters, nitriles, and nitro compounds. The use of a nitroalkane in aldol addition is shown in the following sequence. The use of esters as the donor is discussed further in . Cyclic products can be formed by aldol additions provided the donor carbanion and acceptor carbonyl are part of the molecule. For example, consider how the synthesis of 3-methyl-2-cyclohexanone could be achieved from acyclic substances. The carbon-carbon bond formed in this process of aldol addition closes the ring and ultimately becomes the double bond in the conjugated system when the aldol product undergoes dehydration. Working backwards, we have the sequence and the starting material for the synthesis therefore is 2,6-heptanedione. Because \(\Delta G^0\) for the formation of aldol products is not very favorable, cyclizations involving aldol reactions usually will not proceed to give strained carbocyclic rings. The importance of aldol reactions is in the synthesis of alcohols, especially 1-butanol and 2-ethyl-1-hexanol: Notice that the combination of hydroformylation ( ), aldol addition, dehydration, and hydrogenation takes a simple alkene (propene) to an alcohol with more than twice as many carbons. One of the reactions in the metabolism of carbohydrates by the glycolic pathway is a type of aldol addition. In this reaction \(D\)-fructose (as the 1,6-diphosphate ester) is formed from \(D\)-glyceraldehyde and 1,3-dihydroxypropanone (both as monophosphate esters). The process is readily reversible and is catalyzed by an enzyme known as : It seems likely that this reaction could occur in quite the same way as in the laboratory aldol reactions discussed so far, because the enolate anion of the donor molecule (dihydroxypropanone) is not expected to be formed in significant amount at the pH of living cells. In fact, there is strong evidence that the enzyme behaves as an amino \(\left( \ce{ENH_2} \right)\) compound and reacts with the carbonyl group of dihydroxypropanone to form an imine, analogous to the reactions described in : This implies that the imine form of dihydroxypropanone is a key intermediate in the overall aldol-type addition. How can the imine behave as the carbon in addition to the aldehyde carbonyl of glyceraldehyde 3-phosphate? It is unlikely to do so directly, but it can rearrange to an enamine which, as we will explain in , can act as a carbon nucleophile: Attack of the nucleophilic carbon of the enamine at the aldehyde carbonyl of glyceraldehyde 3-phosphate forms the aldol of the imine which, on hydrolysis, gives the aldol and regenerates the enzyme: By using the neutral enamine as the carbon nucleophile rather than an enolate anion, the biological system avoids the need for strongly basic reaction conditions in aldol addition. \(^2\)This value probably is too large by \(3\) to \(4 \: \text{kcal}\), because resonance stabilization of alkoxyalkanes has been ignored in this calculation. and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases/Gas_Equilibrium_Constants

\(K_c\) and \(K_p\) are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that \(K_c\) is defined by molar concentrations, whereas \(K_p\) is defined by the partial pressures of the gasses inside a closed system. The equilibrium constants do not include the concentrations of single components such as liquids and solid and they may have units depending on the nature of the reaction (although do not). Here are some easy steps on writing gas equilibrium constants (this is the same for finding K , K , K , Q and etc.):. The standard example of writing Gas Equilibrium Constants are: \[ \ aA + bB \; \rightleftharpoons \; cC + dD \nonumber \] \[ K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \nonumber \] \[ K_p = \dfrac{(C)^{c}(D)^{d}} {(A)^{a}(B)^{b}} \nonumber \] Consider the thermal decomposition of \( NH_4SH_{(s)} \): \[ NH_4SH_{(s)} \rightleftharpoons NH_{3 (g)} + H_2S_{(g)} \nonumber \] This also is related to K \[ K_c = \dfrac{[NH_{3},H_{2}S]}{[NH_{4}SH]} \nonumber \] but since \(NH_4SH\) is a solid, we get: \[ K_c = \dfrac{[NH_{3},H_{2}S]}{[1]} \nonumber \] \[ K_c = [NH_{3},H_{2}S] \nonumber \] As for K , it is the same as K , but instead of brackets [ ], K uses parentheses ( ): \[ K_p = \dfrac{(NH_{3})(H_{2}S)}{(NH_{4}SH)} \nonumber \] \[ K_p = \dfrac{(NH_{3})(H_{2}S)}{(1)} \nonumber \] \[ K_p = (NH_{3})(H_{2}S) \nonumber \] Consider the double replacement reaction of hydrogen and iodine gas: \[ H_2 (g) + I_2 (g) \rightleftharpoons 2HI (g) \nonumber \] \[ K_c = \dfrac{[HI]^{2}}{[H_{2},I_{2}]} \nonumber \] \[ K_p = \dfrac{(HI)^{2}}{(H_{2})(I_{2})} \nonumber \] \(K_c\) is an equilibrium constant in terms of molar and is usually defined as: \[ K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \nonumber \] in the general reaction, \[ aA + bB \rightleftharpoons cC + dD\nonumber \] If a large \(K_c\) is formed then there are more products formed. Inversely, a small \(K_c\) indicates that the reaction favors the reactants. \(K_p\) is an equilibrium constant in terms of . and is usually defined as: \[ K_p = \dfrac{(C)^{c}(D)^{d}}{(A)^{a}(B)^{b}} \nonumber \] for the general reaction \[ aA + bB \rightleftharpoons cC + dD \nonumber \] : Reactants/Products all in a single phase. For example: \[A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}\nonumber \] Reactants/Products in more than one phase. For example: \[A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(s)}\nonumber \] The relationship between the two equilibrium constants are: \[ K_p = K_c (RT)^{ \Delta{n}} \nonumber \] or \[K_c = \dfrac{K_p}{(RT)^{ \Delta{n}}}\nonumber \] where, The value of K depends on whether the solution being calculated for is using concentrations or partial pressures. The gas equilibrium constants relate to the equilibrium (K) because they are both derived from (PV = nRT). \(K_c\) is the concentration of the reaction, it is usually shown as: \[ \dfrac{c[C]c[D]}{c[A]c[B]} \nonumber \] \(K_p\) is the amount of partial pressure in the reaction, usually shown as: \[ \dfrac{p(C)p(D)}{p(A)p(B)} \nonumber \] As we have seen above, K = K (RT) ,we can derive this formula from the Ideal Gas Law. We know that K is in terms Molarity \(\left(\dfrac{Moles}{Liters}\right)\), and we can also arrange the Ideal Gas Law (PV = nRT) as: \(\left(\dfrac{n}{L}\right) = \left(\dfrac{P}{RT}\right)\). We know that Partial Pressure is directly proportional to Concentration: \(P = \left(\dfrac{n}{L}\right) * RT\) Pressure can be in units of: Pascal (Pa), Atmosphere (atm), or Torr. Therefore we can replace K with Molarity: the equation become, K = K (RT) (RT) = \(\dfrac{(RT)^{c}(RT)^{d}}{(RT)^{a}(RT)^{b}}\) Also: \(\left(\dfrac{n}{L}\right) = \left(\dfrac{P}{RT}\right)\), can be shown as K = K (RT) K is also written the same as K and K : \(\ aA + bB \rightleftharpoons cC + dD\). \( K = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\) \[ 2 NOBr_{(g)} \rightleftharpoons 2 NO_{(s)} + Br_{2 (g)} \nonumber \] Given: To set up K , it is \(\dfrac{Products}{Reactants}\) \[ K = \dfrac{[NO]^2 \; [Br_{2}]}{[NOBr]^2}\nonumber \] \[ K =\dfrac{[0.1]^2 \; [0.3]}{[0.46]^2}\nonumber \] K = 0.0142 M N O (l) is an important component of rocket fuel, At 25 °C N O is a colorless gas that partially dissociates into NO . The color of an equilibrium mixture of these 2 gasses depends on their relative proportions, which are dependent on temperature. Equilibrium is established in the reaction \( N_2O_4 (g) \rightleftharpoons 2NO_2 (g) \) at 25 °C. Given: What is the K for this reaction? Step 1: Convert grams to moles mol N O = 7.64 g * \( \dfrac{1 mol N_2 O_4}{92.01 g} \) = 8.303 * 10 mol mol NO = 1.56 g * \( \dfrac{1 mol NO_2}{46.01 g} \) = 3.391 * 10 mol Step 2: Convert moles to Molarity (moles/L) [N O ] M \(\dfrac{8.303 * 10^{-2} mol N_{2}O_{4}}{3.00 L}\) = 0.0277 M [NO ] M = \(\dfrac{3.391 * 10^{-2} mol NO_{2}}{3.00 L}\) = 0.0113 M Step 3: Write the Equilibrium constant for K : \[K_{c} = \dfrac{[NO_{2}]^{2}}{[N_{2}O_{4}]} = \dfrac{[0.0113]^{2}}{[0.0277]} = 4.61 \times 10^{-3}\nonumber \] Calculate K for the reaction \[N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber \] Given: When equilibrium is established the mole percent of Nitrogen Oxide (NO) at 1.8% \[N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber \] \(N_2\) X = 0.018 X = (0.79-x)+(.021-x)+(2x)= 1 X = \(\dfrac{2x}{X_{total}}\) 0.018 = \(\dfrac{2x}{1}\) x = 0.009 K = \(\dfrac{p(NO)^{2}}{p(N_{2})p(O_{2})}\) PV = NRT \(P = \dfrac{\dfrac{[n(NO)(RT)]^{2}}{V^{2}}}{\dfrac{n(N_{2})(RT)}{V}\dfrac{n(O_{2})(RT)}{V}}\) [Volume Cancels out] \(P = \dfrac{n(NO)^{2}}{n(N_{2})n(O_{2})}\) \(K_{P} = \dfrac{(2x)^{2}}{(0.79-x)(0.21-x)}\) x = 0.009 K = 2.1x10 The process of finding the Reaction Quotient (Q ) is the same as finding K and K , where the products of the reaction is divided by the reactants of the reaction \(\left(\dfrac{Products}{Reactants}\right)\) at any time not necessarily at equilibrium. If a problem asks you to find which way the reaction will shift in order to achieve equilibrium, and K is given, you would have to calculate for Q and compare the two numbers. When comparing K and Q: A trick to remember to which what the reaction will favor is: Put: K _ Q (in alphabetical order! - or it will not work) K < Q : K \(\leftarrow\) Q The reaction will favor the reactants because reactants are on the left of the equation. K > Q : K \(\rightarrow\) Q The reaction will favor the products because products are on the right of the equation. K = Q : NO CHANGE (See Relationship for more information) \[ CO (g) + H_2O (g) \rightleftharpoons CO_2 (g) + H_2 (g) \nonumber \] Given: K = 1.00 at about 1100 K CO = 1.00 mol H O = 1.00 mol CO = 2.00 mol H = 2.00 mol Compared with their initial amounts, which of the substances will be present in a greater amount and which is in a lesser amount when equilibrium is established? Step 1: Write out the expression for Q \[ Q_c = \dfrac{[CO_2,H_2]}{[CO,H_2O]} \nonumber \] Step 2: Plug in the number of Molarity, since volume is not given, assume it is 1 Liters \[ Q_c = \dfrac{[2.00,2.00]}{[1.00,1.00]} \nonumber \] \[ Q_c = 4.00 \nonumber \] Step 3: Compare K with Q K = 1.00 (unitless) Q = 4.00 (unitless) K < Q 1.00 < 4.00 Therefore, the reaction will shift to the LEFT towards the reactants. A mixture of hydrogen, iodine, and hydrogen iodide, each at 0.0020 M, was introduced into a container heated to 783 Kelvins. At this temperature \(K_c = 46\), Predict if more HI or less will be formed. Step 1: Write out the reaction \[ H_2 (g) + I_2 (g) \leftrightharpoons 2HI (g) \nonumber \] Step 2: Write out the expression for Q \[ Q_c = \dfrac{[HI]^2}{[I_2,H_2]} \nonumber \] Step 3: Plug in the Molarity given Molarity = 0.0020 M \[ Q_c = \dfrac{[0.0020]^{2}}{[0.0020,0.0020]} \nonumber \] \[ Q_{c} = 1.00 \nonumber \] Step 4: Compare K with Q K = 46 (unitless) Q = 1.00 (unitless) K > Q 46 > 1.00 Below are practice problems for K and K : 1. Gaseous Hydrogen Iodide is placed in a closed container at 425 C, Where it partially decomposes to Hydrogen and Iodine: 2HI (g) \(\rightleftharpoons\) H (g) + I (g) The following are given: [HI] = 3.53 * 10 M [H ] = 4.79 * 10 M [I ] = 4.79 * 10 M What is the value of K at this temperature? 2. Write the K for the reaction and state where the reaction is Homogeneous or Heterogeneous. a) N (g) + O (g) \(\rightleftharpoons\) 2NO (g) b) FeO (s) + H (g) \(\rightleftharpoons\) Fe (s) + H O (g) 3. Determine values of K from the K value given: (number 7 from p. 655 in the textbook) 2NO (g) + O (g) \(\rightleftharpoons\) 2NO (g); Kp = 1.48 * 10 at 184 C 1. Write the K for the reaction and state where the reaction is Homogeneous or Heterogeneous. a) 2C H (g) + 2H O (g) \(\rightleftharpoons\) 2C H (g) + O (g) b) Ti (s) + 2Cl (g) \(\rightleftharpoons\) TiCl (g) 2. Determine values of K from the K value given: (number 8 from p. 655 in the textbook) 2H S (g) + CH (g) \(\rightleftharpoons\) 4H (g) + CS (g); K = 5.27 * 10 at 973 K. 3. The two common chlorides of Phosphorus, PCl and PCl , both important in the production of other phosphorous compounds, coexist in equilibrium through: (number 17 from p. 655 in the textbook) PCl (g) + Cl (g) \(\rightleftharpoons\) PCl (g) At 250 C, an equilibrium mixture in a 2.50 L flask contains 0.105 g PCl , 0.220 g PCl , and 2.12 g Cl . What are the values of (a) Kc and (b) Kp for this reaction? 1. 2HI (g) \(\rightleftharpoons\) H (g) + I (g) [HI] = 3.53 * 10 M [H ] = 4.79 * 10 M [I ] = 4.79 * 10 M K = \(\dfrac{[H_{2},I_{2}]}{[HI]^{2}}\) K = \(\dfrac{[4.79 * 10^{-4} M,4.79 * 10^{-4} M]}{[3.53 * 10^{-3} M]^{2}}\) K = \(\dfrac{[2.29441 * 10^{-7}] M^{2}}{[1.24609 * 10^{-5}] M^{2}}\) 2. a) N (g) + O (g) \(\rightleftharpoons\) 2NO (g) K = \(\dfrac{[NO]^{2}}{[N_{2},O_{2}]}\) The reaction is a homogeneous reaction because the reactants/products all have the same phase. b) FeO (s) + H (g) \(\rightleftharpoons\) Fe (s) + H O (g) K = \(\dfrac{[H_{2}O]}{[H_{2}]}\), FeO and Fe are solids so they are no included in equilibrium constants. The reaction is a heterogeneous reaction because the reactants/products have different phases. 3. Converting to K from K (number 7 from p. 655 in the textbook) 2NO (g) + O (g) \(\rightleftharpoons\) 2NO (g) Kp = 1.48 * 10 at 184 C We know that K = K (RT) , we are given K but not K , you can rearrange the equation to: \(K_{c} = \dfrac{K_{p}}{(RT)^{-\Delta{n}}}\) Which can also be written as: K = K (RT) Now that we have your formula, we need to convert 184 C to Kelvin, K = 184 + 273 = 457K K = \(\dfrac{[NO_{2}]^{2}}{[NO]^{2}[O_{2}]}\) -\(\Delta{n}\) = (total number of moles of products) - (total number of moles in reactants) -\(\Delta{n}\) = (2) - (3) = -1 \(\Delta{n}\) = -(-1) R = 0.08206 \(\dfrac{Liter \; Atm}{Mole \; Kelvin}\) Now, plug in all the numbers we found: K = K (RT) K = (1.48 * 10 )[(0.08206)(457K)] [it would be the same if you used this equation: K = K (RT)] 1. a) 2C H (g) + 2H O (g) \(\rightleftharpoons\) 2C H (g) + O (g) K = \(\dfrac{[C {2}H_{6}]^{2}[O_{2}]}{[C_{2}H_{4}]^{2}[H_{2}O]^{2}}\) The reaction is a homogeneous reaction because the reactants/products all have the same phase. b) Ti (s) + 2Cl (g) \(\rightleftharpoons\) TiCl (g) K = \(\dfrac{[TiCl_{4}]}{[Cl_{2}]^{2}}\), Ti is a solid so it is not included in equilibrium constants. The reaction is a heterogeneous reaction because the reactants/products have different phases. 2. Find K , when K is given: (number 8 from p. 655 in the textbook) 2H S (g) + CH (g) \(\rightleftharpoons\) 4H (g) + CS (g) K = 5.27 * 10 at 973 K. K = K (RT) , since Temperature is already converted to Kelvin and R = 0.08206 \(\dfrac{Liter \; Atm}{Mole \; Kelvin}\) We need to find Delta n: K = \(\dfrac{[H {2}]^{4}[CS_{2}]}{[H_{2}S]^{2}[CH_{4}O]}\) \(\Delta{n}\) = (total number of moles of products) - (total number of moles in reactants) \(\Delta{n}\) = (5) - (3) = 2 We can plug in our numbers: K = (5.27 * 10 )[(0.08206)(973)] 3. The two common chlorides of Phosphorus, PCl and PCl , both important in the production of other phosphorous compounds, coexist in equilibrium through: (number 17 from p. 655 in the textbook) PCl (g) + Cl (g) \(\rightleftharpoons\) PCl (g) At 250 C, an equilibrium mixture in a 2.50 L flask contains 0.105 g PCl , 0.220 g PCl , and 2.12 g Cl . What are the values of: (a) K [We need to convert grams to Molarity (mol/L), so we multiply grams with the molar mass and divide by Liters.] PCl = \(\dfrac{0.105 g}{2.50 L}\) \(* \dfrac{1 mol}{137.3 g}\) (Molar Mass) = 2.0173 * 10 M PCl = \(\dfrac{0.220 g}{2.50 L}\) \(* \dfrac{1 mol}{137.3 g}\) (Molar Mass) = 6.4093 * 10 M Cl = \(\dfrac{2.12 g}{2.50 L}\) \(* \dfrac{1 mol}{70.9 g}\) (Molar Mass) = 0.0119605 \( K_{c} = \dfrac{[PCl_{5}]}{[Cl_{2},PCl_{3}]}\) \( K_{c} = \dfrac{[2.0173 * 10^{-4}]}{[0.0119605,6.4093 * 10^{-4}]}\) (b) K K = K (RT) We need to find: \(\Delta{n}\) and K, R = 0.08206 \(\dfrac{L atm}{mol K}\) K = 250 C + 273 = 523 K \(\Delta{n}\) = (1) - (2) = -1 K = (26.32) [(0.08206 )(523)]

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/16%3A_Carbonyl_Compounds_I-_Aldehydes_and_Ketones._Addition_Reactions_of_the_Carbonyl_Group/16.10%3A_Preparative_Methods_for_Aldehydes_and_Ketones

A number of useful reactions for the preparation of aldehydes and ketones, such as ozonization of alkenes and hydration of alkynes, have been considered in previous chapters. These and other methods of preparation are summarized in Tables 16-7 and 16-8 at the end of the chapter. Only a few rather general methods that we have not discussed will be taken up here. Aldehydes and ketones often can be prepared by oxidation of alkenes to 1,2-diols (Sections 11-7C and 11-7D), followed by oxidative cleavage of the 1,2-diols with lead tetraethanoate or sodium periodate. For example, Cleavage of glycols with these reagents proceeds according to the following stoichiometry: In Chapter 15 primary alcohols, \(\ce{RCH_2OH}\), were shown to be readily oxidized to aldehydes, \(\ce{RCHO}\), and secondary alcohols, \(\ce{R_2CHOH}\), to ketones, \(\ce{R_2CO}\), by inorganic reagents such as \(\ce{CrO_3}\) and \(\ce{KMnO_4}\). However, it is a problem to avoid overoxidation with primary alcohols because of the ease with which aldehydes are oxidized to acids, \(\ce{RCHO} \rightarrow \ce{RCO_2H}\). A milder oxidant is methylsulfinylmethane [dimethyl sulfoxide, \(\ce{(CH_3)_2S=O}\)], and this reagent can be used to prepare aldehydes from alcohols by way of an intermediate such as the ester or halide in which the \(\ce{OH}\) group is converted to a better leaving group: Whichever method is employed, the key step is the formation of an alkoxysulfonium salt, \(7\), by a displacement reaction involving dimethyl sulfoxide as an oxygen nucleophile. (Notice that the \(\ce{S=O}\) bond, like the \(\ce{C=O}\) bonds, is strongly polarized as \(\overset{\oplus}{\ce{S}} \ce{-} \overset{\ominus}{\ce{O}}\).) In the examples listed in Equations 16-12 through 16-15, the \(\ce{X}\) group is \(\ce{Br}\), \(\ce{-OSO_2R'}\), \(\ce{-O_2CCF_3}\), and , respectively. In the nest step a sulfur ylide, \(8\), is formed from the reaction of a base with \(7\), but the ylide evidently is unstable and fragments by an internal \(E2\) reaction to form an aldehyde: Conversion of a carboxylic acid to an aldehyde by direct reduction is not easy to achieve, because acids generally are difficult to reduce, whereas aldehydes are easily reduced. Thus the problem is to keep the reaction from going too far. The most useful procedures involve conversion of the acid to a derivative that either is more easily reduced than an aldehyde, or else is reduced to a substance from which the aldehyde can be generated. The so-called involves the first of these schemes; in this procedure, the acid is converted to an acyl chloride, which is reduced with hydrogen over a palladium catalyst to the aldehyde in yields up to \(90\%\). The rate of reduction of the aldehyde to the corresponding alcohol is kept at a low level by poisoning the catalyst with suflur: Metal hydrides, such as lithium aluminum hydride, also can be used to reduce derivatives of carboxylic acids (such as amids and nitriles see Table 16-6) to aldehydes. An example follows: Many carbonyl compounds can be synthesized by acid-catalyzed rearrangements of 1,2-diols (a type of reaction often called the "pinacol-pinacolone" rearrangement). The general characteristics of the reaction are similar to those of carbocation rearrangements (Section 8-9B). The acid assists the reaction by protonating one of the \(\ce{-OH}\) groups to make it a better leaving group. The carbocation that results then can undergo rearrangement by shift of the neighboring \(\ce{R}\) group with its pair of bonding electrions to give a new, thermodynamically more stable species with a carbon-oxygen double bond (see Section 16-7). The prototype of this rearrangement is the conversion of pinacol to pinacolone as follows: An important method of preparing carbonyl (and hydroxy) compounds, especially on an industrial scale, is through rearrangements of alkyl hydroperoxides: The peroxides can be made in some cases by direct air oxidation of hydrocarbons, and in others by sulfuric acid-induced addition of hydrogen peroxide (as \(\ce{H-O_2H}\)) to double bonds: (Notice that hydrogen peroxide in methanoic acid behaves differently toward alkenes in producing addition of \(\ce{HO-OH}\), Section 11-7D.) The direct air oxidation of hydrocarbons is mechanistically similar to that of benzenecarbaldehyde (Section 16-7). The rearrangements of hydroperoxides are acid-catalyzed and are analogous to carbocation rearrangements except that positive oxygen (with only valence electrons) instead of positive carbon is involved in the intermediate stage: In principle, either phenyl or methyl could migrate to the positive oxygen, but only phenyl migration occurs in this case. The rearrangement reaction is closely related to the Baeyer-Villiger reaction (Section 16-7). This reaction is important for a number of reasons. It is an industrial synthesis of aldehydes from alkenes by the addition of carbon monoxide and hydrogen in the presence of a cobalt catalyst. A prime example is the synthesis of butanal from propene, in which 2-methylpropanal also is formed: As you can see, the reaction formally amounts to the addition of methanal as \(\ce{H-CHO}\) to the alkene double bond. Because one additional carbon atom is introduced as a "formyl" \(\ce{CHO}\) group, the reaction often is called , although the older name, , is widely used. Hydroformylation to produce aldehydes is the first step in an important industrial route to alcohols. The intermediate aldehydes are reduced to alcohols by catalytic hydrogenation. Large quantities of \(\ce{C_4}\)-\(\ce{C_8}\) alcohols are prepared by this sequence: The history of the oxo reaction is also noteworthy. It was developed originally in Germany in the years following World War I. At that time, the German chemical industry was faced with inadequate supplies of petroleum. Many German chemists therefore turned to research on ways by which hydrocarbons could be synthesized from smaller building blocks, particularly carbon monoxide and hydrogen derived from coal. The success achieved was remarkable and led to alkane and alkene syntheses known as the : ​​​​​​ This reaction in turn led to the discovery that aldehydes were formed by the further addition of carbon monoxide and hydrogen to alkenes, and was further developed as the oxo process for production of alcohols. The combination \(\ce{CO} + \ce{H_2}\) often is called "synthetic gas". It is prepared by the reduction of water under pressure and at elevated temperatures by carbon (usually coke), methane, or higher-molecular-weight hydrocarbons: The aldehyde synthesis by hydroformylation of alkenes described in the preceding section can be achieved indirectly using boron hydrides. An oversimplified expression of this reaction is The overall reaction is quite complex but involves a rearrangement similar to that described for the hydroboration-oxidation of alkenes (Section 11-6E). The first step is hydroboration of the alkene to a trialkylborane. When the trialkylborane is exposed to carbon monoxide, it reacts (carbonylates) to form a tetracovalent boron, \(9\): The complex \(9\) is unstable and rearranges by transfer of an alkyl group from boron to the electron-deficient carbonyl carbon to give \(10\): Now, if a metal-hydride reducing agent, such as \(\ce{LiAlH_4}\), is present, the carbonyl group of \(10\) is reduced and \(11\) is formed: The reduction product, \(11\), can be converted to an aldehyde by oxidation with aqueous hydrogen peroxide, provided the pH is carefully controlled. (Remember, aldehydes are unstable in strong base.) You may have noticed that only of the three alkyl groups of a trialkylborane is converted to an aldehyde by the carbonylation-reduction-oxidation sequence. To ensure that carbonylation takes the desired course without wasting the starting alkene, hydroboration is achieved conveniently with a hindered borane, such as "9-BBN", \(12\). With \(12\), only the least-hindered alkyl group rearranges in the carbonylation step: Carbonylation of alkylboranes also can produce ketones. The conditions are similar to those in the aldehyde synthesis except that the hydride reducing agent is omitted. By omitting the reducing agent, a second boron-to-carbon rearrangement can occur. Oxidation then produces a ketone: Rearrangement will continue a third time (ultimately to produce a tertiary alcohol) unless movement of the alkyl group remaining on boron in \(13\) is prevented by steric hindrance. and (1977)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/0Group_17%3A_Physical_Properties_of_the_Halogens/Group_17%3A_General_Properties_of_Halogens

The halogens are located on the left of the noble gases on the periodic table. These five toxic, non-metallic elements make up Group 17 of the periodic table and consist of: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Although astatine is radioactive and only has short-lived isotopes, it behaves similar to iodine and is often included in the halogen group. Because the halogen elements have seven valence electrons, they only require one additional electron to form a full octet. This characteristic makes them more reactive than other non-metal groups. Halogens form diatomic molecules (of the form X ​, where X denotes a halogen atom) in their elemental states. The bonds in these diatomic molecules are non-polar covalent single bonds. However, halogens readily combine with most elements and are never seen uncombined in nature. As a general rule, fluorine is the most reactive halogen and astatine is the least reactive. All halogens form Group 1 salts with similar properties. In these compounds, halogens are present as halide anions with charge of -1 (e.g. Cl , Br , etc.). Replacing the -ine ending with an -ide ending indicates the presence of halide anions; for example, Cl is named "chloride." In addition, halogens act as oxidizing agents—they exhibit the property to oxidize metals. Therefore, most of the chemical reactions that involve halogens are oxidation-reduction reactions in aqueous solution. The halogens often form single bonds, when in the -1 oxidation state, with carbon or nitrogen in organic compounds. When a halogen atom is substituted for a covalently-bonded hydrogen atom in an organic compound, the prefix can be used in a general sense, or the prefixes , , , or can be used for specific halogen substitutions. Halogen elements can cross-link to form diatomic molecules with polar covalent single bonds. Chlorine (Cl ) was the first halogen to be discovered in 1774, followed by iodine (I ), bromine (Br ), fluorine (F ), and astatine (At, discovered last in 1940). The name "halogen" is derived from the Greek roots hal- ("salt") and -gen ("to form"). Together these words combine to mean "salt former", referencing the fact that halogens form salts when they react with metals. is the mineral name for rock salt, a natural mineral consisting essentially of sodium chloride (NaCl). Lastly, the halogens are also relevant in daily life, whether it be the fluoride that goes in toothpaste, the chlorine that disinfects drinking water, or the iodine that facilitates the production of thyroid hormones in one's body. \[F < Cl < Br < I < At\] F < Cl < Br < I < At) At < I < Br < Cl < F). At < I < Br < Cl < F). At < I < Br < Cl) The reactivities of the halogens decrease down the group Cl electronegativity A halide is formed when a halogen reacts with another, less electronegative element to form a binary compound. Hydrogen, for example, reacts with halogens to form halides of the form Hydrofluoric acid can etch glass and certain inorganic fluorides over a long period of time. It may seem counterintuitive to say that HF is the weakest hydrohalic acid because fluorine has the highest electronegativity. However,​ the H-F bond is very strong; if The halogens' colors are results of the absorption of visible light by the molecules, which causes electronic excitation. Fluorine absorbs violet light, and therefore appears light yellow. Iodine, on the other hand, absorbs yellow light and appears violet (yellow and violet are complementary colors, which can be determined using a The colors of the halogens grow darker down the group: In closed containers, liquid bromine and solid iodine are in equilibrium with their vapors, which can often be seen as colored gases. One third exception to the rule is this: if a halogen exists in its elemental form (X ), its oxidation state is zero. : Although fluorine is very reactive, it serves many industrial purposes. For example, it is a key component of the plastic (called by the DuPont company) and certain other polymers, often referred to as fluoropolymers. Chlorofluorocarbons (CFCs) are organic chemicals that were used as refrigerants and propellants in aerosols before growing concerns about their possible environmental impact led to their discontinued use. Hydrochlorofluorocarbons (HFCs) are now used instead. Fluoride is also added to toothpaste and drinking water to help reduce tooth decay. Fluorine also exists in the clay used in some ceramics. Fluorine is associated with generating nuclear power as well. In addition, it is used to produce fluoroquinolones, which are antibiotics. Below is a list of some of fluorine's important inorganic compounds. : Chlorine has many industrial uses. It is used to disinfect drinking water and swimming pools. Sodium hypochlorite (NaClO) is the main component of bleach. Hydrochloric acid, sometimes called muriatic acid, is a commonly used acid in industry and laboratories. Chlorine is also present in polyvinyl chloride (PVC), and several other polymers. PVC is used in wire insulation, pipes, and electronics. In addition, chlorine is very useful in the pharmaceutical industry. Medicinal products containing chlorine are used to treat infections, allergies, and diabetes. The neutralized form of hydrochloride is a component of many medications. Chlorine is also used to sterilize hospital machinery and limit infection growth. In agriculture, chlorine is a component of many commercial pesticides: DDT (dichlorodiphenyltrichloroethane) was used as an agricultural insecticide, but its use was discontinued. : Bromine is used in flame retardants because of its fire-resistant properties. It also found in the pesticide methyl bromide, which facilitates the storage of crops and eliminates the spread of bacteria. However, the excessive use of methyl bromide has been discontinued due to its impact on the ozone layer. Bromine is involved in gasoline production as well. Other uses of bromine include the production of photography film, the content in fire extinguishers, and drugs treating pneumonia and Alzheimer's disease. : Iodine is important in the proper functioning of the thyroid gland of the body. If the body does not receive adequate iodine, a goiter (enlarged thyroid gland) will form. Table salt now contains iodine to help promote proper functioning of the thyroid hormones. Iodine is also used as an antiseptic. Solutions used to clean open wounds likely contain iodine, and it is commonly found in disinfectant sprays. In addition, silver iodide is important for photography development. : Because astatine is radioactive and rare, there are no proven uses for this halogen element. However, there is speculation that this element could aid iodine in regulating the thyroid hormones. Also, At has been used in mice to aid the study of cancer.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_13%3A_The_Boron_Family/Z013_Chemistry_of_Aluminum_(Z13)/Case_Study%3A_Conversion_of_Bauxite_Ore_to_Aluminum_______Metal

Aluminum is found in varying amounts in nature as aluminosilicates (contains aluminum, silicon, and oxygen) in various types of clay. As the minerals are weathered they gradually breakdown into various forms of hydrated aluminum oxide, Al O .xH O, known as bauxite. The bauxite is purified by the . First the ore is mixed with a hot concentrated solution of sodium hydroxide. The NaOH will dissolve the oxides of aluminum and silicon but not other impurities such as iron oxides, which remains insoluble. The insoluble materials are removed by filtration. The solution which now contains the oxides of aluminum and silicon are next treated by bubbling carbon dioxide gas through the solution. Carbon dioxide forms a weak acid solution of carbonic acid which neutralizes the sodium hydroxide from the first treatment. This neutralization selectively precipitates the aluminum oxide, but leaves the silicates in solution. Again filtration is used for the separation. After this stage the purified aluminum oxide is heated to evaporate the water. Aluminum in the metal form is very difficult to obtain by using some of the traditional chemical methods involving carbon or carbon monoxide as reducing agents to reduce the aluminum ions to aluminum metal. One of the earliest and costly methods in 1850 was to reduce aluminum chloride with sodium metal to obtain aluminum metal and sodium chloride. (Sodium metal is not easy to obtain either). As a result some of the earliest aluminum metal was made into jewelry. In 1886, Charles Hall, an American (23 yrs. old), and Paul Heroult, a Frenchmen (23 yrs old), simultaneously and independently developed the process still in use today to make aluminum metal. The purified aluminum oxide is mixed with cryolite, a mixture of sodium fluoride and aluminum fluoride, and heated to about 980 degrees Celsius to melt the solids. The mixture melts at a much lower temperature than aluminum oxide would by itself. The hot molten mixture is electrolyzed at a low voltage of 4-5 volts, but a high current of 50,000-150,000 amps. Aluminum ions are reduced to aluminum metal at the cathode (the sides and bottom of the electrolysis cell). At the anode, oxygen is produced from the oxide ions. The anode material is carbon in the form of graphite, which also is oxidized and must be replaced quite frequently. The electricity used to produce aluminum is relatively high. One pound of aluminum requires 6-8 kilowatt-hours of electrical energy. This amount of aluminum can be used to make 23 pop cans or one 300 watt light bulb burning for one hour is required to make one pop can.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.07%3A_Oxidation_Reactions

Most alkenes react readily with ozone \(\left( \ce{O_3} \right)\), even at low temperatures, to yield cyclic peroxidic derivatives known as . For example, These substances, like most compounds with peroxide \(\left( \ce{O-O} \right)\) bonds, may explode violently and unpredictably. Therefore ozonizations must be carried out with appropriate caution. The general importance of these reactions derives not from the ozonides, which usually are not isolated, but from their subsequent products. The ozonides can be converted by hydrolysis with water and reduction, with hydrogen (palladium catalyst) or with zinc and acid, to carbonyl compounds that can be isolated and identified. For example, 2-butene gives ethanal on ozonization, provided the ozonide is destroyed with water and a reducing agent which is effective for hydrogen peroxide: An alternative procedure for decomposing ozonides from di- or trisubstituted alkenes is to treat them with methanol \(\left( \ce{CH_3OH} \right)\). The use of this reagent results in the formation of an aldehyde or ketone and a carboxylic acid: The overall ozonization reaction sequence provides an excellent means for locating the positions of double bonds in alkenes. The potentialities of the method may be illustrated by the difference in reaction products from the 1- and 2-butenes: Ozonization of alkenes has been studied extensively for many years, but there is still disagreement about the mechanism (or mechanisms) involved because some alkenes react with ozone to give oxidation products other than ozonides. It is clear that the ozonide is not formed directly, but by way of an unstable intermediate called a . the molozonide then either isomerizes to the "normal" ozonide or participates in other oxidation reactions. Although the structure of normal ozonides has been established beyond question, that of the molozonide, which is very unstable even at \(-100^\text{o}\), is much less certain. The simplest and most widely accepted mechanism involves formation of a molozonide by a direct of ozone to the double bond.\(^1\) Isomerization of the molozonide appears to occur by a fragmentation-recombination reaction, as shown in Equations 11-7 and 11-8: Several oxidizing reagents react with alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide as \(\ce{HO-OH}\). Of particular importance are alkaline permanganate \(\left( \ce{MnO_4^-} \right)\) and osmium tetroxide \(\left( \ce{OsO_4} \right)\), both of which react in an initial step by a suprafacial cycloaddition mechanism like that postulated for ozone. Each of these reagents produces -1,2-dihydroxy compounds (diols) with cycloalkenes: Osmium tetroxide is superior to permanganate in giving good yields of diol, but its use is restricted because it is a very costly and very toxic reagent. Alkenes can be oxidized with peroxycarboxylic acids, \(\ce{RCO_3H}\), to give oxacyclopropanes (oxiranes, epoxides), which are three-membered cyclic ethers: The reaction, known as , is valuable because the oxacyclopropane ring is cleaved easily, thereby providing a route to the introduction of many kinds of functional groups. In fact, oxidation of alkenes with peroxymethanoic acid (peroxyformic acid), prepared by mixing methanoic acid and hydrogen peroxide, usually does not stop at the oxacyclopropane stage, but leads to ring-opening and the subsequent formation of a diol: This is an alternative scheme for the hydroxylation of alkenes (see ). However, the overall stereochemistry is opposite to that in permanganate hydroxylation. For instance, cyclopentene gives -1,2-cylcopentanediol. First the oxirane forms by suprafacial addition and then undergoes ring opening to give the trans product: The ring opening is a type of \(S_\text{N}2\) reaction. Methanoic acid is sufficiently acidic to protonate the ring oxygen, which makes it a better leaving group, thus facilitating nucleophilic attack by water. The nucleophile always attacks from the side remote from the leaving group: The peroxyacids that are used in the formation of oxacyclopropanes include peroxyethanoic \(\left( \ce{CH_3CO_3H} \right)\), peroxybenzoic \( \left( \ce{C_6H_5CO_3H} \right)\), and trifluoroperoxyethanoic \(\left( \ce{CF_3CO_3H} \right)\) acids. A particularly useful peroxyacid is 3-chloroperoxybenzoic acid, because it is relatively stable and is handled easily as the crystalline solid. The most reactive reagent is trifluoroperoxyethanoic acid, which suggests that the peroxyacid behaves as an electrophile (the electronegativity of fluorine makes the \(\ce{CF_3}\) group strongly electron-attracting). The overall reaction can be viewed as a , in which the proton on oxygen is transferred to the neighboring carbonyl oxygen more or less simultaneously with formation of the three-membered ring: A reaction of immense industrial importance is the formation of oxacyclopropane itself (most often called ethylene oxide) by oxidation fo ethene with oxygen over a silver oxide catalyst at \(300^\text{o}\): Oxacyclopropane is used for many purposes, but probably the most important reaction is ring opening with water to give 1,2-ethanediol (ethylene glycol, bp \(197^\text{o}\)). This diol, mixed with water, is employed widely in automotive cooling systems to provide both a higher boiling and lower freezing coolant than water alone: Propene and higher alkenes are not efficiently epoxidized by oxygen and \(\ce{Ag_2O}\) in the same way as ethene because of competing attack at other than the double-bond carbons. \(^1\)The ozone structure shown here with single electrons having paired spins on the terminal oxygens accords both with the best available quantum mechanical calculations and the low dipole moment of ozone, which is not consonant with the conventional \(\ce{O=} \overset{\oplus}{\ce{O}} - \overset{\ominus}{\ce{O}}\) structure. See W. A. Goddard III, T. H. Dunning, Jr., W. J. Hunt, and P. J. Hay, , 368 (1973). and (1977)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/0Group_17%3A_Physical_Properties_of_the_Halogens/Physical_Properties_of_the_Group_17_Elements

This page discusses the trends in some atomic and physical properties of the Group 17 elements (the halogens): fluorine, chlorine, bromine and iodine. Sections below describe the trends in atomic radius, electronegativity, electron affinity, melting and boiling points, and solubility. There is also a section on the bond enthalpies (and strengths) of halogen-halogen bonds (for example, the Cl-Cl bond) and of hydrogen-halogen bonds (e.g. the H-Cl bond). You can see that the atomic radius increases as you go down the Group. The radius of an atom is governed by As shown in the figure above, electronegativity decreases from fluorine to iodine; the atoms become less effective at attracting bonding pairs of electrons as they grow larger. This can be visualized using dots-and-crosses diagrams for hydrogen fluoride and hydrogen chloride. The bonding electrons between the hydrogen and the halogen experience the same net charge of +7 from either the fluorine or the chlorine. However, in the chlorine case, the nucleus is further away from the bonding pair. Therefore, electrons are not as strongly attracted to the chlorine nucleus as they are to the fluorine nucleus. The stronger attraction to the closer fluorine nucleus makes fluorine is more electronegative. As the halogen atoms get larger, any bonding pair is farther and farther away from the halogen nucleus, and so is less strongly attracted towards it. Hence, the elements become less electronegative as you go down the Group,. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.02%3A_The_Arrhenius_Equation

This page examines rate constant variation with temperature and activation energy, as shown by the Arrhenius equation. The rate equation for a reaction between two substances, A and B, is the following: The rate equation shows the effect of changing the reactant concentrations on the rate of the reaction. All other factors affecting the rate—temperature and catalyst presence, for example—are included in the rate constant, which is only constant if the only change is in the concentration of the reactants. If the temperature is changed or a catalyst is added, for example, the rate constant changes. This is shown mathematically in the Arrhenius equation: The various symbols represent the following: The Arrhenius equation often takes this alternate form, generated by taking the natural logarithm of the standard equation: \[ \large \ln k = \ln A - \dfrac{E_a}{RT} \nonumber \] The Arrhenius equation can be used to determine the effect of a change of temperature on the rate constant, and consequently on the rate of the reaction. If the rate constant doubles, for example, so does the rate of the reaction. What is the kinetic effect of increasing temperature from 20°C to 30°C (293 K to 303 K)? The frequency factor, A, is approximately constant for such a small temperature change. This problem concerns the quantity e , the fraction of molecules with energies equal to or in excess of the activation energy. Let's assume an activation energy of 50 kJ mol , or, equivalently, J mol . The value of the gas constant, R, is 8.31 J K mol . At 20 °C (293 K) the value of the fraction is: Raising the temperature (to 303 K) increases the quantity: You can see that the fraction of the molecules able to react has almost doubled by increasing the temperature by 10°C. The rate of reaction is nearly doubled. A catalyst provides a reaction route with a lower activation energy. Suppose that the catalyzed activation energy is to 25 kJ mol . The calculation is repeated at 293 K: Compared with the corresponding value for an activation energy of 50 kJ mol , there is a significant increase in the fraction of molecules able to react. There are almost 30,000 times more molecules which can react in the presence of the catalyst compared to having no catalyst (using our assumptions about the activation energies). These calculations can also be done in reverse: given the rate of reaction or rate constants at several temperatures, the activation energy can be calculated. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.06%3A_Addition_of_Boron_Hydrides_to_Alkenes._Organoboranes

An especially valuable group of intermediates can be prepared by addition of an compound to carbon-carbon double or triple bonds: The reaction is called and is a versatile synthesis of organoboron compounds. One example is the addition of diborane, \(\ce{B_2H_6}\), to ethene. Diborane behaves as though it is in equilibrium with \(\ce{BH_3}\) \(\left( \ce{B_2H_6} \rightleftharpoons 2 \ce{BH_3} \right)\), and addition proceeds in three stages: The monoalkylborane, \(\ce{RBH_2}\), and the dialkylborane, \(\ce{R_2BH}\), seldom are isolated because they rapidly add to the alkene. These additions amount to reduction of both carbons of the double bond: Organoboranes can be considered to be organometallic compounds. Elemental boron does not have the properties of a metal, and boron-carbon bonds are more covalent than ionic. However, boron is more electropositive than either carbon or hydrogen and when bonded to carbon behaves like most metals in the sense that bonds are polarized with \(\ce{R}\) negative and boron positive: Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes. The simplest borane, \(\ce{BH_3}\), exists as the dimer, \(\ce{B_2H_6}\), or in complexed form with certain ethers or sulfides: Any of these \(\ce{BH_3}\) compounds adds readily to most alkenes at room temperature or lower temperatures. The reactions usually are carried out in ether solvents, although hydrocarbon solvents can be used with the borane-dimethyl sulfide complex. When diborane is the reagent, it can be generated either or externally through the reaction of boron trifluoride with sodium borohydride: \[3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{H_4} + 4 \ce{BF_3} \rightarrow 2 \ce{B_2H_6} + 3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{F_4}\] Hydroborations have to be carried out with some care, because diborane and many alkylboranes are highly reactive and toxic substances; many are spontaneously flammable in air. With unsymmetrical alkenes, hydroboration occurs so that : These additions are : Furthermore, when there is a choice, addition occurs preferentially from the less crowded side of the double bond: If the alkene is a bulky molecule, borane may add only one or two alkene molecules to give either mono- or dialkylborane, \(\ce{RBH_2}\) or \(\ce{R_2BH}\), respectively, as the following reactions show: These bulky boranes still possess \(\ce{B-H}\) bonds and can add further to a multiple bond, but they are highly selective reagents and add only if the alkene or alkyne is unhindered. This selectivity can be useful, particularly to 1-alkynes, which are difficult to stop at the alkenylborane stage when using diborane: With a bulky dialkylborane, such as di-(1,2-dimethylpropyl)borane, further addition to the alkenylborane does not occur. An especially selective hydroborating reagent is prepared from 1,5-cyclooctadiene and borane. The product is a bicyclic compound of structure \(1\) (often abbreviated as 9-BBN), in which the residual \(\ce{B-H}\) bond adds to unhindered alkenes with much greater selectivity than is observed with other hydroborating reagents. It is also one of the few boranes that reacts sufficiently slowly with oxygen that it can be manipulated in air. An example of the difference in selectivity in the hydroboration of -4-methyl-2-pentene with \(\ce{B_2H_6}\) and \(1\) follows: According to the electronegativity chart (Figure 10-11), the boron-hydrogen bond is polarized in the sense \(\overset{\delta \oplus}{\ce{B}} --- \overset{\delta \ominus}{\ce{H}}\). Therefore the direction of addition of \(\ce{B_2H_6}\) to propene is that expected of a mechanism whereby the electrophilic boron atom becomes bonded to the less-substituted carbon of the double bond. However, there is no firm evidence to suggest that a carbocation intermediate is formed through a stepwise electrophilic addition reaction. For this reason, the reaction often is considered to be a . The stepwise formulation explains why boron becomes attached to the less-substituted carbon, but does not account for the fact that the reactions show no other characteristics of carbocation reactions. This could be because of an expected, extraordinarily fast rate of hydride-ion transfer to the carbocation. A more serious objection to the stepwise mechanism is that alkynes react more rapidly than alkenes, something which normally is not observed for stepwise electrophilic additions (cf. ). Some alkylboranes rearrange at elevated temperatures \(\left( 160^\text{o} \right)\) to form more stable isomers. For example, the alkylborane \(2\), produced by hydroboration of 3-ethyl-2-pentene, rearranges to \(3\) on heating: In general, the boron in alkylboranes prefers to be at the of a hydrocarbon chain so it is bonded to a carbon where steric crowding around boron is least severe. Thus rearrangement tends to proceed in the direction Rearrangement is associated with the fact that hydroboration is reversible at elevated temperatures. This makes possible a sequence of elimination-addition reactions in which boron becomes attached to different carbons and ultimately leads to the most stable product that has boron bonded to the carbon at the end of the chain: Rearrangement of alkylboranes can be used to transform alkenes with double bonds in the middle of the chain into 1-alkenes; for example, \(\ce{RCH=CHCH_3} \rightarrow \ce{RCH_2-CH=CH_2}\). The procedure involves hydroboration of the starting alkene in the usual manner; the borane then is isomerized by heating. An excess of 1-decene (bp \(170^\text{o}\)) then is added to the rearranged borane and the mixture is reheated. Heating causes the alkylborane to dissociate into 1-alkene and \(\ce{HBR_2}\); the 1-decene "scavenges" the \(\ce{HBR_2}\) as it forms, thereby allowing a more volatile 1-alkene (bp \(<170^\text{o}\)) to be removed by simple distillation. Thus, for the rearrangement of 3-ethyl-2-pentene to 3-ethyl-1-pentene, Alkylboranes formed in the hydroboration of alkenes and alkynes seldom are isolated; for the most part they are used as reactive intermediates for the synthesis of other substances. In the reactions of alkylboranes, the \(\ce{B-C}\) bond is cleaved in the sense \(\ce{B}^\oplus - \ce{C}^\ominus\) so that carbon is transferred to other atoms, such as \(\ce{H}\), \(\ce{O}\), \(\ce{N}\), and \(\ce{C}\), its bonding electron pair: In the first of these reactions (Equation 11-2), a hydrocarbon is produced by the cleavage of a borane, \(\ce{R_3B}\), with aqueous acid, or better, with anhydrous propanoic acid, \(\ce{CH_3CH_2CO_2H}\). The overall sequence of hydroboration-acid hydrolysis achieves the reduction of a carbon-carbon multiple bond without using hydrogen and a metal catalyst or diimide (Table 11-3): The second reaction (Equation 11-3) achieves the synthesis of a alcohol by the oxidation of the alkylborane with hydrogen peroxide in basic solution. Starting with a 1-alkene, one can prepare a primary alcohol in two steps: This sequence complements the direct hydration of 1-alkenes, which gives alcohols: Hydroboration of an alkene and subsequent reactions of the product trialkylborane, either with hydrogen peroxide or with acid, appear to be highly stereospecific. For example, 1-methylcyclopentene gives exclusively -2-methylcyclopentanol on hydroboration followed by reaction with alkaline hydrogen peroxide. This indicates that, overall, : Hydroboration of an alkyne followed by treatment of the alkenylborane with basic peroxide provides a method of synthesis of aldehydes and ketones. Thus hydroboration of 1-hexyne and oxidation of the 1-hexenylborane, \(4\), with hydrogen peroxide gives hexanal by way of the enol: If \(4\) is treated with deuteriopropanoic acid, replacement of \(\ce{-BR_2}\) by deuterium occurs with of configuration, forming -hexene-1-\(\ce{D_1}\): The stereospecific oxidation of alkylboranes occurs with hydrogen peroxide by an interesting and important general type of rearrangement which, for these reactions, involves migration of an organic group from boron to oxygen. The first step in the oxidation depends on the fact that tricoordinate boron has only six electrons in its valence shell and therefore behaves as if it were electron-deficient. The first step is bond formation at boron by the strongly nucleophilic peroxide anion (from \(\ce{H_2O_2} + \ce{OH}^\ominus \rightleftharpoons ^\ominus \ce{OOH} + \ce{H_2O}\)) to give a tetracovalent boron intermediate: In the second step, an alkyl group moves from boron to the neighboring oxygen and, in so doing, displaces hydroxide ion. Reaction is completed by hydrolysis of the \(\ce{B-O}\) bond: All three groups on boron are replaced in this manner. The rearrangement step (Equation 11-5) is an example of many related rearrangements in which a group, \(\ce{R}\), migrates with its bonding electrons from one atom to an adjacent atom. We already have encountered an example in the rearrangement of carbocations ( ): The difference between the carbocation rearrangement and the rearrangement of Equation 11-5 is that \(\ce{R}\) migrates from boron to oxygen as \(\ce{HO}^\ominus\) departs in what might be considered an internal \(S_\text{N}2\) reaction. We can generalize this kind of reaction of boron with a substance, \(\ce{X-Y}\), as in Equation 11-6: An example of the use of an \(\ce{X-Y}\) reagent is conversion of alkylboranes to primary amines with hydroxylaminesulfonic acid, \(\ce{H_2NOSO_3H}\) (Equation 11-4). The key steps are attack of the nucleophilic nitrogen at boron, followed by rearrangement, and hydrolysis, and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.06%3A_Addition_of_Boron_Hydrides_to_Alkenes._Organoboranes

An especially valuable group of intermediates can be prepared by addition of an compound to carbon-carbon double or triple bonds: The reaction is called and is a versatile synthesis of organoboron compounds. One example is the addition of diborane, \(\ce{B_2H_6}\), to ethene. Diborane behaves as though it is in equilibrium with \(\ce{BH_3}\) \(\left( \ce{B_2H_6} \rightleftharpoons 2 \ce{BH_3} \right)\), and addition proceeds in three stages: The monoalkylborane, \(\ce{RBH_2}\), and the dialkylborane, \(\ce{R_2BH}\), seldom are isolated because they rapidly add to the alkene. These additions amount to reduction of both carbons of the double bond: Organoboranes can be considered to be organometallic compounds. Elemental boron does not have the properties of a metal, and boron-carbon bonds are more covalent than ionic. However, boron is more electropositive than either carbon or hydrogen and when bonded to carbon behaves like most metals in the sense that bonds are polarized with \(\ce{R}\) negative and boron positive: Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes. The simplest borane, \(\ce{BH_3}\), exists as the dimer, \(\ce{B_2H_6}\), or in complexed form with certain ethers or sulfides: Any of these \(\ce{BH_3}\) compounds adds readily to most alkenes at room temperature or lower temperatures. The reactions usually are carried out in ether solvents, although hydrocarbon solvents can be used with the borane-dimethyl sulfide complex. When diborane is the reagent, it can be generated either or externally through the reaction of boron trifluoride with sodium borohydride: \[3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{H_4} + 4 \ce{BF_3} \rightarrow 2 \ce{B_2H_6} + 3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{F_4}\] Hydroborations have to be carried out with some care, because diborane and many alkylboranes are highly reactive and toxic substances; many are spontaneously flammable in air. With unsymmetrical alkenes, hydroboration occurs so that : These additions are : Furthermore, when there is a choice, addition occurs preferentially from the less crowded side of the double bond: If the alkene is a bulky molecule, borane may add only one or two alkene molecules to give either mono- or dialkylborane, \(\ce{RBH_2}\) or \(\ce{R_2BH}\), respectively, as the following reactions show: These bulky boranes still possess \(\ce{B-H}\) bonds and can add further to a multiple bond, but they are highly selective reagents and add only if the alkene or alkyne is unhindered. This selectivity can be useful, particularly to 1-alkynes, which are difficult to stop at the alkenylborane stage when using diborane: With a bulky dialkylborane, such as di-(1,2-dimethylpropyl)borane, further addition to the alkenylborane does not occur. An especially selective hydroborating reagent is prepared from 1,5-cyclooctadiene and borane. The product is a bicyclic compound of structure \(1\) (often abbreviated as 9-BBN), in which the residual \(\ce{B-H}\) bond adds to unhindered alkenes with much greater selectivity than is observed with other hydroborating reagents. It is also one of the few boranes that reacts sufficiently slowly with oxygen that it can be manipulated in air. An example of the difference in selectivity in the hydroboration of -4-methyl-2-pentene with \(\ce{B_2H_6}\) and \(1\) follows: According to the electronegativity chart (Figure 10-11), the boron-hydrogen bond is polarized in the sense \(\overset{\delta \oplus}{\ce{B}} --- \overset{\delta \ominus}{\ce{H}}\). Therefore the direction of addition of \(\ce{B_2H_6}\) to propene is that expected of a mechanism whereby the electrophilic boron atom becomes bonded to the less-substituted carbon of the double bond. However, there is no firm evidence to suggest that a carbocation intermediate is formed through a stepwise electrophilic addition reaction. For this reason, the reaction often is considered to be a . The stepwise formulation explains why boron becomes attached to the less-substituted carbon, but does not account for the fact that the reactions show no other characteristics of carbocation reactions. This could be because of an expected, extraordinarily fast rate of hydride-ion transfer to the carbocation. A more serious objection to the stepwise mechanism is that alkynes react more rapidly than alkenes, something which normally is not observed for stepwise electrophilic additions (cf. ). Some alkylboranes rearrange at elevated temperatures \(\left( 160^\text{o} \right)\) to form more stable isomers. For example, the alkylborane \(2\), produced by hydroboration of 3-ethyl-2-pentene, rearranges to \(3\) on heating: In general, the boron in alkylboranes prefers to be at the of a hydrocarbon chain so it is bonded to a carbon where steric crowding around boron is least severe. Thus rearrangement tends to proceed in the direction Rearrangement is associated with the fact that hydroboration is reversible at elevated temperatures. This makes possible a sequence of elimination-addition reactions in which boron becomes attached to different carbons and ultimately leads to the most stable product that has boron bonded to the carbon at the end of the chain: Rearrangement of alkylboranes can be used to transform alkenes with double bonds in the middle of the chain into 1-alkenes; for example, \(\ce{RCH=CHCH_3} \rightarrow \ce{RCH_2-CH=CH_2}\). The procedure involves hydroboration of the starting alkene in the usual manner; the borane then is isomerized by heating. An excess of 1-decene (bp \(170^\text{o}\)) then is added to the rearranged borane and the mixture is reheated. Heating causes the alkylborane to dissociate into 1-alkene and \(\ce{HBR_2}\); the 1-decene "scavenges" the \(\ce{HBR_2}\) as it forms, thereby allowing a more volatile 1-alkene (bp \(<170^\text{o}\)) to be removed by simple distillation. Thus, for the rearrangement of 3-ethyl-2-pentene to 3-ethyl-1-pentene, Alkylboranes formed in the hydroboration of alkenes and alkynes seldom are isolated; for the most part they are used as reactive intermediates for the synthesis of other substances. In the reactions of alkylboranes, the \(\ce{B-C}\) bond is cleaved in the sense \(\ce{B}^\oplus - \ce{C}^\ominus\) so that carbon is transferred to other atoms, such as \(\ce{H}\), \(\ce{O}\), \(\ce{N}\), and \(\ce{C}\), its bonding electron pair: In the first of these reactions (Equation 11-2), a hydrocarbon is produced by the cleavage of a borane, \(\ce{R_3B}\), with aqueous acid, or better, with anhydrous propanoic acid, \(\ce{CH_3CH_2CO_2H}\). The overall sequence of hydroboration-acid hydrolysis achieves the reduction of a carbon-carbon multiple bond without using hydrogen and a metal catalyst or diimide (Table 11-3): The second reaction (Equation 11-3) achieves the synthesis of a alcohol by the oxidation of the alkylborane with hydrogen peroxide in basic solution. Starting with a 1-alkene, one can prepare a primary alcohol in two steps: This sequence complements the direct hydration of 1-alkenes, which gives alcohols: Hydroboration of an alkene and subsequent reactions of the product trialkylborane, either with hydrogen peroxide or with acid, appear to be highly stereospecific. For example, 1-methylcyclopentene gives exclusively -2-methylcyclopentanol on hydroboration followed by reaction with alkaline hydrogen peroxide. This indicates that, overall, : Hydroboration of an alkyne followed by treatment of the alkenylborane with basic peroxide provides a method of synthesis of aldehydes and ketones. Thus hydroboration of 1-hexyne and oxidation of the 1-hexenylborane, \(4\), with hydrogen peroxide gives hexanal by way of the enol: If \(4\) is treated with deuteriopropanoic acid, replacement of \(\ce{-BR_2}\) by deuterium occurs with of configuration, forming -hexene-1-\(\ce{D_1}\): The stereospecific oxidation of alkylboranes occurs with hydrogen peroxide by an interesting and important general type of rearrangement which, for these reactions, involves migration of an organic group from boron to oxygen. The first step in the oxidation depends on the fact that tricoordinate boron has only six electrons in its valence shell and therefore behaves as if it were electron-deficient. The first step is bond formation at boron by the strongly nucleophilic peroxide anion (from \(\ce{H_2O_2} + \ce{OH}^\ominus \rightleftharpoons ^\ominus \ce{OOH} + \ce{H_2O}\)) to give a tetracovalent boron intermediate: In the second step, an alkyl group moves from boron to the neighboring oxygen and, in so doing, displaces hydroxide ion. Reaction is completed by hydrolysis of the \(\ce{B-O}\) bond: All three groups on boron are replaced in this manner. The rearrangement step (Equation 11-5) is an example of many related rearrangements in which a group, \(\ce{R}\), migrates with its bonding electrons from one atom to an adjacent atom. We already have encountered an example in the rearrangement of carbocations ( ): The difference between the carbocation rearrangement and the rearrangement of Equation 11-5 is that \(\ce{R}\) migrates from boron to oxygen as \(\ce{HO}^\ominus\) departs in what might be considered an internal \(S_\text{N}2\) reaction. We can generalize this kind of reaction of boron with a substance, \(\ce{X-Y}\), as in Equation 11-6: An example of the use of an \(\ce{X-Y}\) reagent is conversion of alkylboranes to primary amines with hydroxylaminesulfonic acid, \(\ce{H_2NOSO_3H}\) (Equation 11-4). The key steps are attack of the nucleophilic nitrogen at boron, followed by rearrangement, and hydrolysis, and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/16%3A_Carbonyl_Compounds_I-_Aldehydes_and_Ketones._Addition_Reactions_of_the_Carbonyl_Group/16.10%3A_Preparative_Methods_for_Aldehydes_and_Ketones

A number of useful reactions for the preparation of aldehydes and ketones, such as ozonization of alkenes and hydration of alkynes, have been considered in previous chapters. These and other methods of preparation are summarized in Tables 16-7 and 16-8 at the end of the chapter. Only a few rather general methods that we have not discussed will be taken up here. Aldehydes and ketones often can be prepared by oxidation of alkenes to 1,2-diols (Sections 11-7C and 11-7D), followed by oxidative cleavage of the 1,2-diols with lead tetraethanoate or sodium periodate. For example, Cleavage of glycols with these reagents proceeds according to the following stoichiometry: In Chapter 15 primary alcohols, \(\ce{RCH_2OH}\), were shown to be readily oxidized to aldehydes, \(\ce{RCHO}\), and secondary alcohols, \(\ce{R_2CHOH}\), to ketones, \(\ce{R_2CO}\), by inorganic reagents such as \(\ce{CrO_3}\) and \(\ce{KMnO_4}\). However, it is a problem to avoid overoxidation with primary alcohols because of the ease with which aldehydes are oxidized to acids, \(\ce{RCHO} \rightarrow \ce{RCO_2H}\). A milder oxidant is methylsulfinylmethane [dimethyl sulfoxide, \(\ce{(CH_3)_2S=O}\)], and this reagent can be used to prepare aldehydes from alcohols by way of an intermediate such as the ester or halide in which the \(\ce{OH}\) group is converted to a better leaving group: Whichever method is employed, the key step is the formation of an alkoxysulfonium salt, \(7\), by a displacement reaction involving dimethyl sulfoxide as an oxygen nucleophile. (Notice that the \(\ce{S=O}\) bond, like the \(\ce{C=O}\) bonds, is strongly polarized as \(\overset{\oplus}{\ce{S}} \ce{-} \overset{\ominus}{\ce{O}}\).) In the examples listed in Equations 16-12 through 16-15, the \(\ce{X}\) group is \(\ce{Br}\), \(\ce{-OSO_2R'}\), \(\ce{-O_2CCF_3}\), and , respectively. In the nest step a sulfur ylide, \(8\), is formed from the reaction of a base with \(7\), but the ylide evidently is unstable and fragments by an internal \(E2\) reaction to form an aldehyde: Conversion of a carboxylic acid to an aldehyde by direct reduction is not easy to achieve, because acids generally are difficult to reduce, whereas aldehydes are easily reduced. Thus the problem is to keep the reaction from going too far. The most useful procedures involve conversion of the acid to a derivative that either is more easily reduced than an aldehyde, or else is reduced to a substance from which the aldehyde can be generated. The so-called involves the first of these schemes; in this procedure, the acid is converted to an acyl chloride, which is reduced with hydrogen over a palladium catalyst to the aldehyde in yields up to \(90\%\). The rate of reduction of the aldehyde to the corresponding alcohol is kept at a low level by poisoning the catalyst with suflur: Metal hydrides, such as lithium aluminum hydride, also can be used to reduce derivatives of carboxylic acids (such as amids and nitriles see Table 16-6) to aldehydes. An example follows: Many carbonyl compounds can be synthesized by acid-catalyzed rearrangements of 1,2-diols (a type of reaction often called the "pinacol-pinacolone" rearrangement). The general characteristics of the reaction are similar to those of carbocation rearrangements (Section 8-9B). The acid assists the reaction by protonating one of the \(\ce{-OH}\) groups to make it a better leaving group. The carbocation that results then can undergo rearrangement by shift of the neighboring \(\ce{R}\) group with its pair of bonding electrions to give a new, thermodynamically more stable species with a carbon-oxygen double bond (see Section 16-7). The prototype of this rearrangement is the conversion of pinacol to pinacolone as follows: An important method of preparing carbonyl (and hydroxy) compounds, especially on an industrial scale, is through rearrangements of alkyl hydroperoxides: The peroxides can be made in some cases by direct air oxidation of hydrocarbons, and in others by sulfuric acid-induced addition of hydrogen peroxide (as \(\ce{H-O_2H}\)) to double bonds: (Notice that hydrogen peroxide in methanoic acid behaves differently toward alkenes in producing addition of \(\ce{HO-OH}\), Section 11-7D.) The direct air oxidation of hydrocarbons is mechanistically similar to that of benzenecarbaldehyde (Section 16-7). The rearrangements of hydroperoxides are acid-catalyzed and are analogous to carbocation rearrangements except that positive oxygen (with only valence electrons) instead of positive carbon is involved in the intermediate stage: In principle, either phenyl or methyl could migrate to the positive oxygen, but only phenyl migration occurs in this case. The rearrangement reaction is closely related to the Baeyer-Villiger reaction (Section 16-7). This reaction is important for a number of reasons. It is an industrial synthesis of aldehydes from alkenes by the addition of carbon monoxide and hydrogen in the presence of a cobalt catalyst. A prime example is the synthesis of butanal from propene, in which 2-methylpropanal also is formed: As you can see, the reaction formally amounts to the addition of methanal as \(\ce{H-CHO}\) to the alkene double bond. Because one additional carbon atom is introduced as a "formyl" \(\ce{CHO}\) group, the reaction often is called , although the older name, , is widely used. Hydroformylation to produce aldehydes is the first step in an important industrial route to alcohols. The intermediate aldehydes are reduced to alcohols by catalytic hydrogenation. Large quantities of \(\ce{C_4}\)-\(\ce{C_8}\) alcohols are prepared by this sequence: The history of the oxo reaction is also noteworthy. It was developed originally in Germany in the years following World War I. At that time, the German chemical industry was faced with inadequate supplies of petroleum. Many German chemists therefore turned to research on ways by which hydrocarbons could be synthesized from smaller building blocks, particularly carbon monoxide and hydrogen derived from coal. The success achieved was remarkable and led to alkane and alkene syntheses known as the : ​​​​​​ This reaction in turn led to the discovery that aldehydes were formed by the further addition of carbon monoxide and hydrogen to alkenes, and was further developed as the oxo process for production of alcohols. The combination \(\ce{CO} + \ce{H_2}\) often is called "synthetic gas". It is prepared by the reduction of water under pressure and at elevated temperatures by carbon (usually coke), methane, or higher-molecular-weight hydrocarbons: The aldehyde synthesis by hydroformylation of alkenes described in the preceding section can be achieved indirectly using boron hydrides. An oversimplified expression of this reaction is The overall reaction is quite complex but involves a rearrangement similar to that described for the hydroboration-oxidation of alkenes (Section 11-6E). The first step is hydroboration of the alkene to a trialkylborane. When the trialkylborane is exposed to carbon monoxide, it reacts (carbonylates) to form a tetracovalent boron, \(9\): The complex \(9\) is unstable and rearranges by transfer of an alkyl group from boron to the electron-deficient carbonyl carbon to give \(10\): Now, if a metal-hydride reducing agent, such as \(\ce{LiAlH_4}\), is present, the carbonyl group of \(10\) is reduced and \(11\) is formed: The reduction product, \(11\), can be converted to an aldehyde by oxidation with aqueous hydrogen peroxide, provided the pH is carefully controlled. (Remember, aldehydes are unstable in strong base.) You may have noticed that only of the three alkyl groups of a trialkylborane is converted to an aldehyde by the carbonylation-reduction-oxidation sequence. To ensure that carbonylation takes the desired course without wasting the starting alkene, hydroboration is achieved conveniently with a hindered borane, such as "9-BBN", \(12\). With \(12\), only the least-hindered alkyl group rearranges in the carbonylation step: Carbonylation of alkylboranes also can produce ketones. The conditions are similar to those in the aldehyde synthesis except that the hydride reducing agent is omitted. By omitting the reducing agent, a second boron-to-carbon rearrangement can occur. Oxidation then produces a ketone: Rearrangement will continue a third time (ultimately to produce a tertiary alcohol) unless movement of the alkyl group remaining on boron in \(13\) is prevented by steric hindrance. and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.05%3A_Chemical_Reactions_of_Alcohols._Reactions_Involving_the_O-H_Bond

Several important chemical reactions of alcohols involve only the oxygen-hydrogen bond and leave the carbon-oxygen bond intact. An important example is salt formation with acids and bases. Alcohols, like water, are both weak bases and weak acids. The acid ionization constant \(\left( K_\text{a} \right)\) of ethanol is about \(10^{-18}\), slightly less than that of water. Ethanol can be converted to its conjugate base by the conjugate base of a weaker acid such as ammonia \(\left( K_\text{a} \sim 10^{-35} \right)\), or hydrogen \(\left( K_\text{a} \sim 10^{-38} \right)\). It is convenient to employ sodium metal or sodium hydride, which react vigorously but controllably with alcohols: The order of acidity of various liquid alcohols generally is water \(>\) \(>\) |(>\) \(\ce{ROH}\). By this we mean that the equilibrium position for the proton-transfer reaction (Equation 15-1) lies more on the side of \(\ce{ROH}\) and \(\ce{OH}^\ominus\) as \(\ce{R}\) is changed from to to ; therefore, -butyl alcohol is considered less acidic than ethanol: \[\ce{ROH} + \ce{OH}^\ominus \rightleftharpoons \ce{RO}^\ominus + \ce{HOH} \tag{15-1}\] However, in the gas phase the order of acidity is reversed, and the equilibrium position for Equation 15-1 lies increasingly on the side of \(\ce{RO}^\ominus\) as \(\ce{R}\) is changed from to to . -Butyl alcohol is therefore acidic than ethanol in the gas phase. This seeming contradiction appears more reasonable when one considers what effect solvation (or the lack of it) has on equilibria expressed by Equation 15-1. In solution, the larger anions of alcohols, known as , probably are less well solvated than the smaller ions, because fewer solvent molecules can be accommodated around the negatively charged oxygen in the larger ions: Acidity of alcohols therefore decreases as the size of the conjugate base increases. However, “naked” gaseous ions are more stable the larger the associated \(\ce{R}\) groups, probably because the larger \(\ce{R}\) groups can stabilize the charge on the oxygen atom better than the smaller \(\ce{R}\) groups. They do this by polarization of their bonding electrons, and the bigger the group, the more polarizable it is. (Also see , which deals with the somewhat similar situation encountered with respect to the relative acidities of ethyne and water.) Alcohols are bases similar in strength to water and accept protons from strong acids. An example is the reaction of methanol with hydrogen bromide to give methyloxonium bromide, which is analogous to the formation of hydroxonium bromide with hydrogen bromide and water: Alkoxide ion formation is important as a means of generating a strong nucleophile that will readily form \(\ce{C-O}\) bonds in \(S_\text{N}2\) reactions. Thus ethanol reacts very slowly with methyl iodide to give methyl ethyl ether, but sodium ethoxide in ethanol solution reacts quite rapidly: In fact, the reaction of alkoxides with alkyl halides or alkyl sulfates is an important general method for the preparation of ethers, and is known as the . Complications can occur because the increase of nucleophilicity associated with the conversion of an alcohol to an alkoxide ion always is accompanied by an even greater increase in eliminating power by the \(E2\) mechanism. The reaction of an alkyl halide with alkoxide then may be one of elimination rather than substitution, depending on the temperature, the structure of the halide, and the alkoxide ( ). For example, if we wish to prepare isopropyl methyl ether, better yields would be obtained if we were to use methyl iodide and isopropoxide ion rather than isopropyl iodide and methoxide ion because of the prevalence of \(E2\) elimination with the latter combination: Potassium -butoxide is an excellent reagent to achieve \(E2\) elimination because it is strongly basic and so bulky as to not undergo \(S_\text{N}2\) reactions readily. An ester may be thought of as a carboxylic acid in which the acidic proton has been replaced by some organic group, \(\ce{R}\), Esters can be prepared from carboxylic acids and alcohols provided an acidic catalyst is present, or they can be prepared from acyl halides and alcohols or carboxylic anhydrides and alcohols: These reactions generally can be expressed by the equation \(+ \ce{ROH} \rightarrow\) \(+ \ce{HX}\) which overall is a nucleophilic displacement of the \(\ce{X}\) group by the nucleophile \(\ce{ROH}\). However, the mechanism of displacement is quite different from the \(S_\text{N}2\) displacements of alkyl derivatives, \(\ce{R'X} + \ce{ROH} \rightarrow \ce{R'OR} + \ce{HX}\), and closely resembles the nucleophilic displacements of activated aryl halides ( ) in being an process. Acyl halides have a rather positive carbonyl carbon because of the polarization of the carbon-oxygen and carbon-halogen bonds. Addition of a nucleophilic group such as the oxygen of an alcohol occurs rather easily. The complex \(1\) contains both an acidic group and a basic group , so that a proton shifts from one oxygen to the other to give \(2\), which then rapidly loses hydrogen chloride by either an \(E1\)- or \(E2\)-type elimination to form the ester. A similar but easily reversible reaction occurs between alcohols and carboxylic acids, which is slow in either direction in the absence of a strong mineral acid. The catalytic effect of acids, such as \(\ce{H_2SO_4}\), \(\ce{HCl}\), and \(\ce{H_3PO_4}\) is produced by protonation of the carbonyl oxygen of the carboxylic acid, thereby giving \(3\). This protonation greatly enhances the affinity of the carbonyl carbon for an electron pair on the oxygen of the alcohol (i.e., \(3 \rightarrow 4\)). Subsequently, a proton is transferred from the \(\ce{OCH_3}\) to an \(\ce{OH}\) group of \(4\) to give \(5\). This process converts the \(\ce{OH}\) into a good leaving group \(\left( \ce{H_2O} \right)\). When \(\ce{H_2O}\) leaves, the product, \(6\), is the conjugate acid of the ester. Transfer of a proton from \(6\) to a base such as \(\ce{H_2O}\) or \(\ce{HSO_4^-}\) completes the reaction, giving the neutral ester and regenerating the acid catalyst. Although a small amount of strong acid catalyst is essential in the preparation of esters from acids and alcohols, the amount of acid catalyst added must not be too large. The reason for the “too much of a good thing” behavior of the catalyst can be understood from the basic properties of alcohols ( ). If too much acid is present, then too much of the alcohol is converted to the oxonium salt: Clearly, formation of the methyloxonium ion can operate only to the nucleophilic reactivity of methanol toward the carbonyl carbon of the carboxylic acid. Another practical limitation of esterification reactions is . If either the acid or the alcohol participants possesses highly branched groups, the positions of equilibrium are less favorable and the rates of esterification are slow. In general, the ease of esterification for alcohols, \(\ce{ROH}\), by the mechanism described is \(\ce{R}\) \(>\) \(\ce{R}\) \(>\) \(\ce{R}\) with a given carboxylic acid. As mentioned, esterification is reversible, and with ethanol and ethanoic acid the equilibrium constant for the liquid phase is about 4 \(\left( \Delta G^0 = -0.8 \: \text{kcal} \right)\) at room temperature, which corresponds to \(66\%\) conversion to ester: The reaction may be driven to completion by removing the ester or water or both as they are formed. The structural unit, possesses both an alkoxyl \(\left( \ce{OR} \right)\) and a hydroxyl \(\left( \ce{OH} \right)\) group on the carbon. This arrangement, although often unstable, is an important feature of carbohydrates such as glucose, fructose, and ribose. When the grouping is of the type , it is called a , and if it is , with no hydrogen attached to the carbon, it is called a : The has two alkoxy \(\left( \ce{OR} \right)\) groups and a hydrogen on the same carbon, , whereas the ketal function has the same structure but with no hydrogen on the carbon. These groupings also are found in carbohydrates and in carbohydrate derivatives, and are called (see ). For our present purposes, we are interested in the ways in which hemiacetals, acetals, hemiketals, and ketals are formed. Hemiacetals and hemiketals can be regarded as products of the addition of alcohols to the carbonyl groups of aldehydes and ketones. Thus methanol adds to ethanal to give a hemiacetal, 1 -methoxyethanol: Acetals and ketals result from substitution of an alkoxy group for the \(\ce{OH}\) group of a hemiacetal or hemiketal. Thus methanol can react with 1-methoxyethanol to form the acetal, 1,1-dimethoxyethane, and water: The reactions of alcohols with aldehydes and ketones are related to the reactions of alcohols with acids (esterification) discussed in the preceding section. Both types involve addition of alcohols to carbonyl groups, and both are acid-catalyzed. Acid catalysis of formation, like ester formation, depends on formation of the conjugate acid of the carbonyl compound. This is expected to enhance the positive (electrophilic) character of the carbonyl carbon so that the nucleophilic alcohol can add readily to it: The hemiacetal can react further, also with the aid of an acidic catalyst. Addition of a proton can occur in two ways, to give \(7\) or \(8\): The first of these, \(7\), has \(\ce{CH_3OH}\) as a leaving group and reverts back to the conjugate acid of ethanal. This is the reverse of acid-catalyzed hemiacetal formation: The second of these,\(8\), has \(\ce{H_2O}\) as a leaving group and can form a new entity, the methoxyethyl cation, \(9\): The ion \(9\) resembles and can be expected to behave similarly by adding a second molecule of alcohol to the electrophilic carbon. The product, \(10\), is then the conjugate acid of the acetal and loses a proton to give the acetal: Formation of hemiacetals and acetals, as well as of hemiketals and ketals, is reversible under acidic conditions, as we already have noted for acid-catalyzed esterification. The reverse reaction is and the equilibrium for this reaction can be made favorable by having an excess of water present: The position of equilibrium in acetal and hemiacetal formation is rather sensitive to steric hindrance. Large groups in either the aldehyde or the alcohol tend to make the reaction less favorable. Table 15-3 shows some typical conversions in acetal formation when 1 mole of aldehyde is allowed to come to equilibrium with 5 moles of alcohol. For ketones, the equilibria are still less favorable than for aldehydes, and to obtain reasonable conversion the water must be removed as it is formed. and (1977)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/0Group_17%3A_Physical_Properties_of_the_Halogens/Atomic_and_Physical_Properties_of_Halogens

This page discusses the trends in the atomic and physical properties of the Group 7 elements (the halogens): fluorine, chlorine, bromine and iodine. Sections below cover the trends in atomic radius, electronegativity, electron affinity, melting and boiling points, and solubility, including a discussion of the bond enthalpies of halogen-halogen and hydrogen-halogen bonds. The figure above shows the increase in atomic radius down the group. The radius of an atom is determined by: Compare the numbers of electrons in each layer of fluorine and chlorine: In each case, the outer electrons feel a net +7 charge from the nucleus. The positive charge on the nucleus is neutralized by the negative inner electrons. This is true for all the atoms in Group 7: the outer electrons experience a net charge of +7.. The only factor affecting the size of the atom is therefore the number of layers of inner electrons surrounding the atom. More layers take up more space due to electron repulsion, so atoms increase in size down the group. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is usually measured on the Pauling scale, on which the most electronegative element (fluorine) is assigned an electronegativity of 4.0. The figure below shows electronegativities for each halogen: Notice that electronegativity decreases down the group. The atoms become less effective at attracting bonding pairs of electrons. This effect is illustrated below using simple dots-and-crosses diagrams for hydrogen fluoride and hydrogen chloride: The bonding pair of electrons between the hydrogen and the halogen experiences the same net pull of +7 from both the fluorine and the chlorine. However, in the chlorine case, the nucleus is farther away from the bonding electrons, which are therefore not as strongly attracted as in the fluorine case. The stronger attraction from the closer fluorine nucleus makes fluorine more electronegative than chlorine. As the halogen atoms increase in size, any bonding pair gets farther away from the halogen nucleus, and so is less strongly attracted toward it. Hence, down the group, the elements become less electronegative. The first electron affinity is the energy released when 1 mole of gaseous atoms each acquire an electron to form 1 mole of gaseous 1- ions. In other words, it is the energy released in the following process: \[ X(g) + e^- \rightarrow X^- (g)\] First electron affinities have negative values by convention. For example, the first electron affinity of chlorine is -349 kJ mol . The negative sign indicates a release of energy. The electron affinity is a measure of the attraction between the incoming electron and the nucleus. There is a positive correlation between attraction and electron affinity. The trend down the group is illustrated below: Notice that the trend down the group is inconsistent. The electron affinities generally decrease (meaning less heat is emitted), but the fluorine value deviates from this trend. In the larger atom, the attraction from the more positive nucleus is offset by the additional screening electrons, so each incoming electron feels the effect of a net +7 charge from the center. As the atom increases in size, the incoming electron is farther from the nucleus and so feels less attraction. The electron affinity therefore decreases down the group. However fluorine is a very small atom, with the incoming electron relatively close to the nucleus, and yet the electron affinity is smaller than expected. Another effect must be considered in the case of fluorine. As the new electron comes approaches the atom, it enters a region of space already very negatively charged because of the existing electrons. The resulting repulsion from these electrons offsets some of the attraction from the nucleus. Because the fluorine atom is very small, its existing electron density is very high. Therefore, the extra repulsion is particularly great and diminishes the attraction from the nucleus enough to lower the electron affinity below that of chlorine. Melting and boiling points increase down the group. As indicated by the graph above, fluorine and chlorine are gases at room temperature, bromine is a liquid and iodine a solid. All the halogens exist as diatomic molecules—F , Cl , and so on. van der Waals dispersion forces are the primary intermolecular attractions between one molecule and its neighbors. Larger molecules farther down the group have more electrons which can move around and form the temporary dipoles that create these forces. The stronger intermolecular attractions down the group require more heat energy for melting or vaporizing, increasing their melting or boiling points. Fluorine reacts violently with water to produce aqueous or gaseous hydrogen fluoride and a mixture of oxygen and ozone; its solubility is meaningless. Chlorine, bromine and iodine all dissolve in water to some extent, but there is again no discernible pattern. The following table shows the solubility of the three elements in water at 25°C: Chlorine dissolved in water produces a pale green solution. Bromine solution adopts a range of colors from yellow to dark orange-red depending on the concentration. Iodine solution in water is very pale brown. Chlorine reacts with water to some extent, producing a mixture of hydrochloric acid and chloric(I) acid (also known as hypochlorous acid). The reaction is reversible, and at any time only about a third of the chlorine molecules have reacted. \[ Cl_2 + H_2O \rightleftharpoons HCl + HClO\] Chloric(I) acid is sometimes symbolized as HOCl, indicating the actual pattern bonding pattern. Bromine and iodine form similar compounds, but to a lesser extent. In both cases, about 99.5% of the halogen remains unreacted. Although iodine is only slightly soluble in water, it dissolves freely in potassium iodide solution, forming a dark red-brown solution. A reversible reaction between iodine molecules and iodide ions gives I ions. These are responsible for the color. In the laboratory, iodine is often produced through oxidation of iodide ions. As long as there are any excess iodide ions present, the iodine reacts to form I . Once the iodide ions have all reacted, the iodine is precipitated as a dark gray solid. The halogens are much more soluble in organic solvents such as hexane than they are in water. Both hexane and the halogens are non-polar molecules, so the only intermolecular forces between them are van der Waals dispersion forces. Because of this, the attractions broken (between hexane molecules and between halogen molecules) are similar to the new attractions made when the two substances mix. Organic solutions of iodine are pink-purple in color. Bond enthalpy is the heat required to break one mole of covalent bonds to produce individual atoms, starting from the original substance in the gas state, and ending with gaseous atoms. For chlorine, Cl , it is the heat energy required for the following reaction, per mole: \[ Cl-Cl (g) \rightarrow 2Cl(g)\] Although bromine is a liquid, the bond enthalpy is defined in terms of gaseous bromine molecules and atoms, as shown below: Covalent bonding is effective because the bonding pair is attracted to both the nuclei at either side of it. It is that attraction which holds the molecule together. The extent of the attraction depends in part on the distances between the bonding pair and the two nuclei. The figure below illustrates such a covalent bond: In all halogens, the bonding pair experiences a net +7 charge from either end of the bond, because the charge on the nucleus is offset by the inner electrons. As the atoms get larger down the group, the bonding pair is further from the nuclei and the strength of the bond should, in theory, decrease, as indicated in the figure below. The question is whether experimental data matches this prediction. As is clear from the figure above, the bond enthalpies of the Cl-Cl, Br-Br and I-I bonds decreases as predicted, but the F-F bond enthalpy deviates. Because fluorine atoms are so small, a strong bond is expected—in fact, it is remarkably weak. There must be another factor for consideration. In addition to the bonding pair of electrons between the two atoms, each atom has 3 lone pairs of electrons in the outer shell. If the bond is very short,as in F-F, the lone pairs on the two atoms are close enough to cause significant repulsion, illustrated below: In the case of fluorine, this repulsion is great enough to counteract much of the attraction between the bonding pair and the two nuclei. This weakens the bond. If the halogen atom is attached to a hydrogen atom, this does not occur; there are no lone pairs on a hydrogen atom. Bond enthalpies for halogen-hydrogen bonds are given below: As larger halogens are involved, the bonding pair is more distant from the nucleus. The attraction is lessened, and the bond should be weaker; this is supported by the data, without exception. This fact has significant implications for the thermal stability of the hydrogen halides— they are easily broken into hydrogen and the halogen on heating. Hydrogen fluoride and hydrogen chloride are thermally very stable under typical laboratory conditions. Hydrogen bromide breaks down to some extent into hydrogen and bromine on heating, and hydrogen iodide is even less stable when heated. Weaker bonds are more easily broken. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Optical_Isomers_in_Inorganic_Complexes/Enantiomers_in_Octahedral_Complexes_with_Bidentate_Ligands

Enantiomers are another kind of isomer that occur in octahedral metal complexes. Like the square planar platinum complexes seen before, these compounds consist of metal ions with other atoms or groups bound to them. More information about the binding of ligands to metals is found in the section on . These groups that bind to metal ions are called ligands. One example of a type of octahedral compounds that can form enantiomers is bidentate complexes. In bidentate complexes (from the Greek two teeth) or chelating complexes (from the Greek crab), a ligand binds very tightly to the metal because it holds onto the metal via more than one atom. Ethylenediamine is one example of a bidentate ligand. Bidentate ligands bind very tightly to a metal because they form two bonds with it, rather than just one. In an octahedral complex, the two donor atoms in a bidentate ligand bind to each other. They reach all the way around the molecule to bind trans to each other. The spatial relationship between the metal and the two atoms connected to it from the same ligand forms a plane. If more than one bidentate ligand is connected to the metal, the relative orientation of one plane to another creates the possibility of mirror images. A complex containing three bidentate ligands can take on the shape of a left-handed propeller or a right-handed propeller. These shapes are alternatively described as a left-handed screw and a right handed screw. If you can picture turning the shape so that it screws into the page behind it, which direction would you turn the screwdriver? If you would twist the screwdriver clockwise, then you have a right handed screw. If you would twist the screwdriver counter-clockwise, then you have a left handed screw. What do we know about enantiomers? These kinds of complexes were historically important in demonstrating how small molecules and ions bound to metal cations. By showing that some metal complexes were chiral and displayed optical activity, early 20th century workers such as Alfred Werner were able to rule out some competing ideas about the structures of metal compounds. Today, we know that metal complexes play important roles in enzymes in biology, and Werner's work on metal complexes laid the groundwork for how we think about these complexes. In addition, stereochemistry in metal complexes became very important in the late 20th century, especially as pharmaceutical companies looked for catalysts that could aid in the production of one enantiomer of a drug, and not the other, in order to maximize pharmaceutical effectiveness and minimize side effects. Identify the relationships between the following pairs of iron(III) oxalate ions. Indicate whether each of the following compounds is the Δ or Λ isomer and draw its enantiomer.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.09%3A_General_Structural_Features_that_Modulate_Ligand_Activity

There are many ways in which ligand affinity may be perturbed (Figure 4.25). It is convenient to divide these into two groups, referred to as distal and proximal effects. effects are associated with the stereochemistry of the metalloporphyrinato moiety and the coordination of the axial base, and thus their influence on O and CO affinity is indirect. effects pertain to noncovalent interactions of the metal-porphyrinato skeleton and the sixth ligand (O , CO, etc.) with neighboring solvent molecules, with substituents, such as pickets or caps, on the porphyrin, and with the surrounding protein chain. The distal groups that hover over the O -binding site engender the most important distal effects. For convenience, the effects of crystal packing and the protein matrix on porphyrin conformation will be discussed among the proximal effects, although as nonbonded interactions they properly are distal effects. To a first approximation, the effects of substituents on the porphyrin ring, as transmitted through bonds to the metal center, do not perturb the ligand binding properties as much as do distal effects. Thus substituents, such as vinyl and propionic-acid groups on protoporphyrin IX and o-pivalamidophenyl pickets, are ignored; one porphyrin is much like another. At the end of this subsection the various ways ligand affinity may be modulated will be summarized in an augmented version of Figure 4.3. Few molecules have had their conformational properties characterized as exhaustively as have metalloporphyrins. The cyclic aromatic 24-atom porphyrinato skeleton offers a tightly constrained metal-binding site. The conformation of least strain is planar, and the radius of the hole of the dianion is close to 2.00 Å, leading to metal-porphyrinato nitrogen-atom separations, M-N , of 2.00 Å if the metal is centered in the square plane defined by the four porphyrinato nitrogen atoms. Small deviations from planarity are generally observed and attributed to crystal packing effects; large deviations may be induced by bulky substituents on the porphyrin skeleton, especially at the positions, by the crystal matrix, or by the highly anisotropic protein matrix. The 2.00 Å radius hole neatly accommodates low-spin (S = 0) and intermediatespin (S = 1) iron(II), low-spin (S = \(\frac{1}{2}\)) iron(III), and cobalt(II) and cobalt(III) ions. * With few exceptions the metal is centered in or above the central hole for mononuclear porphyrin species; only rarely do M-N bonds show a significant (though still small) scatter about their mean value. * In order to accommodate smaller ions, such as nickel(II), the porphyrin skeleton may contract by ruffling, with little loss of aromaticity; like a pleated skirt the pyrrole rings rotate alternately clockwise and counterclockwise about their respective M-N vectors. This distortion leaves the four porphyrinato nitrogen atoms, N , still coplanar. Alternatively, the porphyrin skeleton may buckle to give a saddle conformation; the N atoms may acquire a small tetrahedral distortion in this process. M-N bonds as short as 1.92 Å have been observed. Metals with one or two electrons in their 3d orbital have a radius larger than 2.00 Å. In order to accommodate them in the plane of the porphyrin, the porphyrin skeleton expands. M-N separations as long as 2.07 Å may occur with the metal still centered in the plane of the N atoms. For five-coordinate complexes the magnitude of the displacement of the metal from the plane of the four nitrogen atoms, M • • • porph, is a consequence of the electronic configuration of ML complexes. Of course, the effect is augmented if the 3d orbital (directed along M-N bonds, Figure 4.16) is occupied. Compare a displacement of 0.14 Å for Co(TPP)(1,2-Me Im) (no 3d occupancy) with 0.43 Å for Fe(PF)(2-MeIm) (3d occupied). For six-coordinate complexes where the two axial ligands, L and L , are different, the M • • • porph displacement usually reflects relative influences. Generally, displacement of the metal from the plane of the porphyrinatonitrogen atoms is within 0.04 Å of the displacement from the 24-atom mean plane of the entire porphyrin skeleton. On occasions this second displacement may be much larger, for example in Fe(TPP)(2-MeIm), where it is 0.15 Å larger than it is for Fe(PF)(2-MeIm). This effect is called , and it is usually attributed to crystal packing forces. Interaction of the porphyrin with protein side chains leads to considerable doming or folding of the heme in vertebrate hemoglobins. The metal-axial ligand separations, M-L (when more than one, L denotes the heterocyclic axial base), are dependent on the nature of the ligand, L. When L and L are different, the M-L separations are sensitive to the relative influences of L and L as well as to steric factors. For example, for Fe(TPP)(1-Melm) , the Fe—N bond length is 2.016(5) Å, whereas for Fe(TPP)(1-Melm)(NO) it is 2.180(4) Å. For sterically active ligands, such as 2-methylimidazole compared to 1-methylimidazole (4.34), the longer Co—N bond occurs for the 2-Melm ligand because of steric clash between the 2-methyl group and the porphyrin. It is possible that combinations of intrinsic bonding and steric factors may give rise to a double minimum and two accessible axial ligand conformations (Figure 4.26). This situation seems to occur in the solid state for Fe(PF)(2-Melm)(O )•EtOH, where a short Fe—N and a long Fe—O bond are observed both from the structure revealed by single-crystal x-ray diffraction methods and by EXAFS data. On the other hand, for solvate-free Fe(PF)(2-MeIm)(O ) and for Fe(PF)(1,2-Me Im)(O ), the EXAFS patterns are interpreted in terms of a short Fe—O and long Fe—Im bond. This parameter is the minimum angle that the plane of the axial base (e.g., pyridine, substituted imidazole, etc.) makes with a plane defined by the N , M, and L atoms (Figure 4.25). If there are two axial ligands, e.g., 1-methylimidazole and O , then, as before, the angle the axial base makes is denoted \(\phi_{1}\) and the other angle \(\phi_{2}\). For a linear CO ligand bound perpendicularly to the porphyrin plane, \(\phi_{2}\) is undefined. Note that the orientation of the second ligand is influenced by distal effects. When \(\phi\) = 0, the axial base eclipses a pair of M-N bonds; contacts with the porphyrin are maximized. When \(\phi\) = 45°, contacts are minimized. Unless the axial base has a 2-substituent, however, the contacts are not excessively close for any value of \(\phi\). With a 2-methyl substituent, the contacts are sufficiently severe that the M-N vector is no longer perpendicular to the porphyrin plane, and the imidazole group is rotated so that the M-N vector no longer approximately bisects the imidazole C—N—C bond angle, as illustrated Figure 4.25. Distal effects arise from noncovalent interactions of the coordinated dioxygen, carbon monoxide, or other ligand with its surroundings. The protein matrix, the pickets, and the caps are functionally equivalent to an anisotropic solvent matrix that contains a variety of solutes. The limits of this simplification are illustrated in the following example. The electronically similar cobalt meso-, deutero-, and protoporphyrin IX complexes bind dioxygen with similar affinities under identical solvent conditions. When they are embedded in globin, larger differences in affinity and changes in cooperativity are observed. These effects are attributed to the slightly different nestling of the porphyrin molecules in the cleft in hemoglobin or, in the generalization introduced, to slightly different solvation effects. Interaction of the coordinated O or CO molecule with solvent molecules or with the protein has a profound influence on kinetics and thermodynamics (see Figure 4.24, and Tables 4.2 and 4.5). As discussed earlier, there is accumulation of negative charge on the dioxygen ligand. The possibility then arises for stabilization of coordination through hydrogen bonding or dipolar interactions with solute molecules, porphyrin substituents (such as amide groups in the picket-fence porphyrins and some species of strapped porphyrins ), or with protein residues* (such as histidine). Destabilization of coordinated ligands and lowered affinity can result if the coordinated ligand is unable, through steric clash, to achieve its optimum stereochemistry or if the closest neighboring groups are electronegative, as are the ether and ester linkages on capped porphyrins. We will describe in detail in the next subsection (III.C) the fascinating variety of means by which ligand binding is modulated by distal amino-acid residues. * For CoMbO no change in EPR parameters occurs on substituting D O for H O. No hydrogen bond between O and a distal group comparable in strength to that in whale CoMbO was inferred. Dissimilar systems may show similar affinities for a ligand as a result of a different mix of the proximal and distal effects enumerated above. These effects are not all of equal magnitude, and an attempt is made here to show the increment in free energy that occurs if the effect is manifest in the deoxy or liganded state of Figure 4.3. Increasing the free energy of the deoxy state while holding that of the liganded state constant leads to an increase in affinity. The reference state is gaseous Fe(TPP)(1-MeIm). The magnitude and sign of these effects are shown in Figure 4.27. For the coordination of alkylisocyanide molecules to hemoglobin, the steric effects of different alkyl groups have been quantified. Lowered affinity occurs with increasing alkyl chain length, with the exception of methyl isocyanide.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.02%3A_Hydrogenation_with_Heterogeneous_Catalysts

Addition of hydrogen to a multiple bond is . It is applicable to almost all types of multiple bonds and is of great importance in synthetic chemistry, particularly in the chemical industry. Probably the most important technical example is production of ammonia by the hydrogenation of nitrogen: This may appear to be a simple process, but in fact it is difficult to carry out because the equilibrium is not very favorable. High pressures (\(150\)-\(200 \: \text{atm}\)) are required to get a reasonable conversion, and high temperatures (\(430\)-\(510^\text{o}\)) are necessary to get reasonable reaction rates. A catalyst, usually iron oxide, also is required. The reaction is very important because ammonia is used in ever-increasing amounts as a fertilizer either directly or through conversion to urea or ammonium salts. Production of ammonia requires large quantities of hydrogen, most of which comes from the partial oxidation of hydrocarbons with water or oxygen. A simple and important example is the so-called "methane-steam gas" reaction, which is favorable only at very high temperatures because of the entropy effect in the formation of \(\ce{H_2}\) (see ): Therefore the fertilizer industry is allied closely with the natural gas and petroleum industries, and for obvious reasons ammonia and hydrogen often are produced at the same locations., Alkenes and alkynes add hydrogen much more readily than does nitrogen. For example, ethene reacts rapidly and completely with hydrogen at ordinary pressures and temperatures in the presence of metal catalysts such as nickel, platinum, palladium, copper, and chromium: These reactions are unlike any we have encountered so far. They are reactions, which means that the reacting system consists of two or more phases. Usually, the metal catalyst is present as a finely divided solid suspension in the liquid or solution to be reduced. Alternatively, the metal is deposited on an inert solid support such as carbon, barium sulfate, alumina \(\left( \ce{Al_2O_3} \right)\), or calcium carbonate. Then the mixture of the liquid substrate and solid catalyst is shaken or stirred in a hydrogen atmosphere. However, then actual reaction takes place at the surface of the metal catalyst and is an example of or . The exact mechanisms of heterogeneous reactions are difficult to determine, but much interesting and helpful information has been obtained for catalytic hydrogenation. The metal catalyst is believed to act by binding the reactants at the surface of a crystal lattice. As an example, consider the surface of a nickel crystal (Figure 11-1). The nickel atoms at the surface have fewer neighbors (lower covalency) than the atoms in the interior of the crystal. The surface atoms therefore have residual bonding capacity and might be expected to combine with a variety of substances. It has been shown experimentally that ethene combines exothermically \(\left( \Delta H^0 = -60 \: \text{kcal/mol} \right)\) and reversibly with a metal surface. Although the precise structure of the ethene-nickel complex is unknown, the bonding to nickel must involve the electrons of the double bond because saturated hydrocarbons, such as ethane, combine only weakly with the nickel surface. A possible structure with carbon-nickel \(\sigma\) bonds is shown in Figure 11-1. Hydrogen gas combines with nickel quite readily with dissociation of the \(\ce{H-H}\) bonds and formation of \(\ce{Ni-H}\) bonds (nickel hydride bonds). The overall hydrogenation process is viewed as a series of reversible and sequential steps, as summarized in Figure 11-2. First the reactants, hydrogen and ethene, are adsorbed on the surface of the metal catalyst. The energies of the metal-hydrogen and metal-carbon bonds are such that, in a second step, a hydrogen is transferred to carbon to give an ethyl attached to nickel. This is the halfway point. In the next step, the nickel-carbon bond is broken and the second carbon-hydrogen bond is formed. Hydrogenation is now complete and the product is desorbed from the catalyst surface. Ethane has a low affinity for the metal surface and, when desorbed, creates a vacant space for the adsorption of new ethene and hydrogen molecules. The cycle continues until one of the reagents is consumed or some material is adsorbed that "poisons" the surface and makes it incapable of further catalytic activity. Because the reaction occurs only on the surface, small amounts of a catalyst poison can completely stop the reaction. As might be expected for the postulated mechanism, the spacings of the metal atoms in the crystal lattice are quite important in determining the hydrogenation rates. The mechanism also accounts for the observation that . To illustrate, 1,2-dimethylcyclohexene is reduced to -1,2-dimethylcyclohexane: For maximum catalytic activity, the metal usually is prepared in a finely divided state. This is achieved for platinum and palladium by reducing the metal oxides with hydrogen prior to hydrogenation of the alkene. A specially active form of nickel ("Raney nickel") is prepared from a nickel-aluminum alloy. Sodium hydroxide is added to the alloy to dissolve the aluminum. The nickel remains as a black powder which is pyrophoric (burns in air) if not kept moist: \[2 \ce{Ni-Al} + 2 \ce{OH}^\ominus + 2 \ce{H_2O} \rightarrow 2 \ce{Ni} + 2 \ce{AlO_2^-} + 3 \ce{H_2}\] Highly active platinum, palladium, and nickel catalysts also can be obtained by reduction of metal salts with sodium borohydride \(\left( \ce{NaBH_4} \right)\). As mentioned previously, multiple bonds are not hydrogenated with equal facility. This fact can be used to advantage in carrying out selective reactions. For instance, hydrogenation of a carbon-carbon double bond can be achieved without simultaneously reducing a carbonyl bond in the same molecule. For example the carbon-carbon double bond of the following aldehyde can be reduced selectively: Alkynes are hydrogenated more easily than alkenes mainly because alkynes are adsorbed more readily on the catalyst surface. Hydrogenation proceeds in stages, first to the -alkene and then to the alkane. For example, Normally, it is not possible to stop the hydrogenation of an alkyne at the alkene stage, but if the catalyst is suitably deactivated, addition to the triple bond can be achieved without further addition occurring to the resulting double bond. The preferred catalyst for selective hydrogenation of alkynes is palladium partially "poisoned" with a lead salt ( ). This catalyst shows little affinity for adsorbing alkenes and hence is ineffective in bringing about hydrogenation to the alkane stage: Aromatic hydrocarbons are hydrogenated with considerable difficulty, requiring higher temperatures, higher pressures, and longer reaction times than for alkenes or alkynes: and (1977)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_16%3A_The_Oxygen_Family_(The_Chalcogens)/Z008_Chemistry_of_Oxygen_(Z8)/Ozone/Ozone_Layer_and_Ozone_Hole

Contrary to common misconception, ozone is not in the form of thick layer surrounding atmosphere. Equally untrue is another misconception that a hole is made in this ozone layer. To understand low ozone spread in atmosphere, one needs to know the structure and composition of atmosphere. The distribution of ozone in ionosphere, mesosphere and stratosphere is being depleted. The concentration of ozone is gradually reducing. As the content of ozone is highest in ionosphere and the air itself being very thin, the depletion is negligible in ionosphere. But in mesosphere and stratosphere the air is thicker and ozone content is less. The depletion of ozone is of higher order in these layers. The so called hole in ozone layers simply means that above some continents (specifically Antarctica, Asia and parts of South America) the mesosphere and stratosphere have lost their original level of ozone content. The depletion of ozone layer is a global phenomena both in terms of cause and effect. The geographical limits of countries are not barriers to either dispersal of gases in layers of atmosphere or depletion of gases. The causes for depletion may arise in any country. The effects (in terms of depletion) may arise in any other country. The effects (in terms of ozone depletion) need not be exactly above the country causing the depletion. It is now established that chloroflouro carbon (CFC) chemicals evolved from various refrigerants, coolants and propellants are the primary reasons for depletion of ozone. CFC are a group of chlorine bearing gases of low specific gravity. They rise to stratosphere and mesosphere. Due to ionising solar radiation in these layers, (which is the primary reason for production of ozone) fresh chlorine gas is produced from CFCs. This nascent chlorine gas has the capacity to react with ozone and bring down the level of ozone substantially. The ozone in the ozone layer protects the earth from harmful ultraviolet radiation by trapping these radiations. This ozone layer has been thinning gradually and poses potential health hazards for the future. The thinning of ozone layer has been attributed to the presence of chlorofluorohydrocarbons (e..g, \(\ce{CFCl3}\) and \(\ce{CF2Cl2 }\) in the atmosphere. These chemicals have been used as aerosol propellants and cooling mixtures in refrigerators. How these chemicals affect ozone concentration is illustrated with the equations given below. \[ \ce{CF_2Cl_2 + h\nu \rightarrow CF_2Cl^{\cdot} + Cl^{\cdot} }\label{i}\] Chlorine atom being highly reactive reacts with ozone (\(\ce{O3}\)). \[ \ce{Cl^{\cdot} + O_3 \rightarrow ClO^{\cdot} + O_2}\label{ii}\] The monoxide of chlorine further reacts with another molecule of \(\ce{O3}\) \[ \ce{ClO^{\cdot} + O_3 \rightarrow Cl^{\cdot} + 2O_2 } \label{iii}\] The chlorine atom so obtained reacts with another ozone molecule. Hence, steps \ref{ii} and \ref{iii} are repeated again and again and, leads to the depletion of concentration of ozone.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/16%3A_Carbonyl_Compounds_I-_Aldehydes_and_Ketones._Addition_Reactions_of_the_Carbonyl_Group/16.05%3A_Typical_Carbonyl-Addition_Reactions

We turn now to discuss a few specific addition reactions of the carbonyl groups of aldehydes and ketones. We shall not attempt to provide an extensive catalog of reactions, but will try to emphasize the principles involved with especially important reactions that are useful in synthesis. Grignard reagents, organolithium compounds, and the like generally add to aldehydes and ketones rapidly and irreversibly, but the same is not true of many other reagents; their addition reactions may require acidic or basic catalysts; the adducts may be formed reversibly and with relatively unfavorable equilibrium constants. Also, the initial adducts may be unstable and react further by elimination. (We recommend that you review Section 15-4E to see examples of these points.) To organize this very large number of addition reactions, we have arranged the reactions according to the nucleophile that adds to the carbonyl carbon. The types of nucleophiles considered here form \(\ce{C-C}\), \(\ce{C-O}\), \(\ce{C-N}\), \(\ce{C}\)-halogen, \(\ce{C-S}\), and \(\ce{C-H}\) bonds. A summary is given in Table 16-4. Hydrogen cyanide adds to many aldehydes and ketones to give hydroxylnitriles, usually called "cyanohydrins": The products are useful in synthesis - for example, in the preparation of cyanoalkenes and hydroxy acids: An important feature of cyanohydrin formation is that it requires a basic catalyst. In the absence of base, the reaction does not proceed, or is at best very slow. In principle, the basic catalyst may activate either the carbonyl group or hydrogen cyanide. With hydroxide ion as the base, one reaction to be expected is a reversible addition of hydroxide to the carbonyl group: However, such addition is not likely to facilitate formation of cyanohydrin because it represents a saturation of the carbonyl double bond. Indeed, if the equilibrium constant for this addition were large, an excess of hydroxide ion could inhibit cyanohydrin formation by tying up the ketone as the adduct \(1\). Hydrogen cyanide itself has no unshared electron pair on carbon and does not form a bond to a carbonyl carbon. However, a small amount of a strong base can activate hydrogen cyanide by converting it to cyanide ion, which can function as a carbon nucleophile. A complete sequence for cyanohydrin formation follows: The second step regenerates the cyanide ion. Each step of the reaction is reversible but, with aldehydes and most nonhindered ketones, formation of the cyanohydrin is reasonably favorable. In practical syntheses of cyanohydrins, it is convenient to add a strong acid to a mixture of sodium cyanide and the carbonyl compound, so that hydrogen cyanide is generated . The amount of acid added should be insufficient to consume all the cyanide ion, therefore sufficiently alkaline conditions are maintained for rapid addition. There are a number of rather interesting substances for which we can write important dipolar valence-bond structures of the type . The important factor with these structures is that the negative end of the dipole is . The positive end of the dipole can be several kinds of atoms or groups, the most usual being sulfur, phosphorus, or nitrogen. Some examples (each written here as a single dipolar valence-bond structure) are: The systematic naming of these substances is cumbersome, but they have come to be known as . The genesis of this name may seem obscure, but it is an attempt to reconcile the presence of a \(\ce{C-X}\) \(\sigma\) bond, which is covalent and nonpolar as in alk derivatives, as well as an ionic bond as in metal hal . Hence, the combination .\(^2\) As we might expect from the dipolar structure, ylides can behave as carbon nucleophiles to form carbon-carbon bonds by addition to the carbonyl groups of aldehydes and ketones: However, the further course of reaction depends on the type of ylide used. In the case of phosphorus ylides, the overall reaction amounts to a very useful by the transfer of oxygen to phosphorus and carbon to carbon, as summarized in Equation 16-1. This is called the : Reactions with sulfur ylides proceed differently. The prodcuts are oxacylcopropanes (oxiranes) - not alkenes. The addition step proceeds as with the phosphorus ylides, but the negatively charged oxygen of the dipolar adduct then displaces the sulfonium group as a neutral sulfide. This is an intramolecular \(S_\text{N}2\) reaction similar to the formation of oxacyclopropanes from vicinal chloroalcohols (Section 15-11C): As for the nitrogen ylides, a useful reagent of this type is diazomethane, \(\ce{CH_2N_2}\). Diaxomethane can react with carbonyl compounds in different ways, depending on what happens to the initial adduct \(2\). Oxacyclopropanes are formed if the nitrogen is simply displaced (as \(\ce{N_2}\)) by oxygen (Path \(a\), Equation 16-2). Ketones of rearranged carbon framework result if nitrogen is displaced \(as \(\ce{N_2}\)) by \(\ce{R}^\ominus\) which moves over to the \(\ce{CH_2}\) group (Path \(b\), Equation 16-2): Diazoketones, \(\ce{RCOCHN_2}\), are formed if there is a good leaving group, such as halogen, on the carbonyl (Equation 16-3). Under these circumstances the reactant is an acid halide, not an aldehyde or ketone: We already have discussed additions of alcohols and, by analogy, thiols \(\left( \ce{RSH} \right)\) to carbonyl compounds (see Section 15-4E). We will not repeat this discussion here except to point out that addition of to the carbonyl group of an aldehyde is analogous to hemiacetal formation (Section 15-4E) and is catalyzed both by acids and bases: The equilibrium for hydrate formation depends both on steric and electrical factors. Methanal is \(99.99\%\) hydrated in aqueous solution, ethanal is \(58\%\) hydrated, and 2-propanone is not hydrated significantly. The hydrates seldom can be isolated because they readily revert to the parent aldehyde. The only stable crystalline hydrates known are those having strongly electronegative groups associated with the carbonyl (see Section 15-7). Several carbonyl additions have characteristics similar to those of cyanohydrin formation. A typical example is the addition of sodium hydrogen sulfite, which proceeds readily with good conversion in aqueous solution with most aldehydes, methyl ketones, and unhindered cyclic ketones to form a carbon-sulfur bond. No catalyst is required because sulfite is an efficient nucleophilic agent. The addition step evidently involves the sulfite ion - not hydrogen sulfite ion: The addition products often are nicely crystalline solids that are insoluble in excess concentrated sodium hydrogen sulfite solution. Whether soluble or insoluble, the addition products are useful for separating carbonyl compounds from substances that do not react with sodium hydrogen sulfite. A reaction closely related to acetal formation is the polymerization of aldehydes. Both linear and cyclic polymers are obtained. For example, methanal in water solution polymerizes to a solid ling-chain polymer called paraformaldehyde or "polyoxymethylene": This material, when strongly heated, reverts to methanal; it therefore is a convenient source of gaseous methanal. When heated with dilute acid, paraformaldehyde yields the solid trimer, 1,3,5-trioxycyclohexane (mp \(61^\text{o}\)). The cyclic tetramer is also known. Long-chain methanal polymers have become very important as plastics in recent years. The low cost of paraformaldehyde is highly favorable in this connection, but the instability of the material to elevated temperatures and dilute acids precludes its use in plastics. However, the "end-capping" of polyoxymethylene chains through formation of esters or acetals produces a remarkable increase in stability, and such modified polymers have excellent properties as plastics. Delrin (DuPont) is a stabilized methanal polymer with exceptional strengths and ease of molding. Ethanal (acetaldehyde) polymerizes under the influence of acids to the cyclic trimer, "paraldehyde", and a cyclic tetramer, "metaldehyde". Paraldehyde has been used as a relatively nontoxic sleep-producing drug (hypnotic). Metaldehyde is used as a poison for snails and slugs, "Snarol". Ketones do not appear to form stable polymers like those of aldehydes. A wide variety of substances with \(\ce{-NH_2}\) groups react with aldehydes and ketones by an addition-elimination sequence to give compounds and water. These reactions usually require acid catalysts: Table 16-5 summarizes several important reactions of this type and the nomenclature of the reactants and products. Clearly, if the unshared electron pair on the nitrogen of \(\ce{RNH_2}\) is combined with a proton, Equation 16-4, it cannot attack the carbonyl carbon to give the aminoalkanol as in Equation 16-5. So at high acid concentration (low pH) we expect the rate and the equilibrium for the overall reaction to be unfavorable. In more dilute acid, the rate picks up because there is more free \(\ce{RNH_2}\) in solution. Dehydration of the aminoalkanol (Equation 16-6) is acid catalyzed; this reaction normally is fast at pH values smaller than 3-4. Therefore, the slow step at pH \(<\) 4 is addition of \(\ce{RNH_2}\) to the carbonyl group as per Equation 16-5. As the pH is increased above 4, the addition becomes progressively faster because less \(\ce{RNH_2}\) is tied up as \(\ce{RNH_3^+}\). However, then the dehydration step, Equation 16-6, decreases in rate because it requires an acid catalyst. At pH 6 (remember that going from pH 4 to pH 6 is a 100-fold decrease in \(\ce{H}^\oplus\) concentration), dehydration is the slow step, and at higher pH values it finally becomes too slow to give a useful overall rate of reaction. This sequence of changes in rate and equilibria has been shown to account precisely for rate . pH curves such as in Figure 16-4. Dehydration of \(\ce{(CH_3)_2CHNR(OH)}\) to \(\ce{(CH_3)_2C=NR}\) involves acid catalysis in very much the same way as in acetal formation (Section 15-4E): Ammonia adds readily to many aldehydes. For example, The aldehyde-ammonia adducts usually are not very stable. They readily undergo dehydration and polymerization. 1-Aminoethanol, for example, gives a cyclic trimer composition \(\ce{C_6H_{15}N_3} \cdot 3 \ce{H_2O}\), mp \(97^\text{o}\), with structure \(4\): Methanal and ammonia react by a different course with the formation of a substance known as "hexamethylenetetramine": The product is a high-melting solid (mp \(> \: 230^\text{o}\) d.) and its structure has been established by x-ray diffraction (Section 9-3). In fact, it was the first organic substance whose structure was determined in this way. The high melting point is clearly associated with the considerable symmetry and rigidity of the cage structure: The corresponding all-carbon compound, adamantane (Section 12-8), also has a high melting point \(\left( 268^\text{o} \right)\): Treatment of hexamethylenetetramine with nitric acid gives the high explosive "cyclonite", which often is designated as RDX and was widely used in World War II: Then methanal and ammonia that split off the cage structure during the reaction with nitric acid need not be wasted. In the large-scale manufacture of cyclonite, a combination of nitric acid, ammonium nitrate, and ethanoic anhydride is used, which results in full utilization of the methanal and ammonia: \[\ce{C_6H_{12}N_4} + 4 \ce{HNO_3} + 2 \ce{NH_4NO_3} + 6 \ce{(CH_3CO)_2O} \rightarrow 2 \ce{C_3H_6O_6N_6} + 12 \ce{CH_3CO_2H}\] Secondary amino compounds of the type \(\ce{R_2N-H}\) add to aldehyde and ketone carbonyl groups in an acid-catalyzed reaction in much the same way as do \(\ce{RNH_2}\) compounds - with one important difference. The product contains the structural unit \(\ce{C=C-N}\) rather than \(\ce{C-C=N}\); and because there is a carbon-carbon double bond, such a substance is called an enamine (alk \(+\) ). An example is: The course of this reaction can be understood if we notice that loss of \(\ce{OH}\) from the initial product leads to an immonium ion, \(5\), that cannot lose a proton and form a \(\ce{C=N}\) bond: However, if there is a hydrogen on a carbon attached to the immonium carbon, it is possible for such a hydrogen to be lost as a proton with concurrent formation of the neutral enamine: Enamine formation, like many other carbonyl addition reactions, is readily reversible, and the carbonyl compound can be recovered by hydrolysis with aqueous acids. For this reason, to obtain a good conversion of carbonyl compound to enamine, it usually is necessary to remove the water that is formed by distilling it away from the reaction mixture. Enamines are useful synthetic intermediates for the formation of carbon-carbon bonds, as we will discuss in greater detail in Section 17-4B. Enamines generally are unstable if there is a hydrogen on nitrogen. They rearrange to the corresponding imine. This behavior is analogous to the rearrangement of alkenols to carbonyl compounds (Section 10-5A): Addition of hydrogen halides to carbonyl groups usually is so easily reversible as to preclude isolation of the addition products: However, many aldehydes react with alcohols in the presence of an excess of hydrogen chloride to give \(\alpha\)-chloro ethers: In carrying out laboratory syntheses of \(\alpha\)-chloro ethers, gaseous hydrogen chloride is passed into a mixture of the alcohol and aldehyde. Aqueous \(\ce{HCl}\) is not useful because the excess water gives an unfavorable equilibrium. \(\alpha\)-Chloro ethers are highly reactive compounds that very readily undergo \(S_\text{N}2\) as well as \(S_\text{N}1\) and \(E1\) reactions. Two simple examples, methoxychloromethane (chloromethyl methyl ether) and chloromethoxychloromethane (bis-chloromethyl ether), have been put under severe restrictions as the result of tests that show they have strong chemical carcinogenic properties. Replacement of the carbonyl function by two chlorines occurs with phosphorus pentachloride in ether: This reaction is useful in conjunction with \(E2\) elimination to prepare alkenyl halides, allenes, and alkynes. Cycloalkenyl halides are easily prepared, but because of angle strain the cycloalkynes and cycloallenes with fewer than eight atoms in the ring cannot be isolated (see Section 12-7): Replacement of a carbonyl group by -fluorines\(^3\) can be achieved with molybdenum hexafluoride or sulfur tetrafluoride. Sulfur tetrafluoride converts carboxyl functions to trifluoromethyl groups: In recent years, inorganic hydrides such as lithium aluminum hydride, \(\ce{LiAlH_4}\), and sodium borohydride, \(\ce{NaBH_4}\), have become extremely important as reducing agents of carbonyl compounds. These reagents have considerable utility, especially with sensitive and expensive carbonyl compounds. The reduction of cyclobutanone to cyclobutanol is a good example, and you will notice that the net reaction is the addition of hydrogen across the carbonyl double bond, \(\overset{ 2 \left[ \ce{H} \right]}{\longrightarrow}\) , With the metal hydrides, the key step is transfer of a hydride ion to the carbonyl carbon of the substance being reduced. The hydride transfer is analogous to the transfer of \(\ce{R}^\ominus\) from organometallic compounds to carbonyl groups (Section 14-12A). Lithium aluminum hydride is best handled like a Grignard reagent, because it is soluble in ether and is sensitive to both oxygen and moisture. (Lithium hydride is insoluble in organic solvents and is not an effective reducing agent for organic compounds.) All four hydrogens on aluminum can be utilized: The reaction products must be decomposed with water and acid as with the Grignard complexes. Any excess lithium aluminum hydride is decomposed by water and an acid with evolution of hydrogen: \[\ce{LiAlH_4} + 2 \ce{H_2SO_4} \rightarrow \frac{1}{2} \ce{Li_2SO_4} + \frac{1}{2} \ce{Al_2(SO_4)_3} + 4 \ce{H_2}\] Lithium aluminum hydride usually reduces carbonyl groups without affecting carbon-carbon double bonds. It is, in addition, a good reducing agent for carbonyl groups of carboxylic acids, esters, and other acid derivatives, as will be described in Chapter 18. Sodium borohydride is a milder reducing agent than lithium aluminum hydride and will reduce aldehydes and ketones, but not acids or esters. It reacts sufficiently slowly with water in neutral or alkaline solution that reductions which are reasonably rapid can be carried out in water solution without appreciable hydrolysis of the reagent: \[\ce{NaBH_4} + 4 \ce{CH_2=O} + 3 \ce{H_2O} \rightarrow 4 \ce{CH_3OH} + \ce{NaOB(OH)_2}\] Borane (as \(\ce{BH_3}\) in tetrahydrofuran or dimethyl sulfide) is an even milder reducing agent than \(\ce{BH_4^+}\) for the carbonyl group of aldehydes and ketones. This difference in reactivity can be used to advantage when selective reduction is necessary. For example, borohydride reduces a ketone carbonyl more rapidly than a carbon-carbon double bond, whereas borane reduces the carbon-carbon double bond more rapidly than carbonyl: A useful comparison of the reactivities of boranes and metal hydrides toward various types of multiple bonds is given in Table 16-6. A characteristic reaction of aldehydes without \(\alpha\) hydrogens is the self oxidation-reduction they undergo in the presence of strong base. With methanal as an example, If the aldehyde has \(\alpha\) hydrogens, other reactions usually occur more rapidly. The mechanism of this reaction, usually called the ,\(^4\) combines many features of other processes studied in this chapter. The first step is reversible addition of hydroxide ion to the carbonyl group: A hydrogen can be transferred as hydride ion to methanal from the hydroxyalkoxide ion, thereby reducing the methanal to methanol: Hydride transfer similar to that of the Cannizzaro reaction also may be achieved from a \(\ce{C-H}\) grouping in an alkoxide ion corresponding to a primary or secondary, but not a tertiary, alcohol. This is expected to be a reversible reaction, because the products are another alkoxide and another carbonyl compound: To utilize this equilibrium process as a practical reduction method requires rather special conditions. It is preferable to use an aluminum alkoxide, \(\ce{Al(OR)_3}\), rather than a sodium alkoxide, \(\overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{O}} \ce{R}\), to ensure that the reaction mixture is not too strongly basic. (Carbonyl compounds, particularly aldehydes, are sensitive to strong bases.) The overall reaction may be written for which the alkoxide is derived from 2-propanol. The advantage of this method is that the reaction can be driven essentially to completion by distilling out the 2-propanone as it is formed. The reduction product subsequently can be obtained by acid hydrolysis of the aluminum alkoxide: These have been discussed already in the context of the reverse reactions - oxidation of alcohols (Section 15-6C). \(^2\)Pronounced variously as , , , . The dipolar structures usually written for ylides are an oversimplified representation of the bonding in these substances. \(^3\) is an abbreviation for (twinned) and is a common designation for arrangements having two identical substituents on one carbon. \(^4\)Named after its discoverer, the same Cannizzaro who, in 1860, made an enormous contribution to the problem of obtaining self-consistent atomic weights (Section 1-1). and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/17%3A_Carbonyl_Compounds_II-_Enols_and_Enolate_Anions._Unsaturated_and_Polycarbonyl_Compounds/17.05%3A_Nucleophilic_Substitution_with_Enolate_Anions

The synthetic chemistry of enolate anions is centered on their nucleophilic and basic properties. Accordingly these ions participate in \(S_\text{N}2\) reactions with suitable alkyl compounds: However, there are a number of complicating factors to consider. First, the basic conditions needed to form the enolate ions often lead to side reactions such as aldol addition and \(E2\) elimination of \(\ce{RX}\) compounds. Aldol addition is minimized if the carbonyl compound is a ketone with a structure unfavorable for aldol addition or if of the carbonyl compound is converted to its enolate. To convert all of a simple carbonyl to its enolate usually requires a very strong base, such as \(\ce{NH_2^+}\) in an aprotic solvent or liquid ammonia. Because the enolate anion itself is a strong base, best results are obtained when the halide, \(\ce{RX}\), does not undergo \(E2\) reactions readily. The second complication arises if the alkyl compound reacts with both carbon and oxygen of the nucleophilic enolate anion. The carbon product is the result of "\(\ce{C}\)-alkylation", whereas the oxygen product is the result of "\(\ce{O}\)-alkylation": The possibility of the enolate anion acting as if its charge were effectively concentrated on carbon or on oxygen was discussed previously in connection with aldol addition ( ). However, the situation there was quite different from the one here, because aldol addition is easily reversible, whereas alkylation is not. Furthermore, while the aldol reaction involving \(\ce{C-O}\) bond formation is unfavorable \(\left( \Delta H^0 = + 20 \: \text{kcal mol}^{-1} \right)\) compared to \(\ce{C-C}\) bond formation \(\left( \Delta H^0 = - 4 \: \text{kcal mol}^{-1} \right)\), both \(\ce{O}\)- and \(\ce{C}\)-alkylation of the anion have \(\Delta H^0 < 0\). Whether \(\ce{C}\)- or \(\ce{O}\)-alkylation predominates depends on kinetic control ( ). It is not a simple matter to predict which of the two positions of the enolate will be more nucleophilic, and in fact, mixtures of products often are obtained in distributions that depend on the solvent used, the temperature, the nature of \(\ce{X}\), and the nature of the base employed to form the anion. \(\ce{O}\)-Alkylation tends to occur with ketones of high enol content (which usually means that the enolate anion will have especially high charge density on oxygen) and with alkylating agents possessing a high degree of \(S_\text{N}2\) reactivity. There is another correlation that seems to have validity in many situations, at least where kinetic control is dominant; namely, the (less associated) the ambident anions is from its cation, the is the electrophile to attack the atom of the anion with the negative charge. Thus \(\ce{O}\)-alkylation of the sodium enolate of 2-propanone is favored in aprotic solvents that are good at solvating cations [such as \(\ce{(CH_3)_2SO}\), Section 8-7F]. In the alkylation of unsymmetrical ketones, formation of more than one enolate anion is possible, and when this occurs, mixtures of products are obtained. Thus, However, when one of the possible enolate anions is especially stabilized, either by conjugation or by strong electron-withdrawing groups, that enolate usually is the dominant form and only one product is formed. Thus 2,4-pentanedione is methylated at \(\ce{C3}\), not at \(\ce{C1}\): Enamines ( ), like enolate anions, have two reactive positions and, in principle, can give either \(\ce{N}\)- or \(\ce{C}\)-alkylation. Both products may be formed, but they can be separated readily because, on treatment with dilute acid, only the \(\ce{C}\)-alkylation product hydrolyzes to a ketone. Generally, the alkylated ketone is the desired product: Alkylation of enamines therefore is a feasible, and sometimes much more useful, alternative to the direct alkylation of ketones because it proceeds under less strongly basic conditions. The sequence starts with conversion of a ketone to an enamine, \(\overset{\ce{RNH_2}}{\longrightarrow}\) \(+ \ce{H_2O}\), followed by \(\ce{C}\)-alkylation of the enamine, \(\overset{\ce{RX}}{\longrightarrow}\) \(+ \ce{X}^\ominus\), and ends with hydrolysis to the alkylated ketone, \(+ \ce{H_2O} \rightarrow\) \(+ \ce{R_2NH_2^-}\). A typical example of the use of enamines for alkylation of a ketone follows: Several important biological reactions utilize enamine intermediates as carbon nucleophiles in \(\ce{C-C}\) bond-forming reactions. One example is discussed in . The chemistry of carbanions stabilized by groups other than carbonyl functions is closely analogous to the chemistry of enolate anions. We have seen that \(\ce{C-H}\) acidity of compounds with the structural feature can be significant (p\(K_\text{a}\) of 25 or less) when \(\ce{X}\) is an atom or group that can effectively delocalize the negative charge on carbon in . Typical \(\ce{X}\) groups are \(\ce{C=O}\), \(\ce{C \equiv N}\), \(\ce{PR_3^+}\), \(\ce{SR_2^+}\), \(\ce{SO_2R}\), and \(\ce{SR}\). Consequently, we can expect that carbanions of the type , when formed, will resemble enolate anions and will undergo addition reactions to \(\ce{C=O}\) and \(\ce{C=C}\), and will be alkylated with halides of good \(S_\text{N}2\) reactivity. In fact, the reactions of ylides discussed in are examples of the addition of phosphorus-, sulfur-, and nitrogen-stabilized carbanions to carbonyl groups. Sulfur in its higher oxidation states (e.g., sulfone, \(\ce{-SO_2}-\)) is especially effective in stabilizing adjacent carbanion centers. However, from a synthetic standpoint there are disadvantages to the sulfone grouping in that the better stabilized carbanions also are the least reactive, and subsequent removal of the sulfone grouping can be difficult. A good balance between carbanion stability, carbanion reactivity, and ease of \(\ce{C-S}\) bond cleavage is present in the structures \(\ce{RS-CH_2-SR}\) and . This is illustrated below for a strikingly simple concept for preparing cyclobutanone, in which the ring carbons are derived from methanal and 1,3-dibromopropane: To achieve this synthesis, the methanal first is converted to a thioketal, which then is partially oxidized to give \(13\). Treatment of \(13\) with a strong base converts it to the carbanion, which can be readily alkylated. By using 1,3-dibromopropane and two equivalents of base, a double displacement forms the cyclic product, \(14\). The sulfur groups of \(14\) can be removed easily by acid hydrolysis to give cyclobutanone: and (1977)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Optical_Isomers_in_Inorganic_Complexes/Identifying_Planes_of_Symmetry_in_Octahedral_Complexes

The symmetry of a molecule is determined by the existence of performed with respect to . A symmetry element is a line, a plane or a point in or through an object, about which a rotation or reflection leaves the object in an orientation indistinguishable from the original. A plane of symmetry and the reflection operation is the coincidence of atoms on one side of the plane with corresponding atoms on the other side, as though reflected in a mirror. Identifying planes of symmetry in an octahedral geometry can be daunting in evaluating if a molecule is chiral or not. There is a simple approach, which requires identying and testing six possible planes of symmetry that may exist for this geometry. To demonstrated this, we will consider the \(\ce{[Cr(CO)6]}\) complex in Figure \(\Page {1}\). While you can no doubt recognize at least one plane of symmetry to confirm that this is an achiral complex, let's identify all six of them. First we have the three planes that include the \(Cr\) metal center (as they all do) and four of the attached ligands (Figure \(\Page {2}\)). These are often the easy planes to identify. The next six planes are harder to identify and can be viewed as rotations of those in Figure \(\Page {2}\) by 45°. These planes bisect one angle between ligands, but still have two ligands on the plane (instead of four in the planes of Figure \(\Page {2}\)). The next three planes are even harder to view on the same perspective but are also rotations of those in Figure \(\Page {2}\) by 45° in the perpendicular direction. These planes bisect an angle between ligands, but still have two ligands on the plane. All nine planes, in each of planes are combined in the images below. Now that can you identify these nine planes that may be mirror planes in an octahedral complex, you have to evaluate if they truly are mirror planes. That is simply asking if the ligand fields above the plane is IDENTICAL to the ligand field below. For the \(\ce{[Cr(CO)]}\) complex discussed above, all nine planes are indeed mirror planes. However, in other complexes that may not be the case. Fortunately, to determine if a complex is chiral (optically active), then only one plane needs to be identified. ),

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.3%3A_Entropy_and_Heat_-_Experimental_Basis_of_the_Second_Law_of_Thermodynamics

Experiments show that the magnitude of ΔS is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔS . For instance, ΔS for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure \(\Page {1}\), the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of \(\ce{NaCl}\) and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an \(\ce{NaCl}\) solution, disruption of the crystalline \(\ce{NaCl}\) structure and the hydrogen-bonded interactions in water is quantitatively more important, so \(ΔS_{soln} > 0\). Dissolving \(\ce{NaCl}\) in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive. Predict which substance in each pair has the higher entropy and justify your answer. : amounts of substances and temperature : higher entropy : From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy. : Predict which substance in each pair has the higher entropy and justify your answer. 1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates) a mixture of 3 mol of H (g) and 1 mol of N (g) at 25°C and 1 atm (more molecules of gas are present) A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.11%3A_Nuclear_Magnetic_Resonance_Spectroscopy

Nuclear magnetic resonance (NMR) spectroscopy is extremely useful for identification and analysis of organic compounds. The principle on which this form of spectroscopy is based is simple. The nuclei of many kinds of atoms act like tiny magnets and tend to become aligned in a magnetic field. In NMR spectroscopy, we measure the energy required to change the alignment of magnetic nuclei in a magnetic field. To illustrate the procedure with a simple example, consider the behavior of a proton \(\left( ^1H \right)\) in a magnetic field. There are two possible alignments of this magnetic nucleus with respect to the direction of the applied field, as shown in Figure 9-21. The nuclear magnets can be aligned either with the field direction, or opposed to it. The two orientations are not equivalent, and energy is required to change the more stable alignment to the less stable alignment. A schematic diagram of an NMR instrument is shown in Figure 9-22. When a substance such as ethanol, \(CH_3-CH_2-OH\), the hydrogens of which have nuclei (protons) that are magnetic, is placed in the transmitter coil and the magnetic field is increased gradually, at certain field strengths radio-frequency energy is absorbed by the sample and the ammeter indicates an increase in the flow of current in the coil. The overall result is a spectrum such as the one shown in Figure 9-23. This spectrum is detailed enough to serve as a useful "fingerprint" for ethanol, and also is simple enough that we will be able to account for the origin of each line. It is the purpose of this section to explain how the complexities of spectra such as that of Figure 9-23 can be interpreted in terms of chemical structure. For what kinds of substances can we expect nuclear magnetic resonance absorption to occur? Magnetic properties always are found with nuclei of odd-numbered masses, \(^1H\), \(^{13}C\), \(^{15}N\), \(^{17}O\), \(^{19}F\), \(^{31}P\), and so on, as well as for nuclei of even mass but odd atomic number, \(^2H\), \(^{10}B\), \(^{14}N\), and so on.\(^8\) Nuclei such as \(^{12}C\), \(^{16}O\), and \(^{32}S\), which have even mass and atomic numbers, have magnetic properties and do give nuclear magnetic resonance signals. For various reasons, routine use of NMR spectra in organic chemistry is confined to \(^1H\), \(^{19}F\), \(^{13}C\), and \(^{31}P\). We shall be concerned in this chapter only with NMR spectra of hydrogen (\(^1H\)) and of carbon (\(^{13}C\)). The kind of NMR spectroscopy we shall discuss here is limited in its applications because it can be carried out only with or . Fortunately, the allowable range of solvents is large, from hydrocarbons to concentrated sulfuric acid, and for most compounds it is possible to find a suitable solvent. Nuclear magnetic resonance spectra may be so simple as to have only a single absorption peak, but they also can be much more complex than the spectrum of Figure 9-23. However, it is important to recognize that no matter how complex an NMR spectrum appears to be, in involves just parameters: , , and (reaction-rate) . We shall have more to say about each of these later. First, let us try to establish the relationship of NMR spectroscopy to some of the other forms of spectroscopy we have already discussed in this chapter. Nuclear magnetic resonance\(^9\) spectroscopy involves transitions between possible energy levels of magnetic nuclei in an applied magnetic field (see Figure 9-21). The transition energies are related to the frequency of the absorbed radiation by the familiar equation \(\Delta E - h \nu\). An important difference between nmr and other forms of spectroscopy is that \(\Delta E\) is influenced by the strength of the applied field. This should not be surprising, because if we are to measure the energy of changing the direction of alignment of a magnetic nucleus in a magnetic field, then the stronger the field the more energy will be invloved. Nuclear spin (symbolized as \(I\)) is a quantized property that correlates with nuclear magnetism such that when \(I\) is zero the nucleus has no spin and no magnetic properties. Examples are \(^{12}C\) and \(^{16}O\). Several nuclei of particular interest to organic chemists - \(^1H\), \(^{13}C\), \(^{15}\), \(^{19}\), and \(^{31}P\) - have spin of \(\frac{1}{2}\). With \(I = \frac{1}{2}\) there are only magnetic energy states of the nucleus in a magnetic field. These states are designated with the \(+ \frac{1}{2}\) and \(- \frac{1}{2}\). The difference in energy between these states, \(\Delta E\), is given by \[\Delta E = \gamma h H = h \nu\] or \[\nu = \gamma H\] in which \(h\) is Planck's constant, \(\nu\) is in hertz, \(\gamma\) is a nuclear magnetic constant called the ,\(^{10}\), and \(H\) is the magnetic field strength at the nucleus. In general, \(H\) will not be exactly equal to \(H_\text{o}\), the applied magnetic field and, as we will see, this difference leads to important chemical information. Each of nucleus (\(^1H\), \(^{13}C\), \(^{15}N\), etc.) has its own \(\gamma\) value and, consequently, will undergo transitions at different frequencies at any particular value of \(H\). This should become clearer by study of Figure 9-24. There are several modes of operation of an nmr spectrometer. First and most common, we hold \(\nu\) constant and vary (or "sweep") \(H_\text{o}\). Close to \(\nu = \gamma H\), energy is absorbed by the nuclei and the current flow from the transmitter increases until \(\nu\) is exactly equal to \(\gamma H\). Further increase of \(H_\text{o}\) makes \(\nu < \gamma H_\text{o}\) and the current flow decreases. The form of the energy-absorption curve as a function of \(H_\text{o}\) when \(H_\text{o}\) is changed very slowly is shown in Figure 9-25a. The peak is centered on the point where \(\nu = \gamma H\). When \(H_\text{o}\) is changed more rapidly, transient effects are observed on the peak, which are a consequence of the fact that the nuclei do not revert instantly from the \(- \frac{1}{2}\) to \(+ \frac{1}{2}\) state. The resulting phenomenon is called "ringing" and is shown in Figures 9-25b and 9-25c. Evidence of ringing also will be seen on peaks of Figure 9-23. An alternative method of running an nmr spectrometer is to hold the magnetic field constant and to sweep the transmitter frequency through the resonances. This mode of operation is more like other forms of spectroscopy and gives the same line shapes as sweeping the field (Figure 9-25). What energy is associated with a \(^1H\) nmr transition? The magnitude of this energy may be calculated from the relationship between energy and wavelength (frequency) of the absorbed radiation ( ). That is, \[\Delta E = \frac{28,600}{\lambda} \text{kcal mol}^{-1}\] and \[\lambda = \frac{c}{\nu}\] The frequency \(\nu\) is the operating frequency of the spectrometer, which we will take as \(60 \: \text{MHz}\) or \(6 \times 10^7 \: \text{Hz}\) (cycles \(\text{sec}^{-1}\)), and the velocity of light is \(3 \times 10^8 \: \text{m sec}^{-1}\). Hence \[\lambda = \frac{3 \times 10^8 \times 10^9 \left( \text{nm sec}^{-1} \right)}{6 \times 10^7 \left( \text{Hz} \right)} = 5 \times 10^9 \: \text{nm}\] and \[\Delta E = \frac{28,600}{5 \times 10^9} = 5.7 \times 10^{-6} \: \text{kcal mol}^{-1}\] This is a very small energy difference, which means that only very few more of the nuclei are in the more stable \(+ \frac{1}{2}\) state than in the less stable \(- \frac{1}{2}\) state. The equilibrium constant \(K\) for \(- \frac{1}{2} \rightleftharpoons + \frac{1}{2}\) calculated from Equation 4-2 for \(25^\text{o}\) (298 \: \text{K}\)) and neglecting possible entropy effects is 1.000010! The plot of signal against magnetic field strength for ethanol in Figure 9-23 shows three principal groups of lines corresponding to the three varieties of hydrogen present: methyl (\(CH_3\)), methylene (\(CH_3\)), and hydroxyl (\(OH\)). Differences in the field strengths at which signals are obtained for nuclei of the same kind, such as protons, but located in different molecular environments, are called . Another very important point to notice about Figure 9-23 is that the intensities of the three principal absorptions are in the ratio of 1:2:3, corresponding to the ratio of the number of each kind of proton (\(OH\), \(CH_2\), \(CH_3\)) producing the signal. In general, such as in Figure 9-23 . The areas can be measured by electronic integration and the integral often is displayed on the chart, as it is in Figure 9-23, as a stepped line increasing from left to right. The height of each step corresponds to the relative number of nuclei of a particular kind. Unless special precautions are taken, integrals usually should not be considered accurate to better than about \(5\%\). Why do protons in different molecular environments absorb at different field strengths? The field strength \(H\) at a particular nucleus is than the strength of the external magnetic field \(H_\text{o}\). This is because the valence electrons around a particular nucleus and around neighboring nuclei respond to the applied magnetic field so as to the nucleus from the applied field. The way this shielding occurs is as follows. First, when an atom is placed in a magnetic field, its electrons are forced to undergo a rotation about the field axis, as shown in Figure 9-26. Second, rotation of the electrons around the nucleus is a circulation of charge, and this creates a magnetic field at the nucleus in the direction to \(H_\text{o}\). Third, the magnitude of this \(^{11}\) effect is directly proportional to \(H_\text{o}\) and can be quantified as \(\sigma H_\text{o}\), in which \(\sigma\) is the proportionality constant. It is important to recognize that \(\sigma\) is not a nuclear property but depends on the environment of the atom. Each chemically different proton will have a different value of \(\sigma\) and hence a different chemical shift. The actual field \(H\) at the nucleus will be \(H_\text{o} - \sigma H_\text{o}\). Because \(\sigma\) acts to reduce the strength of the applied field at the nucleus, it is called the . The more shielding there is, the stronger the applied field must be to satisfy the resonance condition, Common usage is: upfield, shielding; downfield, shielding; and you should remember that field-sweep spectra always are recorded with the field from to . The value of nmr spectroscopy in structure determination lies in the fact that chemically different nuclei absorb at different field strengths. In later sections we will be concerned with correlating the chemical shifts with structural features. However, before proceeding furher it is extremely important that you be able to identify the number and kind of nonequivalent protons in a given structure, and therefore the number of chemical shifts to expect. This number is not always self-evident, especially when subtle factors of stereochemistry intervene. For this reason, we suggest that you inspect structures \(3\)-\(5\) to convince yourself that the protons labeled with different letter subscripts in any one molecule are indeed chemically different. One way of checking whether two protons are in equivalent environments is to imagine that each is separately replaced with a different atom or group. If the product of replacing \(H_\text{A}\) is identical with that obtained by replacing \(H_\text{B}\), then \(H_\text{A}\) and \(H_\text{B}\) are chemically equivalent. If the two products are nonidentical, then \(H_\text{A}\) and \(H_\text{B}\) are nonequivalent. For example, replacement of \(H_\text{A}\) or \(H_\text{B}\) in \(3\), \(4\), and \(5\) by an atom \(X\) would give different products. Therefore, \(H_\text{A}\) and \(H_\text{B}\) are nonequivalent in \(3\), \(4\), and \(5\). Matters become more complicated with substances such as \(6\) and \(7\): Notice that \(6\) represents a chiral molecule and if \(H_\text{A}\) and \(H_\text{B}\) each are replaced with \(X\) we get \(8\) and \(9\), which are diastereomers (see ). You can verify this with molecular models if necessary. Diastereomers have different chemical and physical properties; therefore \(H_\text{A}\) and \(H_\text{B}\) in \(6\) are nonequivalent. They often are called hydrogens. What of the two methylene protons in ethanol, \(7\), which we have labeled as \(H_\text{A}\) \(H_\text{A'}\)? Are they identical? In a sense they are not identical because, if each were replaced by \(X\), we would have a pair of enantiomers. Therefore, \(H_\text{A}\) and \(H_\text{A'}\) sometimes are called hydrogens. But, you will recall that enantiomers are chemically indistinguishable unless they are in a chiral environment. Therefore we expect shifts of enantiotopic hydrogens to be identical, unless they are in a chiral environment. To summarize, enantiotopic protons normally will have the same chemical shifts, whereas diastereotopic protons normally will have different chemical shifts. We so far have ignored the relationship of chemical shifts to conformational equilibria. Consider a specific example, 1,2-dibromoethane, for which there are three staggered conformations \(10a\), \(10b\), and \(10c\): Each of these conformations is expected to have its own nmr spectrum. The two gauche forms, \(10a\) and \(10b\), are enantiomers and their spectra should be identical. The hydrogens \(H_\text{A}\) in \(10a\) each are trans to the bromine on the adjacent carbon, while the \(H_\text{B}\) hydrogens are cis to the same bromines (see ). Consequently the \(H_\text{A}\) and \(H_\text{B}\) hydrogens are nonequivalent and would be expected to have different chemical shifts. In contrast, all of the hydrogens of the anti conformer, \(10c\), are equivalent and would have the same chemical shift. Therefore we would expect to observe three chemical shifts arising from \(H_\text{A}\), \(H_\text{B}\), and \(H_\text{C}\) for a mixture of \(10a\), \(10b\), and \(10c\). However, the actual spectrum of 1,2-dibromoethane shows only under ordinary conditions. The reason is that the magnetic nuclei can absorb the exciting radiation. The result is that we observe an chemical shift, which reflects the relative shifts and populations of the three conformers present. If we can go to a sufficiently low temperature to make interconversion of the conformations slow (on the order of 10 times per second), then we will expect to see the three different chemical shifts \(H_\text{A}\), \(H_\text{B}\), and \(H_\text{C}\) with intensities corresponding to the actual populations of the conformations at the sample temperature. This is one example of the effect of rate processes on nmr spectra. Other examples and a more detailed account of how to relate the appearance of the signal to the rates of the exchange processes are given in . Chemical shifts are measured with reference to a standard. For protons or \(^{13}C\) in organic molecules, , \(\left( CH_3 \right)_4 Si\), which gives strong, sharp nmr signals in regions where only a very few other kinds of protons or carbon nuclei absorb. Chemical shifts often are expressed in \(\text{Hz}\) (cycles per second) relative to tetramethylsilane (TMS). These may seem odd units for magnetic field strength but because resonance occurs at \(\nu = \gamma H\), either frequency units (\(\text{Hz}\), radians \(\text{sec}^{-1}\)) or magnetic field units (gauss) are appropriate. Ten years ago, most nmr spectrometers operated for protons with radio-frequency (rf) transmitters set at \(60 \: \text{MHz}\) (\(6 \times 10^7\) cycles per second) but there has been a proliferation of different proton-operating frequencies and now \(30\), \(60\), \(90\), \(100\), \(220\), \(270\), \(300\) and \(360 \: \text{MHz}\) machines are commercially available. The cost of these machines is roughly proportional to the square of the frequency, and one well may wonder why there is such an exotic variety available and what this has to do with the chemical shift. High operating frequencies are desirable because chemical shifts increase with spectrometer frequency, and this makes the spectra simpler to interpret. A 12-fold increase in operating frequency (as from \(30 \: \text{MHz}\) to \(360 \: \text{MHz}\)) means a 12-fold increase in \(H_\text{o}\) at the point of resonance (remember \(\nu = \gamma H\)) and this means also a 12-fold increase in \(\sigma H_\text{o}\). Thus resonances that differ because they correspond to different \(\sigma\) values will be at \(360 \: \text{MHz}\) than at \(30 \: \text{MHz}\). This can produce a dramatic simplification of spectra, as can be seen from Figure 9-27, which shows the effect of almost a factor of four in \(\nu\) on the proton nmr spectrum of 2-methyl-2-butanol.\(^{12}\) To reiterate, chemical shifts are strictly proportional to spectrometer frequency, thus lines \(100 \: \text{Hz}\) apart at \(60 \: \text{MHz}\) will be \(167 \: \text{Hz}\) apart at \(100 \: \text{MHz}\). This might seem to make comparisons of nmr spectra on different spectrometers hopelessly complex but, because of the proportionality of shifts to frequency (or field), (relative to the same standard) for any spectrometer , which are useful for all nmr spectrometers. Nmr shifts reported in \(\text{ppm}\) relative to TMS as zero, as shown in Figure 9-23, are called \(\delta\) (delta) values: \[\delta = \frac{\left( \text{chemical shift downfield in Hz relative to TMS} \right) \times 10^6}{\text{spectrometer frequency in Hz}}\] Thus, if at \(60 \: \text{MHz}\) a proton signal comes \(100 \: \text{Hz}\) downfield relative to tetramethylsilane, it can be designated as being \(\left( +100 \: \text{Hz} \times 10^6 \right)/ 60 \times 10^6 \: \text{Hz} = +1.67 : \text{ppm}\) relative to tetramethylsilane. At \(100 \: \text{MHz}\), the line then will be \(\left( 1.67 \times 10^{-6} \right) \left(100 \times 10^6 \right) = 167 \: \text{Hz}\) downfield from tetramethylsilane. Typical proton chemical shifts relative to TMS are given in Table 9-4.\(^{13}\) The values quoted for each type of proton may, in practice, show variations of \(0.1\)-\(0.3 \: \text{ppm}\). This is not unreasonable, because the chemical shift of a given proton is expected to depend somewhat on the nature of the particular molecule involved, and also on the solvent, temperature, and concentration. A positive \(\delta\) value means a shift to lower field (or lower frequency) with respect to TMS, whereas a negative \(\delta\) signifies a shift to higher field (or higher frequency). The \(\delta\) convention is accepted widely, but you often find in the literature proton shifts with reference to TMS reported as "\(\tau\) values." The \(\tau\) scale has the TMS reference at \(+10 \: \text{ppm}\), so most proton signals fall in the range of \(\tau = 0\) to \(\tau = +10\). A \(\tau\) value can be converted to the appropriate \(\delta\) value by subtracting it from 10. Life with nmr spectra would be simpler if the \(\tau\) scale would just go away. Protonc chemical shifts are very valuable for the determination of structures, but to use the shifts in this way we must know something about the correlations that exist between chemical shift and structural environment of protons in organic compounds. The most important effects arise from differences in electronegativity, types of carbon bonding, hydrogen bonding, and chemical exchange. Consider first the chemical shifts of protons attached to an \(sp^3\) carbon, . The degree of shielding of the proton by the carbon valence electrons depends on the character of the substituent atoms and groups present, and particularly on their electron-attracting power, or electronegativity. For a grouping of the type , the shielding will be as \(\ce{X}\) is more electron withdrawing relative to hydrogen: If \(\ce{X}\) is electron-withdrawing, the proton is deshielded. For example, the proton chemical shifts of the methyl halides (Table 9-4) show decreasing shielding, hence progressively low-field chemical shifts with increasing halogen electronegativity \(\left( \ce{F} > \ce{Cl} > \ce{Br} > \ce{I} \right)\): When two electronegative groups, \(\ce{X}\) and \(\ce{Y}\), are bonded to the carbon, as in \(\ce{XCH_2Y}\), the protons are expected to be less shielded and come into resonance downfield of the methylenes of \(\ce{XCH_2CH_2Y}\). There is an approximate relationship (see below) between the shifts of the \(\ce{XCH_2Y}\) protons and the effective shielding constants \(\left( \sigma \right)\) of \(\ce{X}\) and \(\ce{Y}\) known as . \[\delta = 0.23 + \sigma_x + \sigma_y \tag{9-4}\] Appropriate values of \(\sigma\) for use with this equation are given in Table 9-4. The shifts of the protons of alkanes and cycloalkanes fall in the range of \(0.9\)-\(1.5 \: \text{ppm}\) with \(\ce{C-H}\) protons coming at the low-field end of this range and \(\ce{-CH_3}\) protons coming at the high-field end (see Table 9-4). Alkenic hydrogens (vinyl hydrogens, ) normally are observed between \(4.6\)-\(6.3 \: \text{ppm}\) toward fields than the shifts of protons in alkanes and cycloalkanes. This means that alkenic hydrogens in an organic compound can be easily distinguished from alkane hydrogens. Aromatic protons, such as those in benzene, have shifts at still lower fields and commonly are observed at \(7\)-\(8 \: \text{ppm}\). In contrast, alkynic protons of the type \(\ce{-C \equiv CH}\) give resonances that are of alkenic or aromatic protons and come at \(2\)-\(3 \: \text{ppm}\). Another effect associated with multiple bonds is the large difference in shift between a \(\ce{-CH(OCH_3)_2}\) proton, which normally comes at about \(5.5 \: \text{ppm}\), and aldehyde protons, \(\ce{-CH=O}\), which are much farter at \(9\)-\(11 \: \text{ppm}\). Clearly, the shifts of a proton depend on whether the carbon forms single, double, or triple bonds. In a magnetic field, the circulation of electrons in the \(\pi\) orbitals of multiple bonds induced by the field (Figure 9-26) generates diamagnetic shielding effects in some regions of the multiple bond and paramagnetic deshielding effects in other regions. Apparently, protons attached to double-bonded carbons are in the deshielding zones and thus are downfield while protons attached to triple-bonded carbons are in the shielding zones and are observed at rather high field. When a proton is directly bonded to a strongly electronegative atom such as oxygen or nitrogen its chemical shift is critically dependent on the nature of the solvent, temperature, concentration, and whether acidic or basic impurities are present. The usual variations in chemical shift for such protons are so large (up to \(5 \: \text{ppm}\) for alcohols) that no very useful correlations exist. Hydrogen bonding is the major reason for the variable chemical shifts of \(\ce{OH}\) and \(\ce{NH}\) protons. In general, hydrogen bonding results in , which causes the resonances to move downfield. The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in \(\ce{CCl_4}\) (Figure 9-29). The hydroxyl resonance will be seen to move upfield by hydrogen bonding through equilibria such as Many \(\ce{OH}\) and \(\ce{NH}\) compounds are weak acids and weak bases and can undergo autoprotolysis, which means that a proton can be transferred from one molecule to another. Suppose we have a compound such as 2-aminoethanol, \(\ce{H_2NCH_2CH_2OH}\). This substance normally would be expected to have an \(\ce{NH_2}\) proton resonance at about \(1 \: \text{ppm}\) and an \(\ce{OH}\) proton resonance at about \(3 \: \text{ppm}\). Autoprotolysis equilibria can the protons between the molecules and also from one end to the other as shown below, even if the equilibria are not very favorable. \[\begin{align} \ce{NH_2CH_2CH_2OH} &\overset{\rightarrow}{\longleftarrow} ^\oplus \ce{NH_3CH_2CH_2O}^\ominus \tag{9-5} \\ 2 \ce{NH_2CH_2CH_2OH} &\overset{\rightarrow}{\longleftarrow} ^\oplus \ce{NH_3CH_2CH_2OH} + \ce{N_2CH_2CH_2O}^\ominus \tag{9-6} \end{align}\] Such equilibria can be established very rapidly, especially if traces of a strong acid or a strong base are present. In such circumstances, a average \(\left( \ce{-NH_2}, \: \ce{-OH} \right)\) proton signal is observed, because the excitation of a given proton from its lower-energy magnetic state to its higher-energy magnetic state occurs while it is partly on oxygen and partly on nitrogen. This is the same kind of chemical shift averaging that occurs for rapidly equilibrating conformations (see ). To see how nmr and infrared spectra can be used together for structure determination we shall work through a representative example. The objective is to assign a structure to the compound \(\ce{C_4H_8O_3}\) whose nmr spectrum is shown in Figure 9-30 and whose infrared spectrum shows prominent bands at \(2900 \: \text{cm}^{-1}\), \(1750 \: \text{cm}^{-1}\), \(1000 \: \text{cm}^{-1}\), and \(1100 \: \text{cm}^{-1}\). The infrared spectrum indicates \(\left( 1750 \: \text{cm}^{-1} \right)\), \(\ce{C-H} \: \left( 2900 \: \text{cm}^{-1} \right)\), and \(\ce{C-O} \: \left( 1000 \: \text{cm}^{-1}, \: 1100 \: \text{cm}^{-1} \right)\). The position of the carbonyl band suggests that it is probably an ester, . A carboxylic acid is ruled out because there is no sign of an \(\ce{O-H}\) stretch. The nmr spectrum shows three kinds of signals corresponding to three kinds of protons. The integral shows these are in the ratio of 2:3:3. From this, we can conclude that they are two different kinds of \(\ce{CH_3-}\) groups and a \(\ce{-CH_2-}\) group. The chemical shifts of the presumed \(\ce{CH_3}\) groups are at \(3.70 \: \text{ppm}\) and \(3.35 \: \text{ppm}\). Because the compound contains only \(\ce{C}\), \(\ce{H}\), and \(\ce{O}\), the data of Table 9-4 suggest that these resonances arise from \(\ce{OCH_3}\) groups. The low-field resonance is likely to be (we know from the infrared that there probably is an ester function), while the higher-field resonance is possibly an ether function, \(\ce{-OCH_3}\). If you put all of this information together, you find that \(\ce{CH_3OCH_2CO_2CH_3}\) is the only possible structure. To check whether the \(\ce{CH_2}\) resonance at \(3.9 \: \text{ppm}\) is consistent with the assigned structure we can calculate a shift value from Equation 9-4: \[\begin{align} &\delta = 0.23 + \sigma_{OCH_3} + \sigma_{O=COCH_3} \\ &\delta = 0.23 + 2.36 + 1.55 = 4.14 \: \text{ppm} \end{align}\] The agreement between the calculated and observed shifts is not perfect, but is within the usual range of variation for Equation 9-4. We can be satisfied that the assigned structure is correct. If you look at the nmr spectrum of ethanol, \(\ce{CH_3CH_2OH}\), in Figure 9-23, you will see that the \(\ce{CH_2}\) resonance is actually a group of lines and the \(\ce{CH_3}\) resonance is a group of lines. This three-four line pattern for the grouping \(\ce{CH_3CH_2X} \: \left( \ce{X} \neq \ce{H} \right)\) also is evident in the \(220 \: \text{MHz}\) spectrum of 2-methyl-2-butanol (Figure 9-27) and in the \(60 \: \text{MHz}\) spectrum of ethyl iodide (Figure 9-32). Why do certain proton resonances appear as groups of equally spaced lines rather than single resonances? The facts are that protons on contiguous carbons , such as ethyl derivatives \(\ce{CH_3CH_2X}\), interact magnetically to "split" each other's resonances. This multiplicity of lines produced by the mutual interaction of magnetic nuclei is called " ", and while it complicates nmr spectra, it also provides valuable structural information, as we shall see. An example of a complex proton spectrum is that of ethyl iodide (Figure 9-32). To a first approximation, the two main groups of lines appear as sets of three and four lines, arising from what are called "first-order spin-spin interactions". Matters are further complicated by additional splitting of the "three-four" pattern of ethyl iodide, as also can be seen in Figure 9-32. This additional splitting is called "second-order" splitting. When there are so many lines present, how do we know what we are dealing with? From where to we measure the chemical shift in a complex group of lines? First, the chemical shift normally is at the of the group of lines corresponding to first-order splitting. In ethyl iodide, the chemical shift of the methyl protons is in the center of the quartet: Second, the chemical shift can be recognized by the fact that it is directly proportional to the transmitter frequency, \(\nu\). If we double \(\nu\), the chemical shifts double. In contrast, the first-order spin-spin splittings remain the same. By this we mean that the magnitude (in \(\text{Hz}\)) of the spacing between the lines of a split resonance is independent of the transmitter frequency, \(\nu\). This spacing corresponds to what is called the , or simply the , and is symbolized by \(J\). Third, the second-order splitting tends to disappear with increasing transmitter frequency. For ethyl iodide (Figure 9-32), the second-order splitting at \(60 \: \text{MHz}\) is barely discernible at \(100 \: \text{MHz}\) and disappears at \(200 \: \text{MHz}\). This also can be seen to occur for the three-four splitting pattern of 2-methyl-2-butanol as a function of \(\nu\) (Figure 9-27). The next question is how can we understand and predict what spin-spin splitting patterns will be observed? And how do they give us structural information? The important point is that the . For example, the \(\ce{CH_2}\) resonance of the ethyl group of ethyl iodide is a of lines because of the spin-spin interaction with the neighboring three protons \(\left( n = 3 \right)\) of the methyl group. Likewise, the \(\ce{CH_3}\) group is a of lines because of spin-spin interactions with the two protons \(\left( n = 2 \right)\) of the methylene group. The ratios of the line intensities in the spin-spin splitting patterns of Figure 9-33 usually follow simple rules. A doublet appears as two lines of equal intensity; a triplet as three lines in the ratio 1:2:1; a quartet as four lines in the ratio 1:3:3:1; a quintet as 1:4:6:4:1, and so on. The intensities follow the binomial coefficients for \(\left( x + y \right)^n\), where \(n\) is the number of protons in the splitting group. Thus when \(n = 4\), we have \(x^4 + 4 x^3y + 6 x^2 y^2 + 4 x y^3 + y^4\), or 1:4:6:4:1. The spectrum of \(\ce{(CH_3O)_2CHCH_3}\) (Figure 9-34) provides an excellent example of how nmr shows the presence of contiguous protons. The symmetrical doublet and 1:3:3:1 quartet are typical of the interaction between a single proton and an adjacent group of three, that is, . The methyl protons of the \(\ce{(CH_3O)}\) groups are too far from the others to give demonstrable spin-spin splitting; thus they appear as a single six-proton resonance. In general, the magnitude of the spin-spin splitting effect of one proton on another proton (or group of equivalent protons) depends on the of intervening chemical bonds and on the spatial relationships between the groups. For simple systems without double bonds and with normal bond angles, we usually find for protons (i.e., having different chemical shifts): Where restricted rotation or double- and triple-bonded groups are involved, widely divergent splittings are observed. For double bonds, the between two nonequivalent hydrogens located on one end are characteristically small, while the in \(\ce{-HC=CH}-\) are larger, especially for the trans configuration: Coupling through four or more bonds is significant for compounds with double or triple bonds. Examples of these so-called and some other useful splitting values follow: . A very important characteristic of three-bond proton-proton couplings, \(\ce{H-C-C-H}\), is the way that they depend on the conformation at the \(\ce{C-C}\) bond. Typical values for several particular conformations are You may have wondered why the hydroxyl proton of ethanol produces a single resonance in the spectrum of Figure 9-23. It is quite reasonable to expect that the hydroxyl proton would be split by the neighboring methylene protons because they are only three bonds apart, however, this coupling will not be observed if the hydroxyl protons are exchanging rapidly between the ethanol molecules ( ). When proton exchange is rapid, the spin interactions between the \(\ce{-CH_2}-\) and \(\ce{-OH}\) protons average to zero. At intermediate exchange rates, the coupling manifests itself through line broadening or by actually giving multiple lines. If you look at the several spectra of ethanol in Figure 9-29, you will notice how the shape of the \(\ce{OH}\) resonance varies from a broad singlet to a distinct triplet. Rapid chemical exchange of magnetic nuclei is not the only way that spin-coupling interactions can be averaged to zero. The same effect can be achieved by a technique known as double resonance. To understand how this is done, consider two coupled protons \(\ce{H}_\text{A}\) and \(\ce{H}_\text{B}\) having different chemical shifts. Suppose that \(\ce{H}_\text{A}\) is selectively irradiated at its resonance frequency \(\nu_\text{A}\) while at the same time we observe the resonance signal of \(\ce{H}_\text{B}\). The coupling between \(\ce{H}_\text{A}\) and \(\ce{H}_\text{B}\) disappears, and \(\ce{H}_\text{B}\) shows a resonance. Why is this so? By irradiation of \(\ce{H}_\text{A}\), the \(\ce{H}_\text{A}\) nuclei are changed from the +1/2 state to -1/2 and back again sufficiently rapidly that the neighboring nucleus \(\ce{H}_\text{B}\) effectively “sees” neither one state nor the other. The magnetic interaction between the states therefore averages to zero. This decoupling of magnetic nuclei by double resonance techniques is especially important in \(\ce{^{13}C}\) NMR spectroscopy ( ) but also is used to simplify proton spectra by selectively removing particular couplings. The solution of a typical structural analysis problem by nmr methods utilizes at least four kinds of information obtained directly from the spectrum. They are: chemical shifts \(\left( \delta \right)\), line intensities (signal areas), spin-spin splitting patterns (line mulitplicities), and coupling constants \(\left( J \right)\). We already have shown how chemical shifts are used in the absence of spin-spin splitting. We now will illustrate how more complex spectra may be analyzed. Figure 9-35 shows the proton nmr spectrum for a compound of formula \(\ce{C_3H_6O}\). There are three principal groups of lines at \(9.8\), \(2.4\), and \(1.0 \: \text{ppm}\). Look at the multiplicity of these groups before reading further. There are several ways to approach a problem such as this, but probably the easiest is to start with the integral. The relative heights of the stepped integral for the principal groups of lines can be obtained by a pair of dividers, with a ruler, or with horizontal lines as in Figure 9-35. The integral suggests that one hydrogen is responsible for the resonance at \(9.8 \: \text{ppm}\), two hydrogens at \(2.4 \: \text{ppm}\), and three at \(1.0 \: \text{ppm}\). Three hydrogens in a single group suggest a \(\ce{CH_3}-\) group, and because there is a three-four splitting pattern, it is reasonable to postulate \(\ce{CH_3-CH_2}-\). Subtracting \(\ce{C_2H_5}\) from the given formula \(\ce{C_3H_6O}\) leaves \(\ce{CHO}\), which, with normal valences, has to be \(\ce{-CH=O}\). The spectrum thus appears to be consistent with the structure \(\ce{CH_3CH_2CH=O}\) (propanal) as judged from the molecular formula and the spin-spin splitting pattern, which indicates the \(\ce{CH_3CH_2}-\) grouping. To be sure of the structure, we should check it against of the available information. First, from the shifts (Table 9-4) we see that the single proton at \(9.8 \: \text{ppm}\) fits almost perfectly for \(\ce{RCHO}\), the two-proton \(\ce{-CH_2C=O}\) resonance at \(2.4 \: \text{ppm}\) is consistent with that reported for \(\ce{-CH_2COR}\), while the three-proton line at \(1.0 \: \text{ppm}\) checks with \(0.9 \: \text{ppm}\) for \(\ce{CH_3R}\). What about the couplings? The three-four pattern has a spacing of slightly over \(7 \: \text{Hz}\), which is just right for an ethyl group (compare Figures 9-23 and 9-32). The doubling up (almost obscured by second-order splitting) of the \(\ce{-CH_2}-\) resonance and the splitting of the \(\ce{-CH=O}\) resonance into a 1:2:1 triplet indicate about a \(2\)-\(\text{Hz}\) coupling for the \(\ce{-CH_2-CH=O}\) group. Three-bond couplings between \(\ce{-CHO}\) and adjacent \(\ce{-CH_2}-\) protons appear to be generally smaller than \(\ce{-CH_2-CH_3}\) couplings. We usually would not rely on nmr alone in a structure-analysis problem of this kind, but would seek clues or corroboration from the infrared, electronic, or other spectra, as well as chemical tests. In later chapters we will have many problems that will be facilitated by the use of both nmr and infrared spectra. A further worked example will illustrate the approach. A compound has the composition \(\ce{C_3H_3Br}\) and gives the infrared and nuclear magnetic resonance spectra shown in Figure 9-36. The problem is how to use this information to deduce the structure of the compound. The molecular formula tells us the number and kind of atoms the number of multiple bonds or rings. The formulas of the corresponding \(\ce{C_3}\) hydrocarbon without the bromine would be \(\ce{C_3H_4}\), or hydrogens less than the saturated alkane \(\ce{C_3H_8}\). This means there must be two double bonds or the equivalent - one triple bond or one ring and one double bond.\(^{14}\) Because from the formula we suspect unsaturation, we should check this out with the infrared spectrum. There is a band at \(2120 \: \text{cm}^{-1}\), which is indicative of an unsymmetrically substituted \(\ce{-C \equiv C}-\) group (Table 9-2). The strong, sharp band at \(3300 \: \text{cm}^{-1}\) further tells us that the substance is a 1-alkyne \(\ce{-C \equiv C-H}\). The proton nmr spectrum shows that there are only two principal groups of lines - a two-proton doublet at \(3.85 \: \text{ppm}\) and a one-proton triplet at \(2.45 \: \text{ppm}\). The two-three splitting pattern combined with the 2:1 proton ratio suggests a \(\ce{CH_2}\) group coupled with a \(\ce{CH}\) group. The structure must be a 3-bromo-propyne, \(\ce{BrCH_2C \equiv CH}\). To confirm the assignment, the chemical shifts should be checked (Table 9-4). The \(\ce{\equiv C-H}\) at \(2.45 \: \text{ppm}\) agrees well with the tabulated value of \(2.5 \: \text{ppm}\). There is no tabulated data for \(\ce{-C \equiv C-CH_2Br}\) but the observed shift at \(3.85 \: \text{ppm}\) is at slightly lower fields than the tabulated \(3.33 \: \text{ppm}\) for \(\ce{-CH_2Br}\). This is expected because of the triple bond. The correlation of Equation 9-4 predicts a value of \(4.0 \: \text{ppm}\). Very often, a proton will be spin-coupled to two or more protons, and the couplings are not necessarily the same. When this happens, the resulting spectrum can be quite complex, as our next example shows. A compound \(\ce{C_9H_{10}}\) gives the nmr spectrum of Figure 9-37. There are clearly four kinds of protons in the molecule at \(\delta = 7.28 \: \text{ppm}\), \(5.35 \: \text{ppm}\), \(5.11 \: \text{ppm}\), and \(1.81 \: \text{ppm}\). Although the integral is not shown, the main groups of lines have intensities from the low-field to high-field in the ratio of 5:1:1:3. The five-proton signal at \(7.28 \: \text{ppm}\) is typical of a phenyl group, \(\ce{C_6H_5}\), and the one-proton signals at \(5.35\) and \(5.11 \: \text{ppm}\) are in the region for alkenic protons, . The three-proton signal at \(1.81 \: \text{ppm}\) is typical of a methyl group on a carbon-carbon double bond, . There are only three ways to put together a phenyl ring, , and two \(\ce{HC=}\) protons such that they add up to \(\ce{C_9H_{10}}\). They are Coupling between A and B (designated by the constant \(J_\text{AB}\)) should give four lines, two for A and two for B, as shown in Figure 9-38. Because A and B also are coupled to the three hydrogens of the methyl group (C), each of the four lines corresponding to \(J_\text{AB}\) will be further split (into 1:3:3:1 quartets). If \(J_\text{AC} \neq J_\text{BC}\), then the spacing of the lines in the two sets of quartets will not be the same. According to the foregoing analysis, the maximum number of lines observable for the A and B resonances is sixteen (8 for A and 8 for B). In fact, only eleven are visible (6 for A and 5 for B), which means that some of the sixteen possible lines must overlap. Without examining all possibilities, we can see that the actual situation can be reproduced if \(J_\text{AB} \cong J_\text{BC} = 2J_\text{AC}\). The only structure that is consistent with \(J_\text{AB} = 1.5 \: \text{Hz}\) is \(13\), or 2-phenylpropene; the other possibilities are excluded because \(J_\text{AB}\) should be about \(10 \: \text{Hz}\) for \(12\) and \(16 \: \text{Hz}\) for \(11\). The simple \(n + 1\) rule for predicting the multiplicity of spin-coupled proton signals often breaks down whenever the chemical-shift difference between the protons in different groups becomes comparable to coupling constants for magnetic interaction between the groups. Under these circumstances, you may expect to see more lines, or lines in different positions with different intensities, than predicted from the simple first-order treatment. One example is the effect of changing chemical shift on a two-proton spectrum with \(J = 10 \: \text{Hz}\) (Figure 9-44). We see in Figure 9-44 that even when the shift is 7.5 times larger than the coupling, the outside lines are weaker than the inside lines. This general kind of asymmetry of line intensities also is apparent in the spectrum of ethyl iodide (Figure 9-32), in which the lines of each group are more like 0.7:2.5:3.5:1.3 and 1.2:2.0:0.8, rather than the 1:3:3:1 and 1:2:1 ratios predicted from the first-order treatment. The asymmetry is such that two groups of lines that are connected by spin-spin splitting in effect "point" to one another - the lines on the "inside" of the pattern are stronger than predicted from the first-order treatment, whereas those on the "outside" are weaker. The effect can be put to practical use, as illustrated in the following exercise. To explain the effect of chemical shifts on second-order splitting is beyond the scope of this book. In fact, we haven't really explained first-order splitting, although more on this topic will be found in . But regardless of how many lines appear in a complex nmr spectrum, they can be rationalized in terms of the chemical shifts, coupling constants, and exchange effects. Furthermore, the overall signal intensities remain proportional to the number of protons giving rise to the signals. When there are many hydrogens and small chemical-shift differences, as in alkanes, the proton nmr spectra may have so many closely spaced resonance lines that they merge together to give a series of smooth, more-or-less featureless peaks. The proton spectrum of octane (Figure 9-46a) is an excellent example of this type of spectrum. Useful information often can be obtained from such spectra as to the ratio of \(\ce{CH_3}\) : \(\ce{CH_2}\) : \(\ce{CH}\) by investigation of the integrals over the range of alkane proton absorptions. Figure 9-46 illustrates how this can be done for octane and 2,2,4-trimethylpentane. In recent years \(\ce{^{13}C}\) nmr spectroscopy using \(\ce{^{13}C}\) of natural abundance \(\left( 1.1 \% \right)\) has become an important tool for organic structural analysis. That this did not happen sooner is because \(\ce{^{13}C}\) has a much smaller magnetic moment than \(\ce{^1H}\) and the small moment combined with the small natural abundance means that \(\ce{^{13}C}\) is harder to detect in the nmr than \(\ce{^1H}\) by a factor of 5700. This is a large difference and can be put in the proper context in the following way. Suppose two people are talking in a noisy room and one is trying to hear the other. The common request is "talk louder". If this is not possible then the request is "say it again" or "talk more slowly". Either of the latter requests amounts to an integration of signal versus noise and takes time. Improvement in signal-to-noise for a given communication is achieved as the of the of communication. On the crucial , \(\ce{^{13}C}\) nmr signals require \(\left( 5700 \right)^2 \cong 30,000,000\) times more time to get the same signal-to-noise ratio as in \(\ce{^1H}\) nmr for the same number of nuclei per unit volume. This is a problem. Electronic improvements and use of communication theory, with emphasis on the "say-it-again" technique, have provided the means for obtaining routine \(\ce{^{13}C}\) spectra for even fairly dilute solutions of quite complex molecules. Some of the same kinds of structural effects are important for \(\ce{^{13}C}\) chemical shifts as for proton chemical shifts (Section 9-10E). For example, there is a similar parallel between \(\ce{^{13}C}\) shift differences in compounds of the type \(\ce{CH_3-CH_2-X}\) and electronegativity (Figure 9-47) as between the corresponding proton shifts and electronegativity (Figure 9-28). It is important to notice that \(\ce{^{13}C}\) shifts in \(\text{ppm}\) units are much larger than those of protons. This is because carbon uses \(p\) orbitals in forming bonds, whereas hydrogen uses \(s\) orbitals. We therefore will expect to find the the nuclei of other elements that use \(p\) orbitals in bonding, such as \(\ce{^{15}N}\), \(\ce{^{19}F}\), and \(\ce{^{31}P}\), also will have larger shifts than for protons, as indeed they do. A structural application of \(\ce{^{13}C}\) nmr, which shows its power in an area where \(\ce{^1H}\) nmr is indecisive, is shown in Figure 9-48. Here, we see the high-field \(\ce{^{13}C}\) resonances of a substance known variously as Coumadin, or the sodium salt of warfarin, \(14\), which is used widely as a blood anticoagulant in the treatment of diseases such as phlebitis. It also has substantial utility as a rat poison because of its anticoagulant action. There is no indication of any abnormality in the chemical shifts of carbons 11, 12, and 14 shown in Figure 9-48a. Furthermore, there is a downfield resonance \(216.5 \: \text{ppm}\) from the carbons of TMS (not shown in Figure 9-48a) which is typical of a \(\ce{C=O}\) carbon corresponding to C13. When \(14\) is treated with acid, we expect the product (warfarin) of structure \(15\) to be formed, which should have a \(\ce{^{13}C}\) spectrum much like that shown in Figure 9-48a. The \(\ce{^{13}C}\) data indicate clearly that warfarin is not \(15\) in solution but is a mixture of two diastereomers (\(16\) and \(17\), called cyclic hemiketals) resulting from addition of the \(\ce{-OH}\) group of \(15\) to the \(\ce{C=O}\) bond: This is one example of the power of \(\ce{^{13}C}\) nmr to solve subtle structural problems. \(^8\)Although the principal isotopes of \(Cl\), \(Br\), and \(I\) have magnetic properties, because of the special character of all of these isotopes, they act in organic compounds as though they were . \(^9\) in the sense used here means that the radio-frequency absorption takes place at specified "resonance" frequencies. However, you will see that almost all of the forms of spectroscopy we discuss in this book involve "resonance" absorption in the same sense. \(^{10}\)Here, \(\gamma\) is in \(\text{Hz}\) per gauss; physicists usually define \(\gamma\) in radians per second per gauss. \(^{11}\)From the Greek prefix meaning through, across. The opposite of diamagnetic is ; meaning alongside. We shall use this term later. \(^{12}\)In addition to giving better separation of the lines and clearer spectra, going to higher fields also has the beneficial effect of increasing the proportions of the nuclei in the \(+ \frac{1}{2}\) state, thereby giving more intense, easier-to-detect resonances. \(^{13}\)Many other proton-shift values are available in , Volume 1 and 2, Varian Associates, Palo Alto, Calif., 1962, 1963. \(^{14}\)If two rings were present, this also would give four hydrogens less than the alkane. However, two rings are not possible with only three carbons. and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.11%3A_Nuclear_Magnetic_Resonance_Spectroscopy

Nuclear magnetic resonance (NMR) spectroscopy is extremely useful for identification and analysis of organic compounds. The principle on which this form of spectroscopy is based is simple. The nuclei of many kinds of atoms act like tiny magnets and tend to become aligned in a magnetic field. In NMR spectroscopy, we measure the energy required to change the alignment of magnetic nuclei in a magnetic field. To illustrate the procedure with a simple example, consider the behavior of a proton \(\left( ^1H \right)\) in a magnetic field. There are two possible alignments of this magnetic nucleus with respect to the direction of the applied field, as shown in Figure 9-21. The nuclear magnets can be aligned either with the field direction, or opposed to it. The two orientations are not equivalent, and energy is required to change the more stable alignment to the less stable alignment. A schematic diagram of an NMR instrument is shown in Figure 9-22. When a substance such as ethanol, \(CH_3-CH_2-OH\), the hydrogens of which have nuclei (protons) that are magnetic, is placed in the transmitter coil and the magnetic field is increased gradually, at certain field strengths radio-frequency energy is absorbed by the sample and the ammeter indicates an increase in the flow of current in the coil. The overall result is a spectrum such as the one shown in Figure 9-23. This spectrum is detailed enough to serve as a useful "fingerprint" for ethanol, and also is simple enough that we will be able to account for the origin of each line. It is the purpose of this section to explain how the complexities of spectra such as that of Figure 9-23 can be interpreted in terms of chemical structure. For what kinds of substances can we expect nuclear magnetic resonance absorption to occur? Magnetic properties always are found with nuclei of odd-numbered masses, \(^1H\), \(^{13}C\), \(^{15}N\), \(^{17}O\), \(^{19}F\), \(^{31}P\), and so on, as well as for nuclei of even mass but odd atomic number, \(^2H\), \(^{10}B\), \(^{14}N\), and so on.\(^8\) Nuclei such as \(^{12}C\), \(^{16}O\), and \(^{32}S\), which have even mass and atomic numbers, have magnetic properties and do give nuclear magnetic resonance signals. For various reasons, routine use of NMR spectra in organic chemistry is confined to \(^1H\), \(^{19}F\), \(^{13}C\), and \(^{31}P\). We shall be concerned in this chapter only with NMR spectra of hydrogen (\(^1H\)) and of carbon (\(^{13}C\)). The kind of NMR spectroscopy we shall discuss here is limited in its applications because it can be carried out only with or . Fortunately, the allowable range of solvents is large, from hydrocarbons to concentrated sulfuric acid, and for most compounds it is possible to find a suitable solvent. Nuclear magnetic resonance spectra may be so simple as to have only a single absorption peak, but they also can be much more complex than the spectrum of Figure 9-23. However, it is important to recognize that no matter how complex an NMR spectrum appears to be, in involves just parameters: , , and (reaction-rate) . We shall have more to say about each of these later. First, let us try to establish the relationship of NMR spectroscopy to some of the other forms of spectroscopy we have already discussed in this chapter. Nuclear magnetic resonance\(^9\) spectroscopy involves transitions between possible energy levels of magnetic nuclei in an applied magnetic field (see Figure 9-21). The transition energies are related to the frequency of the absorbed radiation by the familiar equation \(\Delta E - h \nu\). An important difference between nmr and other forms of spectroscopy is that \(\Delta E\) is influenced by the strength of the applied field. This should not be surprising, because if we are to measure the energy of changing the direction of alignment of a magnetic nucleus in a magnetic field, then the stronger the field the more energy will be invloved. Nuclear spin (symbolized as \(I\)) is a quantized property that correlates with nuclear magnetism such that when \(I\) is zero the nucleus has no spin and no magnetic properties. Examples are \(^{12}C\) and \(^{16}O\). Several nuclei of particular interest to organic chemists - \(^1H\), \(^{13}C\), \(^{15}\), \(^{19}\), and \(^{31}P\) - have spin of \(\frac{1}{2}\). With \(I = \frac{1}{2}\) there are only magnetic energy states of the nucleus in a magnetic field. These states are designated with the \(+ \frac{1}{2}\) and \(- \frac{1}{2}\). The difference in energy between these states, \(\Delta E\), is given by \[\Delta E = \gamma h H = h \nu\] or \[\nu = \gamma H\] in which \(h\) is Planck's constant, \(\nu\) is in hertz, \(\gamma\) is a nuclear magnetic constant called the ,\(^{10}\), and \(H\) is the magnetic field strength at the nucleus. In general, \(H\) will not be exactly equal to \(H_\text{o}\), the applied magnetic field and, as we will see, this difference leads to important chemical information. Each of nucleus (\(^1H\), \(^{13}C\), \(^{15}N\), etc.) has its own \(\gamma\) value and, consequently, will undergo transitions at different frequencies at any particular value of \(H\). This should become clearer by study of Figure 9-24. There are several modes of operation of an nmr spectrometer. First and most common, we hold \(\nu\) constant and vary (or "sweep") \(H_\text{o}\). Close to \(\nu = \gamma H\), energy is absorbed by the nuclei and the current flow from the transmitter increases until \(\nu\) is exactly equal to \(\gamma H\). Further increase of \(H_\text{o}\) makes \(\nu < \gamma H_\text{o}\) and the current flow decreases. The form of the energy-absorption curve as a function of \(H_\text{o}\) when \(H_\text{o}\) is changed very slowly is shown in Figure 9-25a. The peak is centered on the point where \(\nu = \gamma H\). When \(H_\text{o}\) is changed more rapidly, transient effects are observed on the peak, which are a consequence of the fact that the nuclei do not revert instantly from the \(- \frac{1}{2}\) to \(+ \frac{1}{2}\) state. The resulting phenomenon is called "ringing" and is shown in Figures 9-25b and 9-25c. Evidence of ringing also will be seen on peaks of Figure 9-23. An alternative method of running an nmr spectrometer is to hold the magnetic field constant and to sweep the transmitter frequency through the resonances. This mode of operation is more like other forms of spectroscopy and gives the same line shapes as sweeping the field (Figure 9-25). What energy is associated with a \(^1H\) nmr transition? The magnitude of this energy may be calculated from the relationship between energy and wavelength (frequency) of the absorbed radiation ( ). That is, \[\Delta E = \frac{28,600}{\lambda} \text{kcal mol}^{-1}\] and \[\lambda = \frac{c}{\nu}\] The frequency \(\nu\) is the operating frequency of the spectrometer, which we will take as \(60 \: \text{MHz}\) or \(6 \times 10^7 \: \text{Hz}\) (cycles \(\text{sec}^{-1}\)), and the velocity of light is \(3 \times 10^8 \: \text{m sec}^{-1}\). Hence \[\lambda = \frac{3 \times 10^8 \times 10^9 \left( \text{nm sec}^{-1} \right)}{6 \times 10^7 \left( \text{Hz} \right)} = 5 \times 10^9 \: \text{nm}\] and \[\Delta E = \frac{28,600}{5 \times 10^9} = 5.7 \times 10^{-6} \: \text{kcal mol}^{-1}\] This is a very small energy difference, which means that only very few more of the nuclei are in the more stable \(+ \frac{1}{2}\) state than in the less stable \(- \frac{1}{2}\) state. The equilibrium constant \(K\) for \(- \frac{1}{2} \rightleftharpoons + \frac{1}{2}\) calculated from Equation 4-2 for \(25^\text{o}\) (298 \: \text{K}\)) and neglecting possible entropy effects is 1.000010! The plot of signal against magnetic field strength for ethanol in Figure 9-23 shows three principal groups of lines corresponding to the three varieties of hydrogen present: methyl (\(CH_3\)), methylene (\(CH_3\)), and hydroxyl (\(OH\)). Differences in the field strengths at which signals are obtained for nuclei of the same kind, such as protons, but located in different molecular environments, are called . Another very important point to notice about Figure 9-23 is that the intensities of the three principal absorptions are in the ratio of 1:2:3, corresponding to the ratio of the number of each kind of proton (\(OH\), \(CH_2\), \(CH_3\)) producing the signal. In general, such as in Figure 9-23 . The areas can be measured by electronic integration and the integral often is displayed on the chart, as it is in Figure 9-23, as a stepped line increasing from left to right. The height of each step corresponds to the relative number of nuclei of a particular kind. Unless special precautions are taken, integrals usually should not be considered accurate to better than about \(5\%\). Why do protons in different molecular environments absorb at different field strengths? The field strength \(H\) at a particular nucleus is than the strength of the external magnetic field \(H_\text{o}\). This is because the valence electrons around a particular nucleus and around neighboring nuclei respond to the applied magnetic field so as to the nucleus from the applied field. The way this shielding occurs is as follows. First, when an atom is placed in a magnetic field, its electrons are forced to undergo a rotation about the field axis, as shown in Figure 9-26. Second, rotation of the electrons around the nucleus is a circulation of charge, and this creates a magnetic field at the nucleus in the direction to \(H_\text{o}\). Third, the magnitude of this \(^{11}\) effect is directly proportional to \(H_\text{o}\) and can be quantified as \(\sigma H_\text{o}\), in which \(\sigma\) is the proportionality constant. It is important to recognize that \(\sigma\) is not a nuclear property but depends on the environment of the atom. Each chemically different proton will have a different value of \(\sigma\) and hence a different chemical shift. The actual field \(H\) at the nucleus will be \(H_\text{o} - \sigma H_\text{o}\). Because \(\sigma\) acts to reduce the strength of the applied field at the nucleus, it is called the . The more shielding there is, the stronger the applied field must be to satisfy the resonance condition, Common usage is: upfield, shielding; downfield, shielding; and you should remember that field-sweep spectra always are recorded with the field from to . The value of nmr spectroscopy in structure determination lies in the fact that chemically different nuclei absorb at different field strengths. In later sections we will be concerned with correlating the chemical shifts with structural features. However, before proceeding furher it is extremely important that you be able to identify the number and kind of nonequivalent protons in a given structure, and therefore the number of chemical shifts to expect. This number is not always self-evident, especially when subtle factors of stereochemistry intervene. For this reason, we suggest that you inspect structures \(3\)-\(5\) to convince yourself that the protons labeled with different letter subscripts in any one molecule are indeed chemically different. One way of checking whether two protons are in equivalent environments is to imagine that each is separately replaced with a different atom or group. If the product of replacing \(H_\text{A}\) is identical with that obtained by replacing \(H_\text{B}\), then \(H_\text{A}\) and \(H_\text{B}\) are chemically equivalent. If the two products are nonidentical, then \(H_\text{A}\) and \(H_\text{B}\) are nonequivalent. For example, replacement of \(H_\text{A}\) or \(H_\text{B}\) in \(3\), \(4\), and \(5\) by an atom \(X\) would give different products. Therefore, \(H_\text{A}\) and \(H_\text{B}\) are nonequivalent in \(3\), \(4\), and \(5\). Matters become more complicated with substances such as \(6\) and \(7\): Notice that \(6\) represents a chiral molecule and if \(H_\text{A}\) and \(H_\text{B}\) each are replaced with \(X\) we get \(8\) and \(9\), which are diastereomers (see ). You can verify this with molecular models if necessary. Diastereomers have different chemical and physical properties; therefore \(H_\text{A}\) and \(H_\text{B}\) in \(6\) are nonequivalent. They often are called hydrogens. What of the two methylene protons in ethanol, \(7\), which we have labeled as \(H_\text{A}\) \(H_\text{A'}\)? Are they identical? In a sense they are not identical because, if each were replaced by \(X\), we would have a pair of enantiomers. Therefore, \(H_\text{A}\) and \(H_\text{A'}\) sometimes are called hydrogens. But, you will recall that enantiomers are chemically indistinguishable unless they are in a chiral environment. Therefore we expect shifts of enantiotopic hydrogens to be identical, unless they are in a chiral environment. To summarize, enantiotopic protons normally will have the same chemical shifts, whereas diastereotopic protons normally will have different chemical shifts. We so far have ignored the relationship of chemical shifts to conformational equilibria. Consider a specific example, 1,2-dibromoethane, for which there are three staggered conformations \(10a\), \(10b\), and \(10c\): Each of these conformations is expected to have its own nmr spectrum. The two gauche forms, \(10a\) and \(10b\), are enantiomers and their spectra should be identical. The hydrogens \(H_\text{A}\) in \(10a\) each are trans to the bromine on the adjacent carbon, while the \(H_\text{B}\) hydrogens are cis to the same bromines (see ). Consequently the \(H_\text{A}\) and \(H_\text{B}\) hydrogens are nonequivalent and would be expected to have different chemical shifts. In contrast, all of the hydrogens of the anti conformer, \(10c\), are equivalent and would have the same chemical shift. Therefore we would expect to observe three chemical shifts arising from \(H_\text{A}\), \(H_\text{B}\), and \(H_\text{C}\) for a mixture of \(10a\), \(10b\), and \(10c\). However, the actual spectrum of 1,2-dibromoethane shows only under ordinary conditions. The reason is that the magnetic nuclei can absorb the exciting radiation. The result is that we observe an chemical shift, which reflects the relative shifts and populations of the three conformers present. If we can go to a sufficiently low temperature to make interconversion of the conformations slow (on the order of 10 times per second), then we will expect to see the three different chemical shifts \(H_\text{A}\), \(H_\text{B}\), and \(H_\text{C}\) with intensities corresponding to the actual populations of the conformations at the sample temperature. This is one example of the effect of rate processes on nmr spectra. Other examples and a more detailed account of how to relate the appearance of the signal to the rates of the exchange processes are given in . Chemical shifts are measured with reference to a standard. For protons or \(^{13}C\) in organic molecules, , \(\left( CH_3 \right)_4 Si\), which gives strong, sharp nmr signals in regions where only a very few other kinds of protons or carbon nuclei absorb. Chemical shifts often are expressed in \(\text{Hz}\) (cycles per second) relative to tetramethylsilane (TMS). These may seem odd units for magnetic field strength but because resonance occurs at \(\nu = \gamma H\), either frequency units (\(\text{Hz}\), radians \(\text{sec}^{-1}\)) or magnetic field units (gauss) are appropriate. Ten years ago, most nmr spectrometers operated for protons with radio-frequency (rf) transmitters set at \(60 \: \text{MHz}\) (\(6 \times 10^7\) cycles per second) but there has been a proliferation of different proton-operating frequencies and now \(30\), \(60\), \(90\), \(100\), \(220\), \(270\), \(300\) and \(360 \: \text{MHz}\) machines are commercially available. The cost of these machines is roughly proportional to the square of the frequency, and one well may wonder why there is such an exotic variety available and what this has to do with the chemical shift. High operating frequencies are desirable because chemical shifts increase with spectrometer frequency, and this makes the spectra simpler to interpret. A 12-fold increase in operating frequency (as from \(30 \: \text{MHz}\) to \(360 \: \text{MHz}\)) means a 12-fold increase in \(H_\text{o}\) at the point of resonance (remember \(\nu = \gamma H\)) and this means also a 12-fold increase in \(\sigma H_\text{o}\). Thus resonances that differ because they correspond to different \(\sigma\) values will be at \(360 \: \text{MHz}\) than at \(30 \: \text{MHz}\). This can produce a dramatic simplification of spectra, as can be seen from Figure 9-27, which shows the effect of almost a factor of four in \(\nu\) on the proton nmr spectrum of 2-methyl-2-butanol.\(^{12}\) To reiterate, chemical shifts are strictly proportional to spectrometer frequency, thus lines \(100 \: \text{Hz}\) apart at \(60 \: \text{MHz}\) will be \(167 \: \text{Hz}\) apart at \(100 \: \text{MHz}\). This might seem to make comparisons of nmr spectra on different spectrometers hopelessly complex but, because of the proportionality of shifts to frequency (or field), (relative to the same standard) for any spectrometer , which are useful for all nmr spectrometers. Nmr shifts reported in \(\text{ppm}\) relative to TMS as zero, as shown in Figure 9-23, are called \(\delta\) (delta) values: \[\delta = \frac{\left( \text{chemical shift downfield in Hz relative to TMS} \right) \times 10^6}{\text{spectrometer frequency in Hz}}\] Thus, if at \(60 \: \text{MHz}\) a proton signal comes \(100 \: \text{Hz}\) downfield relative to tetramethylsilane, it can be designated as being \(\left( +100 \: \text{Hz} \times 10^6 \right)/ 60 \times 10^6 \: \text{Hz} = +1.67 : \text{ppm}\) relative to tetramethylsilane. At \(100 \: \text{MHz}\), the line then will be \(\left( 1.67 \times 10^{-6} \right) \left(100 \times 10^6 \right) = 167 \: \text{Hz}\) downfield from tetramethylsilane. Typical proton chemical shifts relative to TMS are given in Table 9-4.\(^{13}\) The values quoted for each type of proton may, in practice, show variations of \(0.1\)-\(0.3 \: \text{ppm}\). This is not unreasonable, because the chemical shift of a given proton is expected to depend somewhat on the nature of the particular molecule involved, and also on the solvent, temperature, and concentration. A positive \(\delta\) value means a shift to lower field (or lower frequency) with respect to TMS, whereas a negative \(\delta\) signifies a shift to higher field (or higher frequency). The \(\delta\) convention is accepted widely, but you often find in the literature proton shifts with reference to TMS reported as "\(\tau\) values." The \(\tau\) scale has the TMS reference at \(+10 \: \text{ppm}\), so most proton signals fall in the range of \(\tau = 0\) to \(\tau = +10\). A \(\tau\) value can be converted to the appropriate \(\delta\) value by subtracting it from 10. Life with nmr spectra would be simpler if the \(\tau\) scale would just go away. Protonc chemical shifts are very valuable for the determination of structures, but to use the shifts in this way we must know something about the correlations that exist between chemical shift and structural environment of protons in organic compounds. The most important effects arise from differences in electronegativity, types of carbon bonding, hydrogen bonding, and chemical exchange. Consider first the chemical shifts of protons attached to an \(sp^3\) carbon, . The degree of shielding of the proton by the carbon valence electrons depends on the character of the substituent atoms and groups present, and particularly on their electron-attracting power, or electronegativity. For a grouping of the type , the shielding will be as \(\ce{X}\) is more electron withdrawing relative to hydrogen: If \(\ce{X}\) is electron-withdrawing, the proton is deshielded. For example, the proton chemical shifts of the methyl halides (Table 9-4) show decreasing shielding, hence progressively low-field chemical shifts with increasing halogen electronegativity \(\left( \ce{F} > \ce{Cl} > \ce{Br} > \ce{I} \right)\): When two electronegative groups, \(\ce{X}\) and \(\ce{Y}\), are bonded to the carbon, as in \(\ce{XCH_2Y}\), the protons are expected to be less shielded and come into resonance downfield of the methylenes of \(\ce{XCH_2CH_2Y}\). There is an approximate relationship (see below) between the shifts of the \(\ce{XCH_2Y}\) protons and the effective shielding constants \(\left( \sigma \right)\) of \(\ce{X}\) and \(\ce{Y}\) known as . \[\delta = 0.23 + \sigma_x + \sigma_y \tag{9-4}\] Appropriate values of \(\sigma\) for use with this equation are given in Table 9-4. The shifts of the protons of alkanes and cycloalkanes fall in the range of \(0.9\)-\(1.5 \: \text{ppm}\) with \(\ce{C-H}\) protons coming at the low-field end of this range and \(\ce{-CH_3}\) protons coming at the high-field end (see Table 9-4). Alkenic hydrogens (vinyl hydrogens, ) normally are observed between \(4.6\)-\(6.3 \: \text{ppm}\) toward fields than the shifts of protons in alkanes and cycloalkanes. This means that alkenic hydrogens in an organic compound can be easily distinguished from alkane hydrogens. Aromatic protons, such as those in benzene, have shifts at still lower fields and commonly are observed at \(7\)-\(8 \: \text{ppm}\). In contrast, alkynic protons of the type \(\ce{-C \equiv CH}\) give resonances that are of alkenic or aromatic protons and come at \(2\)-\(3 \: \text{ppm}\). Another effect associated with multiple bonds is the large difference in shift between a \(\ce{-CH(OCH_3)_2}\) proton, which normally comes at about \(5.5 \: \text{ppm}\), and aldehyde protons, \(\ce{-CH=O}\), which are much farter at \(9\)-\(11 \: \text{ppm}\). Clearly, the shifts of a proton depend on whether the carbon forms single, double, or triple bonds. In a magnetic field, the circulation of electrons in the \(\pi\) orbitals of multiple bonds induced by the field (Figure 9-26) generates diamagnetic shielding effects in some regions of the multiple bond and paramagnetic deshielding effects in other regions. Apparently, protons attached to double-bonded carbons are in the deshielding zones and thus are downfield while protons attached to triple-bonded carbons are in the shielding zones and are observed at rather high field. When a proton is directly bonded to a strongly electronegative atom such as oxygen or nitrogen its chemical shift is critically dependent on the nature of the solvent, temperature, concentration, and whether acidic or basic impurities are present. The usual variations in chemical shift for such protons are so large (up to \(5 \: \text{ppm}\) for alcohols) that no very useful correlations exist. Hydrogen bonding is the major reason for the variable chemical shifts of \(\ce{OH}\) and \(\ce{NH}\) protons. In general, hydrogen bonding results in , which causes the resonances to move downfield. The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in \(\ce{CCl_4}\) (Figure 9-29). The hydroxyl resonance will be seen to move upfield by hydrogen bonding through equilibria such as Many \(\ce{OH}\) and \(\ce{NH}\) compounds are weak acids and weak bases and can undergo autoprotolysis, which means that a proton can be transferred from one molecule to another. Suppose we have a compound such as 2-aminoethanol, \(\ce{H_2NCH_2CH_2OH}\). This substance normally would be expected to have an \(\ce{NH_2}\) proton resonance at about \(1 \: \text{ppm}\) and an \(\ce{OH}\) proton resonance at about \(3 \: \text{ppm}\). Autoprotolysis equilibria can the protons between the molecules and also from one end to the other as shown below, even if the equilibria are not very favorable. \[\begin{align} \ce{NH_2CH_2CH_2OH} &\overset{\rightarrow}{\longleftarrow} ^\oplus \ce{NH_3CH_2CH_2O}^\ominus \tag{9-5} \\ 2 \ce{NH_2CH_2CH_2OH} &\overset{\rightarrow}{\longleftarrow} ^\oplus \ce{NH_3CH_2CH_2OH} + \ce{N_2CH_2CH_2O}^\ominus \tag{9-6} \end{align}\] Such equilibria can be established very rapidly, especially if traces of a strong acid or a strong base are present. In such circumstances, a average \(\left( \ce{-NH_2}, \: \ce{-OH} \right)\) proton signal is observed, because the excitation of a given proton from its lower-energy magnetic state to its higher-energy magnetic state occurs while it is partly on oxygen and partly on nitrogen. This is the same kind of chemical shift averaging that occurs for rapidly equilibrating conformations (see ). To see how nmr and infrared spectra can be used together for structure determination we shall work through a representative example. The objective is to assign a structure to the compound \(\ce{C_4H_8O_3}\) whose nmr spectrum is shown in Figure 9-30 and whose infrared spectrum shows prominent bands at \(2900 \: \text{cm}^{-1}\), \(1750 \: \text{cm}^{-1}\), \(1000 \: \text{cm}^{-1}\), and \(1100 \: \text{cm}^{-1}\). The infrared spectrum indicates \(\left( 1750 \: \text{cm}^{-1} \right)\), \(\ce{C-H} \: \left( 2900 \: \text{cm}^{-1} \right)\), and \(\ce{C-O} \: \left( 1000 \: \text{cm}^{-1}, \: 1100 \: \text{cm}^{-1} \right)\). The position of the carbonyl band suggests that it is probably an ester, . A carboxylic acid is ruled out because there is no sign of an \(\ce{O-H}\) stretch. The nmr spectrum shows three kinds of signals corresponding to three kinds of protons. The integral shows these are in the ratio of 2:3:3. From this, we can conclude that they are two different kinds of \(\ce{CH_3-}\) groups and a \(\ce{-CH_2-}\) group. The chemical shifts of the presumed \(\ce{CH_3}\) groups are at \(3.70 \: \text{ppm}\) and \(3.35 \: \text{ppm}\). Because the compound contains only \(\ce{C}\), \(\ce{H}\), and \(\ce{O}\), the data of Table 9-4 suggest that these resonances arise from \(\ce{OCH_3}\) groups. The low-field resonance is likely to be (we know from the infrared that there probably is an ester function), while the higher-field resonance is possibly an ether function, \(\ce{-OCH_3}\). If you put all of this information together, you find that \(\ce{CH_3OCH_2CO_2CH_3}\) is the only possible structure. To check whether the \(\ce{CH_2}\) resonance at \(3.9 \: \text{ppm}\) is consistent with the assigned structure we can calculate a shift value from Equation 9-4: \[\begin{align} &\delta = 0.23 + \sigma_{OCH_3} + \sigma_{O=COCH_3} \\ &\delta = 0.23 + 2.36 + 1.55 = 4.14 \: \text{ppm} \end{align}\] The agreement between the calculated and observed shifts is not perfect, but is within the usual range of variation for Equation 9-4. We can be satisfied that the assigned structure is correct. If you look at the nmr spectrum of ethanol, \(\ce{CH_3CH_2OH}\), in Figure 9-23, you will see that the \(\ce{CH_2}\) resonance is actually a group of lines and the \(\ce{CH_3}\) resonance is a group of lines. This three-four line pattern for the grouping \(\ce{CH_3CH_2X} \: \left( \ce{X} \neq \ce{H} \right)\) also is evident in the \(220 \: \text{MHz}\) spectrum of 2-methyl-2-butanol (Figure 9-27) and in the \(60 \: \text{MHz}\) spectrum of ethyl iodide (Figure 9-32). Why do certain proton resonances appear as groups of equally spaced lines rather than single resonances? The facts are that protons on contiguous carbons , such as ethyl derivatives \(\ce{CH_3CH_2X}\), interact magnetically to "split" each other's resonances. This multiplicity of lines produced by the mutual interaction of magnetic nuclei is called " ", and while it complicates nmr spectra, it also provides valuable structural information, as we shall see. An example of a complex proton spectrum is that of ethyl iodide (Figure 9-32). To a first approximation, the two main groups of lines appear as sets of three and four lines, arising from what are called "first-order spin-spin interactions". Matters are further complicated by additional splitting of the "three-four" pattern of ethyl iodide, as also can be seen in Figure 9-32. This additional splitting is called "second-order" splitting. When there are so many lines present, how do we know what we are dealing with? From where to we measure the chemical shift in a complex group of lines? First, the chemical shift normally is at the of the group of lines corresponding to first-order splitting. In ethyl iodide, the chemical shift of the methyl protons is in the center of the quartet: Second, the chemical shift can be recognized by the fact that it is directly proportional to the transmitter frequency, \(\nu\). If we double \(\nu\), the chemical shifts double. In contrast, the first-order spin-spin splittings remain the same. By this we mean that the magnitude (in \(\text{Hz}\)) of the spacing between the lines of a split resonance is independent of the transmitter frequency, \(\nu\). This spacing corresponds to what is called the , or simply the , and is symbolized by \(J\). Third, the second-order splitting tends to disappear with increasing transmitter frequency. For ethyl iodide (Figure 9-32), the second-order splitting at \(60 \: \text{MHz}\) is barely discernible at \(100 \: \text{MHz}\) and disappears at \(200 \: \text{MHz}\). This also can be seen to occur for the three-four splitting pattern of 2-methyl-2-butanol as a function of \(\nu\) (Figure 9-27). The next question is how can we understand and predict what spin-spin splitting patterns will be observed? And how do they give us structural information? The important point is that the . For example, the \(\ce{CH_2}\) resonance of the ethyl group of ethyl iodide is a of lines because of the spin-spin interaction with the neighboring three protons \(\left( n = 3 \right)\) of the methyl group. Likewise, the \(\ce{CH_3}\) group is a of lines because of spin-spin interactions with the two protons \(\left( n = 2 \right)\) of the methylene group. The ratios of the line intensities in the spin-spin splitting patterns of Figure 9-33 usually follow simple rules. A doublet appears as two lines of equal intensity; a triplet as three lines in the ratio 1:2:1; a quartet as four lines in the ratio 1:3:3:1; a quintet as 1:4:6:4:1, and so on. The intensities follow the binomial coefficients for \(\left( x + y \right)^n\), where \(n\) is the number of protons in the splitting group. Thus when \(n = 4\), we have \(x^4 + 4 x^3y + 6 x^2 y^2 + 4 x y^3 + y^4\), or 1:4:6:4:1. The spectrum of \(\ce{(CH_3O)_2CHCH_3}\) (Figure 9-34) provides an excellent example of how nmr shows the presence of contiguous protons. The symmetrical doublet and 1:3:3:1 quartet are typical of the interaction between a single proton and an adjacent group of three, that is, . The methyl protons of the \(\ce{(CH_3O)}\) groups are too far from the others to give demonstrable spin-spin splitting; thus they appear as a single six-proton resonance. In general, the magnitude of the spin-spin splitting effect of one proton on another proton (or group of equivalent protons) depends on the of intervening chemical bonds and on the spatial relationships between the groups. For simple systems without double bonds and with normal bond angles, we usually find for protons (i.e., having different chemical shifts): Where restricted rotation or double- and triple-bonded groups are involved, widely divergent splittings are observed. For double bonds, the between two nonequivalent hydrogens located on one end are characteristically small, while the in \(\ce{-HC=CH}-\) are larger, especially for the trans configuration: Coupling through four or more bonds is significant for compounds with double or triple bonds. Examples of these so-called and some other useful splitting values follow: . A very important characteristic of three-bond proton-proton couplings, \(\ce{H-C-C-H}\), is the way that they depend on the conformation at the \(\ce{C-C}\) bond. Typical values for several particular conformations are You may have wondered why the hydroxyl proton of ethanol produces a single resonance in the spectrum of Figure 9-23. It is quite reasonable to expect that the hydroxyl proton would be split by the neighboring methylene protons because they are only three bonds apart, however, this coupling will not be observed if the hydroxyl protons are exchanging rapidly between the ethanol molecules ( ). When proton exchange is rapid, the spin interactions between the \(\ce{-CH_2}-\) and \(\ce{-OH}\) protons average to zero. At intermediate exchange rates, the coupling manifests itself through line broadening or by actually giving multiple lines. If you look at the several spectra of ethanol in Figure 9-29, you will notice how the shape of the \(\ce{OH}\) resonance varies from a broad singlet to a distinct triplet. Rapid chemical exchange of magnetic nuclei is not the only way that spin-coupling interactions can be averaged to zero. The same effect can be achieved by a technique known as double resonance. To understand how this is done, consider two coupled protons \(\ce{H}_\text{A}\) and \(\ce{H}_\text{B}\) having different chemical shifts. Suppose that \(\ce{H}_\text{A}\) is selectively irradiated at its resonance frequency \(\nu_\text{A}\) while at the same time we observe the resonance signal of \(\ce{H}_\text{B}\). The coupling between \(\ce{H}_\text{A}\) and \(\ce{H}_\text{B}\) disappears, and \(\ce{H}_\text{B}\) shows a resonance. Why is this so? By irradiation of \(\ce{H}_\text{A}\), the \(\ce{H}_\text{A}\) nuclei are changed from the +1/2 state to -1/2 and back again sufficiently rapidly that the neighboring nucleus \(\ce{H}_\text{B}\) effectively “sees” neither one state nor the other. The magnetic interaction between the states therefore averages to zero. This decoupling of magnetic nuclei by double resonance techniques is especially important in \(\ce{^{13}C}\) NMR spectroscopy ( ) but also is used to simplify proton spectra by selectively removing particular couplings. The solution of a typical structural analysis problem by nmr methods utilizes at least four kinds of information obtained directly from the spectrum. They are: chemical shifts \(\left( \delta \right)\), line intensities (signal areas), spin-spin splitting patterns (line mulitplicities), and coupling constants \(\left( J \right)\). We already have shown how chemical shifts are used in the absence of spin-spin splitting. We now will illustrate how more complex spectra may be analyzed. Figure 9-35 shows the proton nmr spectrum for a compound of formula \(\ce{C_3H_6O}\). There are three principal groups of lines at \(9.8\), \(2.4\), and \(1.0 \: \text{ppm}\). Look at the multiplicity of these groups before reading further. There are several ways to approach a problem such as this, but probably the easiest is to start with the integral. The relative heights of the stepped integral for the principal groups of lines can be obtained by a pair of dividers, with a ruler, or with horizontal lines as in Figure 9-35. The integral suggests that one hydrogen is responsible for the resonance at \(9.8 \: \text{ppm}\), two hydrogens at \(2.4 \: \text{ppm}\), and three at \(1.0 \: \text{ppm}\). Three hydrogens in a single group suggest a \(\ce{CH_3}-\) group, and because there is a three-four splitting pattern, it is reasonable to postulate \(\ce{CH_3-CH_2}-\). Subtracting \(\ce{C_2H_5}\) from the given formula \(\ce{C_3H_6O}\) leaves \(\ce{CHO}\), which, with normal valences, has to be \(\ce{-CH=O}\). The spectrum thus appears to be consistent with the structure \(\ce{CH_3CH_2CH=O}\) (propanal) as judged from the molecular formula and the spin-spin splitting pattern, which indicates the \(\ce{CH_3CH_2}-\) grouping. To be sure of the structure, we should check it against of the available information. First, from the shifts (Table 9-4) we see that the single proton at \(9.8 \: \text{ppm}\) fits almost perfectly for \(\ce{RCHO}\), the two-proton \(\ce{-CH_2C=O}\) resonance at \(2.4 \: \text{ppm}\) is consistent with that reported for \(\ce{-CH_2COR}\), while the three-proton line at \(1.0 \: \text{ppm}\) checks with \(0.9 \: \text{ppm}\) for \(\ce{CH_3R}\). What about the couplings? The three-four pattern has a spacing of slightly over \(7 \: \text{Hz}\), which is just right for an ethyl group (compare Figures 9-23 and 9-32). The doubling up (almost obscured by second-order splitting) of the \(\ce{-CH_2}-\) resonance and the splitting of the \(\ce{-CH=O}\) resonance into a 1:2:1 triplet indicate about a \(2\)-\(\text{Hz}\) coupling for the \(\ce{-CH_2-CH=O}\) group. Three-bond couplings between \(\ce{-CHO}\) and adjacent \(\ce{-CH_2}-\) protons appear to be generally smaller than \(\ce{-CH_2-CH_3}\) couplings. We usually would not rely on nmr alone in a structure-analysis problem of this kind, but would seek clues or corroboration from the infrared, electronic, or other spectra, as well as chemical tests. In later chapters we will have many problems that will be facilitated by the use of both nmr and infrared spectra. A further worked example will illustrate the approach. A compound has the composition \(\ce{C_3H_3Br}\) and gives the infrared and nuclear magnetic resonance spectra shown in Figure 9-36. The problem is how to use this information to deduce the structure of the compound. The molecular formula tells us the number and kind of atoms the number of multiple bonds or rings. The formulas of the corresponding \(\ce{C_3}\) hydrocarbon without the bromine would be \(\ce{C_3H_4}\), or hydrogens less than the saturated alkane \(\ce{C_3H_8}\). This means there must be two double bonds or the equivalent - one triple bond or one ring and one double bond.\(^{14}\) Because from the formula we suspect unsaturation, we should check this out with the infrared spectrum. There is a band at \(2120 \: \text{cm}^{-1}\), which is indicative of an unsymmetrically substituted \(\ce{-C \equiv C}-\) group (Table 9-2). The strong, sharp band at \(3300 \: \text{cm}^{-1}\) further tells us that the substance is a 1-alkyne \(\ce{-C \equiv C-H}\). The proton nmr spectrum shows that there are only two principal groups of lines - a two-proton doublet at \(3.85 \: \text{ppm}\) and a one-proton triplet at \(2.45 \: \text{ppm}\). The two-three splitting pattern combined with the 2:1 proton ratio suggests a \(\ce{CH_2}\) group coupled with a \(\ce{CH}\) group. The structure must be a 3-bromo-propyne, \(\ce{BrCH_2C \equiv CH}\). To confirm the assignment, the chemical shifts should be checked (Table 9-4). The \(\ce{\equiv C-H}\) at \(2.45 \: \text{ppm}\) agrees well with the tabulated value of \(2.5 \: \text{ppm}\). There is no tabulated data for \(\ce{-C \equiv C-CH_2Br}\) but the observed shift at \(3.85 \: \text{ppm}\) is at slightly lower fields than the tabulated \(3.33 \: \text{ppm}\) for \(\ce{-CH_2Br}\). This is expected because of the triple bond. The correlation of Equation 9-4 predicts a value of \(4.0 \: \text{ppm}\). Very often, a proton will be spin-coupled to two or more protons, and the couplings are not necessarily the same. When this happens, the resulting spectrum can be quite complex, as our next example shows. A compound \(\ce{C_9H_{10}}\) gives the nmr spectrum of Figure 9-37. There are clearly four kinds of protons in the molecule at \(\delta = 7.28 \: \text{ppm}\), \(5.35 \: \text{ppm}\), \(5.11 \: \text{ppm}\), and \(1.81 \: \text{ppm}\). Although the integral is not shown, the main groups of lines have intensities from the low-field to high-field in the ratio of 5:1:1:3. The five-proton signal at \(7.28 \: \text{ppm}\) is typical of a phenyl group, \(\ce{C_6H_5}\), and the one-proton signals at \(5.35\) and \(5.11 \: \text{ppm}\) are in the region for alkenic protons, . The three-proton signal at \(1.81 \: \text{ppm}\) is typical of a methyl group on a carbon-carbon double bond, . There are only three ways to put together a phenyl ring, , and two \(\ce{HC=}\) protons such that they add up to \(\ce{C_9H_{10}}\). They are Coupling between A and B (designated by the constant \(J_\text{AB}\)) should give four lines, two for A and two for B, as shown in Figure 9-38. Because A and B also are coupled to the three hydrogens of the methyl group (C), each of the four lines corresponding to \(J_\text{AB}\) will be further split (into 1:3:3:1 quartets). If \(J_\text{AC} \neq J_\text{BC}\), then the spacing of the lines in the two sets of quartets will not be the same. According to the foregoing analysis, the maximum number of lines observable for the A and B resonances is sixteen (8 for A and 8 for B). In fact, only eleven are visible (6 for A and 5 for B), which means that some of the sixteen possible lines must overlap. Without examining all possibilities, we can see that the actual situation can be reproduced if \(J_\text{AB} \cong J_\text{BC} = 2J_\text{AC}\). The only structure that is consistent with \(J_\text{AB} = 1.5 \: \text{Hz}\) is \(13\), or 2-phenylpropene; the other possibilities are excluded because \(J_\text{AB}\) should be about \(10 \: \text{Hz}\) for \(12\) and \(16 \: \text{Hz}\) for \(11\). The simple \(n + 1\) rule for predicting the multiplicity of spin-coupled proton signals often breaks down whenever the chemical-shift difference between the protons in different groups becomes comparable to coupling constants for magnetic interaction between the groups. Under these circumstances, you may expect to see more lines, or lines in different positions with different intensities, than predicted from the simple first-order treatment. One example is the effect of changing chemical shift on a two-proton spectrum with \(J = 10 \: \text{Hz}\) (Figure 9-44). We see in Figure 9-44 that even when the shift is 7.5 times larger than the coupling, the outside lines are weaker than the inside lines. This general kind of asymmetry of line intensities also is apparent in the spectrum of ethyl iodide (Figure 9-32), in which the lines of each group are more like 0.7:2.5:3.5:1.3 and 1.2:2.0:0.8, rather than the 1:3:3:1 and 1:2:1 ratios predicted from the first-order treatment. The asymmetry is such that two groups of lines that are connected by spin-spin splitting in effect "point" to one another - the lines on the "inside" of the pattern are stronger than predicted from the first-order treatment, whereas those on the "outside" are weaker. The effect can be put to practical use, as illustrated in the following exercise. To explain the effect of chemical shifts on second-order splitting is beyond the scope of this book. In fact, we haven't really explained first-order splitting, although more on this topic will be found in . But regardless of how many lines appear in a complex nmr spectrum, they can be rationalized in terms of the chemical shifts, coupling constants, and exchange effects. Furthermore, the overall signal intensities remain proportional to the number of protons giving rise to the signals. When there are many hydrogens and small chemical-shift differences, as in alkanes, the proton nmr spectra may have so many closely spaced resonance lines that they merge together to give a series of smooth, more-or-less featureless peaks. The proton spectrum of octane (Figure 9-46a) is an excellent example of this type of spectrum. Useful information often can be obtained from such spectra as to the ratio of \(\ce{CH_3}\) : \(\ce{CH_2}\) : \(\ce{CH}\) by investigation of the integrals over the range of alkane proton absorptions. Figure 9-46 illustrates how this can be done for octane and 2,2,4-trimethylpentane. In recent years \(\ce{^{13}C}\) nmr spectroscopy using \(\ce{^{13}C}\) of natural abundance \(\left( 1.1 \% \right)\) has become an important tool for organic structural analysis. That this did not happen sooner is because \(\ce{^{13}C}\) has a much smaller magnetic moment than \(\ce{^1H}\) and the small moment combined with the small natural abundance means that \(\ce{^{13}C}\) is harder to detect in the nmr than \(\ce{^1H}\) by a factor of 5700. This is a large difference and can be put in the proper context in the following way. Suppose two people are talking in a noisy room and one is trying to hear the other. The common request is "talk louder". If this is not possible then the request is "say it again" or "talk more slowly". Either of the latter requests amounts to an integration of signal versus noise and takes time. Improvement in signal-to-noise for a given communication is achieved as the of the of communication. On the crucial , \(\ce{^{13}C}\) nmr signals require \(\left( 5700 \right)^2 \cong 30,000,000\) times more time to get the same signal-to-noise ratio as in \(\ce{^1H}\) nmr for the same number of nuclei per unit volume. This is a problem. Electronic improvements and use of communication theory, with emphasis on the "say-it-again" technique, have provided the means for obtaining routine \(\ce{^{13}C}\) spectra for even fairly dilute solutions of quite complex molecules. Some of the same kinds of structural effects are important for \(\ce{^{13}C}\) chemical shifts as for proton chemical shifts (Section 9-10E). For example, there is a similar parallel between \(\ce{^{13}C}\) shift differences in compounds of the type \(\ce{CH_3-CH_2-X}\) and electronegativity (Figure 9-47) as between the corresponding proton shifts and electronegativity (Figure 9-28). It is important to notice that \(\ce{^{13}C}\) shifts in \(\text{ppm}\) units are much larger than those of protons. This is because carbon uses \(p\) orbitals in forming bonds, whereas hydrogen uses \(s\) orbitals. We therefore will expect to find the the nuclei of other elements that use \(p\) orbitals in bonding, such as \(\ce{^{15}N}\), \(\ce{^{19}F}\), and \(\ce{^{31}P}\), also will have larger shifts than for protons, as indeed they do. A structural application of \(\ce{^{13}C}\) nmr, which shows its power in an area where \(\ce{^1H}\) nmr is indecisive, is shown in Figure 9-48. Here, we see the high-field \(\ce{^{13}C}\) resonances of a substance known variously as Coumadin, or the sodium salt of warfarin, \(14\), which is used widely as a blood anticoagulant in the treatment of diseases such as phlebitis. It also has substantial utility as a rat poison because of its anticoagulant action. There is no indication of any abnormality in the chemical shifts of carbons 11, 12, and 14 shown in Figure 9-48a. Furthermore, there is a downfield resonance \(216.5 \: \text{ppm}\) from the carbons of TMS (not shown in Figure 9-48a) which is typical of a \(\ce{C=O}\) carbon corresponding to C13. When \(14\) is treated with acid, we expect the product (warfarin) of structure \(15\) to be formed, which should have a \(\ce{^{13}C}\) spectrum much like that shown in Figure 9-48a. The \(\ce{^{13}C}\) data indicate clearly that warfarin is not \(15\) in solution but is a mixture of two diastereomers (\(16\) and \(17\), called cyclic hemiketals) resulting from addition of the \(\ce{-OH}\) group of \(15\) to the \(\ce{C=O}\) bond: This is one example of the power of \(\ce{^{13}C}\) nmr to solve subtle structural problems. \(^8\)Although the principal isotopes of \(Cl\), \(Br\), and \(I\) have magnetic properties, because of the special character of all of these isotopes, they act in organic compounds as though they were . \(^9\) in the sense used here means that the radio-frequency absorption takes place at specified "resonance" frequencies. However, you will see that almost all of the forms of spectroscopy we discuss in this book involve "resonance" absorption in the same sense. \(^{10}\)Here, \(\gamma\) is in \(\text{Hz}\) per gauss; physicists usually define \(\gamma\) in radians per second per gauss. \(^{11}\)From the Greek prefix meaning through, across. The opposite of diamagnetic is ; meaning alongside. We shall use this term later. \(^{12}\)In addition to giving better separation of the lines and clearer spectra, going to higher fields also has the beneficial effect of increasing the proportions of the nuclei in the \(+ \frac{1}{2}\) state, thereby giving more intense, easier-to-detect resonances. \(^{13}\)Many other proton-shift values are available in , Volume 1 and 2, Varian Associates, Palo Alto, Calif., 1962, 1963. \(^{14}\)If two rings were present, this also would give four hydrogens less than the alkane. However, two rings are not possible with only three carbons. and (1977)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/20%3A_Molecular_Spectroscopy_and_Photochemistry/20.2%3A_Vibrations_and_Rotations_of_Molecules%3A_Infrared_and_Microwave_Spectroscopy

Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual , a few of which are illustrated below. The energy of molecular vibration is rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the of the vibration will increase, but the vibrational will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds. Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring that is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 10 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 10 Hz. If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state. The value of E - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state. With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 10 to 10 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side. Our 5.15 x 10 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity. The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light. The power of infrared spectroscopy arises from the observation that . The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 10 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 10 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne. Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. . Such vibrations are said to be . In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls. Some kinds of vibrations are . The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules absorb IR light). Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone. There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called (cm ), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘ ’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you! The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 10 Hz, and a E value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: . Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm ). The jagged peak at approximately 2900-3000 cm is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum. As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm . This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules. In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm . We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance. The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens. Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen. Using the online Spectral Database for Organic Compounds, look up IR spectra for the following compounds, and identify absorbance bands corresponding to those listed in the table above. List actual frequencies for each signal to the nearest cm unit, using the information in tables provided on the site. a) 1-methylcyclohexanol b) 4-methylcyclohexene c) 1-hexyne d) 2-hexyne e) 3-hexyne-2,5-diol It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in table 1 in the tables section at the end of the text. As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol (this type of reaction is discussed in detail in ).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__1%3A_The_Alkali_Metals/2Reactions_of_the_Group_1_Elements/Reactions_of_Group_1_Elements_with_Water

All of —lithium, sodium, potassium, rubidium and cesium react vigorously or even explosively with cold water. In each case, the aqueous metal hydroxide and hydrogen gas are produced, as shown: \[ 2X (s) + 2H_2O (l) \rightarrow 2XOH (aq) + H_2 (g)\] where \(X\) is any Group 1 metal. In each of the following descriptions, a very small portion of the metal is dropped into a large container of water. The sodium moves because it is pushed by the hydrogen produced during the reaction. If the sodium becomes trapped on the side of the container, the hydrogen may catch fire and burn with an orange flame. The color is due to contamination of the normally blue hydrogen flame with sodium compounds. The Group 1 metals become more reactive towards water down the group. It is tempting to conclude that because the reactions get more dramatic down the group, the amount of heat given off increases from lithium to cesium. This is not the case. The table below gives estimates of the enthalpy change for each of the elements undergoing the reaction with water: \[ X (s) + H_2O(l) \rightarrow XOH(aq) + \dfrac{1}{2} H_2 (g) \] There is no consistent pattern in these values; they are all very similar, and counter intuitively, lithium releases the most heat during the reaction. The differences between the reactions are determined at the atomic level. In each case, metal ions in a solid are solvated, as in the reaction below: \[ X(s) \rightarrow X^+(aq) + e^-\] The net enthalpy change for this process can be determined using and breaking it into several theoretical steps with known enthalpy changes. \[ X(s) \rightarrow X(g)\] \[ X(g) \rightarrow X^+(g) + e^-\] \[ X^+(g) \rightarrow X^+(aq)\] These values are tabulated below (all energy values are given in kJ / mol): There is no overall trend in the overall reaction enthalpy, but each of the component input enthalpies (in which energy must be supplied) decreases down the group, while the hydration enthalpies increase: The summation of these effects eliminates any overall pattern. Knowing the atomization energy, the first ionization energy, and the hydration enthalpy, however, reveals useful patterns. Consider the energy input terms: A steady decrease down the group is apparent. From lithium to cesium, energy is required to form a positive ion. This energy will be recovered (and overcompensated) later, but must be initially supplied. This process is related to the of the reaction. The lower the activation energy, the faster the reaction. Although lithium releases the most heat during the reaction, it does so relatively slowly—not in one short, sharp burst. Cesium, on the other hand, has a significantly lower activation energy, and so although it does not release as much heat overall, it does so extremely quickly, causing an explosion. The reactions proceed faster as the energy needed to form positive ions falls. This is in part due to a decrease in ionization energy down the group, and in part to a decrease in atomization energy reflecting weaker metallic bonds from lithium to cesium. This leads to lower activation energies, and therefore faster reactions. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Connecting_Gas_Properties_to_Kinetic_Theory_of_Gases

When examining the ideal gas laws in conjunction with the kinetic theory of gases, we gain insights into the behavior of ideal gas. We can then predict how gas particles behaviors such as gas molecular speed, effusion rates, distances traveled by gas molecules. Graham's Law, which was formulated by the Scottish physical chemist Thomas Graham, is an important law that connects gas properties to the kinetic theory of gases. The Kinetic Molecular Theory states that the average energy of molecules is proportional to absolute temperature as illustrated by the following equation: \[e_K=\dfrac{3}{2}\dfrac{R}{N_A}T\] where Since R and N are constants, this means that the Kelvin temperature (T) of a gas is directly proportional to the average kinetic energy of its molecules. This means that at a given temperature, different gases (for example He or O ) will the same average kinetic energy. Gas molecules move constantly and randomly throughout the volume of the container they occupy. When examining the gas molecules individually, we see that not all of the molecules of a particular gas at a given temperature move at exactly the same speed. This means that each molecule of a gas have slightly different kinetic energy. To calculate the average kinetic energy (e ) of a sample of a gas, we use an average speed of the gas, called the root mean square speed (u ). \[e_K=\dfrac{1}{2}m{u_{rms}^2}\] with The root mean square speed, u , can be determined from the temperature and molar mass of a gas. \[u_{rms}=\sqrt{\dfrac{3RT}{M}}\] with When examining the root mean square speed equation, we can see that the changes in temperature (T) and molar mass (M) affect the speed of the gas molecules. The speed of the molecules in a gas is proportional to the temperature and is inversely proportional to molar mass of the gas. In other words, as the temperature of a sample of gas is increased, the molecules speed up and the root mean square molecular speed increases as a result. Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation: \[\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{(u_rms)_A}{(u_rms)_B}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}\] In according with the Kinetic Molecular Theory, each gas molecule moves independently. However, the net rate at which gas molecules move depend on their average speed. By examining the equation above, we can conclude that the heavier the molar mass of the gas molecules slower the gas molecules move. And conversely, lighter the molar mass of the gas molecules the faster the gas molecules move. Graham's Law can only be applied to gases at low pressures so that gas molecules escape through the tiny pinhole slowly. In addition, the pinhole must be tiny so that no collisions occur as the gas molecules pass through. Since Graham's Law is an extension of the Ideal Gas Law, gases that follows Graham's Law also follows the Ideal Gas Law. The random and rapid motion of tiny gas molecules results in effusion. is the escape of gas molecules through a tiny hole or pinhole. The behavior of helium gas in balloons is an example of effusion. The balloons are made of latex which is porous material that the small helium atom can effuse through. The helium inside a newly inflated balloon will eventually effuse out. This is the reason why balloons will deflate after a period of time. Molecular speeds are also used to explain why small molecules (such as He) diffuse more rapidly than larger molecules (O ). That is the reason why a balloon filled with helium gas will deflate faster than a balloon filled with oxygen gas. The effusion rate, , is inversely proportional to the square root of its molar mass, . \[r\propto\sqrt{\dfrac{1}{M}}\] When there are two different gases the equation for effusion becomes \[\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}\] \(M_A\) is the molar mass of gas A, \(M_B\) is the molar mass of gas B, \(T\) Temperature in Kelvin, \(R\) is the ideal gas constant. From the equation above, The relative rates of effusion of two gases at the same temperature is given as: \[\dfrac{\it{Rate\;of\;effusion\;of\;gas\;}\mathrm 1}{\it{Rate\;of \;effusion\;of\;gas\;}\mathrm 2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}\] Units used to express rate of effusion includes: moles/seconds, moles/minutes, grams/seconds, grams/minutes. The relative distances traveled by the two gases is given as: \[\dfrac{Distance\;traveled\;by\;gas\;1}{Distance\;traveled\;by\;gas\;2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}\] By examine the Graham's law as stated above, we can conclude that a lighter gas will effuse or travel more rapidly than a heavier gas. Mathematically speaking, a gas with smaller molar mass will effuse faster than a gas with larger molar mass under the same condition. Similar to effusion, the process of is the spread of gas molecules through space or through a second substance such as the atmosphere. Diffusion has many useful applications. Here is an example of diffusion that is use in everyday households. Natural gas is odorless and used commercially daily. An undetected leakage can be very dangerous as it is highly flammable and can cause an explosion when it comes in contact with an ignition source. In addition, the long term breathing of natural gas can lead to asphyxiation. Fortunately, chemists have discovered a way to easily detect natural gas occur leak by adding a small quantity of a gaseous organic sulfur compound named methyl mercaptan, CH SH, to the natural gas. When a leak happens, the diffusion of the odorous methyl mercaptan in the natural gas will serve as a sign of warning. The following flowchart outlines the steps in solving quantitative problems involving Graham's Law. It can be used as a general guideline. Calculate the root mean square speed, \(u_{rms}\), in m/s of helium at \(30 ^o C\). Start by converting the molar mass for helium from g/mol to kg/mol. \(M=(4.00\;g/mol)\times\dfrac{1\;kg}{1000\;g}\) \(M=4.00\times10^{-3}\;kg/mol\) Now, using the equation for \(u_{rms}\) substitute in the proper values for each variable and perform the calculation. \[u_{rms}=\sqrt{\dfrac{3RT}{M}}\] \[u_{rms}=\sqrt{\dfrac{3 \times (8.314\; kg\;m^2/s^2*mol*K)(303K)}{4.00 \times 10^{-3}kg/mol}}\] \[u_{rms}=1.37 \times 10^3 \;m/s \] What is the ratio of \(u_{rms}\) values for helium vs. xenon at \(30^oC\). Which is higher and why? There are two approaches to solve this problem: the hard way and the easy way The \(u_{rms}\) speed of helium is calculated from the above example. First convert the molar mass of xenon from g/mol to kg/mol as we did for helium in example 1 \(M_{Xe}=(131.3\;g/mol)\times\dfrac{1\;kg}{1000\;g}\) \(M=0.1313\;kg/mol\) Now, using the equation for the u , insert the given and known values and solve for the variable of interest. \[u_{rms}=\sqrt{\dfrac{3(8.314\; kg\;m^2/s^2*mol*K)(303\;K)}{0.1313\;kg/mol}}\] \[u_{rms}=2.40 \times 10^2\;m/s\] Compare the two values for xenon and helium and decide which is greater. So the ratio of RMS speeds is \[\dfrac{u_{Xe}}{u_{He}} \approx 0.18\] Helium has the higher \(u_{rms}\) speed. This is in according with Graham's Law, because helium atoms are much lighter than xenon atoms. Since the temperature is the same for both gases, only the square root of the ratio of molar mass is needed to be calculated. \[\sqrt{\dfrac{M_{He}}{M_{Xe}}} = \sqrt{\dfrac{4.00 \; g/mol}{131.3\;g/mol }} \approx 0.18\] In either approach, helium has a faster RMS speed than xenon and this is due exclusively to its smaller mass. If oxygen effuses from a container in 5.00 minutes, what is the molecular weight of a gas with the same given quantity of molecules effusing from the same container in 4.00 min? Let oxygen be gas A and its rate is 1/5.00 minutes because it takes that much time for a certain quantity of oxygen to effuse and its molecular weight is 32 grams/mole (O (g)). Let the unknown gas be B and its rate is 1/4.00 minutes and M be the molecular weight of the unknown gas. First, choose the appropriate equation, \(\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}\) Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem. \(\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}\) \(\sqrt{M_B}=\dfrac{(1/5.00\;minutes)(\sqrt{32\;grams/mole})}{1/4.00\;minutes}\) \(\sqrt{M_B}=4.525\) \({M_B}=(4.525)^2=20.5\;grams/mole\) Oxygen, O (g), effuses from a container at the rate of 3.64 mL/sec, what is the molecular weight of a gas effusing from the same container under identical conditions at 4.48 mL/sec? Let label oxygen as gas A. Therefore, rate of effusion of gas A is equaled to 3.64 mL/sec and the molecular weight is 32 grams/mole. The unknown gas is B and its rate of effusion is 4.48 mL/sec. We are solving for the molecular weight of gas B which is labeled as M . First, choose the appropriate equation, \(\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}\) \(\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}\) Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem. \(\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A})(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}\) \(\sqrt{M_B}=\dfrac{(3.64\;mL/sec)(\sqrt{32\;grams/mole})}{4.48\;mL/sec}\) \(\sqrt{M_B}=4.50\) \({M_B}=(4.50)^2=20.2\;grams/mole\) Which answer(s) are true when comparing 1.0 mol O (g) at STP (Standard Temperature and Pressure) and 0.50 mole of S (g) at STP? The two gases have equal through the same orifice, a tiny opening: The average kinetic energy of gas molecules depends on only the Temperature as exemplifies in this equation: \(e_K=\dfrac{3}{2}\dfrac{R}{N_A}T\) In addition, the mass of 0.50 mole of S is the same as that of 1.0 mole of O . The rest of the choices are false. Therefore, the correct answers are (a) and (c).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__1%3A_The_Alkali_Metals/2Reactions_of_the_Group_1_Elements/Reactions_of_Group_I_Elements_with_Oxygen

This page examines the reactions of the Group 1 elements (lithium, sodium, potassium, rubidium and cesium) with oxygen, and the simple reactions of the various oxides formed. Group 1 metals are very reactive, and must be stored out of contact with air to prevent oxidation. Reactivity increases as you go down the group; the less reactive metals (lithium, sodium and potassium) are stored in oil (because of its density, lithium floats in oil, but because it is less reactive than the other metals in the group, the thin coating of oil that results is sufficient to prevent reaction). Rubidium and cesium are typically stored in sealed glass tubes to eliminate contact with air. They are stored either in a vacuum or in an inert gas such as argon and the tubes must be broken open when the metal is used. Depending on the period of the metal, a different type of oxide is formed when the metal is burned . The reactions are the same in oxygen and in air, but oxygen will generate a more violent reaction. Lithium is unique in the group because it also reacts with the nitrogen in the air to form lithium nitride. Lithium burns with a strongly red-tinged flame if heated in air; in pure oxygen, the flame is more intense. It reacts with oxygen in the air to give white lithium oxide: \[ 4Li + O_2 \rightarrow 2Li_2O \label{1}\] Lithium also reacts with the nitrogen in the air to produce lithium nitride and is the only Group 1 element that forms a nitride: \[ 6Li + N_2 \rightarrow 2Li_3N \label{2}\] Small pieces of sodium burn in air with a faint orange glow. Using larger amounts of sodium or burning it in oxygen gives a strong orange flame. The reaction produces a white solid mixture of sodium oxide and sodium peroxide. The equation for the formation of the simple oxide is analogous to the lithium equation: \[ 4Na + O_2 \rightarrow 2Na_2O \label{3}\] The peroxide equation is: \[ 2Na + O_2 \rightarrow Na_2O_2 \label{4}\] Small pieces of potassium heated in air melt and convert instantly into a mixture of potassium peroxide and potassium superoxide without a visible flame. Larger pieces of potassium produce a lilac flame. The equation for the formation of the peroxide is like the sodium equation above: \[ 2K + O_2 \rightarrow K_2O_2 \label{5}\] The superoxide reaction is as follows: \[ K + O_2 \rightarrow KO_2 \label{6}\] Both rubidium and cesium metals ignite in air and produce superoxides, \(RbO_2\) and \(CsO_2\) . The equations for these reactions are analogous to the equivalent potassium superoxide equation (Equation 6): \[ Rb + O_2 \rightarrow RbO_2 \label{7}\] \[ Cs + O_2 \rightarrow CsO_2 \label{8}\] Both superoxides are described as either orange or yellow, but rubidium superoxide can also be dark brown. The oxide forms of each element can be summarized as follows: The more complicated ions are unstable in the presence of a small positive ion. Consider the peroxide ion, \(O_2^{2-}\), which has the following structure: The covalent bond between the two oxygen atoms is relatively weak. Now imagine bringing a small positive ion close to the peroxide ion. Electrons in the peroxide ion will be strongly attracted toward the positive ion. A simple oxide ion can be formed if the oxygen atom on the right "breaks off": Hence, the positive ion polarizes the negative ion. This is most effective if the positive ion is small and highly charged (if it has a high charge density, or a lot of charge packed into a small volume). The values for the various potassium oxides show the same trends. As long as there is enough oxygen, forming the peroxide releases more energy per mole of metal than forming the simple oxide. Forming the superoxide has an even greater enthalpy change. \[ X_2O + H_2O \rightarrow 2X^+_{(aq)} + OH^-_{(aq)} \label{9}\] \[ X_2O + 2HCl \rightarrow 2XCl + H_2O \label{10}\] \[ 2XO_2 + 2H_2O \rightarrow 2XOH + H_2O_2 + O_2 \label{14}\] \[ 2XO_2 + 2HCl \rightarrow 2XCl + H_2O_2 + O_2 \label{15}\] Forming complicated oxides from the metals releases more energy and makes the system more energetically stable. However, this only applies to the lower half of the group, in which the metal ions are large and have a low charge density. At the top of the group, the small ions with a higher charge density tend to polarize the more complicated oxide ions to the point of disintegration. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.11%3A_Nature_of_the_Metal-Dioxygen_Linkage_in_Biological_Systems

Many techniques have been used to probe the geometry and electronic structure of the metal-dioxygen moiety in biological systems and in synthetic models. The results form the basis of any understanding of the factors that determine and modulate oxygen affinity. By resonance Raman techniques, the O—O stretch is observed at 744 cm in oxyhemocycanin and at 844 cm in oxyhemerythrin. Dioxygen is therefore coordinated as peroxo species. By use of unsymmetrically labeled dioxygen, O— O, it was established that dioxygen coordinates symmetrically: \(\tag{4.43}\) Carbon monoxide binds to only one of the Cu centers in deoxyhemocyanin, through the C atom,* apparently blocking the second Cu site. Similar behavior is also seen for the nitrosyl adduct. * A report that CO binds to the copper center through the O atom, an unprecedented mode, has been challenged. On the other hand, in an experiment parallel to that just described for hemocyanin, dioxygen was found to bind asymmetrically in oxyhemerythrin: \(\tag{4.45}\) The existence and location of the proton, which cannot be proven in the crystal structure of deoxyhemerythrin or of oxyhemerythrin, are inferred from a model system for the former that contains a hydroxo group bridging two high-spin, weakly antiferromagnetically coupled Fe centers (J = -10 cm ). For oxyhemerythrin (and for one conformation of methydroxohemerythrin), a small change in the position of the symmetric Fe—O—Fe mode is observed when H O is replaced by D O. The strong antiferromagnetic coupling observed for methemerythrin and oxyhemerythrin (-J ~ 100 cm ) is uniquely consistent with a bridging oxo moiety between a pair of Fe centers. Finally, a Bohr effect (release or uptake of protons) is absent in oxygen binding to hemerythrin. These observations are consistent with a \(\mu\)-oxo group slightly perturbed by hydrogen bonding to a coordinated hydroperoxo species. An important role for the protein in hemerythrin is to assemble an asymmetric diiron(II) species. Only a few of the myriad of known \(\mu\)-oxodiiron(III) complexes are asymmetric, and the synthesis of realistic asymmetric models remains a challenge. Deoxyhemocyanin (Cu d ) and deoxyhemerythrin (Fe d ) are colorless. In the oxygenated derivatives there is considerable charge transfer between the coordinated peroxo groups and the metal centers. This phenomenon makes the essentially d-d metal transitions more intense than those for the simple aquated Fe or Cu ions, and permits facile measurement of oxygen-binding curves. The spectral changes accompanying oxygenation are shown in Figures 4.20 and 4.21. Nitric oxide binds to deoxyhemocyanin, to deoxyhemerythrin, and to the mixed valence Fe • • • Fe semimethemerythrin. Carbon monoxide binds to neither form of hemerythrin: apparently the other ligands have insufficiently strong fields to stabilize the low-spin state for which electron density would be available for back donation into the CO \(\pi\)* orbitals. The O—O stretch that is observed by difference infrared techniques at around 1105 cm for oxyhemoglobin and oxymyoglobin clearly categorizes the dioxygen moiety as a superoxo species; that is, the order of the O—O bond is about 1.5. Considerable ink has been spilled about the nature of the Fe—O fragment since Pauling's original suggestion in 1948 that dioxygen binds to iron in an end-on bent fashion: \(\tag{4.46}\) He subsequently reaffirmed this geometry, and proposed that hydrogen bonding between the coordinated dioxygen and the distal imidazole H—N group was important in stabilizing the Fe—O species. In an alternative model Weiss proposed that a low-spin Fe center (S = \(\frac{1}{2}\)) was very strongly antiferromagnetically coupled to a superoxide anion radical (S = \(\frac{1}{2}\)). A triangular peroxo mode has also been advanced. The problem has been how to resolve the observed diamagnetism of oxyhemoglobin with UV-visible, x-ray absorption, and resonance Raman spectroscopic characteristics that are distinctly different from those of Fe systems (such as carbonmonoxyhemoglobin and low-spin six-coordinated hemochromes, such as Fe(Porph)(Py) ) and from unambiguously Fe systems (such as chloromethemoglobin or cyanomethemoglobin). Any adequate theoretical treatment must also explain how iron-porphyrin systems can bind not only O , but also CO, NO, alkyl isocyanides, and alkyl-nitroso moieties. A simple qualitative model presented by Wayland and coworkers conveniently summarizes ligand-binding geometries of cobalt and iron porphyrins. Although a reasonable quantitative theoretical consensus exists for 1:1 cobalt-dioxygen species, the same cannot be said yet for irondioxygen systems. Why does dioxygen bind to iron and cobalt porphyrins in an end-on bentbond fashion as in (4.37) and (4.46)? Why does carbon monoxide bind in a linear manner (Equation 4.40)? Why are six-coordinate dioxygen and carbonmonoxide adducts more stable than five-coordinate ones? A unified picture of ligand binding that addresses these questions is important in understanding properly the specific case of dioxygen binding to hemoglobin and related systems. The splitting of the metal d orbitals for a four-coordinate metalloporphyrin is shown in the center of Figure 4.22. These orbitals contain some porphyrin character and are antibonding with respect to metal-porphyrin bonds. As shown in Figure 4.16, the primary effect of a single \(\sigma\)-donor axial ligand, such as pyridine or 1-methylimidazole, is to elevate the energy of the antibonding d and lower the energy of the d orbital and hence lead to a high-spin species in place of the intermediate-spin four-coordinate one. Thus, for simplicity in highlighting interaction of the metal center with the diatomic \(\sigma\)-donor: \(\pi\)-acid ligands CO, NO, and O , the perturbations wrought by primarily \(\sigma\)-donor ligands, such as 1-methylimidazole, are omitted. For the corresponding cobalt(II) compound, there is an additional electron. The diatomic ligands of interest share a qualitatively similar molecular orbital scheme. The filling of orbitals for CO is shown on the left-hand side. Dioxygen, which is shown on the right-hand side, has two more electrons than CO; these occupy the doubly degenerate \(\pi\)* orbitals. Quantitative calculations show that the energy of the \(\pi\)* orbitals decreases monotonically from CO to NO to O , indicating increasing ease of reduction of the coordinated molecule, a feature that has not been included in the diagram. Only those interactions of molecular orbitals that have appropriate symmetry and energy to interact significantly with the metal d orbitals are shown. Two extremes are shown in Figure 4.22 for the interaction of a diatomic molecule A-B with the metal center: a linear geometry on the left and a bent geometry on the right. A side-on geometry is omitted for the binding of O to a Co or Fe porphyrin, since this would lead to either an M side-on superoxo or an M peroxo species; both these modes of coordination to these metals are currently without precedent. Linear diatomic metal bonding maximizes the metal-d to ligand-p bonding. When a ligand coordinates in a bent manner, axial symmetry is destroyed, and the degeneracy of the ligand p orbitals is lifted. One p orbital is now oriented to combine with the metal d orbital to form a \(\sigma\) bond, and the other is oriented to combine with d and d orbitals to form a \(\pi\) bond. A bent geometry for the diatomic molecule will result when either or both of the metal d or the ligand p orbitals are occupied, since this geometry stabilizes the occupied d orbital in the five-coordinate complex. Thus O binds in a strongly bent manner to Co and Fe porphyrins; NO binds in a strongly bent manner to Co porphyrins; CO binds in a linear fashion to Fe porphyrins. The interaction of NO with Fe porphyrins and CO with Co porphyrins—the resultant species are formally isoelectronic—is more complicated. The degree of bending seen in Fe (TPP)(NO) is midway between the two extremes. For CO the higher-energy p orbitals lead to a greater mismatch in energy between the d and p orbitals, and less effective \(\sigma\) bonding. In EPR experiments the odd electron is found to be localized in a molecular orbital with about 0.87 metal d character for the five-coordinate Co—CO adduct, as expected for a nearly linear geometry. On the other hand, for the Fe—NO adduct the metal d character of the odd electron is about 0.4 to 0.5; a somewhat bent geometry (140°) is observed in the crystal structure of Fe(TPP)(NO). Because the CO ligand is a very weak \(\sigma\) donor, the Co—CO species exists only at low temperatures. Only qualitative deductions can be made from this model about the extent of electron transfer, if any, from the metal onto the diatomic ligand, especially for dioxygen. The higher in energy the metal d orbital is with respect to the dioxygen p orbitals, the closer the superoxo ligand comes to being effectively a coordinated superoxide anion. With an additional electron, the dioxygen ligand in Co—O complexes can acquire greater electron density than it can in Fe—O complexes. From the diagram it may be inferred that a ligand with very strong \(\pi\)-acceptor properties will lower the energy of the d and d orbitals through strong (d , d )-\(\pi\)* interaction. The resultant energy gap between these two orbitals and the other three metal d orbitals may be sufficient to overcome the energy involved in spin-pairing, and hence lead to five-coordinate low-spin species, as happens for complexes containing phosphines and carbon monosulfide.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_14%3A_The_Carbon_Family/1Group_14%3A_General_Properties_and_Reactions/Carbon_Group_(Group_4)_Trends

The halides, oxides, and sulfides of the M ions become more stable on descending the group. For example SiCl , SiBr , and SiI are all stable. PbCl decomposes at 105 °C and PbI does not exist. Similarly the ease of oxidation of the M halides decreases down the column. PbCl may be converted to PbCl by heating in a stream of chlorine. Similarly PbO is an oxidizing agent, whereas SnO , GeO , and SiO are not. The stabilities of the M and M organometallic derivatives of the elements behave differently. Pb(C H ) can be readily stored and is more stable than Pb(C H ) , which is not isolable as a solid. The compounds in the lower oxidation state are in general more ionic, less likely to form molecular structures, the halides are less readily hydrolyzed and the oxides are less acidic. The M-H and M-C mean bond enthalpies decrease down the column and consequently the hydrides and alkyls become thermodynamically less stable and kinetically more reactive. Carbon, of course, forms a very wide range of hydrides, silicon forms primarily SiH and Si H which are spontaneously inflammable. Higher silanes decompose readily to Si H . The Si-H bond polarities are opposite to C-H. Silanes are strong reducing agents. The germanes GeH , Ge H , and Ge H are less flammable than SiH and are resistant to hydrolysis. SnH decomposes at 0°C to Sn and PbH is extremely unstable. The organometallic derivatives of silicon and germaniu are very similar. They are more reactive than the carbon analogues because the M-C bonds are more polar and the central atom can exand its coordination number more easily. The rates of hydrolysis are in order: Pb >> Sn >> Ge >Si Organotin compounds more readily expand their coordination geometries and more readily form cationic species. Organolead compounds decompose readily at 100-200 °C by free radical processes. The element-element mean bond enthalpies decrease in the order: C-C > Si-Si > Ge-Ge > Sn-Sn > Pb-Pb and therefore the range of ring and polyhedral molecules diminish down the group. Carbon not only forms an extensive range of chain and ring compounds, but also polyhedral molecules such as prismanes, C H , and cubane (C H ). Analogous compounds are known for Si, Ge, and Sn if the hydrogens are replaced by bulky organic substituents. However, few examples exist for Pb, which form compounds containing the anionic Zintl polyhedral anoin Pb analogous to Sn . The ability of the elements to form multiple bonds diminishes in the series: C-C > Si-Si > Ge-Ge > Sn-Sn > Pb-Pb Because the - overlaps become less favorable. This has the following manifestations: Carbon is especially in its diamond and polyhedral forms is a typical non-metal, silicon is a semiconductor, and tin & lead are typical metals. Although tin has one modification (grey tin) which is isostructural with Ge, Si, and diamond. Lead only occurs in close packed structural forms. CO and SiO are acidic oxides, SnO is amphoteric, and GeO is mainly acidic with slight amphoteric character. The Si-O mean bond enthalpies are particularly large and this leads to a wide range of silicates. In general for the elements below carbon the M-O bonds are sufficiently strong that the oxides are susceptible to hydrolysis. For carbon the tetradedral geometry predominates unless multiple bonds formed. For the havier elements the tetrahedral geometry is also widespread bu the larger size of the central atoms leads to the formation of compounds with higher coordination numbers, SiF , Trigonal bipyramidal; SiF , Octahedral; SnCl , Trigonal bipyramidal; Sn(C H )(NO ) (OP(C H ) , Pentgonal bipyramidal (7 coord); Sn(NO ) , Dodecahedral (8 coord) The increased facility to achieve the higher coordination numbers is also reflected in the transition from molecular to ppolymeric, CF , SiF , and GeF ar e molecular, where as SnF amnd PbF have infinite lattices based on octahedral metal centers. Of course all of these compounds with coordination numbers is also reflected in the geometries of the oxo-anions: trig. planar: CO tetrahedral: SiO octahedral: Ge(OH) Sn(OH) Pb(OH) The chlorides of Ge, Sn, and Pb react with aqueous HCl to form the [MCl ] anions, where as SiCl hydrolyses and CCl is unreactive. However, SiF does form [SiF ] with HF. Si, Ge, Sn, and Pb all form oxalato-complexes [M(ox) ] and cationic complexes [M(acac) } . Although C-Cl and Ge-Cl bonds in four valent compounds are reduced to the corresponding hydrides by Zn and HCl, Si-Cl and Sn-Cl bonds are not.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/03%3A_Distributions_Probability_and_Expected_Values/3.06%3A_Continuous_Distribution_Functions_-_the_Envelope_Function_is_the_Derivative_of_the_Area

When we can represent the envelope curve as a continuous function, the envelope curve is the derivative of the cumulative probability distribution function: The cumulative distribution function is \(f\left(u\right)\); the envelope function is \({df\left(u\right)}/{du}\). The envelope function is a , and we will refer to the envelope function, \({df\left(u\right)}/{du}\), as the . The probability density function is the derivative, with respect to the random variable, of the cumulative distribution function. This is an immediate consequence of the fundamental theorem of calculus. If \(H\left(u\right)\) is the anti-derivative of a function \(h\left(u\right)\), we have \({dH\left(u\right)}/{du}=h\left(u\right)\), and the fundamental theorem of calculus asserts that the area under \(h\left(u\right)\), from \(u=a\) to \(u=b\) is \[\begin{aligned} \int_{a}^{b} h(u) d u &=\int_{a}^{b}\left(\frac{d H(u)}{d u}\right) d u \\[4pt] &=H(b)-H(a) \end{aligned}\nonumber \] In the present instance, \(H(u) = f(u)\), so that \[\int_{a}^{b}\left(\frac{d f(u)}{d u}\right) d u=f(b)-f(a)\nonumber \] and \[h(u)=\frac{d f(u)}{du}\nonumber \] The envelope function, \(h(u)\), and \(df(u)\) are the same function. This point is also apparent if we consider the incremental change in the area, \(dA\), under a histogram as the variable increases from \(u\) to \(u + du\). If we let the envelope function be \(h(u)\), we have \[dA=h(u)du\nonumber \] or \[h(u) = \dfrac{dA}{du}\nonumber \] That is, the envelope function is the derivative of the area with respect to the random variable, \(u\). The area is \(f(u)\), so the envelope function is \(h(u)=df(u)/du\). Calling the envelope curve the probability density function emphasizes that it is analogous to a function that expresses the density of matter. That is, for an incremental change in \(u\), the incremental change in probability is \[Δ(probability)= \dfrac{df}{du} Δu\nonumber \] analogous to the incremental change in mass accompanying an incremental change in volume \[Δ(mass)= density \times Δ(volume)\nonumber \] where \[density=\dfrac{d\left(mass\right)}{d\left(volume\right)}.\nonumber \] In this analogy, we suppose that mass is distributed in space with a density that varies from point to point in the space. The mass enclosed in any particular volume is given by the integral of the density function over the volume enclosed; that is, \[mass=\int_V{\left(density\right)dV}.\nonumber \] Conversely, the density at any given point is the limit, as the enclosing volume shrinks to zero, of the enclosed mass divided by the magnitude of the enclosing volume.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/0Group_17%3A_Physical_Properties_of_the_Halogens/Halogen_Group_(Group_17)_Trends

1. All elements are diatomic and molecular and the boiling and melting points increase as a result of the increasing van der Waals interactions between diatomic molecules for the heavier elements. 2. The elements are typical non-metals in their physical and chemical properties. They form anionic compounds based on X- (X = halogen) which is associated with a complete octet. 3. The atoms also form strong covalent bonds with other non-metals. The mean bond enthalpies for E-X bonds are particularly large for fluorine and therefore a wide range of molecular fluorides are known and fluorine is particularly effective at bringing out the highest valencies of the non-metals and highest oxidation states of the metals. 4. The oxidizing ability of the halogens decreases markedly down the group: F2 > Cl2 > Br2 > I2 and only iodine is oxidized by nitric acid. 5. The stabilities of the hydrogen halides decrease down the Group, but their acid strengths increase. 6. Only H-F forms strong hydrogen bonds and this is reflected in the boiling and melting points of the hydrogen halides. 7. The halogens form many interhalogen compounds with the less electronegative halogen surrounded by the more electronegative halogens. Neutral, anionic, and cationic interhalogen compounds are known. ICl and IBr are widely used in organic synthesis and are commercially available. 8. Oxygen fluorides are extremely strong and reactive oxidants and have been explored as potential rocket fuels, the oxides become less reactive down the column and more numerous. Iodine forms a particularly wide range of oxides. 9. The perhalates, EO4-, are only known for Cl, Br, and I. They exhibit and alternation in ther oxidizing abilities and the perbromates are particularly strong oxidizing agents. 10. In the highest oxidation state (+7) the relative oxidizing ability is:

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.02%3A_Changing_Reaction_Rates_with_Temperature

The vast majority of reactions depend on thermal activation, so the major factor to consider is the fraction of the molecules that possess enough kinetic energy to react at a given temperature. According to , a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the . The two distribution plots shown here are for a lower temperature and a higher temperature . The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of that are shown. It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures. Temperature is considered a major factor that affects the rate of a chemical reaction. It is considered a source of energy in order to have a chemical reaction occur. Svante Arrhenius, a Swedish chemist, believed that the reactants in a chemical reaction needed to gain a small amount of energy in order to become products. He called this type of energy the activation energy. The amount of energy used in the reaction is known to be greater than the activation energy in the reaction. Arrhenius came up with an equation that demonstrated that rate constants of different kinds of chemical reactions varied with temperature. This equation indicates a rate constant that has a proportional relationship with temperature. For example, as the rate constant increases, the temperature of the chemical reaction generally also increases. The result is given below: \[\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \nonumber \] This equation is known as Arrhenius' equation. T and T are temperature variables expressed in Kelvin. Some may ask how the temperature actually affects the chemical reaction rate. The answer to this is that this phenomenon is related to the collision theory. Molecules only react if they have sufficient energy for a reaction to take place. When the temperature of a solution increases, the molecular energy levels also increase, causing the reaction to proceedfaster. The graph of ln K vs. 1/T is linear, allowing the calculation of the activation energy needed for the reaction. An alternate form of the Arrhenius equation is given below: \[k = A_e^{-\frac{E_a}{RT}} \nonumber \] Some interesting examples: )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/04%3A_Alkanes/4.05%3A_Halogenation_of_Alkanes._Energies_and_Rates_of_Reactions

The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by . In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place. A simple example of a substitution reaction is the formation of chloromethane and chlorine: \[ \ce{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}\] The equation for the reaction is simple, the ingredients are cheap, and the product is useful. However, if we want to decide in advance whether such a reaction is actually feasible, we have to know more. Particularly, we have to know whether the reaction proceeds in the direction it is written and, if so, whether conditions can be found under which it proceeds at a convenient rate. Obviously, if one were to mix methane and chlorine and find that, at most, only \(1 \%\) conversion to the desired product occurred and that the \(1 \%\) conversion could be achieved only after a day or so of strong heating, this reaction would be both too unfavorable and too slow for an industrial process. One way of visualizing the problems involved is with energy diagrams, which show the energy in terms of some arbitrary that is a measure of progress between the initial and final states (Figure 4-4). Diagrams such as Figure 4-4 may not be familiar to you, and a mechanical analogy may be helpful to provide better understanding of the very important ideas involved. Consider a two-level box containing a number of tennis balls. An analog to an energetically favorable reaction would be to have all of the balls on the upper level where any disturbance would cause them to roll down to the lower level under the influence of gravity, thereby losing energy. If the upper level is modified and a low fence added to hold the balls in place, it will be just as energetically favorable as when the fence is not there for the balls to be at the lower level. The difference is that the process will not occur without some major disturbance. We can say there is an to occurrence of the favorable process. Now, if we shake the box hard enough, the balls on the upper level can acquire enough energy to bounce over the barrier and drop to the lower level. The balls then can be said to acquire enough to surmount the barrier. At the molecular level, the activation energy must be acquired either by collisions between molecules as the result of their thermal motions, or from some external agency, to permit the reactants to get over the barrier and be transformed into products. We shortly will discuss this more, but first we wish to illustrate another important concept with our mechanical analogy, that of and . With gentle shaking of our two-level box, all of the balls on the upper level are expected to wind up on the lower level. There will not be enough activation to have them go from the lower to the upper level. In this circumstance, we can say that the balls are not equilibrated between the lower and upper levels. However, if we shake the box and , no matter whether we start with all of the balls on the lower or upper level, an will be set up with, on the average, most of the balls in the energetically more favorable lower level, but some in the upper level as well. To maintain a constant average fraction of the balls at each level with vigorous and continued shaking, the at which balls go from the upper to the lower level must be equal to the that they go in the opposite direction. The balls now will be between the two levels. At equilibrium, the fraction of the balls on each of the two levels is wholly independent of the height of the barrier, just as long as the activation (shaking) is sufficient to permit the balls to go ways. The diagrams of Figure 4-4 are to be interpreted in the same general way. If thermal agitation of the molecules is sufficient, then equilibrium can be expected to be established between the reactants and the products, whether the overall reaction is energetically favorable (left side of Figure 4-4) or energetically unfavorable (right side of Figure 4-4). But as with our analogy, when equilibrium is established we expect the major portion of the molecules to be in the more favorable energy state. What happens when methane is mixed with chlorine? No measurable reaction occurs when the gases are mixed and kept in the dark at room temperature. Clearly, either the reaction is energetically unfavorable or the energy barrier is high. The answer as to which becomes clear when the mixture is heated to temperatures in excess of \(300^\text{o}\) or when exposed to strong violet or ultraviolet light, whereby a rapid or even explosive reaction takes place. Therefore the reaction is energetically favorable, but the activation energy is greater than can be attained by thermal agitation alone at room temperature. Heat or light therefore must initiate a pathway for the reactants to be converted to products that has a low barrier or activation energy. Could we have predicted the results of this experiment ahead of time? First, we must recognize that there really are several questions here. Could we have decided whether the reaction was energetically favorable? That the dark reaction would be slow at room temperature? That light would cause the reaction to be fast? We consider these and some related questions in detail because they are questions and the answers to them are relevant in one way or another to the study of reactions in organic chemistry. Presumably, methane could react with chlorine to give chloromethane and hydrogen chloride, or chloromethane could react with hydrogen chloride to give methane and chlorine. If conditions were found for which both reactions proceeded at a finite rate, equilibrium finally would be established when the rates of the reactions in each direction became equal: \[ \ce{CH_4 + Cl_2 \rightleftharpoons CH_3Cl + HCl}\] At equilibrium, the relationship among the amounts of reactants and products is given by the equilibrium constant expression \[K_{eq} = \dfrac{[CH_3Cl,HCl]}{[CH_4,Cl_2]} \label{4-1}\] in which \(K_\text{eq}\) is the equilibrium constant. The quantities within the brackets of Equation \(\ref{4-1}\) denote either concentrations for liquid reactants or partial pressures for gaseous substances. If the equilibrium constant \(K_\text{eq}\) is \(1\), then on mixing equal volumes of each of the participant substances (all are gases above \(-24^\text{o}\)), reaction to the will be initially faster than reaction to the left, until equilibrium is established; at this point there will be more chloromethane and hydrogen chloride present than methane and chlorine. However, if the equilibrium constant were \(1\), the reaction initially would proceed faster to the and, at equilibrium, there would be more methane and chlorine present than chloromethane and hydrogen chloride.\(^4\) For methane chlorination, we know from experiment that the reaction goes to the right and that \(K_\text{eq}\) is much greater than unity. Naturally, it would be helpful in planning other organic preparations to be able to estimate \(K_\text{eq}\) in advance. It is a common experience to associate chemical reactions with equilibrium constants greater than one with the evolution of heat, in other words, with negative \(\Delta H^\text{0}\) values. There are, in fact, many striking examples. Formation of chloromethane and hydrogen chloride from methane and chlorine has a \(K_\text{eq}\) of \(10^{18}\) and \(\Delta H^\text{0}\) of \(-24 \: \text{kcal}\) per mole of \(CH_3Cl\) formed at \(25^\text{o}\). Combustion of hydrogen with oxygen to give water has a \(K_\text{eq}\) of \(10^{40}\) and \(\Delta H^\text{0} = -57 \: \text{kcal}\) per mole of water formed at \(25^\text{o}\). However, this correlation between \(K_\text{eq}\) and \(\Delta H^\text{0}\) is neither universal nor rigorous. Reactions are known that absorb heat (are endothermic) and yet have \(K_\text{eq} > 1\). Other reactions have large \(\Delta H^\text{0}\) values and equilibrium constants much less than \(1\). The problem is that the energy change that correlates with \(K_\text{eq}\) is not \(\Delta H^\text{0}\) but \(\Delta G^\text{0}\) (the so-called change of " ")\(^5\), and if we know \(\Delta G^\text{0}\), we can calculate \(K_\text{eq}\) by the equation \[ \Delta G^o =-2.303 RT \log_{10} K_{eq} \label{4-2}\] in which \(R\) is the gas constant and \(T\) is the absolute temperature in degrees Kelvin. For our calculations, we shall use \(R\) as \(1.987 \: \text{cal} \: \text{deg}^{-1} \: \text{mol}^{-1}\) and you should not forget to convert \(\Delta G^\text{0}\) to \(\text{cal}\). Tables of \(\Delta G^\text{0}\) values for formation of particular compounds (at various temperatures and states) from the elements are available in handbooks and the literature. With these, we can calculate equilibrium constants quite accurately. For example, handbooks give the following data, which are useful for methane chlorination: Combining these with proper regard for sign gives and \(\text{log} \: K_\text{eq} = -\left( -24.7 \times 1000 \right)/ \left(2.303 \times 1.987 \times 298.2 \right)\), so \(K_\text{eq} = 1.3 \times 10^{18}\). Unfortunately, insufficient \(\Delta G^\text{0}\) values for formation reactions are available to make this a widely applicable method for calculating \(K_\text{eq}\) values. The situation is not wholly hopeless, because there is a relationship between \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) that also involves \(T\) and another quantity, \(\Delta S^\text{0}\), the standard of the process: \[ \Delta G^o = \Delta H^o -T \Delta S^o \label{4-3}\] This equation shows that \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) are equal when \(\Delta S^\text{0}\) is zero. Therefore the sign and magnitude of \(T \Delta S^\text{0}\) determine how well \(K_\text{eq}\) correlates with \(\Delta H^\text{0}\). Now, we have to give attention to whether we can estimate \(T \Delta S^\text{0}\) values well enough to decide whether the \(\Delta H^\text{0}\) of a given reaction (calculated from bond energies or other information) will give a good or poor measure of \(\Delta G^\text{0}\). To decide whether we need to worry about \(\Delta S^\text{0}\) with regard to any particular reaction, we have to have some idea what physical meaning entropy has. To be very detailed about this subject is beyond the scope of this book, but you should try to understand the physical basis of entropy, because if you do, then you will be able to predict at least qualitatively whether \(\Delta H^\text{0}\) will be about the same or very different from \(\Delta G^\text{0}\). Essentially, the entropy of a chemical system is a measure of its or . Other things being the same, the more random the system is, the more favorable the system is. Different kinds of molecules have different degrees of translational, vibrational, and rotational freedom and, hence, different average degrees of molecular disorder or randomness. Now, if for a chemical reaction the degree of molecular disorder is different for the products than for the reactants, there will be a change in entropy and \(\Delta S^\text{0} \neq 0\). A spectacular example of the effect of molecular disorder in contributing to the difference between \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\) is afforded by the formation of liquid nonane, \(C_9H_{20}\), from solid carbon and hydrogen gas at \(25^\text{o}\): \[\ce{9C(s) + 10H_2(g) \rightarrow C_910_{20}(l)}\] with \(\Delta H^o = -54.7 \, kcal\) and \(\Delta S^o = 5.0 \, kcal\). Equations \(\ref{4-2}\) and \(\ref{4-3}\) can be rearranged to calculate \(\Delta S^\text{0}\) and \(K_\text{eq}\) from \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\):   and \[K_{eq} = 10^{-\Delta G^o/2.303 \,RT} = 10^{-5.900/(2.303 \times 1.987 \times 298.2)} = 4.7 \times 10^{-5}\] These \(\Delta H^\text{0}\), \(\Delta S^\text{0}\), and \(K_\text{eq}\) values can be compared to those for \(H_2 + \frac{1}{2} O_2 \longrightarrow H_2O\), for which \(\Delta H^\text{0}\) is \(-57 \: \text{kcal}\), \(\Delta S^\text{0}\) is \(8.6 \: \text{e.u.}\), and \(K_\text{eq}\) is \(10^{40}\). Obviously, there is something about the entropy change from carbon and hydrogen to nonane. The important thing is that there is a great in the constraints on the atoms on each side of the equation. In particular, hydrogen molecules in the gaseous state have great translational freedom and a high degree of disorder, the greater part of which is lost when the hydrogen atoms become attached to a chain of carbons. This makes for a large \(\Delta S^\text{0}\), which corresponds to a in \(K_\text{eq}\). The differences in constraints of the carbons are less important. Solid carbon has an ordered, rigid structure with little freedom of motion of the individual carbon atoms. These carbons are less constrained in nonane, and this would tend to make \(\Delta S^\text{0}\) more positive and \(\Delta G^\text{0}\) more negative, corresponding to an increase in \(K_\text{eq}\) (Equations \(\ref{4-2}\) and \(\ref{4-3}\)). However, this is a effect on \(\Delta S^\text{0}\) compared to the enormous difference in the degree of disorder of hydrogen between hydrogen gas and hydrogen bound to carbon in nonane. Negative entropy effects usually are observed in ring-closure reactions such as the formation of cyclohexane from 1-hexene, which occur with substantial loss of rotational freedom (disorder) about the \(C-C\) bonds: There is an even greater loss in entropy on forming cyclohexane from ethene because substantially more freedom is lost in orienting three ethene molecules to form a ring: For simple reactions, with the same number of molecules on each side of the equation, with no ring formation or other unusual changes in the constraints between the products and reactants, \(\Delta S^\text{0}\) usually is relatively small. In general, for such processes, we know from experience that \(K_\text{eq}\) \(\Delta H^\text{0}\) \(-15 \: \text{kcal}\) \(\Delta H^\text{0}\) \(+15 \: \text{kcal}\). We can use this as a "rule of thumb" to predict whether \(K_\text{eq}\) should be greater or less than unity for vapor-phase reactions involving simple molecules. Some idea of the degree of success to be expected from this rule may be inferred from the examples in Table 4-5, which also contains a further comparison of some experimental \(\Delta H^\text{0}\) values with those calculated from bond energies. Suppose \(\Delta G^\text{0}\) is positive, what hope do we have of obtaining a useful conversion to a desired product? There is no simple straightforward and general answer to this question. When the reaction is reversible the classic procedure of removing one or more of the products to prevent equilibrium from being established has many applications in organic chemistry, as will be seen later. When this approach is inapplicable, a change in reagents is necessary. Thus, iodine does not give a useful conversion with 2,2-dimethylpropane, \(1\), to give 1-iodo-2,2-dimethylpropane, \(2\), because the position of equilibrium is too far to the left (\(K_\text{eq} \cong 10^{-5}\)): Alternative routes with favorable \(\Delta G^\text{0}\) values are required. Development of ways to make indirectly, by efficient processes, what cannot be made directly is one of the most interesting and challenging activities of organic chemists. To reach an understanding of why methane and chlorine do not react in the dark, we must consider the details of the reaction occurs - that is, the . The simplest mechanism would be for a chlorine molecule to collide with a methane molecule in such a way as to have chloromethane and hydrogen chloride formed directly as a result of a breaking of the \(Cl-Cl\) and \(C-H\) bonds and making of the \(C-Cl\) and \(H-Cl\) bonds (see Figure 4-5). The failure to react indicates that there must be an energy barrier too high for this mechanism to operate. Why should this be so? First, this mechanism involves a very precisely oriented "four-center" collision between chlorine and methane that would have a low probability of occurrence (i.e., a large decrease in entropy because a precise orientation means high molecular ordering). Second, it requires pushing a chlorine molecule sufficiently deeply into a methane molecule so one of the chlorine atoms comes close enough to the carbon to form a bond and yield chloromethane. Generally, to bring nonbonded atoms to near-bonding distances (\(1.2 \: \text{A}\) to \(1.8 \: \text{A}\)) requires a large expenditure of energy, as can be seen in Figure 4-6. Interatomic repulsive forces increase rapidly at short distances, and pushing a chlorine molecule into a methane molecule to attain distances similar to the \(1.77\)-\(\text{A}\) carbon-chlorine bond distance in chloromethane would require a considerable amount of compression (see Figure 4-7). Valuable information about interatomic repulsions can be obtained with space-filling models of the CPK type ( ), which have radii scaled to correspond to actual atomic interference radii, that is, the interatomic distance at the point where curves of the type of Figure 4-6 start to rise steeply. With such models, the degree of atomic compression required to bring the nonbonded atoms to within near-bonding distance is more evident than with the ball-and-stick models. It may be noted that four-center reactions of the type postulated in Figure 4-5 are encountered only rarely. If the concerted four-center mechanism for formation of chloromethane and hydrogen chloride from chlorine and methane is discarded, all the remaining possibilities are . A slow stepwise reaction is dynamically analogous to the flow of sand through a succession of funnels with different stem diameters. The funnel with the smallest stem will be the most important bottleneck and, if its stem diameter is much smaller than the others, it alone will determine the flow rate. Generally, a multistep chemical reaction will have a slow (analogous to the funnel with the small stem) and other relatively , which may occur either before or after the slow step. A possible set of steps for the chlorination of methane follows: Reactions (1) and (2) involve dissociation of chlorine into chlorine atoms and the breaking of a \(C-H\) bond of methane to give a methyl radical and a hydrogen atom. The methyl radical, like chlorine and hydrogen atoms, has one election not involved in bonding. Atoms and radicals usually are highly reactive, so formation of chloromethane and hydrogen chloride should proceed readily by Reactions (3) and (4). The crux then will be whether Steps (1) and (2) are reasonable under the reaction conditions. In the absence of some , only collisions due to the usual thermal motions of the molecules can provide the energy needed to break the bonds. At temperatures below \(100^\text{o}\), it is very rare indeed that the thermal agitation alone can supply sufficient energy to break any significant number of bonds stronger than \(30\) to \(35 \: \text{kcal mol}^{-1}\). The \(Cl-Cl\) bond energy from Table 4-3 is \(58.1 \: \text{kcal}\), which is much too great to allow bond breaking from thermal agitation at \(25^\text{o}\) in accord with Reaction (1). For Reaction (2) it is not advisable to use the \(98.7 \: \text{kcal} \: C-H\) bond energy from Table 4-3 because this is one fourth of the energy required to break all four \(C-H\) bonds ( ). More specific are given in Table 4-5, and it will be seen that to break one \(C-H\) bond of methane requires \(104 \: \text{kcal}\) at \(25^\text{o}\), which again is too much to be gained by thermal agitation. Therefore we can conclude that Reactions (1)-(4) can not be an important mechanism for chlorination of methane at room temperature. One might ask whether dissociation into ions would provide viable mechanisms for methane chlorination. Part of the answer certainly is: Not in the vapor phase, as the following thermochemical data show: Ionic dissociation simply does not occur at ordinarily accessible temperatures by collisions between molecules in the vapor state. What is needed for formation of ions is either a highly energetic external stimulus, such as bombardment with fast-moving electrons, or an ionizing solvent that will assist ionization. Both of these processes will be discussed later. The point here is that ionic dissociation is not a viable step for the vapor-phase chlorination of methane. First, we should make clear that the light does more than provide energy merely to lift the molecules of methane and chlorine over the barrier of Figure 4-4. This is evident from the fact that very little light is needed, far less than one light photon per molecule of chloromethane produced. The light could activate either methane or chlorine, or both. However, methane is colorless and chlorine is yellow-green. This indicates that chlorine, not methane, interacts with visible light. A photon of near-ultraviolet light, such as is absorbed by chlorine gas, provides more than enough energy to split the molecule into two chlorine atoms: Once produced, a chlorine atom can remove a hydrogen atom from a methane molecule and form a methyl radical and a hydrogen chloride molecule. The bond-dissociation energies of \(CH_4\) (\(104 \: \text{kcal}\)) and \(HCl\) (\(103.1 \: \text{kcal}\)) suggest that this reaction is endothermic by about \(1 \: \text{kcal}\): Use of bond-dissociation energies gives a calculated \(\Delta H^\text{0}\) of \(-26 \: \text{kcal}\) for this reaction, which is certainly large enough, by our rule of thumb, to predict that \(K_\text{eq}\) will be greater than 1. Attack of a methyl radical on molecular chlorine is expected to require somewhat more oriented collision than for a chlorine atom reacting with methane (the chlorine molecule probably should be endwise, not sidewise, to the radical) but the interatomic repulsion probably should not be much different. The net result of \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) and \(CH_3 \cdot + Cl_2 \longrightarrow CH_3Cl + Cl \cdot\) is formation of chloromethane and hydrogen chloride from methane and chlorine. Notice that the chlorine atom consumed in the first step is replaced by another one in the second step. This kind of sequence of reactions is called a because, in principle, one atom can induce the reaction of an infinite number of molecules through operation of a "chain" or cycle of reactions. In our example, chlorine atoms formed by the action of light on \(Cl_2\) can induce the chlorination of methane by the : In practice, chain reactions are limited by so-called processes. In our example, chlorine atoms or methyl radicals are destroyed by reacting with one another, as shown in the following equations: Chain reactions may be considered to involve three phases: First, must occur, which for methane chlorination is activation and conversion of chlorine molecules to chlorine atoms by light. Second, steps convert reactants to products with no net consumption of atoms or radicals. The propagation reactions occur in competition with steps, which result in destruction of atoms or radicals. Putting everything together, we can write: The chain-termination reactions are expected to be exceedingly fast because atoms and radicals have electrons in unfilled shells that normally are bonding. As a result, bond formation can begin as soon as the atoms or radicals approach one another closely, without need for other bonds to begin to break. The evidence is strong that bond-forming reactions between atoms and radicals usually are , that there is almost no barrier or activation energy required, and the rates of combination are simply the rates at which encounters between radicals or atoms occur. If the rates of combination of radicals or atoms are so fast, you might well wonder how chain propagation ever could compete. Of course, competition will be possible if the propagation reactions themselves are fast, but another important consideration is the fact that the . Suppose that the concentration of \(Cl \cdot\) is \(10^{-11} \: \text{M}\) and the \(CH_4\) concentration \(1 \: \text{M}\). The probability of encounters between two \(Cl \cdot\) atoms will be proportional to \(10^{-11} \times 10^{-11}\), and between \(CH_4\) and \(Cl \cdot\) atoms it will be \(10^{-11} \times 1\). Thus, other things being the same, \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) (propagation) would be favored over \(2Cl \cdot \longrightarrow Cl_2\) (termination) by a factor of \(10^{11}\). Under favorable conditions, the methane-chlorination chain may go through 100 to 10,000 cycles before termination occurs by radical or atom combination. Consequently the efficiency (or ) of the reaction is very high in terms of the amount of chlorination that occurs relative to the amount of the light absorbed. The overall rates of chain reactions usually are slowed very much by substances that can combine with atoms or radicals and convert them into species incapable of participating in the chain-propagation steps. Such substances are called , or . Oxygen acts as an inhibitor in the chlorination of methane by rapidly combining with a methyl radical to form the comparatively stable (less reactive) peroxymethyl radical, \(CH_3OO \cdot\). This effectively terminates the chain: To a considerable degree, we can predict reactivities, provided we use common sense to limit our efforts to reasonable situations. In the preceding section, we argued that reactions in which atoms or radicals combine can well be expected to be extremely fast because each entity has a potentially bonding electron in an outer unfilled shell, and bringing these together to form a bond does not require that other bonds be broken: The difference between the average energy of the reactants and the energy of the transition state is called the (Figure 4-4). We expect this energy to be smaller (lower barrier) if a weak bond is being broken and a strong bond is being made. The perceptive reader will notice that we are suggesting a parallel between reaction rate and \(\Delta H^\text{0}\) because \(\Delta H^\text{0}\) depends on the difference in strengths of the bonds being broken and formed. Yet previously ( ), we pointed out that the energy barrier for a reaction need bear no relationship to how energetically feasible the reaction is, and this is indeed true for complex reactions involving many steps. But our intuitive parallel between rate and \(\Delta H^\text{0}\) usually works quite well for the rates of steps. This is borne out by experimental data on rates of removal of a hydrogen atom from methane by atoms or radicals (\(X \cdot\)), such as \(F \cdot\), \(Cl \cdot\), \(Br \cdot\), \(HO \cdot\), \(H_2N \cdot\), which generally parallel the strength of the new bond formed: Similarly, if we look at the \(H-C\) bond-dissociation energies of the hydrocarbons shown in Table 4-6, we would infer that \(Cl \cdot\) would remove a hydrogen most rapidly from the carbon forming the weakest \(C-H\) bond and, again, this is very much in accord with experience. For example, the chlorination of methylbenzene (toluene) in sunlight leads to the substitution of a methyl hydrogen rather than a ring hydrogen for the reason that the methyl \(C-H\) bonds are weaker and are attacked more rapidly than the ring \(C-H\) bonds. This can be seen explicitly in the \(\Delta H^\text{0}\) values for the chain-propagation steps calculated from the bond-dissociation energies of Table 4-6. The \(\Delta H^\text{0}\) of ring-hydrogen abstraction is unfavorable by \(+7 \: \text{kcal}\) because of the high \(C-H\) bond energy (\(110 \: \text{kcal}\)). Thus this step is not observed. It is too slow in comparison with the more favorable reaction at the methyl group even though the second propagation step is energetically favorable by \(-37 \: \text{kcal}\) and presumably would occur very rapidly. Use of bond-dissociation energies to predict relative reaction rates becomes much less valid when we try to compare different kinds of reactions. To illustrate, ethane might react with \(F \cdot\) to give fluoromethane or hydrogen fluoride: It is not a good idea to try to predict the relative rates of these two reactions on the basis of their overall \(\Delta H^\text{0}\) values because the nature of the bonds made and broken is too different. Faced with proposing a mechanism for a reaction that involves overall making or breaking of more than two bonds, the beginner almost invariably tries to concoct a process wherein, with a step, all of the right bonds break and all of the right bonds form. Such mechanisms, called , have three disadvantages. First, they are almost impossible to prove correct. Second, prediction of the relative rates of reactions involving concerted mechanisms is especially difficult. Third, concerted mechanisms have a certain sterility in that one has no control over what happens while they are taking place, except an overall control of rate by regulating concentrations, temperature, pressure, choice of solvents, and so on. To illustrate, suppose that methane chlorination appeared to proceed by way of a one-step concerted mechanism: At the instant of reaction, the reactant molecules in effect would disappear into a dark closet and later emerge as product molecules. There is no way to prove experimentally that all of the bonds were made and formed simultaneously. All one could do would be to use the most searching possible tests to probe for the existence of discrete steps. If these tests fail, the reaction still would not be concerted because other, still more searching tests might be developed later that would give a different answer. The fact is, once you accept that a particular reaction is concerted, you, in effect, accept the proposition that further work on its is futile, no matter how important you might feel that other studies would be regarding the factors affecting the reaction rate. The experienced practitioner in reaction mechanisms accepts a concerted mechanism for a reaction involving the breaking and making of more than two bonds as a last resort. He first will try to analyze the overall transformation in terms of discrete steps that are individually simple enough surely to be concerted and that also involves energetically reasonable intermediates. Such an analysis of a reaction in terms of discrete mechanistic steps offers many possibilities for experimental studies, especially in development of procedures for detecting the existence, even if highly transitory, of the proposed intermediates. We shall give many examples of the fruitfulness of this kind of approach in subsequent discussions. \(^4\)If calculations based on chemical equilibrium constants are unfamiliar to you, we suggest you study one of the general chemistry texts listed for supplemental reading at the end of Chapter 1. \(^5\)Many books and references use \(\Delta F^\text{0}\) instead of \(\Delta G^\text{0}\). The difference between standard Gibbs energy \(\Delta G^\text{0}\) and the Gibbs energy \(\Delta G\) is that \(\Delta G^\text{0}\) is defined as the value of the free energy when all of the participants are in standard states. The free energy for \(\Delta G\) for a reaction \(\text{A} + \text{B} + \cdots \longrightarrow \text{X} + \text{Y} + \cdots\) is equal to \(\Delta G^\text{0} - 2.303 RT \: \text{log} \: \frac{\left[ \text{X} \right] \left[ \text{Y} \right] \cdots}{\left[ \text{A} \right] \left[ \text{B} \right] \cdots}\) where the products, \(\left[ \text{X} \right], \left[ \text{Y} \right] \cdots\), and the reactants, \(\left[ \text{A} \right], \left[ \text{B} \right] \cdots\), do not have to be in standard states. We shall use only \(\Delta G^\text{0}\) in this book. \(^6\)The entropy unit \(\text{e.u.}\) has the dimensions calorie per degree or \(\text{cal deg}^{-1}\). and (1977)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.2%3A_Bonding_in_Simple_Molecules_That_Contain_Transition_Metals

The oxidation state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Almost all of the have multiple oxidation states experimentally observed. Filling atomic orbitals requires a set number of electrons. The s-block is composed of elements of Groups I and II, the alkali and alkaline earth metals (sodium and calcium belong to this block). Groups XIII through XVIII comprise of the p-block, which contains the nonmetals, halogens, and noble gases (carbon, nitrogen, oxygen, fluorine, and chlorine are common members). Transition metals reside in the d-block, between Groups III and XII. If the following table appears strange, or if the orientations are unclear, please review the section on . The key thing to remember about electronic configuration is that the most stable noble gas configuration is ideal for any atom. Forming bonds are a way to approach that configuration. In particular, the transition metals form more lenient bonds with anions, cations, and neutral complexes in comparison to other elements. This is because the d orbital is rather diffused (the f orbital of the lanthanide and actinide series more so). Counting through the periodic table is an easy way to determine which electrons exist in which orbitals. As mentioned before, by counting protons (atomic number), you can tell the number of electrons in a neutral atom. Organizing by block quickens this process.  1s (H, He), 2s (Li, Be), 2p (B, C, N, O, F, Ne), 3s (Na, Mg), 3p (Al, Si, P, S, Cl, Ar), 4s (K, Ca), 3d (Sc, Ti, V). If you do not feel confident about this counting system and how electron orbitals are filled, please see the section on . Referring to the periodic table below confirms this organization. We have three elements in the 3d orbital. Therefore, we write in the order the orbitals were filled. 1s 2s 2p 3s 3p 4s 3d or [Ar] 4s 3d . The neutral atom configurations of the fourth period transition metals are in Table \(\Page {2}\). Chromium and copper appear anomalous. Take a brief look at where the element (atomic number 24) lies on the Periodic Table (Figure \(\Page {1}\)). The electronic configuration for chromium is not [Ar] 4s 3d but instead it is [Ar] 4s 3d . This is because the half-filled 3d manifold (with one 4s electron) is more stable than a partially filled d-manifold (and a filled 4s manifold). You will notice from Table \(\Page {2}\) that the copper exhibits a similar phenomenon, although with a fully filled d-manifold. When considering ions, we add or subtract negative charges from an atom. Keeping the atomic orbitals when assigning oxidation numbers in mind helps in recognizing that transition metals pose a special case, but not an exception to this convenient method. An atom that accepts an electron to achieve a more stable configuration is assigned an oxidation number of -1. The donation of an electron is then +1. When a transition metal loses electrons, it tends to lose it's s orbital electrons before any of its d orbital electrons. For more discussion of these compounds form, see . Write the electronic configurations of: The atomic number of iron is 26 so there are 26 protons in the species. Determine the more stable configuration between the following pair: Most transition metals have multiple oxidation states, since it is relatively easy to lose electron(s) for transition metals compared to the alkali metals and alkaline earth metals. Alkali metals have one electron in their valence s-orbital and their ions almost always have oxidation states of +1 (from losing a single electron). Similarly, alkaline earth metals have two electrons in their valences s-orbitals, resulting in ions with a +2 oxidation state (from losing both). However, transitions metals are more complex and exhibit a range of observable oxidation states due primarily to the removal of d-orbital electrons. The following chart describes the most common oxidation states of the period 3 elements. Scandium is one of the two elements in the first transition metal period which has only one oxidation state (zinc is the other, with an oxidation state of +2). All the other elements have at least two different oxidation states. Manganese, which is in the middle of the period, has the highest number of oxidation states, and indeed the highest oxidation state in the whole period since it has five unpaired electrons (see table below). To help remember the stability of higher oxidation states for transition metals it is important to know the trend: the stability of the higher oxidation states progressively increases down a group. For example, in group 6, (chromium) Cr is most stable at a +3 oxidation state, meaning that you will not find many stable forms of Cr in the +4 and +5 oxidation states. By contrast, there are many stable forms of molybdenum (Mo) and tungsten (W) at +4 and +5 oxidation states. What makes zinc stable as Zn ? What makes scandium stable as Sc ? Zinc has the neutral configuration [Ar]4s 3d . Losing 2 electrons does not alter the complete d orbital. Neutral scandium is written as [Ar]4s 3d . Losing 3 electrons brings the configuration to the noble state with valence 3p . Why is iron almost always Fe or Fe ? Iron is written as [Ar]4s 3d . Losing 2 electrons from the s-orbital (3d ) or 2 s- and 1 d-orbital (3d ) electron are fairly stable oxidation states. Write manganese oxides in a few different oxidation states. Which ones are possible and/or reasonable? Although Mn is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn , Mn(II) as MnO, Mn(II,III) as Mn O , Mn(IV) as MnO , or manganese dioxide, Mn(VII) in the permanganate ion MnO , and so on. When given an ionic compound such as \(\ce{AgCl}\), you can easily determine the oxidation state of the transition metal. In this case, you would be asked to determine the oxidation state of silver (Ag). Since we know that chlorine (Cl) is in the group of the periodic table, we then know that it has a charge of -1, or simply Cl . In addition, by seeing that there is no overall charge for \(\ce{AgCl}\), (which is determined by looking at the top right of the compound, i.e., AgCl # represents Determine the oxidation state of cobalt in \(\ce{CoBr2}\). Similar to chlorine, bromine (\(\ce{Br}\)) is also a halogen with an oxidation charge of -1 (\(\ce{Br^{-}}\)). Since there are two bromines each with a charge of -1. In addition, we know that \(\ce{CoBr2}\) has an overall neutral charge, therefore we can conclude that the cation (cobalt), \(\ce{Co}\) must have an oxidation state of +2 to neutralize the -2 charge from the two bromine anions. What is the oxidation state of zinc in \(\ce{ZnCO3}\). (Note: the \(\ce{CO3}\) anion has a charge state of -2) \(\ce{CO3}\) s us \(\ce{Zn^{2+}}\) an \(\ce{CO3^{-2}}\) Consider the manganese (\(\ce{Mn}\)) atom in the permanganate (\(\ce{MnO4^{-}}\)) ion. Since has an oxidation state of -2 and we know there are four oxygen atoms. In addition, this compound has an overall charge of -1; therefore the overall charge is not neutral in this example. Thus, since the oxygen atoms in the ion contribute a total oxidation state of -8, and since the overall charge of the ion is -1, the sole manganese atom must have an oxidation state of +7. This gives us \(\ce{Mn^{7+}}\) and \(\ce{4 O^{2-}}\), which will result as \(\ce{MnO4^{-}}\). This example also shows that manganese atoms can have an oxidation state of +7, which is the highest possible oxidation state for the fourth period transition metals. is widely studied because it is an important reducing agent in chemical analysis and is also studied in biochemistry for catalysis and in metallurgy in fortifying alloys. In plants, manganese is required in trace amounts; stronger doses begin to react with enzymes and inhibit some cellular function. Due to manganese's flexibility in accepting many oxidation states, it becomes a good example to describe general trends and concepts behind electron configurations. Electron configurations of unpaired electrons are said to be and respond to the proximity of magnets. Fully paired electrons are and do not feel this influence. Manganese, in particular, has paramagnetic and diamagnetic orientations depending on what its oxidation state is. \(\ce{Mn2O3}\) is manganese(III) oxide with manganese in the +3 state. 4 unpaired electrons means this complex is paramagnetic. \[\ce{[Ar]} 4s^{0} 3d^{4}\nonumber\] \(\ce{MnO2}\) is manganese(IV) oxide, where manganese is in the +4 state. 3 unpaired electrons means this complex is less paramagnetic than Mn . \[\ce{[Ar]} 4s^{0} 3d^{3}\nonumber\] \(\ce{KMnO4}\) is potassium permanganate, where manganese is in the +7 state with no electrons in the 4s and 3d orbitals. \[\ce{[Ar]} 4s^{0} 3d^{0}\nonumber\] Since the 3p orbitals are all paired, this complex is diamagnetic. Oxidation states of transition metals follow the general rules for most other ions, except for the fact that the d orbital is degenerated with the s orbital of the higher quantum number. Transition metals achieve stability by arranging their electrons accordingly and are oxidized, or they lose electrons to other atoms and ions. These resulting cations participate in the or synthesis of other compounds. Determine the oxidation states of the transition metals found in these neutral compounds. Note: The transition metal is underlined in the following compounds.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/09%3A_Separation_Purification_and_Identification_of_Organic_Compounds/9.10%3A_Electronic_Spectra_of_Organic_Molecules

A year after Herschel discovered infrared radiation, Johann Ritter discovered radiation beyond the violet end of the visible spectrum. This radiation came to be known as and soon was recognized as being especially effective in causing chemical reactions. Absorption of light in the ultraviolet and visible regions produces changes in the electronic energies of molecules associated with excitation of an electron from a stable to an unstable orbital. Because the energy required to excite the valence-shell electrons of molecules is comparable to the strengths of chemical bonds, absorption may lead to chemical reactions. We discussed this briefly in Chapter 4 in connection with photochemical halogenation of alkanes; a more detailed account of photochemistry is given in . The transition of an electron from the ground state, \(E_1\), to an excited electronic state, \(E_2\), is accompanied by vibrational and rotational changes in the molecule, as shown in Figure 9-17. In condense phase samples, it usually is not possible to resolve the resulting absorption well enough to see the fine structure due to vibration-rotation transitions. Consequently, absorptions due to electronic excitation are relatively broad. The ultraviolet spectrum of 2-propanone (acetone) is shown in Figure 9-18. The weak absorption, which peaks (i.e., has \(\lambda_\text{max}\) at \(280 \: \text{nm}\), is the result of excitation of one of the unshared electrons on oxygen to a higher energy level. This is called an \(n \rightarrow \pi^*\) (often \(N \rightarrow A\)) transition, in which \(n\) denotes that the excited electron is one of the unshared \(n\) electrons on oxygen and \(\pi^*\) (pi star) denotes that the excited electron goes to a high-energy orbital of the carbon-oxygen double bond (cf. and ). The same kind of \(n \rightarrow \pi^*\) transition occurs at about the same wavelength and intensity for many simple compounds of the type \(\ce{R_2C=O}\) and \(\ce{RCH=O}\), in which \(R\) is an alkyl group. In a very schematic way, we can write The \(\pi \rightarrow \pi^*\) transition for ethene has \(\lambda_\text{max} = 175 \: \text{nm}\) and \(\epsilon_\text{max} = 10,000\). It would be expected that an alkadiene would give an absorption spectrum similar to that of ethene but with a larger \(\epsilon\), because there are more double bonds per mole to absorb radiation. This expectation is more or less realized for compounds such as 1,5-hexadiene and 1,3-dimethylenecyclobutane, which have isolated double bonds, but not for 1,3-butadiene or ethenylbenzene, which have conjugated double bonds ( ): In general, This means that the difference in energy between the normal and excited states of conjugated systems is than for isolated systems of double bonds. For 1,3-butadiene and 1,5-hexadiene we can calculate from Equation 9-2 that the transition energy is about \(23 \: \text{kcal}\) less for the conjugated system. The ground state of 1,3-butadiene is stabilized by perhaps \(3 \: \text{kcal}\) relative to a nonconjugated system of double bonds, which means that the excited state must be much more stabilized than this if the transition energy is to be \(23 \: \text{kcal}\) less than for 1,5-hexadiene. Why is the excited state of a conjugated system of double bonds stabilized more, relative to the ground state, than for a nonconjugated system? Resonance theory provides an explanation (see Section 6-5). Of the several conventional valence-bond structures that can be written for 1,3-butadiene, four of which are shown here, \(2a\)-\(2d\), only structure \(2a\) has a low enough energy to be dominant for the ground state of 1,3-butadiene: Now, when the molecule is excited to the extent of \(132 \: \text{kcal mol}^{-1}\) by \(217 \: \text{nm}\) ultraviolet light, its energy is so large that pairing schemes such as \(2b\), \(2c\), and \(2d\), which are too unfavorable to contribute very much to the ground state, can be very important for the excited state. Thus the stabilization energy of the excited state, which has a multiplicity of nearly equal-energy pairing schemes, is expected to be greater than that of the ground state with one dominant pairing scheme. The more double bonds in the conjugated system, the smaller the energy difference between the normal and excited states. The diphenylpolyenes of formula \(C_6H_5-(CH=CH)_n-C_6H_5\) absorb radiation at progressively wavelengths as \(n\) is increased. This is apparent from the colors of the compounds, which range from colorless with \(n = 1\), to orange with \(n = 2-7\), to red with \(n = 8\), as \(\lambda_\text{max}\) goes from the ultraviolet into the visible region of the electromagnetic spectrum. Similar effects are found with conjugated \(\ce{C=O}\) and \(\ce{C=N}\) double bonds. For example, the electronic spectra of 2-butanone and 3-buten-2-one are shown in Figure 9-20. Conjugation also can influence infrared spectra. Transitions arising from \(C=C\) and \(C=O\) stretching vibrations generally are more intense and are shifted to slightly lower frequencies (longer wavelengths) for conjugated compounds relative to nonconjugated compounds. Thus the \(C=C\) stretching of 1-butene occurs at \(1650 \: \text{cm}^{-1}\), whereas that of 1,3-butadiene is observed at \(1597 \: \text{cm}^{-1}\). Alkanes and cycloalkanes have no low-energy electronic transitions comparable to conjugated systems or molecules with nonbonding electrons. Therefore alkanes and cycloalkanes show no absorption above \(200 \: \text{nm}\) and are good solvents to use for electronic spectroscopy. How do we use electronic spectroscopy in chemical analysis? The two principal applications are structure determinations and quantitative analysis. The position and intensity of an electronic absorption band provides information as to chemical structure. Such absorptions normally are not as useful as infrared absorptions because they do not give as detailed information. For our purposes here, the main points to remember are: If we are dealing with compounds for which the wavelengths and the molar intensities of the absorption bands are known, then we can use the degree of absorption for quantitative analysis with the aid of the (see Table 9-3 for definitions): \[A = \epsilon cl\] By measuring the absorbance \(A\) of a sample of known \(\epsilon\) in a cell of known path length \(I\), the concentration \(c\) may be determined. Because changes in absorbance reflect changes in concentration, it is possible to use absorbance measurements to follow rates of chemical reactions, to determine equilibrium constants (such as the dissociation constants of acids and bases), and to follow conformational changes in bio-organic molecules such as proteins and nucleic acids. The capacity of electronic spectroscopy for performing qualitative and quantitative of the elements in chemical compounds has been done and applications have been made to elements of low atomic number, such as carbon and oxygen. Electronic spectra has been developed in the study of this elements and it’s compounds. and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/03%3A_Distributions_Probability_and_Expected_Values/3.13%3A_The_Expected_Value_of_a_Function_of_Several_Variables_and_the_Central_Limit_Theorem

We can extend the idea of an expected value to a function of multiple random variables. Let and be distributions whose random variables are \(u\) and \(v\), respectively. Let the probability density functions for these distributions be \({df_u\left(u\right)}/{du}\) and \({df_v\left(v\right)}/{dv}\). In general, these probability density functions are different functions; that is, \(U\) and \(V\) are different distributions. Let \(g\left(u,v\right)\) be some function of these random variables. The probability that an observation made on \(U\) produces a value of \(u\) in the range \(u^*<u<u^*+du\) is \[P\left(u^*<u<u^*+du\right)\ =\frac{df_u\left(u^*\right)}{du}du\nonumber \] and the probability that an observation made on \(V\) produces a value of \(v\) in the range \(v^*<v<v^*+dv\) is \[P\left(v^*<v<v^*+dv\right)=\frac{df_v\left(v^*\right)}{dv}dv\nonumber \] The probability that making one observation on each of these distributions produces a value of \(u\) that lies in the range \(u^*<u<u^*+du\) and a value of \(v\) that lies in the range \(v^*<v<v^*+dv\) is \[\frac{df_u\left(u^*\right)}{du}\frac{df_v\left(v^*\right)}{dv}du\,dv\nonumber \] In a straightforward generalization, we define the expected value of \(g\left(u,v\right)\), \(\left\langle g\left(u,v\right)\ \right\rangle\), as \[\left\langle g\left(u,v\right)\ \right\rangle =\int^{\infty }_{v=-\infty }{\int^{\infty }_{u=-\infty }{g\left(u,v\right)}}\frac{df_u\left(u\right)}{du}\frac{df_v\left(v\right)}{dv}dudv\nonumber \] If \(g\left(u,v\right)\) is a sum of functions of independent variables, \(g\left(u,v\right)=h\left(u\right)+k\left(v\right)\), we have \[\left\langle g\left(u,v\right)\right\rangle =\int^{\infty }_{-\infty }{\int^{\infty }_{-\infty }{\left[h\left(u\right)+k\left(v\right)\right]\frac{df_u\left(u\right)}{du}\frac{df_v\left(v\right)}{dv}}dudv}=\int^{\infty }_{-\infty }{h\left(u\right)\frac{df_u\left(u\right)}{du}}du+\int^{\infty }_{-\infty }{k\left(v\right)\frac{df_v\left(v\right)}{dv}}dv=\ \left\langle h\left(u\right)\ \right\rangle +\left\langle k\left(v\right)\ \right\rangle\nonumber \] If \(g\left(u,v\right)\) is a product of independent functions, \(g\left(u,v\right)=h\left(u\right)k\left(v\right)\), we have \[\left\langle g\left(u,v\right)\right\rangle =\int^{\infty }_{-\infty }{\int^{\infty }_{-\infty }{h\left(u\right)k\left(v\right)\frac{df_u\left(u\right)}{du}\frac{df_v\left(v\right)}{dv}}dudv}\ \ \ \ =\int^{\infty }_{-\infty }{h\left(u\right)\frac{df_u\left(u\right)}{du}}du\times \int^{\infty }_{-\infty }{k\left(v\right)\frac{df_v\left(v\right)}{dv}}d=\ \left\langle h\left(u\right)\right\rangle \ \left\langle k\left(v\right)\right\rangle\nonumber \] We can extend these conclusions to functions of the random variables of any number of distributions. If \(u_i\) is the random variable of distribution \(U_i\) whose probability density function is \({df_i\left(u_i\right)}/{du_i}\), the expected value of \[g\left(u_1,\dots ,u_i,\dots ,u_N\right)=h_1\left(u_1\right)+\dots +h_i\left(u_i\right)+\dots +h_N\left(u_N\right)\nonumber \] becomes \[\left\langle g\left(u_1,\dots ,u_i,\dots ,u_N\right)\right\rangle =\sum^N_{i=1}{\left\langle h_i\left(u_i\right)\right\rangle }\nonumber \] and the expected value of \[g\left(u_1,\dots ,u_i,\dots ,u_N\right)=h_1\left(u_1\right)\dots h_i\left(u_i\right)\dots h_N\left(u_N\right)\nonumber \] becomes \[\left\langle g\left(u_1,\dots ,u_i,\dots ,u_N\right)\right\rangle =\ \ \prod^N_{i=1}{\left\langle h_i\left(u_i\right)\right\rangle }\nonumber \] We are particularly interested in expected values for repeated trials made on the same distribution. We consider distributions for which the outcome of one trial is independent of the outcome of any other trial. The probability density function is the same for every trial, so we have \(f\left(u\right)=f_1\left(u_1\right)=\dots =f_i\left(u_i\right)=\dots =f_N\left(u_N\right)\). Let the values obtained for the random variable in a series of trials on the same distribution be \(u_1\), , \(u_i\), , \(u_N\). For each trial, we have \[\left\langle h_i\left(u_i\right)\right\rangle \ =\ \ \int^{\infty }_{-\infty }{h_i\left(u_i\right)\frac{df_i\left(u_i\right)}{du_i}}du_i\nonumber \] If we consider the special case of repeated trials in which the functions \(h_i\left(u_i\right)\) are all the same function, so that \(h\left(u\right)=h_1\left(u_1\right)=\dots =h_i\left(u_i\right)=\dots =h_N\left(u_N\right)\), the expected value of \[g\left(u_1,\dots ,u_i,\dots ,u_N\right)\nonumber \] \[=h_1\left(u_1\right)+\dots +h_i\left(u_i\right)+\dots +h_N\left(u_N\right)\nonumber \] becomes \[\left\langle g\left(u_1,\dots ,u_i,\dots ,u_N\right)\right\rangle =\ \sum^N_{i=1}{\left\langle h_i\left(u_i\right)\right\rangle \ }=N\left\langle h\left(u\right)\right\rangle\nonumber \] and the expected value of \[g\left(u_1,\dots ,u_i,\dots ,u_N\right)=h_1\left(u_1\right)\dots h_i\left(u_i\right)\dots h_N\left(u_N\right)\nonumber \] becomes \[\left\langle g\left(u_1,\dots ,u_i,\dots ,u_N\right)\right\rangle =\ \prod^N_{i=1}{\left\langle h_i\left(u_i\right)\right\rangle }={\left\langle h\left(u\right)\right\rangle }^N\nonumber \] Now let us consider \(N\) independent trials on the same distribution and let \(h_i\left(u_i\right)=h\left(u_i\right)=u_i\). Then, the expected value of \[g\left(u_1,\dots ,u_i,\dots ,u_N\right)= h_1\left(u_1\right)+\dots +h_i\left(u_i\right)+\dots +h_N\left(u_N\right)\nonumber \] becomes \[\ \left\langle u_1+\dots +u_i+\dots +u_N\right\rangle =\sum^N_{i=1}{\left\langle u_i\right\rangle }=N\left\langle u\right\rangle =N\mu\nonumber \] By definition, the average of \(N\) repeated trials is \({\overline{u}}_N={\left(u_1+\dots +u_i+\dots +u_N\right)}/{N}\), so that the expected value of the mean of a distribution of an average-of- \(N\) repeated trials is \[\left\langle {\overline{u}}_N\right\rangle =\frac{\left\langle u_1+\dots +u_i+\dots +u_N\right\rangle }{N}=\mu\nonumber \] This proves one element of the : The mean of a distribution of averages-of- \(N\) values of a random variable drawn from a parent distribution is equal to the mean of the parent distribution. The variance of these averages-of- \(N\) is \[\sigma^2_N=\left\langle {\left({\overline{u}}_N-\mu \right)}^2\right\rangle =\left\langle {\left[\left(\frac{1}{N}\sum^N_{i=1}{u_i}\right)-\mu \right]}^2\ \right\rangle =\left\langle {\left[\left(\frac{1}{N}\sum^N_{i=1}{u_i}\right)-\frac{N\mu }{N}\right]}^2\right\rangle =\ \frac{1}{N^2}\ \left\langle \ {\left[\left(\sum^N_{i=1}{u_i}\right)-N\mu \right]}^2\right\rangle =\ \frac{1}{N^2}\ \left\langle {\left[\left(\sum^N_{i=1}{\left(u_i-\mu \right)}\right)\right]}^2\ \right\rangle =\frac{1}{N^2\ }\ \left\langle \ \sum^N_{i=1}\left(u_i-\mu \right)^2 \right\rangle +\frac{2}{N^2} \left\langle \sum^{N-1}_{i=1} \left(u_i-\mu \right)\sum^N_{j=i+1} \left(u_j-\mu \right) \right \rangle =\frac{1}{N^2}\sum^N_{i=1} \left\langle \left(u_i-\mu \right)^2 \right\rangle +\frac{2}{N^2} \left\langle \sum^{N-1}_{i=1} \left(u_i-\mu \right) \right\rangle \ \left\langle \sum^N_{j=i+1} \left(u_j-\mu \right) \right\rangle\nonumber \] Where the last term is zero, because \[\left\langle \sum^{N-1}_{i=1}{\left(u_i-\mu \right)}\ \right\rangle = \sum^{N-1}_{i=1}{\left\langle \ \left(u_i-\mu \right)\ \right\rangle }\] and \[\left\langle \ \left(u_i-\mu \right)\ \right\rangle \ =0\nonumber \] By definition, \(\sigma^2=\ \left\langle \ {\left(u_i-\mu \right)}^2\ \right\rangle\), so that we have \[\sigma^2_N=\frac{N\sigma^2}{N^2}=\frac{\sigma^2}{N}\nonumber \] This proves a second element of the central limit theorem: The variance of an average of \(N\) values of a random variable drawn from a parent distribution is equal to the variance of the parent distribution divided by \(N\).

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/17%3A_Carbonyl_Compounds_II-_Enols_and_Enolate_Anions._Unsaturated_and_Polycarbonyl_Compounds/17.02%3A_Enolization_of_Aldehydes_and_Ketones

Transformation of a carbonyl compound to an enol at a useful rate normally requires either a basic catalyst or an acidic catalyst and, of course, at least one hydrogen on the \(\alpha\) carbon. The features of each type of catalysis follow. With a basic catalyst such as hydroxide ion, the first step in enolization is removal of a proton from the \(\alpha\) position to give the enolate anion \(1\): Normally, \(\ce{C-H}\) bonds are highly resistant to attack by basic reagents, but removal of a proton to a carbonyl group results in the formation of a considerably stabilized anion with a substantial proportion of the negative charge on oxygen, as represented by the valence-bond structure \(1a\). Carbonyl compounds such as 2-propanone therefore are weak acids, only slightly weaker than alcohols (compare the p\(K_\text{a}\) values for some representative compounds in Table 17-1).\(^1\) carbonyl groups greatly increase the acidity. For example, 2,4-pentanedione (acetylacetone, \(2\)) has a p\(K_\text{a} \cong\) 9, which is comparable to the \(\ce{O-H}\) acidity of phenols (see Table 17-1). The reason is that the enolate anion \(3\) has the charge largely shared by the two oxygen atoms (cf. \(3b\) and \(3c\)). As a result, the enolate anion \(3\) is stabilized more with respect to the ketone than the enolate anion from 2-propanone is stabilized relative to 2-propanone: You will notice from Structures \(1a\) and \(1b\) that because the negative charge of the enolate anion is distributed on both oxygen and carbon, the ion can, in principle, combine with a proton at either site. If the enolate ion adds a proton to oxygen, the enol is formed; if it adds a proton to carbon, the ketone is formed: Ions of this type, which can react at either of two different sites, often are called . In fact, enolate anions add a proton at oxygen at least \(10^{10}\) times faster than at carbon; the proton also is removed from oxygen much faster than from carbon. Thus the enolate anion of 2-propanone is in rapid equilibrium with the enol, but is converted back and forth to the ketone only slowly (Equation 17-1). Another important point is that, although enolization by way of enolate anions requires a basic catalyst, both an acid and a base are necessary: a base to form the enolate anion; an acid to donate a proton to the anion to form the enol. If there is no acid available that is strong enough to donate a proton to the anion, then only the enolate anion is formed: Catalysis of the enolization of 2-propanone by acids involves first, oxonium-salt formation and second, removal of an \(\alpha\) proton with water or other proton acceptor (base): This sequence differs from enolization induced by basic catalysis (as discussed in ) in that the enol is formed directly and not subsequent to the formation of the enolate anion. The proton addition to the carbonyl oxygen greatly facilitates proton removal from the \(\alpha\) carbon because of the electron-attracting power of the positively charged oxygen. Nevertheless, this last step is the for enolization in acid solution. The equilibrium position between a simple ketone and its enol usually lies far on the side of the ketone (see Table 17-2). However, there are some interesting and important exceptions to this generalization. For instance, the influence of carbonyl groups on the enol content is very striking, as we can see from the fact that \(85\%\) of 2,4-pentanedione is the enol form at equilibrium: The enol form of 2,4-pentanedione (and of related dicarbonyl compounds of the type ) not only is stabilized by electron-delocalization, as shown in Structures \(4a\) and \(4b\), but by hydrogen-bonding of the acidic hydrogen between the two oxygens: Of course, such stabilization is not possible for the keto form. An extreme example of the stabilization of an enol by electron delocalization is benzenol (phenol), which exists \(100\%\) in the enol form. In this case the extra stability of the benzene ring is the important factor: In the succeeding sections of this chapter we will discuss several important reactions that take place by way of enols or enolate anions. \(^1\)The important difference between 2-propanone and ethanol as acids is that the of establishment of equilibrium with 2-propanone or similar compounds where ionization involves breaking a \(\ce{C-H}\) bond is very much than the corresponding reaction with \(\ce{O-H}\) bonds. and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Ion_-_Induced_Dipole_Interactions

The charges on ions and the charge separation in polar molecules explain the fairly strong interactions between them, with very strong ion - ion interactions, weaker ion - dipole interactions, and considerably weaker . Even in a non-polar molecule, however, the valence electrons are moving around and there will occasionally be instances when more are on one side of the molecule than on the other. This gives rise to fluctuating or instantaneous dipoles: These instantaneous dipoles may be induced and stabilized as an ion or a polar molecule approaches the non-polar molecule. , Professor of Chemistry, University of Missouri-Rolla

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/20%3A_Molecular_Spectroscopy_and_Photochemistry/20.3%3A_Excited_Electronic_States%3A_Electronic_Spectroscopy_of_Molecules

While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo . What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital. Let’s take as our first example the simple case of molecular hydrogen, H . As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding MO, and a higher energy antibonding * MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding * orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO). If the molecule is exposed to light of a wavelength with energy equal to E, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the to the * orbital. This is referred to as a . E for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm. When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a Because - * energy gaps are narrower than gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen. The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated \(\pi\) systems. In these groups, the energy gap for - * transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called . Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2p atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding. Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm. As conjugated pi systems become larger, the energy gap for a - * transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the - * transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a E of 111 kcal/mol. In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange. The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a - * transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a * antibonding MO: This is referred to as an . The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an \(n \rightarrow \pi^*\) transition is smaller that that of a - * transition – and thus the n - * peak is at a longer wavelength. In general, n - * transitions are weaker (less light absorbed) than those due to * transitions. What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the of -carotene? Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below. Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA ( -aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O. We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred. (Image from ) Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD . This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems. You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm as is the convention in IR spectroscopy. Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is \(\lambda_{max}\), which is the wavelength at maximal light absorbance. As you can see, NAD has \(\lambda_{max} = 260\;nm\). We also want to record how much light is absorbed at \(\lambda_{max}\). Here we use a unitless number called , abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength it passes through the sample (I ), divides this value by the intensity of the same wavelength it passes through the sample (I), then takes the log of that number: \[A = \log \dfrac{I_0}{I} \tag{4.3.1}\] You can see that the absorbance value at 260 nm (A ) is about 1.0 in this spectrum. Here is the absorbance spectrum of the common food coloring Red #3: Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1: Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue. UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance. If we divide the observed value of A at by the concentration of the sample ( , in mol/L), we obtain the , or ( ), which is a characteristic value for a given compound. \[\epsilon = \dfrac{A}{c} \tag{4.3.2}\] The absorbance will also depend, of course, on the - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L mol cm . If we look up the value of e for our compound at , and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD the literature value of at 260 nm is 18,000 L mol cm . In our NAD spectrum we observed A = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10 M. The bases of DNA and RNA are good chromophores: Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of = 0.020 ng ×mL for double-stranded DNA at its of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology). Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’). As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other. Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD , the compound whose spectrum we saw earlier in this section) to NADH: Both NAD and NADH absorb at 260 nm. However NADH, unlike NAD , has a second absorbance band with = 340 nm and = 6290 L mol cm . The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis: By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction. UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores. Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.04%3A_Electrophilic_Additions_to_Alkenes

The reactions of alkanes discussed in are processes, which means that the bonds are made and broken through radical or atomic intermediates. In contrast, the \(S_\text{N}\) and \(E\) reactions of alkyl halides, considered in Chapter 8, involve heterolytic bond cleavage and ionic reagents or products. An especially important factor contributing to the differences between the reactions of the alkanes and alkyl halides is the slight ionic character of \(\ce{C-H}\) compared to \(\ce{C}\)-halide bonds (see ). The alkenes are like the alkanes in being nonpolar compounds ( ) and it may come as a surprise that many important reactions of alkenes are heterolytic reactions. Why should this be so? No doubt because the electrons in the alkene double bonds are more exposed and accessible than the electrons in an alkane \(\ce{C-C}\) bond. This is evident from the atomic-orbital models of ethene described in . The electrons of the double bond are pushed outward by their mutual repulsions, and their average positions are considerably farther from the bond axis than the electron positions of a single bond (Figure 10-6). In such circumstances, electrophilic reagents, which act to acquire electrons in chemical reactions ( ), are expected to be particularly reactive. This is actually the case. Furthermore, reagents that are primarily nucleophilic (electron-donating) are notoriously poor for initiating reactions at carbon-carbon double bonds. Exceptions occur when the double bonds carry substituents with a sufficiently high degree of electron-attracting power to reduce the electron density in the double bond enough to permit attack by a nucleophilic agent. Example of electrophilic reagents that normally add to carbon-carbon double bonds of alkenes to give saturated compounds include halogens (\(\ce{Cl_2}\), \(\ce{Br_2}\), and \(\ce{I_2}\)), hydrogen halides (\(\ce{HCl}\) and \(\ce{HBr}\)), hypohalous acids (\(\ce{HOCl}\) and \(\ce{HOBr}\)), water, and sulfuric acid: The mechanisms of these reactions have much in common and have been studied extensively from this point of view. They also have very considerable synthetic utility. The addition of water to alkenes (hydration) is particularly important for the preparation of a number of commercially important alcohols. Thus ethanol and 2-propanol (isopropyl alcohol) are made on a very large scale by the hydration of the corresponding alkenes (ethene and propene) using sulfuric or phosphoric acids as catalysts. The nature of this type of reaction will be described later. We shall give particular attention here to the addition of bromine to alkenes because this reaction is carried out very conveniently in the laboratory and illustrates a number of important points about electrophilic addition reactions. Much of what follows applies to addition of the other halogens, except fluorine. A significant observation concerning bromine addition is that it and many of the other reactions listed above proceed in the dark and are influenced by radical inhibitors. This is evidence against a radical-chain mechanism of the type involved in the halogenation of alkanes ( ). However, it does not preclude the operation of radical-addition reactions under other conditions, and, as we shall see later in this chapter, bromine, chlorine, and many other reagents that commonly add to alkenes by ionic mechanisms also can add by radical mechanisms. One alternative to a radical-chain reaction for bromine addition to an alkene would be the simple four-center, one-step process shown in Figure 10-7. The mechanism of Figure 10-7 cannot be correct for bromine addition to alkenes in solution for two important reasons. First, notice that this mechanism requires that the two \(\ce{C-Br}\) bonds be formed on the side of the double bond, and hence produce . However, there is much evidence to show that bromine and many other reagents add to alkenes to form products (Figure 10-8). Cyclohexene adds bromine to give -1,2-dibromocyclohexane: The cis isomer is not formed at all. To give the trans isomer, the two new \(\ce{C-Br}\) bonds have to be formed on of the double bond by antarafacial addition. But this is impossible by a one-step mechanism because the \(\ce{Br-Br}\) bond would have to stretch too far to permit the formation of both \(\ce{C-Br}\) bonds at the same time. The second piece of evidence against the mechanism of Figure 10-7 is that bromine addition reactions carried out in the presence of more than one nucleophilic reagent usually give mixtures of products. Thus the addition of bromine to an alkene in methanol solution containing lithium chloride leads not only to the expected dibromoalkane, but also to products resulting from attack by chloride ions and by the solvent: The intervention of extraneous nucleophiles suggests a mechanism in which the nucleophiles compete for a reactive intermediate formed in one of the steps. A somewhat oversimplified two-step mechanism that accounts for most of the foregoing facts is illustrated for the addition of bromine to ethene. [In the formation shown below, the curved arrows are not considered to have real mechanistic significance, but are used primarily to show which atoms can be regarded as nucleophilic (donate electrons) and which as electrophilic (accept electrons). The arrowheads always should be drawn to point to the atoms that are formulated as accepting a pair of electrons.] The first step (which involves electrophilic attack by bromine on the double bond) produces a bromide ion and a carbocation, as shown in Equation 10-1.\(^1\) As we know from our study of \(S_\text{N}1\) reactions ( ), carbocations react readily with nucleophilic reagents. Therefore in the second step of the bromine-addition mechanism, shown in Equation 10-2, the bromoethyl cation is expected to combine rapidly with bromide ion to give the dibromo compound. However, if other nucleophiles, such as \(\ce{Cl}^\ominus\) or \(\ce{CH_3OH}\), are present in solution, they should be able to compete with bromide ion for the cation, as in Equations 10-3 and 10-4, and mixtures of products will result: To account for the observation that all of these reactions result in antarafacial addition, we must conclude that the first and second steps take place from . The simple carbocation intermediate of Equation 10-1 does not account for formation of the antarafacial-addition product. The results with \(S_\text{N}1\) reactions (Section 8-6) and the atomic-orbital representation (see ) predict that the bonds to the positively charged carbon atom of a carbocation should lie in a plane. Therefore, in the second step of addition of bromine to cycloalkenes, bromide ion could attack either side of the planar positive carbon to give a mixture of - and -1,2-dibromocyclohexanes. Nonetheless, antarafacial addition occurs exclusively: To account for the stereospecificity of bromine addition to alkenes, it has been suggested that in the initial electrophilic attack of bromine a cyclic intermediate is formed that has bromine bonded to carbons of the double bond. Such a "bridged" ion is called a because the bromine formally carries the positive charge: An \(S_\text{N}2\)-type of attack of bromide ion, or other nucleophile, at carbon on the side to the bridging group then results in formation of the antarafacial-addition product: We may seem to have contradicted ourselves because Equation 10-1 shows a carbocation to be formed in bromine addition, but Equation 10-5 suggests a bromonium ion. Actually, the formulation of intermediates in alkene addition reactions as "open" ions or as cyclic ions is a controversial matter, even after many years of study. Unfortunately, it is not possible to determine the structure of the intermediate ions by any direct physical method because, under the conditions of the reaction, the ions are so reactive that they form products more rapidly than they can be observed. However, it is possible to generate stable bromonium ions, as well as the corresponding chloronium and iodonium ions. The technique is to use low temperatures in the absence of any strong nucleophiles and to start with a 1,2-dihaloalkane and antimony pentafluoride in liquid sulfur dioxide: The \(\ce{C_2H_4Br}^\oplus\) ions produced in this way are relatively stable and have been shown by nmr to have the cyclic halonium ion structure. There is a further aspect of polar additions to alkenes that we should consider, namely, that electrophilic reagents form loose complexes with the \(\pi\) electrons of the double bonds of alkenes to reaction by addition. Complexes of this type are called (or ). Formation of a complex between iodine and cyclohexene is demonstrated by the fact that iodine dissolves in cyclohexene to give a solution, whereas its solutions in cyclohexane are . The brown solution of iodine in cyclohexene slowly fades as addition occurs to give colorless -1,2-diiodocyclohexane. Precise Lewis structures cannot be written for charge-transfer complexes, but they commonly are represented as with the arrow denoting that electrons of the double bond are associated with the electrophile. These complexes probably represent the first stage in the formation of addition products by a sequence such as the following for bromine addition: We have seen that electrophiles can react with alkenes to form carbon-halogen bonds by donating positive halogen, \(\ce{Br}^\oplus\), \(\ce{Cl}^\oplus\), or \(\ce{I}^\oplus\). Likewise, carbon-hydrogen bonds can be formed by appropriately strong proton donors, which, of course, are typically strong proton acids. These acids are more effective in the absence of large amounts of water because water can compete with the alkene as a proton acceptor (also see ). Hydrogen chloride addition to ethene occurs by way of a proton-transfer step to give the ethyl cation and a chloride ion (Equation 10-6) followed by a step in which the nucleophilic chloride ion combines with the ethyl cation (Equation 10-7): All of the hydrogen halides \(\ce{HF}\), \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\)) will add to alkenes. Addition of hydrogen fluoride, while facile, is easily reversible. However, a solution of \(70\%\) anhydrous hydrogen fluoride and \(30\%\) of the weak organic base, pyridine, which is about 1/10,000 times as strong as ammonia, works better, and with cyclohexene gives fluorocyclohexane. With hydrogen iodide, care must be taken to prevent \(\ce{I_2}\) addition products resulting from iodine formed by oxidation reactions such as \[4 \ce{HI} + \ce{O_2} \rightarrow 2 \ce{I_2} + 2 \ce{H_2O}\] With hydrogen bromide, radical-chain addition may intervene unless the reaction conditions are controlled carefully (this will be discussed in ). The stereochemistry of addition depends largely on the structure of the alkene, but for simple alkenes and cycloalkenes, addition occurs predominantly in an antarafacial manner. For example, hydrogen bromide reacts with 1,2-dimethylcyclohexene to give the antarafacial addition product: We mentioned previously that the hydration of alkenes required a strong acid as a catalyst, because water itself is too weak an acid to initiate the proton-transfer step. However, if a small amount of a strong acid such as sulfuric acid is present, hydronium ions, \(\ce{H_3O}^\oplus\), are formed in sufficient amount to protonate reasonably reactive alkenes, although by no means as effectively as does concentrated sulfuric acid. The carbocation formed then is attacked rapidly by a nucleophilic water molecule to give the alcohol as its conjugate acid,\(^2\) which regenerates hydronium ion by transferring a proton to water. The reaction sequence follows for 2-methylpropene: In this sequence, the acid acts as a because the hydronium ion used in the proton addition step is regenerated in the final step. Sulfuric acid (or phosphoric acid) is preferred as an acid catalyst for addition of water to alkenes because the conjugate base, \(\ce{HSO_4-}\) (or \(\ce{H_2PO_4-}\)), is a poor nucleophile and does not interfere in the reaction. However, if the water concentration is kept low by using concentrated acid, addition occurs to give sulfate (or phosphate) esters. The esters formed with sulfuric acid are either alkyl acid sulfates \(\ce{R-OSO_3H}\) or dialkyl sulfates \(\ce{(RO)_2SO_2}\). In fact, this is one of the major routes used in the commercial production of ethanol and 2-propanol. Ethen and sulfuric acid give ethyl hydrogen sulfate, which reacts readily with water in a second step to give ethanol: One of the more confusing features of organic chemistry is the multitude of conditions that are used to carry out a given kind of reaction, such as the electrophilic addition of proton acids to different alkenes. Strong acids, weak acids, water, no water - Why can't there be a standard procedure? The problem is that alkenes have very different tendencies to accept protons. In the vapor phase, \(\Delta H^0\) for addition of a proton to ethene is about \(35 \: \text{kcal}\) more positive than for 2-methylpropene, and although the difference should be smaller in solution, it still would be large. Therefore we can anticipate (and we find) that a much more powerful proton donor is needed to initiate addition of an acid to ethene than to 2-methylpropene. But why not use in all cases a strong enough acid to protonate alkene one might want to have a proton acid add to? Two reasons: First, strong acids can induce undesirable side reactions, so that one usually will try not to use a stronger acid than necessary; second, very strong acid may even prevent the desired reaction from occurring! In elementary chemistry, we usually deal with acids in more or less dilute aqueous solution and we think of sulfuric, hydrochloric, and nitric acids as being similarly strong because each is essentially completely disassociated in dilute water solution: \[\ce{HCl} + \ce{H_2O} \overset{\longrightarrow}{\leftarrow} \ce{H_3O}^\oplus + \ce{Cl}^\ominus\] This does not mean they actually are equally strong acids. It means only that each of the acids is sufficiently strong to donate all of its protons to water. We can say that water has a "leveling effect" on acid strengths because as long as an acid can donate its protons to water, the solution has but one acid "strength" that is determined by the \(\ce{H_3O}^\oplus\) concentration, because \(\ce{H_3O}^\oplus\) is where the protons are. Now, if we use poorer proton acceptors as solvent we find the proton-donating powers of various "strong" acids begin to spread out immensely. Furthermore, new things begin to happen. For example, ethene is not hydrated appreciably by dilute aqueous acid; it just is too hard to transfer a proton from hydronium ion to ethene. So we use concentrated sulfuric acid, which is strong enough to add a proton to ethene. But now we don't get hydration, because any water that is present in concentrated sulfuric acid is virtually all converted to \(\ce{H_3O}^\oplus\), which is non-nucleophilic! \[\ce{H_2SO_4} + \ce{H_2O} \rightarrow \ce{H_3O}^\oplus + \ce{HSO_4-}\] However, formation of \(\ce{H_3O}^\oplus\) leads to formation of \(\ce{HSO_4-}\), which has enough nucleophilic character to react with the \(\ce{CH_3CH_2+}\) to give ethyl hydrogen sulfate and this is formed instead of the conjugate acid of ethanol ( ). The epitome of the use of stronger acid and weaker nucleophile is with liquid \(\ce{SO_2}\) (bp \(\sim 10^\text{o}\)) as the solvent and \(\ce{HBF_6}\) as the acid. This solvent is a very poor proton acceptor (which means that its conjugate acid is a very good proton donor) and \(\ce{SbF_6-}\) is an extremely poor nucleophile. If we add ethene to such a solution, a stable solution of \(\ce{CH_3CH_2+} \ce{SbF_6-}\) is formed. The reason is that there is no better proton acceptor present than \(\ce{CH_2=CH_2}\) and no nucleophile good enough to combine with the cation. The conversion of fumaric acid to malic acid is an important biological hydration reaction. It is one of a cycle of reactions (Krebs citric acid cycle) involved in the metabolic combustion of fuels (amino acids and carbohydrates) to \(\ce{CO_2}\) and \(\ce{H_2O}\) in a living cell. \(^1\)An alternative to Equation 10-1 would be to have \(\ce{Br_2}\) ionize to \(\ce{Br}^\oplus\) and \(\ce{Br}^\ominus\), with a subsequent attack of \(\ce{Br}^\oplus\) on the double bond to produce the carbocation. The fact is that energy required for such an ionization of \(\ce{Br_2}\) is prohibitively large even in water solution \(\left( \Delta H^0 \geq 80 \: \text{kcal} \right)\). One might well wonder why Equation 10-1 could possibly be more favorable. The calculated \(\Delta H^0\) for \(\ce{CH_2=CH_2} + \ce{Br_2} \rightarrow \cdot \ce{CH_2-CH_2Br} + \ce{Br} \cdot\) is \(+41 \: \text{kcal}\), which is only slightly more favorable than the \(\Delta H^0\) for \(\ce{Br_2} \rightarrow 2 \ce{Br} \cdot\) of \(46.4 \: \text{kcal}\). However, available thermochemical data suggest that the ease of transferring an electron from \(\cdot \ce{CH_2CH_2Br}\) to \(\ce{Br} \cdot\) to give \(^\oplus \ce{CH_2CH_2Br} + \ce{Br}^\ominus\) is about \(80 \: \text{kcal}\) more favorable than \(2 \ce{Br} \cdot \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus\). Thus the overall \(\Delta H^0\) of Equation 10-1 is likely to be about \(85 \: \text{kcal}\) more favorable than \(\ce{Br_2} \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus\). \(^2\)The terms and are very convenient to designate substances that are difficult to name simply as acids, bases, or salts. The conjugate acid of a compound \(\ce{X}\) is \(\ce{XH}^\oplus\) and the conjugate base of \(\ce{HY}\) is \(\ce{Y}^\ominus\). Thus \(\ce{H_3O}^\oplus\) is the conjugate acid of water, while \(\ce{OH}^\ominus\) is its conjugate base. Water itself is then both the conjugate base of \(\ce{H_3O}^\oplus\) and the conjugate acid of \(\ce{OH}^\ominus\) and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.03%3A_Conformations_of_Cycloalkanes

The equilibria (relative stabilities) and equilibration (rate of interconversion) of the rotational conformations of ethane and butane were discussed in Section 5-2. If you review this material, it will be clear that forming a ring from a hydrocarbon chain will greatly reduce the number of possible staggered and eclipsed conformations. We will begin our discussion with cyclohexane because of its special importance, proceed to smaller rings, then give a brief exposition of the conformations of the larger rings. If the carbons of a cyclohexane ring were placed at the corners of a regular hexagon, all the \(\ce{C-C-C}\) bond angles would have to be \(120^\text{o}\). Because the expected normal \(\ce{C-C-C}\) bond angle should be near the tetrahedral value of \(109.5^\text{o}\), the suggested planar configuration of cyclohexane would have at each of the carbons, and would correspond to less stable cyclohexane molecules than those with more normal bond angles. The actual normal value for the \(\ce{C-C-C}\) bond angle of an open-chain \(\ce{-CH_2-CH_2-CH_2}-\) unit appears to be about \(112.5^\text{o}\), which is \(3^\text{o}\) greater than the tetrahedral value. From this we can conclude that the angle strain at each carbon of a planar cyclohexane would be \(\left( 120^\text{o} - 112.5^\text{o} \right) = 7.5^\text{o}\). Angle strain is not the whole story with regard to the instability of the planar form, because in addition to having \(\ce{C-C-C}\) bond angles different from their normal values, the planar structure also has its carbons and hydrogens in the unfavorable arrangement, as shown in Figure 12-2. If the carbon valence angles are kept near the tetrahedral value, you will find that you can construct ball-and-stick models of the cyclohexane six-carbon ring with two quite different conformations. These are known as the "chair" and "boat" conformations (Figure 12-3). It has not been possible to separate cyclohexane at room temperature into pure isomeric forms that correspond to these conformations, and actually the two forms appear to be rapidly interconverted. The chair conformation is considerably more stable and comprises more than \(99.9\%\) of the equilibrium mixture at room temperature.\(^1\) Why is the boar form less stable than the chair form, if both have normal \(\ce{C-C-C}\) bond angles? The answer is that the boat form has unfavorable nonbonded interactions between the hydrogen atoms around the ring. If we make all of the bond angles normal and orient the carbons to give the "extreme boat" conformation of Figure 12-4, a pair of 1,4 hydrogens (the so-called "flagpole" hydrogens) have to be very close together \(\left( 1.83 \: \text{Å} \right)\). Hydrogens this close together would be on the rising part of a repulsion potential energy curve, such as Figure 4-6, for hydrogen-hydrogen nonbonded interactions. This at an \(\ce{H-H}\) distance of \(1.83 \: \text{Å}\) corresponds to a repulsion energy of about \(3 \: \text{kcal mol}^{-1}\). There is still another factor that makes the extreme boat unfavorable; namely, that the eight hydrogens along the "sides" of the boat are eclipsed, which brings them substantially closer together than they would be in a staggered arrangement (about \(2.27 \: \text{Å}\) compared with \(2.50 \: \text{Å}\)). This is in striking contrast with the chair form (Figure 12-5), for which adjacent hydrogens are in staggered positions with respect to one another all around the ring. Therefore the chair form is expected to be more stable than the boat form because it has less repulsion between the hydrogens. You should make and inspect models such as those in Figure 12-3 to see the rather striking difference between the chair and boat conformations that is not obvious from the diagrams. You will find that the chair structure is quite rigid, and rotation does occur around the \(\ce{C-C}\) bonds with interconversion to the boat structure. In contrast, the boat form is quite flexible. Rotation about the \(\ce{C-C}\) bonds permits the ring to twist one way or the other from the extreme boat conformation to considerably more stable, equal-energy conformations, in which the flagpole hydrogens move farther apart and the eight hydrogens along the sides become largely but not completely staggered. These arrangements are called the (sometimes ) conformations (see Figure 12-6) and are believed to be about \(5 \: \text{kcal mol}^{-1}\) less stable than the chair form. It is possible to measure the spectral properties of the twist-boat form by a very elegant technique employed by F. A. L. Anet. Because the equilibrium constant for conversion of chair to boat increases with temperature, a considerable proportion of the molecules exist as the twist-boat form in the vapor at \(800^\text{o}\). If such vapor is allowed to impinge on a surface cooled to \(20 \: \text{K}\), the film condensate contains about \(25\%\) of the twist-boat form. At this low temperature, the twist-boat form is converted to the more stable chair form at a very slow rate. Infrared spectra can be taken of the boat-chair mixture at \(10 \: \text{K}\). If the mixture is allowed to warm to \(75 \: \text{K}\), the normal equilibrium favoring the chair form is established in a short time. The spatial arrangement (stereochemistry) of cyclohexane and other organic compounds are studied conveniently with the aid of , which are made with standard bond angles and scaled bond distances. The bonds have stainless-steel rods that make a snap-fit into stainless-steel sleeves. Rotation is smooth about the bonds and there is sufficient flexibility to accommodate some angle strain. Dreding models of the conformations of cyclohexane are shown in Figure 12-7. Notice that these models correspond closely to the sawhorse representations in Figures 12-4, 12-5, and 12-6. Figure 12-5 shows that there are two distinct kinds of hydrogen in the chair form of cyclohexane - six that are close to the "average" plane of the ring (called hydrogens) and three above and three below this average plane (called hydrogens). This raises interesting questions in connection with substituted cyclohexanes: For example, is the methyl group in methylcyclohexane equatorial or axial? Since only methylcyclohexane is known, the methyl group must be exclusively equatorial \(\left( e \right)\), exclusively axial \(\left( a \right)\), or the two forms must be interconverted so rapidly that they cannot be separated into isomeric forms. It appears that the latter circumstance prevails, with the ring changing rapidly from one chair form to another by flipping one end of the chair up and the other end down: Such a change would cause a substituent in an axial position to go to an equatorial position and . This process is called and its rate often is called the . With cyclohexane, inversion is so fast at room temperature that, on average, the molecules, flip about 100,000 times per second, over an energy barrier of about \(11 \: \text{kcal mol}^{-1}\). You will understand this flipping process if you make a model of a cyclohexane ring carrying a single substituent. By manipulating the model you can discover some of the different ways the process can occur. The simplest route is simply to flip up one corner of the ring to convert the chair into a boat and then flip down the opposite carbon: Because of the flexibility of the boat conformation, it is possible to transform it to other boat conformations whereby carbons other than the one indicated flip down and complete the interconversion. At room temperature the conformation of methylcyclohexane with the methyl equatorial is more stable than the one with the methyl axial by \(1.7 \: \text{kcal mol}^{-1}\). The same is true of all monosubstituted cyclohexanes to a greater or lesser degree. Reasons for this can be seen from space-filling models (Figure 12-8), which show that a substituent group has more room when the substituent is equatorial than when it is axial. In the axial position the substituent is considerably closer to the two axial hydrogens on the same side of the ring than to other hydrognes, even hydrogens on adjacent carbons when the substituent is in the equatorial position (Figure 12-8). For example, when the substituent is bromine, which has a \(\ce{C-Br}\) bond length of \(1.94 \: \text{Å}\), the distance from axial bromine to the axial hydrogen at \(\ce{C_3}\) or \(\ce{C_5}\) on the same side of the ring is about \(2.7 \: \text{Å}\). In contrast, the distance from equatorial bromine to any of the hydrogens on the adjacent carbons is about \(3.1 \: \text{Å}\). There is a very important general aspect of the difference between these two nonbonded \(\ce{H} \cdot \cdot \cdot \ce{Br}\) interactions at \(2.7 \: \text{Å}\) and \(3.1 \: \text{Å}\). Whenever two nonbonded atoms are brought close together, and before the massive repulsion sets in (which is so evident in Figure 4-6), there is a slight in the energy curve corresponding to .\(^2\) For nonbonded \(\ce{H} \cdot \cdot \cdot \ce{Br}\) interactions the bottom of the dip comes at about \(3.1 \: \text{Å}\) (Figure 12-9), and the resulting attraction between the atoms will provide some stabilization of the equatorial conformation relative to the axial conformation. Weak attractive forces between nonbonded atoms are called or , and are of great importance in determining the properties of liquids. They also can be expected to play a role in determining conformation equilibria whenever the distances between the atoms in the conformations correspond to the so-called . Table 12-2 shows the contribution made by various substituents to the free-energy change from the axial to the equatorial orientations of the substituent. Thus, for bromine, the free-energy change, \(\Delta G^0\), is \(-0.5 \: \text{kcal mol}^{-1}\), which means that at \(25^\text{o}\), the equilibrium constant, \(K\), for the axial \(\rightleftharpoons\) equatorial equilibrium is about 2.3 (from \(-2.303 RT \: \text{log} \: K = \Delta G^0\); see Section 4-4A). From many studies it is known that the interconversion of conformations with the substituent in the equatorial and the axial positions occurs about 100,000 times per second, which corresponds to a transition-state energy (activation energy) of about \(11 \: \text{kcal mol}^{-1}\) above the ground-state energy. The rate decreases as the temperature is lowered. If one cools chlorocyclohexane to its melting point \(\left( -44^\text{o} \right)\), the substance crystallizes to give the pure equatorial isomer. The crystals then can be cooled to \(-150^\text{o}\) and dissolved at this temperature in a suitable solvent. At \(-150^\text{o}\) it would take about 130 days for half of the equatorial form to be converted to the axial form. However, when the solution is warmed to \(-60^\text{o}\) the equatorial conformation is converted to the equilibrium mixture in a few tenths of a second. The cis-trans isomerism of cyclohexane derivatives (Section 5-1A) is complicated by conformational isomerism. For example, 4- -butylcyclohexyl chloride theoretically could exist in four stereoisomeric chair forms, \(1\), \(2\), \(3\), and \(4\). trans cis Use of \(\ce{^{13}C}\) nmr spectroscopy to determine whether a substituent is in an axial or equatorial position is well illustrated with - and -4- -butylcyclohexanols, \(5\) and \(6\): The five \(\ce{-CH_2}-\) groups of cyclopentane theoretically could form a regular planar pentagon (internal angles of \(108^\text{o}\)) with only a little bending of the normal \(\ce{C-C-C}\) bond angles. Actually, cyclopentane molecules are flat. The planar structure has completely eclipsed hydrogens, which makes it less stable by about \(10 \: \text{kcal mol}^{-1}\) than if there were no eclipsed hydrogens. The result is that each molecule assumes a puckered conformation that is the best compromise between distortion of bond angles and eclipsing of hydrogens. The best compromise conformations have the ring twisted with one or two of the \(\ce{-CH_2}-\) groups bent substantially out of a plane passed through the other carbons (Figure 12-14). The flexibility of the ring is such that these deformations move rapidly around the ring. Formation of a four-membered ring of carbon atoms can be achieved only with substantial distortion of the normal valence angles of carbon, regardless of whether the ring is planar or nonplanar. In cyclobutane, for example, if the valence bonds are assumed to lie along straight lines drawn between the carbon nuclei, each \(\ce{C-C-C}\) bond angle will be \(19.5^\text{o}\) smaller than the \(109.5^\text{o}\) tetrahedral value: The three carbon atoms of the cyclopropane ring lie in a plane. Therefore the angle strain is expected to be considerable because each \(\ce{C-C-C}\) valence angle must be deformed \(49.5^\text{o}\) from the tetrahedral value. It is likely that some relief from the strain associated with the eclipsing of the hydrogens of cyclopropane is achieved by distortion of the \(\ce{H-C-H}\) and \(\ce{H-C-C}\) bond angles: If one is willing to consider a carbon-carbon double bond as a two-membered ring, then ethene, \(\ce{C_2H_4}\), is the simplest possible cycloalkane ("cycloethane"). As such, \(\ce{C_2H_4}\) has \(\ce{C-C-C}\) valence angles of \(0^\text{o}\) and therefore an angle strain of \(109.5^\text{o}\) at each \(\ce{CH_2}\) group compared to the tetrahedral value: and (1977) \(^1\)Pioneering work on the conformations of cyclohexane and its derivatives was carried out by O. Hassel (Norway) and D. H. R. Barton (United Kingdom) for which they shared a Nobel Prize in 1969. \(^2\)The vertical scale of Figure 4-6 does not permit seeing the dip in the curve resulting from attractive forces between neon atoms. It is deepest when \(r\) is about \(3.12 \: Å\) and amounts to \(0.070 \: \text{kcal mol}^{-1}\). \(^3\)After F. London, who developed a quantum-mechanical theory of the origin of these forces and also pioneered many quantum calculations of great consequence to chemistry, including bonding in \(\ce{H_2}\), which will be discussed in Section 21-1.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.2%3A_Entropy_and_Spontaneity_-_A_Molecular_Statistical_Interpretation

We know that the first law of thermodynamics governs changes in the state function we called (\(U\)) and that changes in the internal energy (\(ΔU\)) are closely related to changes in the enthalpy (\(ΔH\)), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of . This information, however, does not tell us whether a particular process or reaction will occur spontaneously. Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved. Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously. For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH NO , NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., \(ΔH_{soln} > 0\)). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure \(\Page {1}\). Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (\(S\)), a thermodynamic property of all substances that is proportional to their degree of disorder. thus the cannot predict the direction of a natural process Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (\(ΔS > 0\)) or a decrease in entropy (\(ΔS < 0\)), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: \[ΔS = S_f − S_i\] When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy. How can we express disorder quantitatively? From the example of coins, you can probably see that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number of all possible arrangements of ten coins. Using the language of molecular statistics, we say that a collection of coins in which a given fraction of its members are heads-up constitutes a of the system. Since we do not care which coins are heads-up, there are clearly numerous configurations of the individual coins which can result in this “macrostate”. Each of these configurations specifies a of the system. The greater the number of microstates that correspond to a given macrostate (or configuration), the greater the probability of that macrostate. To see what this means, consider the possible outcomes of a toss of four coins (Table ): A toss of four coins will yield one of the five outcomes (macrostates) listed in the leftmost column of the table. The second column gives the number of “ways”— that is, the number of head/tail configurations of the set of coins (the number of microstates)— that can result in the macrostate. The probability of a toss resulting in a particular macrostate is proportional to the number of microstates corresponding to the macrostate, and is equal to this number, divided by the total number of possible microstates (in this example, 2 =16). An important assumption here is that all microstates are equally probable; that is, the toss is a “fair” one in which the many factors that determine the trajectory of each coin operate in an entirely random way. The greater the number of that correspond to a given , the greater the probability of that . Following the work of Carnot and Clausius, Ludwig developed a molecular-scale statistical model that related the entropy of a system to the number of possible for the system. A (\(\Omega\)) is a specific configuration of the locations and energies of the atoms or molecules that comprise a system like the following: \[S=k \ln \Omega \label{Eq2}\] Here is the Boltzmann constant and has a value of 1.38 × 10 J/K. As for other state functions, the change in entropy for a process is the difference between its final (\(S_f\)) and initial (\(S_i\)) values: \[\begin{align} ΔS &= S_\ce{f}−S_\ce{i} \\[4pt] &=k \ln \Omega_\ce{f} − k \ln \Omega_\ce{i} \\[4pt] &= k \ln\dfrac{\Omega_\ce{f}}{\Omega_\ce{i}} \label{Eq2a} \end{align}\] For processes involving an increase in the number of microstates, \(\Omega_f > \Omega_i\), the entropy of the system increases, \(ΔS > 0\). Conversely, processes that reduce the number of microstates, \(\Omega_f < \Omega_i\), yield a decrease in system entropy, \(ΔS < 0\). This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs. Consider the general case of a system comprised of particles distributed among boxes. The number of microstates possible for such a system is . For example, distributing four particles among two boxes will result in 2 = 16 different microstates as illustrated in Figure \(\Page {2}\). Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called . The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates (Equation \ref{Eq2}), . For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is \[\dfrac{6}{16} = \dfrac{3}{8} \nonumber\] The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of \[\dfrac{1}{16} \nonumber\] The probability of finding all particles in only one box (either the left box or right box) is then \[\left(\dfrac{1}{16}+\dfrac{1}{16}\right)=\dfrac{2}{16} = \dfrac{1}{8} \nonumber\] As you add more particles to the system, the number of possible microstates increases exponentially (2 ). A macroscopic (laboratory-sized) system would typically consist of moles of particles ( ~ 10 ), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations. is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. The spontaneous process whereby the gas contained initially in one flask expands to fill both flasks equally therefore yields an increase in entropy for the system. A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of energy (represented as “*”) in Figure \(\Page {3}\). The hot object is comprised of particles and and initially contains both energy units. The cold object is comprised of particles and , which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. And so, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is \(\frac{3}{10}\). More likely is the flow of heat to yield one of the other two distribution, the combined probability being \(\frac{7}{10}\). The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being \(\frac{4}{10}\). As for the previous example of matter dispersal, extrapolating this treatment to macroscopic collections of particles dramatically increases the probability of the uniform distribution relative to the other distributions. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects’ temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy. Consider the system shown here. What is the change in entropy for a process that converts the system from distribution (a) to (c)? We are interested in the following change: The initial number of microstates is one, the final six: \[\begin{align*} ΔS &=k \ln\dfrac{\Omega_\ce{c}}{\Omega_\ce{a}}\nonumber \\[4pt] &= 1.38×10^{−23}\:J/K × \ln\dfrac{6}{1} \\[4pt] &= 2.47×10^{−23}\:J/K \end{align*} \] The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive. Consider the system shown in Figure \(\Page {3}\). What is the change in entropy for the process where the energy is transferred from the hot object ( ) to the cold object ( )? 0 J/K A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔS > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔS > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0. Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy. Thus coins and cards tend to assume random configurations when tossed or shuffled, and socks and books tend to become more scattered about a teenager’s room during the course of daily living. But there are some important differences between these large-scale mechanical, or macro systems, and the collections of sub-microscopic particles that constitute the stuff of chemistry, and which we will refer to here generically as . Molecules, unlike macro objects, are capable of accepting, storing, and giving up energy in tiny amounts (quanta), and act as highly efficient carriers and spreaders of thermal energy as they move around. Thus, , The importance of these last two points is far greater than you might at first think, but to fully appreciate this, you must recall the various ways in which thermal energy is stored in molecules. The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure \(\Page {4}\). In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system’s component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, \(S_{liquid} > S_{solid}\) and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, \(ΔS > 0\). By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, \(ΔS < 0\). Now consider the vapor or gas phase. The atoms or molecules occupy a greater volume than in the liquid phase; therefore each atom or molecule can be found in many more locations than in the liquid (or solid) phase. Consequently, for any substance, \[S_{gas} \gg S_{liquid} > S_{solid}\] and the processes of vaporization and sublimation likewise involve increases in entropy, \(ΔS > 0\). Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, \(ΔS < 0\). According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure \(\Page {5}\) ). The entropy of a substance is influenced by structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle’s mass and the spacing of quantized translational energy levels (which is a topic beyond the scope of our treatment). For molecules, greater numbers of atoms (regardless of their masses) increase the ways in which the molecules can vibrate and thus the number of possible microstates and the system entropy. Finally, variations in the types of particles affect the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, \(ΔS > 0\). Considering the various factors that affect entropy allows us to make informed predictions of the sign of \(ΔS\) for various chemical and physical processes as illustrated in Example \(\Page {2}\). Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions. Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction. Positive; The solid dissolves to give an increase of mobile ions in solution. Negative; The liquid becomes a more ordered solid. Positive; The relatively ordered solid becomes a gas. Positive; There is a net production of one mole of gas. Entropy (\(S\)) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system. For a given substance, \(S_{solid} < S_{liquid} \ll S_{gas}\) in a given physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions may be reliably predicted. )   ).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.5%3A_Crystal_Field_Theory%3A_Optical_and_Magnetic_Properties

One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes. CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion ( ) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands. We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure \(\Page {1a}\)). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure \(\Page {1b}\), the d and d orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the e orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (d , d , and d , collectively called the t orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge. The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δ ), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δ .) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two e orbitals increase in energy by 0.6Δ , whereas the three t orbitals decrease in energy by 0.4Δ . Thus the total change in energy is \[2(0.6Δ_o) + 3(−0.4Δ_o) = 0. \nonumber \] Crystal field splitting does not change the total energy of the d orbitals. Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure \(\Page {1a}\)). We can use the d-orbital energy-level diagram in Figure \(\Page {1}\) to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure 24.6.2, for d –d systems—such as [Ti(H O) ] , [V(H O) ] , and [Cr(H O) ] , respectively—the electrons successively occupy the three degenerate t orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H O) ] , for example, by saying that the chromium ion has a d electron configuration or, more succinctly, Cr is a d ion. When we reach the d configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty e orbitals or one of the singly occupied t orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the . If Δ is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty e orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H O) ] ion, is called a high-spin complex. Conversely, if Δ is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN) ] ion, is called a low-spin complex. Similarly, metal ions with the d , d , or d electron configurations can be either high spin or low spin, depending on the magnitude of Δ . In contrast, only one arrangement of d electrons is possible for metal ions with d –d electron configurations. For example, the [Ni(H O) ] ion is d with two unpaired electrons, the [Cu(H O) ] ion is d with one unpaired electron, and the [Zn(H O) ] ion is d with no unpaired electrons. If Δ is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δ is greater, a low-spin configuration forms. The magnitude of Δ dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δ (i.e., Δ > P) yield a low-spin complex, whereas small values of Δ (i.e., Δ < P) produce a high-spin complex. As we noted, the magnitude of Δ depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δ for some representative transition-metal complexes are given in Table \(\Page {1}\). Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994). Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δ increases as the charge on the metal ion increases. Typically, Δ for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H O) ] , Δ = 11,800 cm ; for [V(H O) ] , Δ = 17,850 cm . For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δ increases with increasing principal quantum number: Δ (3d) < Δ (4d) < Δ (5d). The data for hexaammine complexes of the trivalent group 9 metals illustrate this point: [Rh(NH ) ] : Δ = 34,100 cm The increase in Δ with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions. Experimentally, it is found that the Δ observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δ decreases as the size of the donor atom increases. For example, Δ values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F , interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH , results in significantly larger Δ values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F . The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δ : The values of Δ listed in Table \(\Page {1}\) illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand. The largest Δ splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons. The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition (Figure 24.6.3). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δ . Recall that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δ , which depends on the structure of the complex. For example, the complex [Cr(NH ) ] has strong-field ligands and a relatively large Δ . Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H O) ] ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color. We can now understand why emeralds and rubies have such different colors, even though both contain Cr in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al O in rubies and Be Al Si O in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δ relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si O ] silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δ . Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion. Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d complex such as [Ti(H O) ] is located in one of the t orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δ . The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t orbitals by the energy of those orbitals (−0.4Δ ), multiplying the number of electrons in e orbitals by the energy of those orbitals (+0.6Δ ), and summing the two. Table \(\Page {2}\) gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d configuration. CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences. Octahedral d and d complexes and low-spin d , d , d , and d complexes exhibit large CFSEs. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. : complexes structure, high spin versus low spin, and the number of unpaired electrons B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d electron configuration. C Because of the weak-field ligands, we expect a relatively small Δ , making the compound high spin. D In a high-spin octahedral d complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t orbitals, giving four unpaired electrons. B C Because rhodium is a second-row transition metal ion with a d electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δ , making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order d d , and d (these are degenerate); d ; and d . D The eight electrons occupy the first four of these orbitals, leaving the d . orbital empty. Thus there are no unpaired electrons. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δ ), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δ is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δ is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δ , whereas weak-field ligands interact more weakly and give a smaller Δ . The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/16%3A_Carbonyl_Compounds_I-_Aldehydes_and_Ketones._Addition_Reactions_of_the_Carbonyl_Group/16.05%3A_Typical_Carbonyl-Addition_Reactions

We turn now to discuss a few specific addition reactions of the carbonyl groups of aldehydes and ketones. We shall not attempt to provide an extensive catalog of reactions, but will try to emphasize the principles involved with especially important reactions that are useful in synthesis. Grignard reagents, organolithium compounds, and the like generally add to aldehydes and ketones rapidly and irreversibly, but the same is not true of many other reagents; their addition reactions may require acidic or basic catalysts; the adducts may be formed reversibly and with relatively unfavorable equilibrium constants. Also, the initial adducts may be unstable and react further by elimination. (We recommend that you review Section 15-4E to see examples of these points.) To organize this very large number of addition reactions, we have arranged the reactions according to the nucleophile that adds to the carbonyl carbon. The types of nucleophiles considered here form \(\ce{C-C}\), \(\ce{C-O}\), \(\ce{C-N}\), \(\ce{C}\)-halogen, \(\ce{C-S}\), and \(\ce{C-H}\) bonds. A summary is given in Table 16-4. Hydrogen cyanide adds to many aldehydes and ketones to give hydroxylnitriles, usually called "cyanohydrins": The products are useful in synthesis - for example, in the preparation of cyanoalkenes and hydroxy acids: An important feature of cyanohydrin formation is that it requires a basic catalyst. In the absence of base, the reaction does not proceed, or is at best very slow. In principle, the basic catalyst may activate either the carbonyl group or hydrogen cyanide. With hydroxide ion as the base, one reaction to be expected is a reversible addition of hydroxide to the carbonyl group: However, such addition is not likely to facilitate formation of cyanohydrin because it represents a saturation of the carbonyl double bond. Indeed, if the equilibrium constant for this addition were large, an excess of hydroxide ion could inhibit cyanohydrin formation by tying up the ketone as the adduct \(1\). Hydrogen cyanide itself has no unshared electron pair on carbon and does not form a bond to a carbonyl carbon. However, a small amount of a strong base can activate hydrogen cyanide by converting it to cyanide ion, which can function as a carbon nucleophile. A complete sequence for cyanohydrin formation follows: The second step regenerates the cyanide ion. Each step of the reaction is reversible but, with aldehydes and most nonhindered ketones, formation of the cyanohydrin is reasonably favorable. In practical syntheses of cyanohydrins, it is convenient to add a strong acid to a mixture of sodium cyanide and the carbonyl compound, so that hydrogen cyanide is generated . The amount of acid added should be insufficient to consume all the cyanide ion, therefore sufficiently alkaline conditions are maintained for rapid addition. There are a number of rather interesting substances for which we can write important dipolar valence-bond structures of the type . The important factor with these structures is that the negative end of the dipole is . The positive end of the dipole can be several kinds of atoms or groups, the most usual being sulfur, phosphorus, or nitrogen. Some examples (each written here as a single dipolar valence-bond structure) are: The systematic naming of these substances is cumbersome, but they have come to be known as . The genesis of this name may seem obscure, but it is an attempt to reconcile the presence of a \(\ce{C-X}\) \(\sigma\) bond, which is covalent and nonpolar as in alk derivatives, as well as an ionic bond as in metal hal . Hence, the combination .\(^2\) As we might expect from the dipolar structure, ylides can behave as carbon nucleophiles to form carbon-carbon bonds by addition to the carbonyl groups of aldehydes and ketones: However, the further course of reaction depends on the type of ylide used. In the case of phosphorus ylides, the overall reaction amounts to a very useful by the transfer of oxygen to phosphorus and carbon to carbon, as summarized in Equation 16-1. This is called the : Reactions with sulfur ylides proceed differently. The prodcuts are oxacylcopropanes (oxiranes) - not alkenes. The addition step proceeds as with the phosphorus ylides, but the negatively charged oxygen of the dipolar adduct then displaces the sulfonium group as a neutral sulfide. This is an intramolecular \(S_\text{N}2\) reaction similar to the formation of oxacyclopropanes from vicinal chloroalcohols (Section 15-11C): As for the nitrogen ylides, a useful reagent of this type is diazomethane, \(\ce{CH_2N_2}\). Diaxomethane can react with carbonyl compounds in different ways, depending on what happens to the initial adduct \(2\). Oxacyclopropanes are formed if the nitrogen is simply displaced (as \(\ce{N_2}\)) by oxygen (Path \(a\), Equation 16-2). Ketones of rearranged carbon framework result if nitrogen is displaced \(as \(\ce{N_2}\)) by \(\ce{R}^\ominus\) which moves over to the \(\ce{CH_2}\) group (Path \(b\), Equation 16-2): Diazoketones, \(\ce{RCOCHN_2}\), are formed if there is a good leaving group, such as halogen, on the carbonyl (Equation 16-3). Under these circumstances the reactant is an acid halide, not an aldehyde or ketone: We already have discussed additions of alcohols and, by analogy, thiols \(\left( \ce{RSH} \right)\) to carbonyl compounds (see Section 15-4E). We will not repeat this discussion here except to point out that addition of to the carbonyl group of an aldehyde is analogous to hemiacetal formation (Section 15-4E) and is catalyzed both by acids and bases: The equilibrium for hydrate formation depends both on steric and electrical factors. Methanal is \(99.99\%\) hydrated in aqueous solution, ethanal is \(58\%\) hydrated, and 2-propanone is not hydrated significantly. The hydrates seldom can be isolated because they readily revert to the parent aldehyde. The only stable crystalline hydrates known are those having strongly electronegative groups associated with the carbonyl (see Section 15-7). Several carbonyl additions have characteristics similar to those of cyanohydrin formation. A typical example is the addition of sodium hydrogen sulfite, which proceeds readily with good conversion in aqueous solution with most aldehydes, methyl ketones, and unhindered cyclic ketones to form a carbon-sulfur bond. No catalyst is required because sulfite is an efficient nucleophilic agent. The addition step evidently involves the sulfite ion - not hydrogen sulfite ion: The addition products often are nicely crystalline solids that are insoluble in excess concentrated sodium hydrogen sulfite solution. Whether soluble or insoluble, the addition products are useful for separating carbonyl compounds from substances that do not react with sodium hydrogen sulfite. A reaction closely related to acetal formation is the polymerization of aldehydes. Both linear and cyclic polymers are obtained. For example, methanal in water solution polymerizes to a solid ling-chain polymer called paraformaldehyde or "polyoxymethylene": This material, when strongly heated, reverts to methanal; it therefore is a convenient source of gaseous methanal. When heated with dilute acid, paraformaldehyde yields the solid trimer, 1,3,5-trioxycyclohexane (mp \(61^\text{o}\)). The cyclic tetramer is also known. Long-chain methanal polymers have become very important as plastics in recent years. The low cost of paraformaldehyde is highly favorable in this connection, but the instability of the material to elevated temperatures and dilute acids precludes its use in plastics. However, the "end-capping" of polyoxymethylene chains through formation of esters or acetals produces a remarkable increase in stability, and such modified polymers have excellent properties as plastics. Delrin (DuPont) is a stabilized methanal polymer with exceptional strengths and ease of molding. Ethanal (acetaldehyde) polymerizes under the influence of acids to the cyclic trimer, "paraldehyde", and a cyclic tetramer, "metaldehyde". Paraldehyde has been used as a relatively nontoxic sleep-producing drug (hypnotic). Metaldehyde is used as a poison for snails and slugs, "Snarol". Ketones do not appear to form stable polymers like those of aldehydes. A wide variety of substances with \(\ce{-NH_2}\) groups react with aldehydes and ketones by an addition-elimination sequence to give compounds and water. These reactions usually require acid catalysts: Table 16-5 summarizes several important reactions of this type and the nomenclature of the reactants and products. Clearly, if the unshared electron pair on the nitrogen of \(\ce{RNH_2}\) is combined with a proton, Equation 16-4, it cannot attack the carbonyl carbon to give the aminoalkanol as in Equation 16-5. So at high acid concentration (low pH) we expect the rate and the equilibrium for the overall reaction to be unfavorable. In more dilute acid, the rate picks up because there is more free \(\ce{RNH_2}\) in solution. Dehydration of the aminoalkanol (Equation 16-6) is acid catalyzed; this reaction normally is fast at pH values smaller than 3-4. Therefore, the slow step at pH \(<\) 4 is addition of \(\ce{RNH_2}\) to the carbonyl group as per Equation 16-5. As the pH is increased above 4, the addition becomes progressively faster because less \(\ce{RNH_2}\) is tied up as \(\ce{RNH_3^+}\). However, then the dehydration step, Equation 16-6, decreases in rate because it requires an acid catalyst. At pH 6 (remember that going from pH 4 to pH 6 is a 100-fold decrease in \(\ce{H}^\oplus\) concentration), dehydration is the slow step, and at higher pH values it finally becomes too slow to give a useful overall rate of reaction. This sequence of changes in rate and equilibria has been shown to account precisely for rate . pH curves such as in Figure 16-4. Dehydration of \(\ce{(CH_3)_2CHNR(OH)}\) to \(\ce{(CH_3)_2C=NR}\) involves acid catalysis in very much the same way as in acetal formation (Section 15-4E): Ammonia adds readily to many aldehydes. For example, The aldehyde-ammonia adducts usually are not very stable. They readily undergo dehydration and polymerization. 1-Aminoethanol, for example, gives a cyclic trimer composition \(\ce{C_6H_{15}N_3} \cdot 3 \ce{H_2O}\), mp \(97^\text{o}\), with structure \(4\): Methanal and ammonia react by a different course with the formation of a substance known as "hexamethylenetetramine": The product is a high-melting solid (mp \(> \: 230^\text{o}\) d.) and its structure has been established by x-ray diffraction (Section 9-3). In fact, it was the first organic substance whose structure was determined in this way. The high melting point is clearly associated with the considerable symmetry and rigidity of the cage structure: The corresponding all-carbon compound, adamantane (Section 12-8), also has a high melting point \(\left( 268^\text{o} \right)\): Treatment of hexamethylenetetramine with nitric acid gives the high explosive "cyclonite", which often is designated as RDX and was widely used in World War II: Then methanal and ammonia that split off the cage structure during the reaction with nitric acid need not be wasted. In the large-scale manufacture of cyclonite, a combination of nitric acid, ammonium nitrate, and ethanoic anhydride is used, which results in full utilization of the methanal and ammonia: \[\ce{C_6H_{12}N_4} + 4 \ce{HNO_3} + 2 \ce{NH_4NO_3} + 6 \ce{(CH_3CO)_2O} \rightarrow 2 \ce{C_3H_6O_6N_6} + 12 \ce{CH_3CO_2H}\] Secondary amino compounds of the type \(\ce{R_2N-H}\) add to aldehyde and ketone carbonyl groups in an acid-catalyzed reaction in much the same way as do \(\ce{RNH_2}\) compounds - with one important difference. The product contains the structural unit \(\ce{C=C-N}\) rather than \(\ce{C-C=N}\); and because there is a carbon-carbon double bond, such a substance is called an enamine (alk \(+\) ). An example is: The course of this reaction can be understood if we notice that loss of \(\ce{OH}\) from the initial product leads to an immonium ion, \(5\), that cannot lose a proton and form a \(\ce{C=N}\) bond: However, if there is a hydrogen on a carbon attached to the immonium carbon, it is possible for such a hydrogen to be lost as a proton with concurrent formation of the neutral enamine: Enamine formation, like many other carbonyl addition reactions, is readily reversible, and the carbonyl compound can be recovered by hydrolysis with aqueous acids. For this reason, to obtain a good conversion of carbonyl compound to enamine, it usually is necessary to remove the water that is formed by distilling it away from the reaction mixture. Enamines are useful synthetic intermediates for the formation of carbon-carbon bonds, as we will discuss in greater detail in Section 17-4B. Enamines generally are unstable if there is a hydrogen on nitrogen. They rearrange to the corresponding imine. This behavior is analogous to the rearrangement of alkenols to carbonyl compounds (Section 10-5A): Addition of hydrogen halides to carbonyl groups usually is so easily reversible as to preclude isolation of the addition products: However, many aldehydes react with alcohols in the presence of an excess of hydrogen chloride to give \(\alpha\)-chloro ethers: In carrying out laboratory syntheses of \(\alpha\)-chloro ethers, gaseous hydrogen chloride is passed into a mixture of the alcohol and aldehyde. Aqueous \(\ce{HCl}\) is not useful because the excess water gives an unfavorable equilibrium. \(\alpha\)-Chloro ethers are highly reactive compounds that very readily undergo \(S_\text{N}2\) as well as \(S_\text{N}1\) and \(E1\) reactions. Two simple examples, methoxychloromethane (chloromethyl methyl ether) and chloromethoxychloromethane (bis-chloromethyl ether), have been put under severe restrictions as the result of tests that show they have strong chemical carcinogenic properties. Replacement of the carbonyl function by two chlorines occurs with phosphorus pentachloride in ether: This reaction is useful in conjunction with \(E2\) elimination to prepare alkenyl halides, allenes, and alkynes. Cycloalkenyl halides are easily prepared, but because of angle strain the cycloalkynes and cycloallenes with fewer than eight atoms in the ring cannot be isolated (see Section 12-7): Replacement of a carbonyl group by -fluorines\(^3\) can be achieved with molybdenum hexafluoride or sulfur tetrafluoride. Sulfur tetrafluoride converts carboxyl functions to trifluoromethyl groups: In recent years, inorganic hydrides such as lithium aluminum hydride, \(\ce{LiAlH_4}\), and sodium borohydride, \(\ce{NaBH_4}\), have become extremely important as reducing agents of carbonyl compounds. These reagents have considerable utility, especially with sensitive and expensive carbonyl compounds. The reduction of cyclobutanone to cyclobutanol is a good example, and you will notice that the net reaction is the addition of hydrogen across the carbonyl double bond, \(\overset{ 2 \left[ \ce{H} \right]}{\longrightarrow}\) , With the metal hydrides, the key step is transfer of a hydride ion to the carbonyl carbon of the substance being reduced. The hydride transfer is analogous to the transfer of \(\ce{R}^\ominus\) from organometallic compounds to carbonyl groups (Section 14-12A). Lithium aluminum hydride is best handled like a Grignard reagent, because it is soluble in ether and is sensitive to both oxygen and moisture. (Lithium hydride is insoluble in organic solvents and is not an effective reducing agent for organic compounds.) All four hydrogens on aluminum can be utilized: The reaction products must be decomposed with water and acid as with the Grignard complexes. Any excess lithium aluminum hydride is decomposed by water and an acid with evolution of hydrogen: \[\ce{LiAlH_4} + 2 \ce{H_2SO_4} \rightarrow \frac{1}{2} \ce{Li_2SO_4} + \frac{1}{2} \ce{Al_2(SO_4)_3} + 4 \ce{H_2}\] Lithium aluminum hydride usually reduces carbonyl groups without affecting carbon-carbon double bonds. It is, in addition, a good reducing agent for carbonyl groups of carboxylic acids, esters, and other acid derivatives, as will be described in Chapter 18. Sodium borohydride is a milder reducing agent than lithium aluminum hydride and will reduce aldehydes and ketones, but not acids or esters. It reacts sufficiently slowly with water in neutral or alkaline solution that reductions which are reasonably rapid can be carried out in water solution without appreciable hydrolysis of the reagent: \[\ce{NaBH_4} + 4 \ce{CH_2=O} + 3 \ce{H_2O} \rightarrow 4 \ce{CH_3OH} + \ce{NaOB(OH)_2}\] Borane (as \(\ce{BH_3}\) in tetrahydrofuran or dimethyl sulfide) is an even milder reducing agent than \(\ce{BH_4^+}\) for the carbonyl group of aldehydes and ketones. This difference in reactivity can be used to advantage when selective reduction is necessary. For example, borohydride reduces a ketone carbonyl more rapidly than a carbon-carbon double bond, whereas borane reduces the carbon-carbon double bond more rapidly than carbonyl: A useful comparison of the reactivities of boranes and metal hydrides toward various types of multiple bonds is given in Table 16-6. A characteristic reaction of aldehydes without \(\alpha\) hydrogens is the self oxidation-reduction they undergo in the presence of strong base. With methanal as an example, If the aldehyde has \(\alpha\) hydrogens, other reactions usually occur more rapidly. The mechanism of this reaction, usually called the ,\(^4\) combines many features of other processes studied in this chapter. The first step is reversible addition of hydroxide ion to the carbonyl group: A hydrogen can be transferred as hydride ion to methanal from the hydroxyalkoxide ion, thereby reducing the methanal to methanol: Hydride transfer similar to that of the Cannizzaro reaction also may be achieved from a \(\ce{C-H}\) grouping in an alkoxide ion corresponding to a primary or secondary, but not a tertiary, alcohol. This is expected to be a reversible reaction, because the products are another alkoxide and another carbonyl compound: To utilize this equilibrium process as a practical reduction method requires rather special conditions. It is preferable to use an aluminum alkoxide, \(\ce{Al(OR)_3}\), rather than a sodium alkoxide, \(\overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{O}} \ce{R}\), to ensure that the reaction mixture is not too strongly basic. (Carbonyl compounds, particularly aldehydes, are sensitive to strong bases.) The overall reaction may be written for which the alkoxide is derived from 2-propanol. The advantage of this method is that the reaction can be driven essentially to completion by distilling out the 2-propanone as it is formed. The reduction product subsequently can be obtained by acid hydrolysis of the aluminum alkoxide: These have been discussed already in the context of the reverse reactions - oxidation of alcohols (Section 15-6C). \(^2\)Pronounced variously as , , , . The dipolar structures usually written for ylides are an oversimplified representation of the bonding in these substances. \(^3\) is an abbreviation for (twinned) and is a common designation for arrangements having two identical substituents on one carbon. \(^4\)Named after its discoverer, the same Cannizzaro who, in 1860, made an enormous contribution to the problem of obtaining self-consistent atomic weights (Section 1-1). and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/Number_of_Vibrational_Modes_in_a_Molecule

The Heisenberg uncertainty principle argues that all atoms in a molecule are constantly in motion (otherwise we would know position and momentum accurately). For molecules, they exhibit three general types of motions: translations (external), rotations (internal) and vibrations (internal). A diatomic molecule contains only a single motion., while polyatomic molecules exhibit more complex vibrations, known as A molecule has translational and rotational motion as a whole while each atom has it's own motion. The vibrational modes can be IR or Raman active. For a mode to be observed in the IR spectrum, changes must occur in the permanent dipole (i.e. not diatomic molecules). Diatomic molecules are observed in the Raman spectra but not in the IR spectra. This is due to the fact that diatomic molecules have one band and no permanent dipole, and therefore one single vibration. An example of this would be \(\ce{O2}\) or \(\ce{N2}\). However, unsymmetric diatomic molecules (i.e. \(\ce{CN}\)) do absorb in the IR spectra. Polyatomic molecules undergo more complex vibrations that can be summed or resolved into normal modes of vibration. The normal modes of vibration are: asymmetric, symmetric, wagging, twisting, scissoring, and rocking for polyatomic molecules. Degree of freedom is the number of variables required to describe the motion of a particle completely. For an atom moving in 3-dimensional space, three coordinates are adequate so its degree of freedom is three. Its motion is purely translational. If we have a molecule made of N atoms (or ions), the degree of freedom becomes 3N, because each atom has 3 degrees of freedom. Furthermore, since these atoms are bonded together, all motions are not translational; some become rotational, some others vibration. For non-linear molecules, all rotational motions can be described in terms of rotations around 3 axes, the rotational degree of freedom is 3 and the remaining 3N-6 degrees of freedom constitute vibrational motion. For a linear molecule however, rotation around its own axis is no rotation because it leave the molecule unchanged. So there are only 2 rotational degrees of freedom for any linear molecule leaving 3N-5 degrees of freedom for vibration. The degrees of vibrational modes for can be calculated using the formula: \[3N-5 \label{1}\] The degrees of freedom for can be calculated using the formula: \[3N-6 \label{2}\] \(n\) is equal to the number of atoms within the molecule of interest. The following procedure should be followed when trying to calculate the number of vibrational modes: How many vibrational modes are there in the linear \(\ce{CO_2}\) molecule ? There are a total of \(3\) atoms in this molecule. It is a linear molecule so we use Equation \ref{1}. There are \[3(3)-5 = 4 \nonumber\] vibrational modes in \(\ce{CO_2}\). Would \(\ce{CO_2}\) and \(\ce{SO_2}\) have a different number for degrees of vibrational freedom? Following the procedure above, it is clear that \(\ce{CO_2}\) is a linear molecule while \(\ce{SO_2}\) is nonlinear. \(\ce{SO_2}\) contains a lone pair which causes the molecule to be bent in shape, whereas, \(\ce{CO_2}\) has no lone pairs. It is key to have an understanding of how the molecule is shaped. Therefore, \(\ce{CO_2}\) has 4 vibrational modes and \(\ce{SO_2}\) has 3 modes of freedom. Would \(\ce{CO_2}\) and \(\ce{SO_2}\) have a different number for degrees of vibrational freedom? Following the procedure above, it is clear that \(\ce{CO_2}\) is a linear molecule while \(\ce{SO_2}\) is nonlinear. \(\ce{SO_2}\) contains a lone pair which causes the molecule to be bent in shape, whereas, \(\ce{CO_2}\) has no lone pairs. It is key to have an understanding of how the molecule is shaped. Therefore, \(\ce{CO_2}\) has 4 vibrational modes and \(\ce{SO_2}\) has 3 modes of freedom. How many vibrational modes are there in the tetrahedral \(\ce{CH_4}\) molecule ? In this molecule, there are a total of 5 atoms. It is a nonlinear molecule so we use Equation \ref{2}. There are \[3(5)-6 = 9\nonumber\] vibrational modes in \(\ce{CH_4}\). How many vibrational modes are there in the nonlinear \(\ce{C_{60}}\) molecule ? In this molecule, there are a total of 60 carbon atoms. It is a nonlinear molecule so we use Equation \ref{2}. There are \[3(60)-6 = 174\nonumber\] vibrational modes in \(\ce{C_60}\).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/Z020_Chemistry_of_Calcium_(Z20)/Calcium_Hydroxide

Calcium hydroxide, Ca(OH) , forms colorless crystals (resulting in white powder) and is obtained by mixing calcium oxide with water (calcium hydroxide is also called ). Calcium hydroxide is produced commerically in enormous quantities by thermal decomposition of limestone and subsequent exothermic reaction of the calcium oxide with water: CaCO CaO + CO CaO + H O Ca(OH) (+65.2 kJ) The exothermic reaction with water yields enough energy to evaporate the water. In the laboratory calcium hydroxide can be prepared by mixing aqueous solutions of calcium chloride and sodium hydroxide: CaCl + 2 NaOH Ca(OH) + 2 NaCl When heating calcium hydroxide to 512 °C the calcium hydroxide decomposes into calcium oxide and water: Ca(OH) CaO + H O Ca(OH) is only slightly soluble in water (0.16g Ca(OH) /100g water at 20°C) forming a basic solution called . The solubility decreases with increasing temperature. The suspension of calcium hydroxide particles in water is called . Lime water turns milky in the presence of carbon dioxide due to formation of calcium carbonate: Ca(OH) + CO CaCO + H O Calcium hydroxide is used in the construction industry as part of mortar, since its reaction with carbon dioxide of the air binds the particles of sand and gravel by forming calcium carbonate. Another major applications are the usage of calcium hydroxide as a flocculant in water and sewage treatment, and the recovery of in the paper industry.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/Z004_Chemistry_of_Beryllium_(Z4)/Beryllium_Fluoride

Beryllium fluoride (\(BeF_2\)) is a white solid being used as the principal precursor for the manufacture of beryllium metal. The production of beryllium is done by the reduction of\( BeF_2\) at 1300°C with magnesium in a graphite crucible: \[BeF_2 + Mg \rightarrow Be + MgF_2\] In order to obtain \(BeF_2\) beryllium ores (mainly bertrandite) are converted into \(Be(OH)_2\). The impure \(Be(OH)_2\) then reacts with ammonium bifluoride to give ammonium tetrafluorberyllate: \[Be(OH)_2 + 2 (NH_4)HF_2 \rightarrow (NH_4)_2BeF_4 + 2 H_2O\] Tetrafluorberyllate is a robust ion, which allows its purification by precipitation of various impurities as their hydroxides. Heating purified \((NH_4)_2BeF_4\) gives the desired product: \[(NH_4)_2BeF_4 \rightarrow 2 NH_3 + 2 HF + BeF_2\]

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/1b_Properties_of_Transition_Metals/Metallurgy/An_Introduction_to_the_Chemistry_of_Metal_Extraction

This page looks at the various factors which influence the choice of method for extracting metals from their ores, including reduction by carbon, reduction by a reactive metal (like sodium or magnesium), and by electrolysis. Details for the extraction of aluminum, copper, iron and titanium are given in separate pages in this section. An ore is any naturally-occurring source of a metal that you can economically extract the metal from. Aluminum, for example, is the most common metal in the Earth's crust, occurring in all sorts of minerals. However, it isn't economically worthwhile to extract it from most of these minerals. Instead, the usual ore of aluminum is bauxite - which contains from 50 - 70% of aluminum oxide. Copper is much rarer, but fortunately can be found in high-grade ores (ones containing a high percentage of copper) in particular places. Because copper is a valuable metal, it is also worth extracting it from low-grade ores as well. Ores are commonly oxides, for example: . . . or sulfides, for example: This simply means getting rid of as much of the unwanted rocky material as possible before the ore is converted into the metal. In some cases this is done chemically. For example, pure aluminum oxide is obtained from bauxite by a process involving a reaction with sodium hydroxide solution. This is described in detail on the aluminum page in this section. Some copper ores can be converted into copper(II) sulfate solution by leaving the crushed ore in contact with dilute sulphuric acid for a long time. Copper can then be extracted from the copper(II) sulfate solution. But, in many cases, it is possible to separate the metal compound from unwanted rocky material by physical means. A common example of this involves froth flotation. The ore is first crushed and then treated with something which will bind to the particles of the metal compound that you want and make those particles hydrophobic. "Hydrophobic" literally means "water fearing". In concentrating copper ores, for example, pine oil is often used. The pine oil binds to the copper compounds, but not to the unwanted rocky material. The treated ore is then put in a large bath of water containing a foaming agent (a soap or detergent of some kind), and air is blown through the mixture to make a lot of bubbles. Because they are water-repellent, the coated particles of the metal compound tend to be picked up by the air bubbles, float to the top of the bath, and are allowed to flow out over the sides. The rest of the rocky material stays in the bath. At its simplest, where you are starting from metal oxides, the ore is being reduced because oxygen is being removed. However, if you are starting with a sulfide ore, for example, that's not a lot of help! It is much more helpful to use the definition of reduction in terms of addition of electrons. To a reasonable approximation, you can think of these ores as containing positive metal ions. To convert them to the metal, you need to add electrons - reduction. There are various economic factors you need to think about in choosing a method of reduction for a particular ore. These are all covered in detail on other pages in this section under the extractions of particular metals. What follows is a quick summary. You need to consider: There may be various environmental considerations as well - some of which will have economic costs. Carbon (as coke or charcoal) is cheap. It not only acts as a reducing agent, but it also acts as the fuel to provide heat for the process. However, in some cases (for example with aluminum) the temperature needed for carbon reduction is too high to be economic - so a different method has to be used. Carbon may also be left in the metal as an impurity. Sometimes this can be removed afterwards (for example, in the extraction of iron); sometimes it can't (for example in producing titanium), and a different method would have to be used in cases like this. Other more reactive metals can be used to reduce the ore. Titanium is produced by reducing titanium(IV) chloride using a more reactive metal such as sodium or magnesium. As you will see if you read the page about titanium extraction, this is the only way of producing high purity metal. \[ TiCl_4 + 4Na \rightarrow Ti + 4NaCl\] The more reactive metal sodium releases electrons easily as it forms its ions: \[ 4Na \rightarrow 4Na^+ + 4e^-\] These electrons are used to reduce the titanium(IV) chloride: \[ TiCl_4 + 4e^- \rightarrow Ti + 4Cl^-\] The downside of this is expense. You have first to extract (or to buy) the sodium or magnesium. The more reactive the metal is, the more difficult and expensive the extraction becomes. That means that you are having to use a very expensive reducing agent to extract the titanium. As you will see if you read the page about titanium extraction, there are other problems in its extraction which also add to the cost. This is a common extraction process for the more reactive metals - for example, for aluminum and metals above it in the electrochemical series. You may also come across it in other cases such as one method of extracting copper and in the purification of copper. During electrolysis, electrons are being added directly to the metal ions at the cathode (the negative electrode). The downside (particularly in the aluminum case) is the cost of the electricity. An advantage is that it can produce very pure metals. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.04%3A_Relaxation_Methods/2.1.4.01%3A_Flash_Photolysis

In 1967 the Nobel Prize in Chemistry was awarded to Manfred Eigen, Ronald George, George Porter and Wreyford Norrish for their co-discovery of in 1949. Flash Photolysis is used extensively to study reactions that happen extremely quickly, even down to the femtosecond depending on the laser that is used. The technique was born out of cameras developed during and shortly after WWII, which were used to take pictures of fast moving planes, rockets, and missiles. Since then the technology of lasers and optics has progressed allowing faster and faster reactions to be studied. Flash Photolysis is often used to study reactions that are light dependent such as photosynthesis and reactions in the cones on the retina of the our eye, but the meathod can also be applied to other reactions. The light in the form of a laser excites a molecule into a reactive state, usually in the form of a free radical. From there it is possible to measure the reaction spectroscopically, using the exitory flash as a light source to measure absorbance. The laser pulse must be aproximatly half the length of the reaction, and of sufficient energy to induce the reaction to take place. Further the flash must cover the spectrum of frequencies which are being studied because not only is the flash producing intermediates of the reaction that are usually not observed, it is also producing the source for spectroscopic analysis. Intermediates of most reactions are rarely observed, this techniques isolates even low concentrations of otherwise unobservable portions of reactions allowing research into synthetic, biochemical, and photo-sensitive reactions.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Calculating_an_Equilibrium_Constant_Using_Partial_Pressures

The equilibrium constant is known as \(K_{eq}\). A common example of \(K_{eq}\) is with the reaction: \[aA + bB \rightleftharpoons cC + dD\] \[K_{eq} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\] where: All the equilibrium constants tell the relative amounts of products and reactants at equilibrium. For any reversible reaction, there can be constructed an equilibrium constant to describe the equilibrium conditions for that reaction. Since there are many different types of reversible reactions, there are many different types of equilibrium constants: Referring to equation: \[aA + bB \rightleftharpoons cC + dD\] \[K_p = \dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\] Partial Pressures: In a mixture of gases, it is the pressure an individual gas exerts. The partial pressure is independent of other gases that may be present in a mixture. According to the ideal gas law, partial pressure is inversely proportional to volume. It is also directly proportional to moles and temperature. At equilibrium in the following reaction at room temperature, the partial pressures of the gases are found to be \(P_{N_2}\) = 0.094 atm, \(P_{H_2}\) = 0.039 atm, and \(P_{NH_3}\) = 0.003 atm. \[\ce{N_2 (g) + 3 H_2 (g) \rightleftharpoons 2 NH_3 (g)} \nonumber \] What is the \(K_p\) for the reaction? First, write \(K_{eq}\) (equilibrium constant expression) in terms of activities. \[K = \dfrac{(a_{NH_3})^2}{(a_{N_2})(a_{H_2})^3} \nonumber\] Then, replace the activities with the partial pressures in the equilibrium constant expression. \[K_p = \dfrac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \nonumber\] Finally, substitute the given partial pressures into the equation. \[K_p = \dfrac{(0.003)^2}{(0.094)(0.039)^3} = 1.61 \nonumber\] At equilibrium in the following reaction at 303 K, the total pressure is 0.016 atm while the partial pressure of \(P_{H_2}\) is found to be 0.013 atm. \[\ce{3 Fe_2O_3 (s) + H_2 (g) \rightleftharpoons 2 Fe_3O_4 (s) + H_2O (g)} \nonumber\] What is the \(K_p\) for the reaction? First, calculate the partial pressure for \(\ce{H2O}\) by subtracting the partial pressure of \(\ce{H2}\) from the total pressure. \[ \begin{align*} P_{H_2O} &= {P_{total}-P_{H_2}} \\[4pt] &= (0.016-0.013) \; atm \\[4pt] &= 0.003 \; atm \end{align*}\] Then, write K (equilibrium constant expression) in terms of activities. Remember that solids and pure liquids are ignored. \[K = \dfrac{(a_{H_2O})}{(a_{H_2})}\nonumber\] Then, replace the activities with the partial pressures in the equilibrium constant expression. \[K_p = \dfrac{(P_{H_2O})}{(P_{H_2})}\nonumber\] Finally, substitute the given partial pressures into the equation. \[K_p = \dfrac{(0.003)}{(0.013)} = 0.23 \nonumber\] A flask initially contained hydrogen sulfide at a pressure of 5.00 atm at 313 K. When the reaction reached equilibrium, the partial pressure of sulfur vapor was found to be 0.15 atm. \[\ce{2 H_2S (g) \rightleftharpoons 2 H_2 (g) + S_2 (g) } \nonumber\] What is the \(K_p\) for the reaction? For this kind of problem, are used. Now, set up the equilibrium constant expression, \(K_p\). \[K_p = \dfrac{(P_{H_2})^2(P_{S_2})}{(P_{H_2S})^2} \nonumber\] Finally, substitute the calculated partial pressures into the equation. \[ \begin{align*} K_p &= \dfrac{(0.3)^2(0.15)}{(4.7)^2} \\[4pt] &= 6.11 \times 10^{-4} \end{align*} \]

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/17%3A_Carbonyl_Compounds_II-_Enols_and_Enolate_Anions._Unsaturated_and_Polycarbonyl_Compounds/17.06%3A_-Unsaturated_Aldehydes_and_Ketones

The most generally useful preparation of \(\alpha\),\(\beta\)-unsaturated carbonyl compounds is by dehydration of aldol addition products, as described in . Conjugation of the carbonyl group and double bond has a marked influence on spectroscopic properties, particularly on ultraviolet spectra, as the result of stabilization of the excited electron states, which for \(\pi \rightarrow \pi^*\) transitions can be described in terms of important contributions of polar resonance structures (see and ): Such resonance is much less important in the ground state but is still sufficiently important to account for the moderate differences in dipole moments between saturated and \(\alpha\),\(\beta\)-unsaturated aldehydes and ketones; for example, The effect of conjugation also is reflected in infrared carbonyl frequencies ( ) and NMR spectra. With respect to the latter, it is found that the protons on the \(\beta\) carbon of \(\alpha\),\(\beta\)-unsaturated carbonyl compounds usually come at \(0.7\) to \(1.7 \: \text{ppm}\) fields than ordinary alkenic protons. The effect is smaller for the \(\alpha\) protons. There are many addition reactions of \(\alpha\),\(\beta\)-unsaturated aldehydes, ketones, and related compounds that are the same as the carbonyl addition reactions described previously. Others are quite different and result in addition to the double bond. Organometallic compounds are examples of nucleophilic reagents that can add to either the alkene or the carbonyl bonds of conjugated ketones ( ). Hydrogen cyanide behaves likewise and adds to the carbon-carbon double bond of 3-butene-2-one, but to the carbonyl group of 2-butenal: All of these reactions may be classified as nucleophilic additions, but when addition occurs at the alkene bond, the orientation always is such that the nucleophile adds at the \(\beta\) carbon. An example is the addition of methanol catalyzed by sodium methoxide: Nucleophilic reagents normally do not attack carbon-carbon double bonds ( ). The adjacent carbonyl function therefore must greatly enhance the reactivity of the double bond toward such reagents. This enhancement is not surprising when it is realized that the attack of a nucleophile produces a stabilized enolate anion: The products are formed from the enolate intermediate by proton transfer to either carbon or oxygen. If the proton adds to oxygen the enol is formed, which is unstable with respect to the ketone and ultimately will rearrange: Reactions of this type are referred to in a variety of terms, many of which are rather confusing and nondescriptive. They sometimes are classified as , implying that addition occurs across the terminal protons of the conjugated system. A synonymous term is . When the nucleophile is a carbanion, the reaction is called a . Thus, by this definition, Equation 17-7 represents a Michael addition. Another, perhaps more typical, example is the addition of an enolate to a conjugated ketone: Michael-type additions, like aldol additions, are useful for the formation of carbon-carbon bonds. Electrophilic addition of hydrogen halides to \(\alpha\),\(\beta\)-unsaturated aldehydes and ketones places the halogen on the \(\beta\) carbon. This orientation is opposite to that observed for related additions to conjugated dienes: and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Dipole-Dipole_Interactions

Dipole-Dipole interactions result when two dipolar molecules interact with each other through space. When this occurs, the partially negative portion of one of the polar molecules is attracted to the partially positive portion of the second polar molecule. This type of interaction between molecules accounts for many physically and biologically significant phenomena such as the elevated boiling point of water. occur due to the unequal sharing of electrons between atoms in a molecule. Those atoms that are more electronegative pull the bonded electrons closer to themselves. The buildup of electron density around an atom or discreet region of a molecule can result in a molecular dipole in which one side of the molecule possesses a partially negative charge and the other side a partially positive charge. Molecules with dipoles that are not canceled by their molecular geometry are said to be polar. In Figure 1 above, the more electronegative Oxygen atoms pull electron density towards themselves as demonstrated by the arrows. Carbon Dioxide is not polar however because of its linear geometry. A molecule's overall dipole is directional, and is given by the vector sum of the dipoles between the atoms. If we imagined the Carbon Dioxide molecule centered at 0 in the XY coordinate plane, the molecule's overall dipole would be given by the following equation: \[\mu \cos(0) + -\mu \cos(0) = 0. \] Where \(μ\) is the dipole moment of the bond (given by μ=Q x r where Q is the charge and r is the distance of separation). Therefore, the two dipoles cancel each other out to yield a molecule with no net dipole. In contrast, figure 2 demonstrates a situation where a molecular dipole does result. There is no opposing dipole moment to cancel out the one that is shown above. If we were to imagine the hydrogen flouride molecule placed so that the Hydrogen sat at the origin in the XY coordinate plane, the dipole would be given by \(\mu \cos(0)=\mu\). Potential energy is the maximum energy that is available for an object to do work. In physics, work is a quantity that describes the energy expended as a force operates over a distance. Potential energy is positional because it depends on the forces acting on an object at its position in space. For instance, we could say that an object held above the ground has a potential energy equal to its mass x acceleration due to gravity x its height above the ground (i.e., \(mgh\)). This potential energy that an object has as a result of its position can be used to do work. For instance we could use a pulley system with a large weight held above the ground to hoist a smaller weight into the air. As we drop the large weight it converts its potential energy to kinetic energy and does work on the rope which lifts the smaller weight into the air. It is important to remember that due to the second law of thermodynamics, the amount of work done by an object can never exceed (and is often considerably less) than the objects potential energy. On a subatomic level, charged atoms have an electric potential which allows them to interact with each other. Electric potential refers to the energy held by a charged particle as a result of it's position relative to a second charged particle. Electric potential depends on charge polarity, charge strength and distance. Molecules with the same charge will repel each other as they come closer together while molecules with opposite charges will attract. For two positively charged particles interacting at a distance r, the potential energy possessed by the system can be defined using Coulomb's Law: \[V = \dfrac{kQq}{r} \label{1}\] where The above equation can also be used to calculate the distance between two charged particles (\(r\)) if we know the potential energy of the system. While Coulomb's law is important, it only gives the potential energy between two point particles. Since molecules are much larger than point particles and have charge concentrated over a larger area, we have to come up with a new equation. The potential energy possessed by two polar atoms interacting with each other depends on the dipole moment, μ, of each molecule, the distance apart, r, and the orientation in which the two molecules interact. For the case in which the partially positive area of one molecule interacts only with the partially negative area of the other molecule, the potential energy is given by: \[V(r) = -\dfrac{2\mu_{1}\mu_{2}}{4\pi\epsilon_{o}r^{3}} \label{2}\] where \(\epsilon_o\) is the permeability of space. If it is not the case that the molecular dipoles interact in this straight end to end manor, we have to account mathematically for the change in potential energy due to the angle between the dipoles. We can add an angular term to the above equation to account for this new parameter of the system: \[V (r) =-\dfrac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r_{12}^{3}}{(\cos\theta_{12}- 3\cos\theta_{1}\cos\theta_{2})} \label{3}\] In this formula \(\theta_{12}\) is the angle made by the two oppositely charged dipoles, and \(r_{12}\) is the distance between the two molecules. Also, \(\theta_{1}\) and \(\theta_{2}\) are the angles formed by the two dipoles with respect to the line connecting their centers. It is also important to find the potential energy of the dipole moment for more than two interacting molecules. An important concept to keep in mind when dealing with multiple charged molecules interacting is that like charges repel and opposite charges attract. So for a system in which three charged molecules (2 positively charged molecules and 1 negatively charged molecule) are interacting, we need to consider the angle between the attractive and repellant forces. The potential energy for the dipole interaction between multiple charged molecules is: \[V = \dfrac{kp \cos\theta}{r^{2}} \label{4}\] where Calculate the potential energy of the dipole-dipole interaction between 2 \(\ce{HF}\) molecules oriented along the x axis in an XY coordinate plane whose area of positive charge is separated by 5.00 Angstroms from the area of negative charge of the adjacent molecule: The Dipole moment of the HF molecules can be found in many tables, μ=1.92 D. Assume the molecules exist in a vacuum such that \(\epsilon_{0}=8.8541878\times10^{-12}C^2N^{-1}m^{-2}\) \[\mu=1.92D\times3.3356\times10^{-30}\dfrac{Cm}{D}=6.4044\times10^{-30}Cm\] Now use Equation \ref{2} to calculate the interaction energy \[\begin{align*}V&=-\dfrac{2(6.4044\times10^{-30}Cm)^2}{4(8.8541878\times10^{-12}C^2N^{-1}m^{-2})(5.0\times10^{-10})^3} \\[4pt] &=1.4745\times10^{-19}Nm \end{align*}\] Now imagine the same two HF molecules in the following orientation: Given: \(\theta_{1}=\dfrac{3\pi}{4}\), \(\theta_{2}=\dfrac{\pi}{3}\) and \(\theta_{12}=\dfrac{5\pi}{12}\) \[\begin{align*} V&=-\dfrac{(6.4044\times10^{-30}Cm)^2}{4\pi(8.8541878\times10^{-12}C^2N^{-1}m^{-2}(5.00\times10^{-10}m)^3}(\cos\dfrac{5\pi}{12}-3\cos\dfrac{3\pi}{4}\cos\dfrac{\pi}{3}) \\[4pt] &=-9.73\times10^{-20}Nm=9.73\times10^{-20}J\end{align*}\] It would seem, based on the above discussion, that in a system composed of a large number of dipolar molecules randomly interacting with one another, V should go to zero because the molecules adopt all possible orientations. Thus the negative potential energy of two molecular dipoles participating in a favorable interaction would be cancelled out by the positive energy of two molecular dipoles participating in a high potential energy interaction. Contrary to our assumption, in bulk systems, it is more probable for dipolar molecules to interact in such a way as to minimize their potential energy (i.e., dipoles form less energetic, more probable configurations in accordance with the ). For instance, the partially positive area of a molecular dipole being held next to the partially positive area of a second molecular dipole is a high potential energy configuration and few molecules in the system will have sufficient energy to adopt it at room temperature. Generally, the higher potential energy configurations are only able to be populated at elevated temperatures. Therefore, the interactions of dipoles in a bulk Solution are not random, and instead adopt more probable, lower energy configurations. The following equation takes this into account: \[V=-\dfrac{2\mu_{A}^2\mu_{B}^2}{3(4\pi\epsilon_{0})^2r^6}\dfrac{1}{k_{B}T} \label{5}\] Looking at Equation \ref{5}, what happens to the potential energy of the interaction as temperature increases. The potential energy of the dipole-dipole interaction decreases as T increases. This can be seen from the form of the above equation, but an explanation for this observation is relatively simple to come by. As the temperature of the system increases, more molecules have sufficient energy to occupy the less favorable configurations. The higher, less favorable, configurations are those that give less favorable interactions between the dipoles (i.e., higher potential energy configurations). Calculate the average energy of HF molecules interacting with one another in a bulk Solution assuming that the molecules are 4.00 Angstroms apart in room temperature Solution. Using Equation \ref{5} to calculate the bulk potential energy: \[\begin{align*} V&=-\dfrac{2}{3}\dfrac{(6.4044\times10^{-30}Cm)^4}{(4\pi(8.8541878\times10^{-12}C^2N^{-1}m^{-2})^2(4.00\times10^{-10}m)^6}\dfrac{1}{(1.381\times10^{-23}Jk^{-1})(298k)} \\[4pt] &=-5.46\times10^{-21}J\end{align*}\] What is the amount of energy stabilization that is provided to the system when 1 mole of HF atoms interact through dipole-dipole interactions. Since we have already calculated above the average potential energy of the HF dipole-dipole interaction this problem can be easily solved. \[\begin{align*} V &=-5.46\times10^{-21}J\times(6.022\times10^{23}mol^{-1}) \\[4pt] &=-3288\dfrac{J}{mol}=3.29\dfrac{kJ}{mol} \end{align*}\] The potential energy from dipole interactions is important for living organisms. The biggest impact dipole interactions have on living organisms is seen with protein folding. Every process of protein formation, from the binding of individual amino acids to secondary structures to tertiary structures and even the formation of quaternary structures is dependent on dipole-dipole interactions. A prime example of quaternary dipole interaction that is vital to human health is the formation of erythrocytes. , commonly known as red blood cells are the cell type responsible for the gas exchange (i.e. respiration). Inside the erythrocytes, the molecule involved in this crucial process, is 'hemoglobin', formed by four protein subunits and a heme group'. For an heme to form properly, multiple steps must occur, all of which involve dipole interactions. The four protein subunits—two alpha chains, two beta chains—and the heme group, interact with each other through a series of dipole-dipole interactions which allow the erythrocyte to take its final shape. Any mutation that destroys these dipole-dipole interactions prevents the erythrocyte from forming properly, and impairs their ability to carry oxygen to the tissues of the body. So we can see that without the dipole-dipole interactions, proteins would not be able to fold properly and all life as we know it would cease to exist.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/19%3A_The_Distribution_of_Outcomes_for_Multiple_Trials/19.03%3A_Distribution_of_Results_for_Multiple_Trials_with_Many_Possible_Outcomes

It is now easy to extend our results to multiple trials with any number of outcomes. Let the outcomes be \(A\), \(B\), \(C\), ., \(Z\), for which the probabilities in a single trial are \(P_A\), \(P_B\), \(P_C\), \(P_Z\). We again want to write an equation for the total probability after \(n\) trials. We let \(n_A\), \(n_B\), \(n_C\), \(n_Z\) be the number of \(A\), \(B\), \(C\), , \(Z\) outcomes exhibited in \(n_A+n_B+n_C+...+n_Z=n\) trials. If we do not care about the order in which the outcomes are obtained, the probability of \(n_A\), \(n_B\), \(n_C\), , \(n_Z\) outcomes in \(n\) trials is \[C\left(n_A,n_B,n_C,\dots ,n_Z\right)P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\] and the total probability sum is \[1={\left(P_A+P_B+P_C+\dots +P_Z\right)}^n=\sum_{n_I}{C\left(n_A,n_B,n_C,\dots ,n_Z\right)P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z}\] where the summation is to be carried out over all combinations of integer values for \(n_A\), \(n_B\), \(n_C\), , \(n_Z\) consistent with \(n_A+n_B+n_C+...+n_Z=n\). Let one of the terms for \(n_A\) \(A\)-outcomes, \(n_B\) \(B\)-outcomes, \(n_C\) \(C\)-outcomes, , \(n_Z\)\({}_{\ }\)\(Z\)-outcomes, be \[\left(P_{A,a}P_{A,b}\dots P_{A,f}\right)\left(P_{B,g}P_{B,h}\dots P_{B,m}\right)\times \left(P_{C,p}P_{C,q}\dots P_{C,t}\right)\dots \left(P_{Z,u}P_{Z,v}\dots P_{Z,z}\right)\] where there are \(n_A\) indices in the set \(\{a,\ b,\ \dots ,\ f\}\), \(n_B\) indices in the set \(\{g,\ h,\ \dots ,\ m\}\), \(n_C\) indices in the set \(\{p,\ q,\ \dots ,\ t\}\), , and \(n_Z\) indices in the set \(\{u,\ v,\ \dots ,\ z\}\). There are \(n_A!\) ways to order the \(A\)-outcomes, \(n_B!\) ways to order the \(B\)-outcomes, \(n_C!\) ways to order the \(C\)-outcomes, , and \(n_Z!\) ways to order the \(Z\)-outcomes. So, there are \(n_A!n_B!n_C!\dots n_Z!\) ways to order \(n_A\) \(A\)-outcomes, \(n_B\) \(B\)-outcomes, \(n_C\) \(C\)-outcomes, , and \(n_Z\) \(Z\)-outcomes. The same is true for any other distinguishable combination; for every distinguishable combination belonging to the population set \(\{n_A\), \(n_B\), \(n_C\), , \(n_Z\}\) there are \(n_A!n_B!n_C!\dots n_Z!\) indistinguishable permutations. Again, we can express this result as the general relationship: so that \[n!=n_A!n_B!n_C!\dots n_Z!C\left(n_A,n_B,n_C,\dots ,n_Z\right)\] and \[C\left(n_A,n_B,n_C,\dots ,n_Z\right)=\frac{n!}{n_A!n_B!n_C!\dots n_Z!}\] Equivalently, we can construct a sum, \(T\), in which we add up all of the \(n!\) permutations of \(P_{A,a}\) factors for \(n_A\) \(A\)-outcomes, \(P_{B,b}\) factors for \(n_B\) \(B\)-outcomes, \(P_{C,c}\) factors for \(n_C\) \(C\)-outcomes, , and \(P_{Z,z}\) factors for \(n_Z\) \(Z\)-outcomes. The value of each term in \(T\) will be \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\). So we have \[T=n!P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\] \(T\) will contain all \(C\left(n_A,n_B,n_C,\dots ,n_Z\right)\) of the \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\)-valued products (distinguishable combinations) that are a part of the total-probability sum. Moreover, \(T\) will also include all of the \(n_A!n_B!n_C!\dots n_Z!\) indistinguishable permutations of each of these \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\)-valued products. Then we also have \[T=n_A!n_B!n_C!\dots n_Z!C\left(n_A,n_B,n_C,\dots ,n_Z\right)\] \[\times P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\] Equating these two expressions for\(\ T\) gives us the number of \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\)-valued products \[n!P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z=n_A!n_B!n_C!\dots n_Z!\] \[\times C\left(n_A,n_B,n_C,\dots ,n_Z\right)P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\] and hence, \[C\left(n_A,n_B,n_C,\dots ,n_Z\right)=\frac{n!}{n_A!n_B!n_C!\dots n_Z!}\] In the special case that \(P_A=P_B=P_C=\dots =P_Z\), all of the products \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\) have the same value. Then, the probability of any set of outcomes, \(\{n_A,n_B,n_C,\dots ,n_Z\}\), is proportional to \(C\left(n_A,n_B,n_C,\dots ,n_Z\right)\).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/Z056_Chemistry_of_Barium_(Z56)/Barium_Hydroxide

Barium hydroxide, \(Ba(OH)_2\), is also known as baryta. It can be prepared either by dissolving barium oxide (\(BaO\)) or barium sulfide (\(BaS\)) in water: \[BaO + H_2O \rightarrow Ba(OH)_2 + H_2O\] \[BaS + H_2O \rightarrow Ba(OH)_2 + H_2S\] Barium hydroxide crystallizes as the octahydrate, \(Ba(OH)_2 \cdot 8H_2O\), which can be converted to the monohydrate by heating in air. The white granular monohydrate is the usual commercially available form. Barium hydroxide decomposes to barium oxide when heated to 800°C. Barium hydroxide forms an alkaline solution in water ("baryta water") which can be used to titrate weak acids, since its clear aqueous solution is guaranteed to be free of carbonate (barium carbonate is insoluble in water).

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Selection_Rules_for_Electronic_Spectra_of_Transition_Metal_Complexes/Derivation_of_Laporte_Rule

The Laporte Rule is a selection rule in electron absorption spectroscopy that applies to centrosymmetric molecules. It says that transitions between states of the same symmetry with respect to inversion are forbidden. Using the mathematical concept of even and odd functions, the Laporte Rule can be derived and summarized as follows: Electronic transitions from waveunfunctions with g symmetry to wavefunctions with symmetry are forbidden, as are transitions from wavefunctions with symmetry to wavefunctions with symmetry. Transitions from to and to , where the symmetry switches, may be (but are not necessarily) allowed. \[ \int_{- \infty}^{ \infty} \Psi_{el} \hat{ M} \Psi^{ex}_{el} d \tau = \left | \left \langle i \left | \hat{M} \right | f \right \rangle \right | \label{1}\] (Ψ is the wavefunction, M is the transition dipole moment operator, i refers to the initial state and f refers to the final state). The Laporte Rule tells us that if the integrand of the transition moment integral does not contain the totally symmetric representation, then the transition is forbidden. \[ \mu = \int_{- \infty }^{\infty} \Psi \hat{ \mu} \Psi^{ex} \label{2}\] The transition moment dipole given above includes the wave functions of the final and initial states, which contain both electronic and nuclear wave functions. \[ \Psi = \Psi_{nuc} \Psi_{el} \label{3}\] The Born–Oppenheimer approximation tells us that electronic transitions happen on a faster time scale than nuclear transitions so the two can be separated, and when dealing with electronic transitions we can effectively ignore any nuclear motion. Thus, in discussing electronic spectroscopy and the "allowedness" of certain electronic transitions, we consider only the electronic integral. \[ \int_{- \infty}^{ \infty} \Psi_{el} H \Psi_{el}^{ex}\label{4} \] Where is the time-dependent Hamiltonian. We can make this integral easier to work with by invoking , which allows us to replace with , the electric dipole moment operator , which is part of the time-dependent Hamiltonian. \[ \hat{ M} = e \sum_{ i} \vec{ r_{ i}} \label{5}\] We arrive at the following transition moment integral to describe electronic transitions from a ground state to an excited state. \[ \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \left | \left \langle i \left | \hat{M} \right | f \right \rangle \right | \label{6}\] The transition moment integral describes the probability of a transition taking place. The integral must be non-zero in order for a transition to occur. In other words, the two electronic states must overlap in order for a transition to occur. To determine if a transition is allowed, the point group of the molecule is determined, and the corresponding character table gives the symmetry representation of each electronic wavefunction. The character table is also used to determine the electric dipole moment operator, which is contained in the transition moment integral and has three parts that transform with the Cartesian x, y, and z axes. \[ \hat{ M} = \begin {vmatrix} \hat {M_{x}}\\ \hat {M_{y}}\\ \hat {M_{z}} \end{vmatrix}\label{7}\] For calculating the integrand of the transition moment integral, the symmetry representations corresponding to x, y, and z are used for the electric dipole moment operator. The integrand of the transition moment integral is then given by the following direct products: \[ \Psi_{el} \otimes \begin {vmatrix} \hat {M_{x}}\\ \hat {M_{y}}\\ \hat {M_{z}} \end{vmatrix} \otimes \Psi_{el}^{ex} \label{8}\] The component of the electric dipole moment operator that gives the totally symmetric representation corresponds to the direction of polarization of the electronic transition. For a point group of low symmetry like C , computing the integrand for all possible transitions is a relatively short task. However, when considering a point group with a high degree of symmetry, like O , computing the integrand for all possible transitions in a point group becomes a long and tedious task. If we examine these computations through the lens of mathematics, this problem can be simplified. The Laporte Rule applies to centrosymmetric molecules, those containing a center of inversion. All states have a symmetry with respect to the inversion center, and all representations for a given point group have either a (gerade) or (ungerade) designation with respect to the center of inversion. Designations of and refer to even and odd symmetries with respect to the inversion center. Odd and even with respect to orbital symmetry correspond to the mathematical definitions of odd and even functions. M is an odd function. For the derivation of the Laporte Rule, allow states to be represented by the even function cos(x), states to be represented by the odd function sin(x), and M to be represented by the odd function x. \[ F(x) = \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \int \cos{x} \times{ x} \times{ \cos{x}} dx \label{9}\] The integrand is given by Evaluate the integrand Evaluate the integrand \[ G(x) = - G(-x)\] Thus, \[ G(x) = x \times{ \cos^2 {x}} \] is an odd function, and the transition between two even states is forbidden. Similarly, a transition between two odd states is given by the integral \[ F(x) = \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \int \sin{x} \times{x} \times{ \sin{x}} dx \label{10} \] The integrand is given by \[ G(x) = x \times{ \sin^2 {x}} \] Evaluate the integrand \[ G(x = \frac{- \pi}{4}) = (\frac{- \pi}{4}) \times{ \sin^2 {( \frac{- \pi}{4})}} = \frac {- \pi}{8}\] \[ G(x) = - G(-x)\] Thus, \[ G(x) = x \times{ \sin^2 {x}} \] is an odd function, and the transition between two odd states is forbidden. A transition between an odd and an even state is given by the integral \[ F(x) = \int_{- \infty}^{ \infty} \Psi_{el} \hat {M} \Psi_{el}^{ex} = \int \cos{x} \times{x} \times{ \sin{x}} dx \label{11}\] Evaluate the integrand \[ G(x = \frac{ \pi}{4}) = \sin{( \frac{ \pi}{4})} \times{( \frac{ \pi}{4})} \times{ \cos {( \frac{ \pi}{4})}} = \frac{ \pi}{8} \] Evaluate the integrand \[ G(x = \frac{- \pi}{4}) = \sin{( \frac{- \pi}{4})} \times{( \frac{- \pi}{4})} \times{ \cos {( \frac{- \pi}{4})}} = \frac{ \pi}{8} \] \[ G(x) = G(-x) \] Thus, \[ G(x) = x \times{ \sin{x}} \times{ \cos{x}} \] is an even function, and the transition between an even and an odd state is not immediately forbidden. May be Allowed May be Allowed It is shown above that the Laporte Rule expressly states that transitions between two even states or two odd states are forbidden. It should be emphasized, however, that the Laporte Rule does state that transitions between an even and an odd state are allowed. The integrand of the transition moment integral for the transition between an odd and an even state is even, but there are multiple states with even inversion symmetry that are not the totally symmetric representation. Only an integrand that contains the totally symmetric representation is allowed. To determine which of the even transitions are actually allowed, the cross product of the representations of the initial and final electronic states with the electric dipole moment operator must be computed. All resulting cross products will be even; only those which contain the totally symmetric representation will be allowed. O (octahedral) point group \[ M_x, M_y, M_z = T_{1u} \] Equation (3) Equation (1) shows a forbidden even-even transition. Equation (2) shows a forbidden odd-odd transition. Equations (3) and (4) demonstrate the subtlety of the Laporte Rule. Both are even transitions between an odd and an even state. However, (3) contains the totally symmetric representation and is an allowed transition, while (4) does not contain the totally symmetric representation and is thereby forbidden. 1. The C point group contains the representations A , B , E , A , B , and E . Via the Laporte rule: g to g and u to u transitions are orbitally forbidden. For g to u and u to g transitions, the integrand of the transition moment integral must contain the totally symmetric representation, which is A in this case. \[ A_g \otimes{ A_u} \] \[ A_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} A_{u} = \begin {vmatrix} A_{g}\\ E_{u} \end{vmatrix} \] \[ A_g \otimes{ B_u} \] \[ A_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} B_{u} = \begin {vmatrix} E_{g}\\ B_{g} \end{vmatrix} \] \[ A_g \otimes{ E_u} \] \[ A_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} E_{u} = \begin {vmatrix} A_{g} + A_{u} + 2B_{g}\\ E_{g} \end{vmatrix} \] \[ B_g \otimes{ A_u} \] \[ B_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} A_{u} = \begin {vmatrix} E_{g}\\ B_{g} \end{vmatrix} \] \[ B_g \otimes{ B_u} \] \[ B_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} B_{u} = \begin {vmatrix} E_{g}\\ B_{g} \end{vmatrix} \] \[ B_g \otimes{ E_u} \] \[ B_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} E_{u} = \begin {vmatrix} A_{g} + A_{u} + 2B_{g}\\ E_{g} \end{vmatrix} \] \[ E_g \otimes{ A_u} \] \[ E_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} A_{u} = \begin {vmatrix} A_{g} + A_{u} + 2B_{g}\\ E_{g} \end{vmatrix} \] \[ E_g \otimes{ B_u} \] \[ E_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} B_{u} = \begin {vmatrix} B_{u} + B_{g} + 2A_{g}\\ E_{g} \end{vmatrix} \] \[ E_g \otimes{ E_u} \] \[ E_{g} \begin {vmatrix} E_{u}\\ A_{u} \end{vmatrix} E_{u} = \begin {vmatrix} E_{g} + E_{u} + 2E_{g}\\ A_{g} +A_{u} + 2B_{g} \end{vmatrix} \] \[ A_u \otimes{ A_u} \] 2. Laporte Forbidden 3. Using the character table, we see that d-orbitals in octahedral have the following symmetries: Possible transitions are \[ E_{g} \times{ E_{g}} \] \[E_{g} \times{ T_{2g}} \] \[T_{2g} \times{ T_{2g}} \] All possible transitions are g to g, and are thus forbidden by the Laporte Rule. P-orbitals have the following symmetries: All possible transitions are u to u, and are thus forbidden by the Laporte Rule. P-orbital to d-orbital transitions would all be u to g, and are not forbidden by the Laporte Rule. 4.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/Z020_Chemistry_of_Calcium_(Z20)/Calcium_Sulfide

Calcium sulfide, CaS, crystallizes in cubes like rock salt and occurs as oldhamite in nature. CaS has an odor of rotten eggs, which stems from H S formed by hydrolysis of the calcium sulfide. CaS is produced by reduction of calcium sulfate with charcoal: CaSO + 2 C CaS + 2 CO It also occurs as a by-product of the , which was in heavy use in the 19 century before the invention of the . CaS can be oxidized by calcium sulfate to form calcium oxide and sulfur dioxide: 3 CaSO + CaS 4 CaO + 4 SO Calcium sulfide decomposes upon contact with water: CaS + H O Ca(SH)(OH) Ca(SH)(OH) + H O Ca(OH) + H S In the presence of traces of heavy metals CaS becomes phosphorescent (like many other sulfides such as zinc sulfide or cadmium sulfide).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/Z020_Chemistry_of_Calcium_(Z20)/Calcium_Sulfate

Calcium sulfate, CaSO , is a common laboratory and industrial chemical and an often used material in the building trade. It occurs naturally in various forms, which differ in their crystal water content: gypsum (CaSO ·2H O), the hemihydrate (CaSO ·0.5H O) also known as plaster of Paris, and the anhydrite. Despite its name the anhydrite is not always entirely devoid of water, the water content ranges from 0.0 to 0.05 mol-percent. The main sources of calcium sulfate are naturally occurring minerals (gypsum and anhydrite). World production of natural gypsum is about 100 million tones per annum. Besides the natural sources, calcium sulfate is also produced as a by-product, mainly from the desulfurization of exhaust gases of fossil-fuel power stations.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.6%3A_Optical_Properties_and_the_Spectrochemical_Series

Electromagnetic radiation is a form of energy that is produced by oscillating electric and magnetic disturbance, or by the movement of electrically charged particles traveling through a vacuum or matter. and have certain characteristics, including amplitude, wavelength, and frequency (Figure \(\Page {1}\)). General properties of all electromagnetic radiation include: The energy of a photon is expressed by Planck's law in terms of the frequency (\( u\)) of the photon \[E=h u \label{24.5.1} \] since \(\lambda u =c\) for all light Plancks law can be also expressed in terms of the wavelength of the photon \[E = h u = \dfrac{hc}{\lambda} \label{24.5.2} \] If white light is passed through a prism, it splits into all the colors of the rainbow (Figure \(\Page {2}\)). Visible light is simply a small part of an electromagnetic spectrum most of which we cannot see - gamma rays, X-rays, infra-red, radio waves and so on. Each of these has a particular wavelength, ranging from 10 meters for gamma rays to several hundred meters for radio waves. Visible light has wavelengths from about 400 to 750 nm (1 nanometer = 10 meters). If white light (ordinary sunlight, for example) passes through copper(II) sulfate solution, some wavelengths in the light are absorbed by the solution. Copper(II) ions in solution absorb light in the red region of the spectrum. The light which passes through the solution and out the other side will have all the colors in it except for the red. We see this mixture of wavelengths as pale blue (cyan). The diagram gives an impression of what happens if you pass white light through copper(II) sulfate solution. Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. You would not have thought that all the other colors apart from some red would look cyan, for example. Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors. The process of absorption involves the excitation of the valence electrons in the molecule typically from the low lying level called the Highest Occupied Molecular Orbital (HOMO) into a higher lying state called the the Lowest Unoccupied Molecular Orbital ( ). When this and LUMO transition (Figure \(\Page {3}\)) involves the absorption of visible light, the sample is colored. The HOMO-LUMO energy difference \[\Delta E = E_{HOMO} - E_{LUMO} \label{24.5.3A} \] depends on the nature of the molecule and can be connected to the wavelength of the light absorbed \[\Delta E = h u = \dfrac{hc}{\lambda} \label{24.5.3B} \] Equation \(\ref{24.5.3B}\) is the most important equation in the field of light-matter interactions (spectroscopy). As Example \(\Page {1}\) demonstrated, when white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other (Figure \(\Page {5}\)). Thus, absorption of 420-430 nm light renders a substance yellow, and absorption of 500-520 nm light makes it red. Green is unique in that it can be created by absorption close to 400 nm as well as absorption near 800 nm. Colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light. What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. Copper(II) sulfate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum and cyan is the complementary color of red (Table \(\Page {1}\)). We often casually talk about the transition metals as being those in the middle of the Periodic Table where d orbitals are being filled, but these should really be called rather than transition elements (or metals). The definition of a transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. Zinc with the electronic structure [Ar] 3d 4s does not count as a transition metal whichever definition you use. In the metal, it has a full 3d level. When it forms an ion, the 4s electrons are lost - again leaving a completely full 3d level. At the other end of the row, scandium ([Ar] 3d 4s ) does not really counts as a transition metal either. Although there is a partially filled d level in the metal, when it forms its ion, it loses all three outer electrons. The Sc ion does not count as a transition metal ion because its 3d level is empty. The diagrams show the approximate colors of some typical hexaaqua metal ions, with the formula [ M(H O) ] . The charge on these ions is typically 2+ or 3+.   The corresponding transition metal ions are colored. Some, like the hexaaquamanganese(II) ion (not shown) and the hexaaquairon(II) ion, are quite faintly colored - but they are colored. So, what causes transition metal ions to absorb wavelengths from visible light (causing color) whereas non-transition metal ions do not? And why does the color vary so much from ion to ion? This is discussed in the next sections. The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Only the first two are introduced below. refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, \(O_2\) is a good example of (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet: As shown in the video, molecular oxygen (\(O_2\) is paramagnetic and is attracted to the magnet. , Molecular nitrogen, \(N_2\), however, has no unpaired electrons and it is diamagnetic (this concept is discussed below); it is therefore unaffected by the magnet. There are some exceptions to the rule; these concern some transition metals, in which the unpaired electron is not in a d-orbital. Examples of these metals include \(Sc^{3+}\), \(Ti^{4+}\), \(Zn^{2+}\), and \(Cu^+\). These metals are the not defined as paramagnetic: they are considered diamagnetic because all d-electrons are paired. Paramagnetic compounds sometimes display bulk magnetic properties due to the clustering of the metal atoms. This phenomenon is known as ferromagnetism, but this property is not discussed here. Diamagnetic substances are characterized by paired electrons—except in the previously-discussed case of transition metals, there are no unpaired electrons. According to the which states that no two identical electrons may take up the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly by a magnetic field. In fact, diamagnetic substances are weakly by a magnetic field as demonstrated with the carbon sheet in Figure \(\Page {6}\). Figure \(\Page {6}\): Levitating pyrolytic carbon: A small (~ ) piece of graphite levitating over a permanent neodymium magnet array ( cubes on a piece of steel). Note that the poles of the magnets are aligned vertically and alternate (two with north facing up, and two with south facing up, diagonally). from . The magnetic form of a substance can be determined by examining its electron configuration: if it shows unpaired electrons, then the substance is paramagnetic; if all electrons are paired, the substance is diamagnetic. This process can be broken into four steps: Are chlorine atoms paramagnetic or diamagnetic? For Cl atoms, the electron configuration is 3s 3p Ignore the core electrons and focus on the valence electrons only. There is one unpaired electron. Since there is an unpaired electron, Cl atoms are paramagnetic (but is quite weak). Example 2: Zinc Atoms For Zn atoms, the electron configuration is 4s 3d There are no unpaired electrons. Because there are no unpaired electrons, Zn atoms are diamagnetic. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/Z056_Chemistry_of_Barium_(Z56)/Barium_Sulfate

Barium sulfate, \(BaSO_4\), is a white crystalline solid that is insoluble in water (solubility product \(K_{sp} = 1.1 \times 10^{-10}\)). The main commercial source of barium sulfate is the mineral baryte, which is normally highly impure. Baryte is reduced by heating with coke to barium sulfide, \(BaS\): \[BaSO_4 + 4 C \rightarrow BaS + 4 CO\] Barium sulfide is soluble in water can be easily converted to oxide, carbonate, and the halides. In order to produce highly pure barium sulfate, barium sulfide or barium chloride is treated with sulfuric acid: \[BaS + H_2SO_4 \rightarrow BaSO_4 + H_2S\] \[BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2 HCl\] Barium sulfate belongs to the most insoluble sulfates. Its low solubility results in many important applications, such as contrast agent in x-ray imaging, detection of \(Ba^{2+}\) ions, or the detoxification of barium salt solutions. Most of the synthetic barium sulfate is used as a white pigment (together with \(TiO_2\) or \(ZnS\)) for paints. One particular inorganic pigment called lithopone is a combination of barium sulfate and zinc sulfide.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Lennard-Jones_Potential

Proposed by Sir John Edward Lennard-Jones, the Lennard-Jones potential describes the potential energy of interaction between two non-bonding atoms or molecules based on their distance of separation. The potential equation accounts for the difference between attractive forces ( , dipole-induced dipole, and ) and repulsive forces. Imagine two rubber balls separated by a large distance. Both objects are far enough apart that they are not interacting. The two balls can be brought closer together with minimal energy, allowing interaction. The balls can continuously be brought closer together until they are touching. At this point, it becomes difficult to further decrease the distance between the two balls. In order to bring the balls any closer together, increasing amounts of energy must be added. This is because eventually, as the balls begin to invade each other’s space, they repel each other; the force of repulsion is far greater than the force of attraction. This scenario is similar to that which takes place in neutral atoms and molecules and is often described by the . The Lennard-Jones model consists of two 'parts'; a steep repulsive term, and smoother attractive term, representing the London dispersion forces. Apart from being an important model in itself, the Lennard-Jones potential frequently forms one of 'building blocks' of many force fields. It is worth mentioning that the 12-6 Lennard-Jones model is not the most faithful representation of the potential energy surface, but rather its use is widespread due to its computational expediency. The Lennard-Jones Potential is given by the following equation: \[ V(r)= 4 \epsilon \left [ {\left (\dfrac{\sigma}{r} \right )}^{12}-{\left (\dfrac{\sigma}{r} \right )}^{6} \right] \label{1}\] or is sometimes expressed as: \[ V(r) = \frac{A}{r^{12}}- \dfrac{B}{r^6} \label{2}\] where The \(\epsilon\) and \(\sigma\) values for Xenon (Xe) are found to be 1.77 kJ/mol and 4.10 Angstroms, respectively. Determine the van der Waals radius for the Xenon atom. Recall that the van der Waals radius is equal to one-half of the internuclear distance between nonbonding particles. Because \(\sigma\) gives a measure of how close two non-bonding particles can be, the van der Waals radius for Xenon (Xe) is given by: r = \(\sigma\)/2 = 4.10 Angstroms/2 = 2.05 Angstroms The Lennard-Jones potential is a function of the distance between the centers of two particles. When two non-bonding particles are an infinite distance apart, the possibility of them coming together and interacting is minimal. For simplicity's sake, their bonding potential energy is considered zero. However, as the distance of separation decreases, the probability of interaction increases. The particles come closer together until they reach a region of separation where the two particles become bound; their bonding potential energy decreases from zero to a negative quantity. While the particles are bound, the distance between their centers continue to decrease until the particles reach an equilibrium, specified by the separation distance at which the minimum potential energy is reached. If the two bound particles are further pressed together, past their equilibrium distance, repulsion begins to occur: the particles are so close together that their electrons are forced to occupy each other’s orbitals. Repulsion occurs as each particle attempts to retain the space in their respective orbitals. Despite the repulsive force between both particles, their bonding potential energy increases rapidly as the distance of separation decreases. Calculate the intermolecular potential between two Argon (Ar) atoms separated by a distance of 4.0 Angstroms (use \(\epsilon\)=0.997 kJ/mol and \(\sigma\)=3.40 Angstroms). To solve for the intermolecular potential between the two Argon atoms, we use equation 2.1 where V is the intermolecular potential between two non-bonding particles. \[ V= 4 \epsilon \left [ {\left (\dfrac{\sigma}{r} \right )}^{12}-{\left (\dfrac{\sigma}{r} \right )}^{6} \right] \nonumber\] The data given are \(\epsilon\) , and the distance of separation, We plug these values into equation 2.1 and solve as follows: \[V = 4(0.997\;\text{kJ/mol}) \left[\left(\dfrac{3.40\;\text{Angstroms}}{4.0\;\text{Angstroms}}\right)^{12}-\left(\dfrac{3.40 \;\text{Angstroms}}{4.0 \;\text{Angstroms}}\right)^6\right] \nonumber\] V = 3.988(0.14-0.38) V = 3.988(-0.24) V = -0.96 kJ/mol Like the bonding potential energy, the stability of an arrangement of atoms is a function of the Lennard-Jones separation distance. As the separation distance decreases below equilibrium, the potential energy becomes increasingly positive (indicating a repulsive force). Such a large potential energy is energetically unfavorable, as it indicates an overlapping of atomic orbitals. However, at long separation distances, the potential energy is negative and approaches zero as the separation distance increases to infinity (indicating an attractive force). This indicates that at long-range distances, the pair of atoms or molecules experiences a small stabilizing force. Lastly, as the separation between the two particles reaches a distance slightly greater than σ, the potential energy reaches a minimum value (indicating zero force). At this point, the pair of particles is most stable and will remain in that orientation until an external force is exerted upon it. Two molecules, separated by a distance of 3.0 angstroms, are found to have a \(\sigma\) value of 4.10 angstroms. By decreasing the separation distance between both molecules to 2.0 angstroms, the intermolecular potential between the molecules becomes more negative. Do these molecules follow the Lennard-Jones potential? Why or why not? Recall that \(\sigma\) is the distance at which the bonding potential between two particles is zero. On a graph of the Lennard-Jones potential, then, this value gives the x-intersection of the graph. According to the Lennard-Jones potential, any value of \(r\) greater than \(\sigma\) should yield a negative bonding potential and any value of r smaller than \(\sigma\) should yield a positive bonding potential. In this scenario, as the separation between the two molecules decreases from 3.0 angstroms to 2.0 angstroms, the bonding potential is becomes more negative. In essence however, because the starting separation (3.0 angstroms) is already less than \(\sigma\) (4.0 angstroms), decreasing the separation even further (2.0 angstroms) should result in a more positive bonding potential. Therefore, these molecules do not follow the Lennard-Jones potential. From section 2.1, , dipole-induced dipole, and interactions are all attractive forces. See Figure C. A species will have a repulsive force acting on it when r is less than the equilibrium distance between the particles. A species will have an attractive force acting on it when r is greater than the equilibrium distance between the particles. Lastly, when \(r\) is equal to t he equilibrium distance between both particles, the species will have no force acting upon it.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.8%3A_Distribution_of_a_Single_Species_between_Immiscible_Phases_-_Extraction_and_Separation

A partitioning of a compound exist between a mixture of two immiscible phases at equilibrium, which is a measure of the difference in solubility of the compound in these two phases. If one of the solvents is a gas and the other a liquid, the "gas/liquid partition coefficient" is the same as the dimensionless form of the constant. A solute can partition when one or both solvents is a solid (e.g., solid solution). We call the process of moving a species from one phase to another phase an . Simple extractions are particularly useful for separations where only one component has a favorable partition coefficient. Several important separation techniques are based on a simple extraction, including liquid–liquid, liquid–solid, solid–liquid, and gas–solid extractions. The most important group of separation techniques uses a selective partitioning of the solute between two immiscible phases. If we bring a phase containing a solute, S, into contact with a second phase, the solute partitions itself between the two phases, as shown by the following equilibrium reaction. \[\textrm S_\textrm{phase 1}\rightleftharpoons \textrm S_\textrm{phase 2}\label{7.20}\] The equilibrium constant for this equilbrium is \[K_\textrm D=\mathrm{\dfrac{[S_{phase\;2}]}{[S_{phase\;1}]}}\] is called the distribution constant or . If \(K_D\) is sufficiently large, then the solute moves from phase 1 to phase 2. The solute remains in phase 1 if the partition coefficient is sufficiently small. When we bring a phase containing two solutes into contact with a second phase, if is favorable for only one of the solutes a separation of the solutes is possible. The physical states of the phases are identified when describing the separation process, with the phase containing the sample listed first. For example, if the sample is in a liquid phase and the second phase is a solid, then the separation involves liquid–solid partitioning. It often happens that two immiscible liquid phases are in contact, one of which contains a solute. How will the solute tend to distribute itself between the two phases? One’s first thought might be that some of the solute will migrate from one phase into the other until it is distributed equally between the two phases, since this would correspond to the maximum dispersion (randomness) of the solute. This, however, does not take into the account the differing solubilities the solute might have in the two liquids; if such a difference does exist, the solute will preferentially migrate into the phase in which it is more soluble (Figure \(\Page {1}\) ). Figure \(\Page {1}\): Scheme for a simple liquid–liquid extraction in which the solute’s partitioning depends only on the K equilibrium. The transport of substances between different phases is of immense importance in such diverse fields as pharmacology and environmental science. For example, if a drug is to pass from the aqueous phase with the stomach into the bloodstream, it must pass through the lipid (oil-like) phase of the epithelial cells that line the digestive tract. Similarly, a pollutant such as a pesticide residue that is more soluble in oil than in water will be preferentially taken up and retained by marine organism, especially fish, whose bodies contain more oil-like substances; this is basically the mechanism whereby such residues as DDT can undergo as they become more concentrated at higher levels within the food chain. For this reason, environmental regulations now require that oil-water distribution ratios be established for any new chemical likely to find its way into natural waters. The standard “oil” phase that is almost universally used is octanol, C H OH. In preparative chemistry it is frequently necessary to recover a desired product present in a reaction mixture by extracting it into another liquid in which it is more soluble than the unwanted substances. On the laboratory scale this operation is carried out in a as shown below. If the distribution ratio is too low to achieve efficient separation in a single step, it can be repeated; there are automated devices that can carry out hundreds of successive extractions, each yielding a product of higher purity. In these applications our goal is to exploit the Le Chatelier principle by repeatedly upsetting the phase distribution equilibrium that would result if two phases were to remain in permanent contact. The distribution ratio for iodine between water and carbon disulfide is 650. Calculate the concentration of I remaining in the aqueous phase after 50.0 mL of 0.10M I in water is shaken with 10.0 mL of CS . The equilibrium constant is \[K_d = \dfrac{C_{CS_2}}{C_{H_2O}} = 650 \nonumber\] Let and represent the numbers of millimoles of solute in the water and CS layers, respectively. can then be written as ( /10 mL) ÷ ( /50 mL) = 650. The number of moles of solute is (50 mL) × (0.10 mmol mL ) = 5.00 mmol, and mass conservation requires that 5.00 mmol, so = (5.00 – ) mmol and we now have only the single unknown . The equilibrium constant then becomes \[ \dfrac{(5.00 – m_1)\, mmol / 10\, mL}{ m_1\, mmol / 50\, mL} = 650 \nonumber\] Simplifying and solving for \(m_1\) yields \[ \dfrac{(0.50 – 0.1)m_1}{0.02\, m_1} = 650 \nonumber\] with = 0.0382 mmol. The concentration of solute in the water layer is (0.0382 mmol) / (50 mL) = , showing that almost all of the iodine has moved into the CS layer. In an extraction, the sample is one phase and we extract the analyte or the interferent into a second phase. We also can separate the analyte and interferents by continuously passing one sample-free phase, called the mobile phase, over a second sample-free phase that remains fixed or stationary. The sample is injected into the mobile phase and the sample’s components partition themselves between the mobile phase and the stationary phase. Those components with larger partition coefficients are more likely to move into the stationary phase, taking a longer time to pass through the system. This is the basis of all chromatographic separations. Chromatography provides both a separation of analytes and interferents, and a means for performing a qualitative or quantitative analysis for the analyte. Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure \(\Page {3}\) provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced at the top of the column as a narrow band. Ideally, the solute’s initial concentration profile is rectangular (Figure \(\Page {1a}\)). As the sample moves down the column the solutes begin to separate (Figures \(\Page {1b,c}\)), and the individual solute bands begin to broaden and develop a Gaussian profile. If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure \(\Page {1d}\)). We can follow the progress of the separation either by collecting fractions as they elute from the column (Figure \(\Page {4}\)), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a (Figure \(\Page {5}\)), and consists of a peak for each solute. )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/The_Acidity_of_the_Hydrogen_Halides

This page discusses the acidity of the hydrogen halides: hydrogen fluoride, hydrogen chloride, hydrogen bromide and hydrogen iodide. It begins by describing their physical properties and synthesis and then explains what happens when they react with water to make acids such as hydrofluoric acid and hydrochloric acid. The hydrogen halides are colorless gases at room temperature, producing steamy fumes in moist air. The boiling points of these compounds are shown in the figure below: Hydrogen fluoride has an abnormally high boiling point for a molecule of its size(293 K or 20°C), and can condense under cool conditions. This is due to the fact that hydrogen fluoride can form hydrogen bonds. Because fluorine is the most electronegative of all the elements, the fluorine-hydrogen bond is highly polarized. The hydrogen atom carries a high partial positive charge (δ+); the fluorine is fairly negatively charged (δ-). In addition, each fluorine atom has 3 lone pairs of electrons. Fluorine's outer electrons are at the n=2 level, and the lone pairs represent small, highly charged regions of space. Hydrogen bonds form between the δ+ hydrogen on one HF molecule and a lone pair on the fluorine of another one.The figure below illustrates this association: The other hydrogen halides do not form hydrogen bonds because the larger halogens are not as electronegative as fluorine; therefore, the bonds are less polar. In addition, their lone pairs are at higher energy levels, so the halogen does not carry such an intensely concentrated negative charge; therefore, other hydrogen atoms are not attracted as strongly. There are several ways of synthesizing hydrogen halides; the method considered here is the reaction between an ionic halide, like sodium chloride, and an acid like concentrated phosphoric(V) acid, H PO , or concentrated sulfuric acid. When concentrated sulfuric acid is added to sodium chloride under cold conditions, the acid donates a proton to a chloride ion, forming hydrogen chloride. In the gas phase, it immediately escapes from the system. \[ Cl^- + H_2SO_4 \rightarrow HCl + HSO_4^-\] The full equation for the reaction is: \[ NaCl + H_2SO_4 \rightarrow HCl + NaHSO_4\] Sodium bisulfate is also formed in the reaction. Concentrated phosphoric(V) acid reacts similarly, according to the following equation: \[ Cl^- + H_3PO_4 \rightarrow HCl + H_2PO_4^-\] The full ionic equation showing the formation of the salt, sodium biphosphate(V), is given below: \[ NaCl + H_3PO_4 \rightarrow HCl + NaH_2PO_4\] All hydrogen halides can be formed by the same method, using concentrated phosphoric(V) acid. Concentrated sulfuric acid, however, behaves differently. Hydrogen fluoride can be made with sulfuric acid, but hydrogen bromide and hydrogen iodide cannot. The problem is that concentrated sulfuric acid is an oxidizing agent, and as well as producing hydrogen bromide or hydrogen iodide, some of the halide ions are oxidized to bromine or iodine. Phosphoric acid does not have this ability because it is not an oxidant. By the Bronsted-Lowry definition of an acid as a proton donor, hydrogen chloride is an acid because it transfers protons to other species. Consider its reaction with water. Hydrogen chloride gas is soluble in water; its solvated form is hydrochloric acid. Hydrogen chloride fumes in moist air are caused by hydrogen chloride reacting with water vapor in the air to produce a cloud of concentrated hydrochloric acid. A proton is donated from the hydrogen chloride to one of the lone pairs on a water molecule. A coordinate ( ) bond is formed between the oxygen and the transferred proton. The equation for the reaction is the following: \[ H_2O + HCl \rightarrow H_3O^+ + Cl^-\] The H O ion is the hydroxonium ion (also known as the hydronium ion or the oxonium ion). This is the normal form of protons in water; sometimes it is shortened to the proton form, H (aq), for brevity. When hydrogen chloride dissolves in water (to produce hydrochloric acid), almost all the hydrogen chloride molecules react in this way. Hydrochloric acid is therefore a strong acid. An acid is strong if it is fully ionized in solution. Hydrogen bromide and hydrogen iodide dissolve in (and react with) water in exactly the same way as hydrogen chloride does. Hydrogen bromide forms hydrobromic acid; hydrogen iodide gives hydriodic acid. Both of these are also strong acids. By contrast, although hydrogen fluoride dissolves freely in water, hydrofluoric acid is only a weak acid; it is similar in strength to organic acids like methanoic acid. The complicated reason for this is discussed below. Because the fluorine atom is so small, the bond enthalpy (bond energy) of the hydrogen-fluorine bond is very high. The chart below gives values for all the hydrogen-halogen bond enthalpies: In order for ions to form when the hydrogen fluoride reacts with water, the H-F bond must be broken. It would seem reasonable to say that the relative reluctance of hydrogen fluoride to react with water is due to the large amount of energy needed to break that bond, but this explanation does not hold. The energetics of this sequence are of interest: All of these terms are involved in the overall enthalpy change as you convert HX into its ions in water. However, the terms involving the hydrogen are the same for every hydrogen halide. Only the values for the red terms in the diagram need be considered. The values are tabulated below: There is virtually no difference in the total HF and HCl values. The large bond enthalpy of the H-F bond is offset by the large hydration enthalpy of the fluoride ion. There is a very strong attraction between the very small fluoride ion and the water molecules. This releases a lot of heat (the hydration enthalpy) when the fluoride ion becomes wrapped in water molecules. The energy terms considered previously have concerned HX molecules in the gas phase. To reach a more correct explanation, the molecules must first be considered as unreacted aqueous HX molecules. The equation for this is given below: The equation is incorporated into an improved energy cycle as follows: Unfortunately, values for the first step in the reaction are not readily available. However, in each case, the initial separation of the HX from water molecules is endothermic. Energy is required to break the intermolecular attractions between the HX molecules and water. That energy is much greater for hydrogen fluoride because it forms hydrogen bonds with water. The other hydrogen halides experience only the weaker van der Waals dispersion forces or dipole-dipole attractions. The overall enthalpy changes (including all the stages in the energy cycle) for the reactions are given in the table below: \[ HX(aq) + H_2O (l) \rightarrow H_3O^+ (aq) + X^- (aq)\] The enthalpy change for HF is much smaller in magnitude than that for the other three hydrogen halides, but it is still negative exothermic change. Therefore, more information is needed to explain why HF is a weak acid. The free energy change, not the enthalpy change, determines the extent and direction of a reaction. Free energy change is calculated from the enthalpy change, the temperature of the reaction and the entropy change during the reaction. For simplicity, entropy can be thought of as a measure of the amount of disorder in a system. Entropy is given the symbol . If a system becomes more disordered, then its entropy increases. If it becomes more ordered, its entropy decreases. The key equation is given below: In simple terms, for a reaction to happen, the free energy change must be negative. But more accurately, the free energy change can be used to calculate a value for the equilibrium constant for a reaction using the following expression: The term K is the equilibrium constant for the reaction below: \[ HX(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + X^-(aq)\] The values for TΔS (needed to calculate ΔG) for the four reactions at a temperature of 298 K are tabulated below: Notice that at the top of the group, the systems become more ordered when the HX reacts with the water. The entropy of the system (the amount of disorder) decreases, particularly for the hydrogen fluoride. The reason for this is that the very strong attraction between H O and F imposes a lot of order on the system, as does the attraction between the water molecules and the various ions present. These attractions are each greatest for the small fluoride ions. The total effect on the free energy change, and therefore the value of the equilibrium constant, can now be considered. These values are calculated in the following table: The values for these estimated equilibrium constants for HCl, HBr and HI are so high that the reaction can be considered "one-way". The ionization is virtually 100% complete. These are all strong acids, increasing in strength down the group. By contrast, the estimated K for hydrofluoric acid is small. Hydrofluoric acid only ionizes to a limited extent in water. Therefore, it is a weak acid. The estimated value for HF in the table can be compared to the experimental value: These values differ by an order of magnitude, but because of the logarithmic relationship between the free energy and the equilibrium constant, a very small change in ΔG has a very large effect on K . To have the values in close agreement, ΔG would must increase from +16 to +18.5 kJ mol . Given the uncertainty in the values used to calculate ΔG, the difference between the calculated value and the experimental value could easily fall within this range. The two main factors are: Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Connecting_Electronic_Configurations_to_the_Periodic_Table

Electron configuration can be described as how electrons are assembled within the orbitals shells and subshells of an atom. It is important to understand what an electron is in order to fully understand the electron configuration. An electron is a sub atomic particle that is associated with a negative charge. Electrons are found outside of the nucleus, as opposed to neutrons (particles with neutral charge,) and protons (particles with positive charge.) Furthermore, electrons are associated with energy, more specifically quantum energy, and exemplify wave-like and particle-like characteristics. The word configuration simply means the arrangement of something. Therefore electron configuration in straightforward language means the arrangement of electrons. In general when filling up the electron diagram, it is customary to fill the lowest energies first and work your way up to the higher energies. Principles and rules such as the Pauli exclusion principle, Hund’s rule, and the Aufbau process are used to determine how to properly configure electrons. The Pauli exclusion rule basically says that at most, 2 electrons are allowed to be in the same orbital. Hund’s rule explains that each orbital in the subshell must be occupied with one single electron first before two electrons can be in the same orbital. Lastly, the Aufbau process describes the process of adding electron configuration to each individualized element in the periodic table. Fully understanding the principles relating to electron configuration will promote a better understanding of how to design them and give us a better understanding of each element in the periodic table. How the periodic table was formed has an intimate correlation with electron configuration. After studying the relationship between electron configuration and the period table, it was pointed out by Niels Bohr that electron configurations are similar for elements within the same group in the periodic table. Groups occupy the vertical rows as opposed to a period which is the horizontal rows in the table of elements. Now consider elements. These elements also will also have similar electron configurations to each another because they are in the same group; these elements have 6 valence electrons. Answer: False. Elements in the same GROUP have similar electron configurations. 2. Question: What element has the electron configuration [Ar] 4s 3d 4p ? Answer: Bromine 3. Question: What element has the electron configuration [Xe] 4f 5d 6s 6p Answer: Bismuth 4. Question: Demonstrate how elements in a group share similar characteristics by filling in the electron configurations for the Group 18 elements: Answer: 1s , [He]2s 2p , [Ne]3s 3p , [Ar]3d 4s 4p , [Kr]4d 5s 5p , [Xe]4f 5d 6s 6p 5. Question: How many valence electrons are there in Iodine? Answer: Iodine, z=53, group 17. This means there are seven valence electrons. 6. Question: What is the highest number of electrons a 4p subshell can hold? Answer: 6! Each 3 p orbital can hold 2 electrons so if they are all filled, the answer is 6. You get this by multiplying the three orbitals by 2 electrons per orbital, so 3 multiplied by 2 equals 6. Make up some practice problems for the future readers.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Ion_-_Ion_Interactions

The interactions between ions (ion - ion interactions) are the easiest to understand: like charges repel each other and opposite charges attract. These Coulombic forces operate over relatively long distances in the gas phase. The force depends on the product of the charges (\(Z_1\), \(Z_2\)) divided by the square of the distance of separation (\(d^2\)): \[F \propto \dfrac{- Z_1Z_2}{d^2} \tag{1}\] Two oppositely-charged particles flying about in a vacuum will be attracted toward each other, and the force becomes stronger and stronger as they approach until eventually they will stick together and a considerable amount of energy will be required to separate them. They form an , a new particle which has a positively-charged area and a negatively-charged area. There are fairly strong interactions between these ion pairs and free ions, so that these the clusters tend to grow, and they will eventually fall out of the gas phase as a liquid or solid (depending on the temperature). , Professor of Chemistry, University of Missouri-Rolla

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_14%3A_The_Carbon_Family/Z014_Chemistry_of_Silicon_(Z14)/Silicon_and_Group_14_Elements

Group 14 elements play more important roles in our lives and our civilization than elements of any other group. Thus, every educated person should know something about them. Carbon exists as diamond, graphite, fullerenes, and charcoal. Their structures are interesting; so are their properties. You probably know a lot about diamond and graphite, but the fullerenes were discovered after 1970, and this discovery has opened a door for a lot of interesting research. Read about them in books, magazines and journals. You might find yourself working with them some day. Among the fullerenes, one of the most common "molecules" has 60 \(\ce C\) atoms, and is represented by \(\ce{C60}\); a diagram is shown here. If you connect the 60 carbon atoms with bonds, the structure looks like a cage with 5- and 6- member rings. Synthesis, bonding, symmetry and stability of the cagelike fullerenes have already attracted a lot of attention, and their properties are even more fascinating. Regarding carbon compounds, you already know something about \(\ce{CO2}\) and \(\ce{CO}\) including their roles in the environment. The hard carbides such as \(\ce{Fe3C}\), \(\ce{WC}\), and \(\ce{TiC}\) are more interesting to material scientists and engineers for their application in cutting tools. The calcium carbide \(\ce{CaC2}\) produced by reducing \(\ce{CaO}\) by carbon was a valuable commodity at one time due to its reaction with water to give acetylene gas: \(\ce{CaC_2 + 2 H_2O \rightarrow Ca(OH)_2 + C_2H_{2\large{(g)}}}\) Acetylene is an important industrial gas, for the manufacture of polymers. Do you know that: By weight, silicon is the most abundant element in the Earth's crust. It usually exists in the form of oxide, \(\ce{SiO2}\). This formula does not do justice to represent so many different materials we call silicates, but these substances are indeed \(\ce{SiO2}\). Some of the minerals contain impurities. In pure form, \(\ce{SiO2}\) is quartz. Small particles of quartz are sand. They are hard. In the structure at the atomic level, every silicon atom is bonded to 4 oxygen atoms, and every oxygen is bonded to two \(\ce{Si}\) atoms. The four \(\ce{Si-O}\) bonds point towards the corners of a tetrahedron, as do the \(\ce{C-C}\) bonds in the diamond structure. When impurity is present, the quartz may be colored. Due to various arrangements of the \(\ce{Si-O-Si}\) bonds, the same substance appears in many forms. A basic unit of silicate structures is \(\ce{SiO4^4-}\). The gemstone zircon has a formula \(\ce{ZrSiO4}\), and olivine has a chemical formula of \(\ce{(MgFe)2SiO4}\). Two \(\ce{SiO4^4-}\) units combine to give the pyrosilicate unit \(\ce{Si2O7^6-}\), and it appears in akermanite, \(\ce{Ca2MgSi2O7}\). When the number of units increase, the tetrahedral units combine to form rings, chains, layers and 3-dimensional networks. Thus, the structure and classification of silicate is a major part of . (This site has some interesting pictures.) Elemental silicon can be obtained from reduction of silicates. The reduction of sand, \(\ce{SiO2}\) by carbon at 3300 K in the reaction, \(\ce{SiO_2 + 2 C \rightarrow Si_{\large{(l)}} + 2 CO} \;\;\; \textrm{at 3300 K}\) gives liquid silicon. The silicon so obtained is usually not pure, and for the computer industry, the element must be purified. Crystal growth and silicon fabrication dominate the industry in the 1980s and 1990s, and perhaps into the next century, and the production of the element is only the beginning of the process. If a more reactive element, \(\ce{Mg}\), is used in the reduction, \(\ce{Mg2Si}\) is formed, \(\ce{SiO_2 + 4 Mg \rightarrow Mg_2Si + 2 MgO}\) \(\ce{Mg2Si}\) is a compound, and it reacts with water to form silane. Silane, \(\ce{SiH4}\), can be produced by reacting \(\ce{Mg2Si}\) with acids \(\ce{Mg_2Si + 4 H_2O \rightarrow 2 Mg(OH)_2 + SiH_4}\) and \(\ce{SiH4}\) is ignited when it contacts air, much more reactive than methane, \(\ce{SiH_4 + O_2 \rightarrow SiO_2 + H_2O}\) In a basic solution, \(\ce{SiH4}\) reacts with water to give \(\ce{SiO(OH)3-}\), \(\ce{SiH_4 + OH^- + 3H_2O \rightarrow SiO(OH)_3^- + 4 H_2}\) Silicon tetrafluoride is formed when glass (\(\ce{SiO2}\)) is exposed to \(\ce{HF}\), and when \(\ce{Si}\) reacts with \(\ce{F2}\), \(\ce{SiO2 + 4 HF_{\large{(aq)}} \rightarrow 2 H2O + SiF_{4\large{(g)}}}\) \(\ce{Si + 2 F2 \rightarrow SiF_{4\large{(g)}}}\) When chlorine passes through hot sand (\(\ce{SiO2}\)) and carbon, \(\ce{SiCl4}\) is produced, \(\ce{SiO2 + Cl + 2 C2 \rightarrow 2CO + SiCl_{4\large{(g)}}}\) \(\ce{SiCl4}\) and \(\ce{SiF4}\) react with water to give silicic acid, \(\ce{SiCl4 + 4H2O \rightarrow 4 HCl + Si(OH)_{4\large{(aq)}}}\), \(\ce{SiF4 + 4H2O \rightarrow 4 HF + Si(OH)_{4\large{(aq)}}}\). Silicones are polymers with general formula \(\mathrm{(R_2SiO_2)_{\large n}}\) or \(\mathrm{(RSiO_3)_{\large n}}\), (\(\ce{R}\) = \(\ce{CH3}\), \(\ce{C2H5}\), \(\ce{C6H5}\), etc). The chain is held together by \(\ce{Si-O-Si}\) bonds. A simple one is \(\mathrm{((CH_3)_2SiO_2)_{\large n}}\), Of course, the 4 bonds around the \(\ce{Si}\) atoms point to the corners of a tetrahedron. These siloxane polymers are widely used as sealants, adhesives, additives, flame retardants, and lubricants. They have a wide application in industries. Depending on the organic group attached to silicon, the inorganic polymer Silicones has been an important class of materials. diamond Diamond is the hardest thing in the world. Fullerenes are large molecules consisting of 40 to hundreds of carbon atoms, with \(\mathrm{C_{60}}\) being the most common. \(\ce{CaC2}\) The reaction to produce acetylene gas is \(\ce{CaC2 + H2O \rightarrow C2H2 + Ca(OH)2}\) Acetylene is still an important industrial gas, as raw material for polymers. \(\ce{C}\) Hint... Carbon or coke is used for silicon metal, because \(\ce{Mg2Si}\) is formed if \(\ce{Mg}\) is used. methane is stable Methane is the major component of natural gas, and it will not react with air until ignited, whereas silane ignites explosively as soon as it contacts air. silicones Silicon polymers are an important class of materials invented not too long ago.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Structural_Isomers%3A_Coordination_Isomerism_in_Transition_Metal_Complexes

Coordination isomerism occurs in compounds containing complex anionic and cationic parts and can be viewed as the interchange of one or more ligands between the cationic complex ion and the anionic complex ion. For example, \(\ce{[Co(NH3)6,Cr(CN)6]}\) is a coordination isomer with \(\ce{[Cr(NH3)6,Co(CN)6]}\). Alternatively, coordination isomers may be formed by switching the metals between the two complex ions like \(\ce{[Zn(NH3)4,CuCl4]}\) and \(\ce{[Cu(NH3)4,ZnCl4]}\). Are \(\ce{[Cu(NH3)4,PtCl4]}\) and \(\ce{[Pt(NH3)4,CuCl4]}\) coordination isomers? Here, both the cation and anion are complex ions. In the first isomer, \(\ce{NH3}\) is attached to the copper and the \(\ce{Cl^{-}}\) are attached to the platinum. In the second isomer, they have swapped. Yes, they are coordination isomers. What is one coordination isomer of \(\ce{[Co(NH3)6] [Cr(C2O4)3]}\)? Coordination isomers involve swapping the species from the inner coordination sphere to one metal (e.g, cation) to inner coordination sphere of a different metal (e.g., the anion) in the compound. One isomer is completely swapping the ligand sphere, e.g, \(\ce{[Co(C2O4)3] [Cr(NH3)6]}\). Alternative coordination isomers are \(\ce{ [Co(NH3)4(C2O4)] [Cr(NH3)2(C2O4)2]}\) and \(\ce{ [Co(NH3)2(C2O4)2] [Cr(NH3)4(C2O4)]}\).

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.03%3A_Rate_vs._Concentration_Proportionalities

To determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form A → products in which the rate is – [A]/ , we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A]. Use the tabulated experimental data to determine the order of the reaction \[2 N_2O_5 \rightarrow 4 NO_2 + O_2 \nonumber \] The ideal gas law can be used to convert the partial pressures of \(N_2O_5\) to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of \([N_2O_5]\) and \([N_2O_5]^2\). It is apparent that the rates are directly proportional to \([N_2O_5]^1\), indicating that this is a first-order reaction. When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found. The slope of this plot gives the value of the rate constant (Figure 1; right). = (5.2 × 10 ) [N2O5] mol L s A study of the gas-phase reduction of nitric oxide by hydrogen \[2 NO + 2 H_2 \rightarrow N_2 + 2 H_2O \nonumber \] yielded the following initial-rate data (all pressures in torr): Find the order of the reaction with respect to each component. In looking over this data, take note of the following: : Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide. : Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen. The rate law is thus \[ \text{rate} = k[NO]^2[H_2]. \nonumber \] It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is \[\ce{A + B + C → products} \nonumber \] and we need to find the order with respect to [B] in the rate law. If we set [B] to 0.020 M and let [A] = [C] = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively the one in which we are interested. )

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al)/09%3A_The_Electronic_States_of_the_Multielectron_Atoms/9.07%3A_The_Self-Consistent_Field_Approximation_(Hartree-Fock_Method)

In this section we consider a method for finding the best possible one-electron wavefunctions that was published by Hartree in 1948 and improved two years later by Fock. For the Schrödinger equation to be solvable, the variables must be separable. The variables are the coordinates of the electrons. In order to separate the variables in a way that retains information about electron-electron interactions, the Coulomb repulsion term, e.g. \(\dfrac {e^2}{4 \pi \epsilon _0 r_{12}}\) for helium, must be approximated so it depends only on the coordinates of one electron. Such an approximate Hamiltonian can account for the interaction of the electrons in an average way. The exact one-electron eigenfunctions of this approximate Hamiltonian then can be found by solving the Schrödinger equation. These functions are the best possible one-electron functions. The best possible one-electron wavefunctions, by definition, will give the lowest possible total energy for a multi-electron system when combined into a Slater determinant and used with the complete multielectron Hamiltonian to calculate the expectation value for the total energy of the system. These wavefunctions are called the Hartree-Fock wavefunctions and the calculated total energy is the Hartree-Fock energy of the system. Application of the variational method to the problem of minimizing the total energy leads to the following set of Schrödinger-like equations called Hartree-Fock equations, \[ \hat {F} \varphi _i = \epsilon _i \varphi _i \label {9-51}\] where \(\hat {F}\) is called the Fock operator. The Fock operator is a one-electron operator and solving a Hartree-Fock equation gives the energy and Hartree-Fock orbital for one electron. For a system with 2N electrons, the variable i will range from 1 to N; i.e there will be one equation for each orbital. The reason for this is that only the spatial wavefunctions are used in Equation \(\ref{9-51}\). Since the spatial portion of an orbital can be used to describe two electrons, each of the energies and wavefunctions found by solving \(\ref{9-51}\) will be used to describe two electrons. The nature of the Fock operator reveals how the Hartree-Fock (HF) or Self-Consistent Field (SCF) Method accounts for the electron-electron interaction in atoms and molecules while preserving the idea of atomic and molecular orbitals. The full antisymmetrized wavefunction written as a Slater determinant of spin-orbitals is necessary to derive the form of the Fock operator, which is \[\hat {F} = \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j ) \label {9-52}\] As shown by the expanded version on the far right, the first term in this equation, \(\hat {H}^0\), is the familiar hydrogen-like operator that accounts for the kinetic energy of an electron and the potential energy of this electron interacting with the nucleus. For electron 1 in helium, for example, \[\hat {H}^0 (1) = - \dfrac {\hbar ^2}{2m} \nabla ^2_1 - \dfrac {2e^2}{4 \pi \epsilon _0 r_1} \label {9-53}\] The second term in Equation \(\ref{9-52}\), \(\sum _{j=1}^N (2 \hat {J} _j - \hat {K} _j )\), accounts for the potential energy of one electron in an average field created by all the other electrons in the system. The Fock operator is couched in terms of the coordinates of the one electron whose perspective we are taking (which we’ll call electron 1 throughout the following discussion), and the average field created by all the other electrons in the system is built in terms of the coordinates of a generic “other electron” (which we’ll call electron 2) that is considered to occupy each orbital in turn during the summation over the N spatial orbitals. The operators \(\hat {j}\) and \(\hat {K}\) result from the electron-electron repulsion terms in the full Hamiltonian for a multi-electron system. These operators involve the one-electron orbitals as well as the electron-electron interaction energy, \(\dfrac {e^2}{4 \pi \epsilon _0 r_{12}}\), which in atomic units simplifies to 1/r12. Atomic units are used in the rest of this discussion to simplify the notation by removing fundamental constants. The operators \(\hat {J}\) and \(\hat {K}\) are most conveniently defined by examining how they operate on a wavefunction, \(\varphi _i\), which describes electron 1. \[ \hat {J}_j (1) \varphi _i (1) = \left [ \int \varphi ^*_i (2) \dfrac {1}{r_{12}} \varphi _i (2) d \tau _2 \right ] \varphi _i (1) \label {9-54}\] \[ \hat {K}_j (1) \varphi _i (1) = \left [ \int \varphi ^*_j (2) \dfrac {1}{r_{12}} \varphi _i (2) d \tau _2 \right ] \varphi _j (1) \label {9-55}\] \(\hat {J}\) is called a Coulomb operator. As mentioned above, the specific coordinates 1 and 2 are used here to underline the fact that \(\hat {J}\) operates on a function of one electron in an orbital (here, electron 1 in \(\varphi _i\) using the results of an expectation value integral over the coordinates of a different electron (electron 2 in \(\varphi _j\). The second electron can be described by the same spatial orbital (if i = j) or by a different spatial orbital (if i ≠ j). \(\hat {J}\) takes the complex conjugate of an orbital describing electron 2, \(\varphi ^*_j (2) \varphi _j (2) \), multiplies by 1/r12, and integrates over the coordinates of electron 2. The quantity \(d \tau _2 \varphi ^*_j (2) \varphi _j (2) \) represents the charge distribution in space due to electron 2 in orbital j. The quantity \(d \tau _2 \varphi ^*_j (2) \varphi _j (2) \dfrac {1}{r_{12}}\) thus represents the potential energy at \(r_1\) due to the charge density at \(r_2\) where \(r_{12}\) is the distance between \(r_1\) and \(r_2\). Evaluation of the integral gives the total potential energy at r1 due to the overall, or average, charge density produced by electron 2 in orbital j. Since the part of the Fock operator containing \(\hat {J}\) involves a sum over all the orbitals, and a multiplicative factor of 2 to account for the presence of two electrons in each orbital, solution of the Hartree-Fock equation produces a spatial orbital \(\varphi _i\) that is determined by the average potential energy or Coulomb field of all the other electrons. The other operator under the summation in the Fock operator is \(\hat {K}\), the exchange operator. Equation \(\ref{9-55}\) reveals that this operator involves a change in the labels on the orbitals. In analogy with the Coulomb operator, \(d \tau _2 \varphi ^*_j (2) \varphi _i (2) \dfrac {1}{r_{12}}\) represents the potential energy at \(r_1\) due to the overlap charge distribution at \(r_2\) associated with orbitals i and j. The integral is the potential energy due to the total overlap charge density associated with electron 2. The term exchange operator is used because the electron is exchanged between the two orbitals i and j. This overlap contribution to the charge density and potential energy is a quantum mechanical effect. It is a consequence of the wave-like properties of electrons. Wave-like properties means the electrons are described by wavefunctions. While difficult to understand in a concrete physical way, the effects of the exchange operator are important contributors to the total energy of the orbitals and the system as a whole. There is no classical analog to this interaction energy, and a classical theory is unable to calculate correctly the energies of multi-electron systems. For the ground state of helium, electrons 1 and 2 are both described by spatial orbital \(\varphi _i\), so N = 1 and the sum in Equation \(\ref{9-52}\) includes only j = 1. Furthermore, since i = j = 1, the exchange and Coulomb integrals are identical in this case. As a result, the summation in the Fock operator takes a very simple form and the complete Fock operator for electron 1 in helium is given by \[ \hat {F} (1) = -\dfrac {\hbar ^2}{2m} \nabla ^2_1 - \dfrac {2e^2}{4 \pi \epsilon _0 r_1} + \hat {U} (1) \label {9-56}\] where \(\hat {U} (1) \) is given by the summation in the Fock operator. \[ \hat {U} (1) = \sum ^1_{j=1} (2 \hat {j} _j - \hat {K} _j ) = \int \varphi _1 (2) \dfrac {1}{r_{12}} \varphi _1 (2) d \tau _2 \label {9-57}\] Show that \(\sum ^1_{j=1} (2 \hat {j} _j - \hat {K} _j ) = \int \varphi _1 (2) \dfrac {1}{r_{12}} \varphi _1 (2) d \tau _2\) by substituting the definitions of \(\hat {J} (1)\) and \(\hat {K} (1) \) for helium into the summation and evaluating the summation over the only occupied spatial orbital. The interaction of electron 1 with electron 2 is averaged over all positions of electron 2 to produce \(\hat {U} (1) \). By integrating over the coordinates of electron 2, the explicit dependence of the potential energy on the coordinates of electron 2 is removed. This approach makes it possible to account for the electron-electron repulsion in terms of the spatial distribution of the two electrons using only single-electron terms in the Fock operator and one-electron wavefunctions. Since both electrons in helium are described by the same spatial wavefunction, the Fock equation given by \(\ref{9-58}\) and \(\ref{9-56}\) describes either electron equally well. Solving the Fock equation therefore will give us the spatial wavefunction and the one-electron energy associated with either of the electrons. The energy of an electron in the spatial orbital \(\varphi _1\) can be calculated either by solving the Fock equation \[ \hat {F} (1) \varphi _1 (1) = \epsilon _1 (1) \varphi _1 (1) \label {9-58}\] or by using an expectation value expression. \[ \epsilon _1(1) = \int \varphi ^*_1 (r_1) \hat {F} (r_1) \varphi _1 (r_1) d \tau \label {9-59}\] In order to solve either of these equations for the energy \(\epsilon _1\), we need to evaluate the potential energy function \(\hat {U} (1) \) that is part of the Fock operator \(\hat {F} (1) \). In order to evaluate \(\hat {U} (1) \), the forms of all of the occupied spatial orbitals, \(\varphi _i (2) \) must be known. For the simple case of helium, only the \(\varphi _1 (2) \) function is required, but for larger multi-electron systems, the forms of occupied orbitals \(\epsilon _1, \epsilon _2\) etc. will be needed to specify \(\hat {U} (1) \). For helium, we know that \(\varphi _1 (2)\) will have the same form as \(\varphi _1 (1)\), and the \(\varphi _1\) functions can be obtained by solving the Fock equation (9-58). However, we are now caught in a circle because the Fock operator depends upon the \(\varphi _1\) function. The problem with solving Equation \(\ref{9-58}\) to obtain the Fock orbitals is that the Fock operator, as we have seen, depends on the Fock orbitals. In other words, we need to know the solution to this equation in order to solve the equation. We appear to be between a rock and a hard place. A procedure has been invented to wiggle out of this situation. One makes a guess at the orbitals, e.g. one inserts some adjustable parameters into hydrogenic wavefunctions, for example. These orbitals are used to construct the Fock operator that is used to solve for new orbitals. The new orbitals then are used to construct a new Fock operator, and the process is repeated until no significant change in the orbital energies or functions occurs. At this end point, the orbitals produced by the Fock operator are the same as the orbitals that are used in the Fock operator to describe the average Coulomb and overlap (or exchange) potentials due to the electron-electron interactions. The solution therefore is self-consistent, and the method therefore is called the self-consistent field (SCF) method. The objective of the Hartree-Fock method is to produce the best possible one-electron wavefunctions for use in approximating the exact wavefunction for a multi-electron system, which can be an atom or a molecule. So what kind of guess functions should we write to get the best possible one-electron wavefunctions? Answers to this question have spawned a huge area of research in computation chemistry over the past 40 years, including a Nobel Prize in 1998. In Chapter 10 we examine in detail the various alternatives for constructing one-electron wavefunctions from basis functions. Write a paragraph without using any equations that describes the essential features of the Hartree-Fock method. Create a block diagram or flow chart that shows the steps involved in the Hartree-Fock method. The expectation value of the Fock operator gives us the energy of an electron in a particular orbital. \[ \epsilon _i (1) = \int d\tau _1 \varphi ^*_i (1) \hat {F} (1) \varphi _i (1) \label {9-60}\] Using the definition of the Fock operator and representing the integrals with gives \[ \epsilon _i (1) = \left \langle H^0_i \right \rangle + \sum ^N_{j=1} (2 J_{ij} - K_{ij}) \label {9-61}\] where \[\left \langle H^0_i \right \rangle = \left \langle \varphi _i | - \dfrac {1}{2} \nabla ^2_1 | \varphi _i \right \rangle - \left \langle \varphi _i | \dfrac {Z}{r_1} | \varphi _i \right \rangle \label {9-62}\] The kinetic and potential energy terms in the operator \(\hat {H} ^0\), defined in Equation \(\ref{9-53}\), are written here in atomic units for simplicity of notation. The sum involving the Coulomb and exchange integrals, \(J\) and \(K\), accounts for the electron-electron interaction energy between the electron in orbital i and all the other electrons in the system. We now want to examine the meaning and the nature of the sum over all the orbitals in Equation \(\ref{9-61}\). Describe the contributions to the orbital energy or single-electron energy \( epsilon\) in words as represented by Equation \(\ref{9-61}\). For the case j = k one has for \(2 J_{ik} - K_{ik}\) \[ 2 \left \langle \varphi _i (1) \varphi _k (2) | \dfrac {1}{r_{12}} |\varphi _i (1) \varphi _k (2) \right \rangle - \left \langle \varphi _i (1) \varphi _k (2) | \dfrac {1}{r_{12}} |\varphi _k (1) \varphi _i (2) \right \rangle \label {9-63}\] The first term with the factor of 2 is the average potential energy due to the charge distribution caused by an electron in orbital i with the charge distribution caused by the two electrons in orbital k. The factor of 2 accounts for the two electrons in orbital k. The second term is the average potential energy due to the overlap charge distribution caused by electrons 1 and 2 in orbitals i and k. The second term appears only once, i.e. without a factor of 2, because only one of the electrons in orbital k has the same spin as the electron in orbital i and can exchange with it. The minus sign results from the wavefunction being antisymmetric with respect to electron exchange. Rewrite Equation \(\ref{9-63}\) including the spin functions \(\alpha\) and \(\beta\) explicitly for each electron. Since these two spin functions form an orthonormal set and factor out of the spatial integrals, show that that the exchange integral is zero if the two electrons have different spin. For the case j = i one has \[ 2 \left \langle \varphi _i (1) \varphi _i (2) | \dfrac {1}{r_{12}} |\varphi _i (1) \varphi _i (2) \right \rangle - \left \langle \varphi _i (1) \varphi _i (2) | \dfrac {1}{r_{12}} |\varphi _i (1) \varphi _i (2) \right \rangle \label {9-64}\] revealing that \(J_{ii} = K_{ii}\), and \(2 J_{ii} - K_{ii} = J_{ii}\), which corresponds to the between the two electrons in orbital i. Because these electrons have opposite spin, there is no exchange energy. This was the case for our helium example, above. If the single-electron orbital energies are summed to get the total electronic energy, the Coulomb and exchange energies for each pair of electrons are counted twice: once for each member of the pair. These additional \(2J - K\) contributions to the single-electron energies must be subtracted from the sum of the single-electron energies to get the total electronic energy, as shown in Equation \(\ref{9-65}\). The factor of 2 accounts for the fact that two electrons occupy each spatial orbital, and is the energy of a single electron in a spatial orbital. \[E_{elec} = \sum ^N_{i=1} \left [ 2 \epsilon _i - \sum ^N_{j=1} (2J_{ii} - K_{ij}) \right ] \label {9-65}\] Use Equations \(\ref{9-61}\) and \(\ref{9-65}\) to show that the total electronic energy also can be expressed in the following \[E_{elec} = \sum ^N_{i=1} \left [ 2\left \langle H^0_i \right \rangle - \sum ^N_{j=1} (2J_{ii} - K_{ij}) \right ] \] \[E_{elec} = \sum ^N_{i=1} \left ( \epsilon _i + \left \langle H^0_i \right \rangle \right ) \] forms. Write out all the terms in Equation \(\ref{9-65}\) for the case of 4 electrons in 2 orbitals with different energies. As we increase the flexibility of wavefunctions by adding additional parameters to the guess orbitals used in Hartree-Fock calculations, we expect to get better and better energies. The variational principle says that any approximate energy calculated using the exact Hamiltonian is an upper bound to the exact energy of a system, so the lowest energy that we calculate using the Hartree-Fock method will be the most accurate. At some point, the improvements in the energy will be very slight. This limiting energy is the lowest that can be obtained with a single Slater determinant wavefunction. This limit is called the Hartree-Fock limit, the energy is the Hartree-Fock energy, the orbitals producing this limit are by definition the best single-electron orbitals that can be constructed and are called Hartree-Fock orbitals, and the Slater determinant is the Hartree-Fock wavefunction. ")

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Calculating_the_pH_of_the_Solution_of_a_Polyprotic_Base%2F%2FAcid

Thus far, we have been discussing problems and answers in equilibria--perhaps the most popular type of problem being how to find the of a weak solution given a certain concentration of a molecule. However, those problems in particular usually only involve what is called a monoprotic acid. “Mono” in the word “monoprotic” indicates that there is only one ionizeable hydrogen atom in an acid when immersed in water, whereas the concept of allows for two or more ionizeable hydrogen atoms. Consider the following chemical equation as the molecule acetic acid equilibrates in the solution: \[\ce{ CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}} \nonumber \] Although acetic acid carries a four hydrogen atoms, only a single becomes ionized. Not to get into too much detail between monoprotic and polyprotic acids, but if you desire to find the pH given a concentration of a weak acid (in this case, acetic acid), you would create and complete an adjusting for how much acetic acid disassociates. However, if you wish to find the pH of a solution after a disassociates, there are extra steps that would need to be done. Let's first take a look at a unique example: What is the pH of 0.75 M sulfuric acid? In sulfuric acid (H SO ), there are two ionizable hydrogen atoms. What makes this molecule interesting is that its ionization constant for the first hydrogen (\(K_{a1}\)) ionized is significantly larger than is the second ionization constant (\(K_{a2}\)). The \(K_{a1}\) constant for sulfuric acid is conveniently dubbed “very large” while the \(K_{a2}\) constant is 1.1 x 10 . As such, the sulfuric acid will completely disassociate into HSO and H O ions (as a strong acid). \[ \ce{H2SO4 (aq) + H2O -> HSO4^{-} (aq) + H_3O^{+}} \nonumber \] Since the sulfuric acid completely disassociates in the solution, we can skip the ICE table process for sulfuric acid, and assert that the concentration \(\ce{HSO4^{-}}\) and \(\ce{H3O^{+}}\) are the same as that of H SO , that is 0.75 M. (This neglects the background concentration of \(\ce{H_3O^{+}}\) in water of \(1 \times 10^{-7}M\)). Equation: \[ \ce{HSO4^{-} (aq) + H2O <=> SO4^{2-} (aq) + H_3O^{+}} \nonumber \] ICE Table: \(K_{a2} = \dfrac{[SO_4^{2-},H_3O^+]}{[HSO_4^-]} = 1.1 \times 10^{-2} = 0.011 = \dfrac{x(0.75+x)}{0.75 - x}\) Assume \(x\) in the denominator is negligible. Therefore, \[x = 0.011 M = [SO_4^{2-}] \nonumber \] Since we know the value of x, we can use the equation from the ICE table to find the value of [HSO ]. \[ [HSO_4^-] = 0.75 \; M - x = 0.75 - 0.011 = 0.74 \; M \nonumber \] We can also find [H O ] using the equation from the ICE table. \[[H_3O+] = 0.75 \; M + x = 0.75 + 0.011 = 0.76 \; M \nonumber \] We can then find the pH from the calculated [H O ] value. \[ pH = -log[H_3O^+] = -log0.76 = 0.119 \nonumber \] Let's say our task is to find the pH given a polyprotic which gains protons in water. Thankfully, the process is essentially the same as finding the pH of a polyprotic acid except in this case we deal with the concentration of OH instead of H O . Let's take a look at how to find the pH of C H O N , a diprotic base with a concentration of 0.00162 M, and a \(K_{b1}\) of 10 and a \(K_{b2}\) of 10 . Equation: \(C_{20}H_{24}O_2N \; (aq) + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H^+ + OH^- \) \[K_{b1} = \dfrac{[C_{20}H_{24}O_2N_2H^+,OH^-]}{[C_{20}H_{24}O_2N_2]} = 10^{-6} = 0.011 = \dfrac{(x)(x)}{0.00162 - x} \nonumber \] Again, assume x in the denominator is negligible. Therefore, \(0.011 \approx \dfrac{x^2}{0.00162}\) Then, \(x \approx 4 \times 10^{-5}\) We can then find the pH. \(pOH = -log(4 \times 10^{-5}) = 4.4\) \(pH = 14 - 4.4 = 9.6\) As we determine the pH of the solution, we realize that the OH gained using the second ionization constant is so insignificant that it does not impact the final pH value. For good measure, the following is the process to determine the pH in case the second use of the ICE table would indeed make a difference. Equation: \(C_{20}H_{24}O_2N_2H^+ + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H_2^{2+} + OH^-\) ICE Table: \(K_{b2} = \dfrac{[C_{20}H_{24}O_2N_2H_2^{2^+},OH^-]}{[C_{20}H_{24}O_2N_2H^+]} = 10^{-9.8}\) \[10^{-9.8} = \dfrac{(4 \times 10^{-5} + x)(x)}{(4 \times 10^{-5} - x)} \nonumber \] \[10^{-9.8} = \dfrac{0.00004 \; x + x^2}{0.00004 - x} \nonumber \] \[10^{-9.8}(0.00004 - x) = 0.00004 x + x^2 \nonumber \] \[x^2 + (4 \times 10^{-5})x - 6.3 \times 10^{-15} = 0 \nonumber \] \[ \begin{align*} x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[4pt] &= \dfrac{-4 \times 10^{-5} \pm \sqrt{(4 \times 10^{-5})^2 - 4(1)(6.3 \times 10^{-15})}}{2(1)} \\[4pt] & = 0 \end{align*} \]

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Interpreting_Infrared_Spectra/IR3._Subtle_Points_of_IR_Spectroscopy

show two sets of peaks in the IR spectrum. Alkanes contain two kinds of bonds: C-C bonds and C-H bonds. However, these two facts are not related. The reasons are explained through bond polarity and molecular vibrations. Bond polarity can play a role in IR spectroscopy. Hooke's Law states: Hooke's Law in IR spectroscopy means: The reasons explaining why C-H bending vibrations are at lower frequency than C-H stretching vibrations are also related to Hooke's Law. An H-C-H bending vibration involves three atoms, not just two, so the mass involved is greater than in a C-H stretch. That means lower frequency. Also, it turns out that the "stiffness" of a bond angle (analogous to the strength of a spring) is less than the "stiffness" of a bond length; the angle has a little more latitude to change than does the length. Both factors lead to a lower bending frequency.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Ionization_Energy/Decomposing_the_Standard_Reduction_Potential

Ionization energy is the amount of energy required to remove an electron from a given chemical species. Reduction Potential is a measurement of the amount of force required for a chemical species to gain electrons. The ionization energy is a single step process and follows a constant trend by decreasing down a period within a group. Standard reduction potential is more closely related to a multistep process known as solvation. Solvation upon how easily a solute can break its own bonds, how easily the solute can break its bonds, and how much easily the solute can attract the solvent toward it self in order to form and ionic compound. is depend and upon the polarity of the molecules within the solution. As we will see, some species will have lower ionization energies as well as having higher reduction potential when compared to species of the same period.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_14%3A_The_Carbon_Family/1Group_14%3A_General_Chemistry/Oxides_of_Group_4_Elements

This page briefly examines the oxides of carbon, silicon, germanium, tin and lead. It concentrates on the structural differences between carbon dioxide and silicon dioxide, and on the trends in acid-base behavior of the oxides down . The physical properties of carbon dioxide differ significantly from those of silicon dioxide (also known as silicon(IV) oxide or silica). Carbon dioxide is a gas whereas silicon dioxide is a hard, high-melting solid. The other dioxides in Group 4 are also solids, making the structure of carbon dioxide the anomaly. The fact that carbon dioxide is a gas indicates that it consists of small, simple molecules. Carbon can form these molecules because it can form double bonds with oxygen. None of the other elements in Group 4 form double bonds with oxygen, so their oxides adopt completely different structures. When carbon forms bonds with oxygen, it promotes one of its 2s electrons into the empty 2p level. This produces 4 unpaired electrons. These electrons are rearranged by hybridizing the 2s electron and one of the 2p electrons to make two sp hybrid orbitals of equal energy. The other 2p electrons are unaffected during this process. The figure below illustrates this: Notice that the two green lobes are two different hybrid orbitals, arranged as far from each other as possible (although the two hybrid orbitals have an arrangement similar to a p orbital, it is important not to confuse the two). Oxygen's electronic structure is 1s 2s 2p 2p 2p , and its orbitals must also hybridize. In this case, sp hybrids are formed from the s orbital and two of the p orbitals, rearranging to form 3 orbitals of equal energy, leaving a temporarily unaffected p orbital. As shown below, two of the sp hybrid orbitals contain lone pairs of electrons. In the figure below, the carbon and oxygen atoms are arranged in pre-bonding position: The green hybrid orbitals overlap end-to-end, forming covalent bonds. These are called sigma bonds, and are shown as orange in the next diagram. The sigma bonding brings the p orbitals close enough to overlap. This overlap between the two sets of p orbitals produces two \(pi\) bonds, similar to the \(pi\) bond found in ethene. These \(pi\) bonds are twisted at 90° to each other in the final molecule. To form a carbon-oxygen double bond, it is necessary for the lobes of the p orbitals on the carbon and the oxygen to overlap correctly. Silicon does not double bond with oxygen. Because silicon atoms are larger than carbon atoms, silicon-oxygen bonds are longer than carbon-oxygen bonds. Consider a hypothetical silicon-oxygen double bond, analogous to the carbon-oxygen double bond discussed above. Because silicon-oxygen bonds are longer than carbon-oxygen equivalents, the p orbitals on silicon and oxygen cannot overlap enough to form a stable pi bond. Therefore, only single bonds are formed. There are several structures for silicon dioxide. One of the simplest is shown below: This is similar to the structure of diamond, with each of the silicon atoms bridged to its four neighbors via an oxygen atom, forming a large network covalent structure. Strong bonds in three dimensions make silicon dioxide a hard, high melting point solid. The oxides of the elements at the top of Group 4 are acidic, but this acidity decreases down the group. Toward the bottom of the group the oxides are more basic, but do not lose their acidic character completely. A compound with both acidic and basic properties is called . The trend, therefore, ranges from oxides at the top of the group toward at the bottom. Carbon monoxide is usually treated as a neutral oxide, but it is slightly acidic. It does not react with water, but it can react with hot concentrated sodium hydroxide solution to give a solution of sodium methanoate. \[ NaOH + CO \rightarrow HCOONa\] The reaction of carbon monoxide with basic hydroxide ions displays its acidic character. These compounds are both weakly acidic. The chemical equation for CO is given below: \[ H_2O (l) + CO_2(aq) \rightleftharpoons H^+(aq) + HCO_3^- (aq)\] The solution of carbon dioxide in water is sometimes known as carbonic acid, but in fact only about 0.1% of the carbon dioxide actually reacts. The equilibrium lies well to the left. At low temperatures, carbon dioxide reacts with sodium hydroxide to form solutions of either sodium carbonate or sodium bicarbonate, depending on the proportions used. Equations for these reactions are given below: \[ 2NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O\] \[NaOH + CO_2 \rightarrow NaHCO_3\] At high temperatures and concentrations, silicon dioxide also reacts with sodium hydroxide, forming sodium silicate as shown: \[2NaOH + SiO_2 \rightarrow Na_2SiO_3 + H_2O\] Another familiar reaction occurs in the Blast Furnace ; in this process, calcium oxide (from limestone, one of the raw materials) reacts with silicon dioxide to produce a liquid slag of calcium silicate. This is another example of acidic silicon dioxide reacting with a base. \[ CaO(s) + SiO_2(s) \rightarrow CaSiO_3 (l)\] These elements form amphoteric oxides. These oxides react with acids to form salts. An example of this reaction, using hydrochloric acid, is given below: \[ XO(s) + 2HCl(aq) \rightarrow XCl_2(aq) + H_2O (l)\] where \ (X\) represents Ge or Sn; a modification is required for lead. When lead(II) oxide is mixed with hydrochloric acid, an insoluble layer of lead(II) chloride forms over the oxide; this stops the reaction from proceeding. A more accurate chemical equation from that given above is the following: \[ PbO(s) + 2HCl(aq) \rightarrow PbCl_2 (s) + H_2O\] However, if a high enough concentration of HCl is used, the large excess of chloride ions can react with lead(II) chloride to produce soluble complexes such as PbCl . These ionic complexes are soluble in water, so the reaction continues. The equation for this process is given below: \[ PbCl_2 (s) + 2Cl^-(aq) \rightarrow PbCl_4^{2-} (aq)\] These oxides also react with bases like sodium hydroxide, without exception, as follows: \[XO(s) + 2OH^-(aq) \rightarrow XO_2^{2-}(aq) + H_2O(l)\] Lead(II) oxide, for example, reacts to form PbO (plumbate(II)) ions. These dioxides are amphoteric, as mentioned before. These dioxides react first with concentrated hydrochloric acid to form compounds of the type XCl : \[ XO_2 + 4HCl \rightarrow XCl_4 + 2H_2O\] These XCl compounds react with excess chloride ions in the hydrochloric acid to form complexes such as XCl . \[ XCl_4 + 2Cl^- \rightarrow XCl_6^{2-}\] In the case of lead(IV) oxide, the reaction requires ice-cold hydrochloric acid. If the reaction is carried out at warmer temperatures, the lead(IV) chloride decomposes to give lead(II) chloride and chlorine gas. This is because the preferred oxidation state of lead is +2 rather than +4. The dioxides react with hot concentrated sodium hydroxide, forming soluble complexes of the form [X(OH) ] . \[XO_2(s) + 2OH^-(aq) + 2H_2O (l) \rightarrow [X(OH)_6]^{2-} (aq)\] Some sources suggest that lead(IV) oxide requires molten sodium hydroxide to react. In that case, the equation is modified as follows: \[ PbO_2 (s) + 2NaOH (l) \rightarrow Na_2PbO_3(s) + H_2O (g)\] Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/21%3A_Resonance_and_Molecular_Orbital_Methods/21.03%3A_Comparison_of_the_Resonance_and_Molecular-Orbital_Methods

In this section, we will sketch the similarities and differences in the resonance (or valence-bond, VB) and molecular-orbital (MO) approaches for electron-pair bonds. Both methods normally start with atomic orbitals \(1s\), \(2s\), \(2p\), and so on, of the types discussed in . Where the methods differ is in how these orbitals are used. For a bond between two atoms, the combines (or mixes) , one from each atom, with proper account of orbital phase ( ) to obtain , one of low energy and one of higher energy. The atomic orbitals can be pure or hybrid orbitals ( and ). In Figure 21-2, we show the results of combining the \(1s\) orbitals of hydrogen. The calculation for the most stable state proceeds by determining the energy of the system when two electrons are in the low-energy molecular orbital. The binding energy is the difference between the energy so calculated and the energies of the separated atoms. Because the molecular orbital extends over atoms, the bonding electrons must be associated with both atoms. Remember, the MO method first combines the atomic orbitals to give molecular orbitals, then populates the molecular orbitals with electrons (no more than two electrons per orbital). This part of the procedure is similar to the way electrons are allocated to atomic orbitals ( ). The starts with the same atomic orbitals but assigns one electron to each orbital. For an electron-pair bond between two hydrogen atoms, the VB treatment in its simplest form considers two . One of these has electron 1 in the orbital of hydrogen 1 and electron 2 in the orbital of hydrogen 2, \(\left( 1 \right)\). The other configuration, \(2\), has electron 2 in the orbital of hydrogen 1 and electron 1 in the orbital of hydrogen 2: The calculation then proceeds to predict a low-energy state and a high-energy state. These states can be regarded as of \(1\) and \(2\). The low-energy state, which is the one of more interest to us, usually is called a . In the VB method, of the electrons becomes associated with atoms through the mixing of the two configurations. A very important point here is that the calculation that mixes \(1\) and \(2\) leads to a six times greater binding energy than calculated for \(1\) and \(2\) alone. Thus in the VB treatment we combine (here \(1\) and \(2\), \(\leftrightarrow\) symbolizing mixing), whereas in the MO treatment we combine to get low- and high-energy molecular orbitals. The forces that hold atoms together through chemical bonds are electrostatic, that is, the attraction of positively charged nuclei for negatively charged electrons. But the energy calculated for a single configuration, such as \(1\), only accounts for about one sixth of the total binding. In either the VB or the MO method the electrons in an electron pair between two nuclei brought to within bonding distances are equivalent and indistinguishable. That is, we are unable to identify one electron any more than the other with a given atom. The significance of the pairing of the electrons is that is permits each electron to have maximum possible freedom to move through the orbitals of the two-atom system rather than being "localized" on particular atoms. Quantum-mechanical calculations tell us that freedom of motion of the electrons is very important. Thus, using the VB method, we calculate that fully five sixths of the binding of the hydrogen molecule is associated with the "delocalization" of the electrons between the two nuclei. There are many compounds with structures in which electrons are delocalized over . Such molecules should be more stable than would be expected for molecules with the same geometry but with electron pairs constrained to be associated with just one or two atoms. We will shortly discuss some specific examples, but because most of these examples involve the delocalization of \(\pi\) electrons, it is expedient to first discuss ethene as a prototype, using both the MO and VB methods. The atomic-orbital picture of ethene (Figure 6-14) formulates the \(\pi\) bond as resulting from overlap of two adjacent \(p\) atomic orbitals, one from each of two \(sp^2\) hybridized carbons. The \(p\) orbitals are directed perpendicularly to the plane defined by the hybrid orbitals of the \(\sigma\) bonds, and to a first approximation, we assume that exchange of the \(\pi\) and \(\sigma\) electrons between their respective orbitals does not affect the energy of the molecule. If this assumption is valid, \(\pi\) bonding can be treated independently of \(\sigma\) bonding. Although undoubtedly oversimplified, the VB and MO methods have been remarkably successful using this assumption. In our subsequent discussions, we shall treat the \(\pi\) electrons separately from localized \(\sigma\) electrons. The \(\pi\) bond of the ethene molecule can be formulated very much like the bond in the hydrogen molecule ( ), with the difference that the bonding is achieved by the overlap of two \(2p\) atomic orbitals of carbon rather than two \(1s\) atomic orbitals of hydrogen. In the MO method the mixing of the two \(2p\) atomic orbitals gives two molecular orbitals. The details of the mathematics of the mixing process to give an optimum set of molecular orbitals are well beyond the scope of this book,\(^1\) but the results are shown in Figure 21-3. The two \(\pi\) electrons of ethene are taken as occupying the low-energy bonding orbital, while the high-energy antibonding orbital normally is empty. How much more stable is the bonding molecular orbital relative to a pair of noninteracting \(p\) atomic orbitals? It is difficult to provide a numerical answer in \(\text{kcal mol}^{-1}\) that is meaningful, but we can describe the energy in symbolic terms. First, the energy of one electron in the \(p\) atomic orbital of an \(sp^2\)-hybridized carbon, as in \(3\), is taken as a standard quantity, \(\alpha\), often called the : Thus, if there were no \(\pi\) bonding in ethene and no repulsion between the electrons, the energy of the two electrons (one in each of the two adjacent \(p\) orbitals of the carbons) would be twice the Coulomb energy, or \(2 \alpha\). This would be the situation for two carbons such as \(3\) that are widely separated. The MO calculation shows that the bonding molecular orbital of ethene is more stable (of lower energy) than the nonbonding level, \(\alpha\), by a quantity, \(\beta\), where \(\beta\) is a energy term (Figure 21-3). Likewise, the antibonding level is destabilized by an amount \(-\beta\). For paired electrons in the bonding molecular orbital, the \(\pi\)-electron energy of ethene is calculated to be \(2 \left( \alpha + \beta \right) = 2 \alpha + 2 \beta\). In the valence-bond approach, the \(\pi\) bond of ethene is considered to be a hybrid of all reasonable electronic configurations of two indistinguishable paired electrons distributed between two \(p\) orbitals. Each of the configurations that can be written, \(4a\), \(4b\), \(4c\), and \(4d\), have identical locations of the atomic nuclei in space: The four valence-bond structures or configurations, \(4a\)-\(d\), are combined mathematically to give four hybrid states, and of these, the lowest-energy one corresponds approximately to the normal state of the molecule. The calculation shows that the structures \(4a\) and \(4b\), which have one electron in each \(p\) orbital, are the major contributors to the "hybrid" of ethene. The valence-bond structures, \(4c\) and \(4d\), are , which correspond to the conventional formulas, \(4e\) and \(4f\): These valence-bond structures are not important to the \(\pi\) bond of the ground state of ethene, although they are important for carbonyl bonds ( ). \(^1\)There are many excellent books that cover this subject in great detail; however, the simplest introductory work is J. D. Roberts; , W. A. Benjamin, Inc., Menlo Park, Calif., 1961. and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colligative_Properties/Anomalous_Colligative_Properties_(Real_Solutions)

Anomalous colligative properties are that deviate from the norm. Chemist Jacobus van 't Hoff was the first to describe anomalous colligative properties, but it was Svante Arrhenius who succeeded in explaining anomalous values of colligative properties. are the properties of solutions that rely only on the number (concentration) of the solute particles, and on the identity/type of solute particles, in an ideal solution (e.g., vapor pressure lowering, freezing point depression, boiling point elevation and osmotic pressure). There is a direct relationship between the concentration and the effect that is recorded. Therefore, the colligative properties are helpful when characterizing the nature of a solute after it is dissolved in a solvent. There are some solutes that produce a greater effect on colligative properties than what is expected. Arrhenius explained this by using the following equation: \[\Delta{T_f} = -K_f \times m = -1.86 \; ^{\circ}C \; m^{-1} \times 0.0100 \; m = -0.0186 \; ^{\circ}C \tag{1}\] The expected freezing point for this solution would be: - 0.0186 \(^{\circ}C\). Lets say that this solution was that of urea, the measured freezing point is close to -0.0186 \(^{\circ}C\). If it were to be a solution of NaCl, then the measured freezing point would then be -0.0361 \(^{\circ}C\). According to Van't Hoff the factor, \(i\) is the ratio of the measured value of a colligative property to that of the expected value if the solute is a nonelectrolyte. Now, for 0.0100 NaCl, it would be: \[i = \dfrac{\text{measured} \; \Delta{T_f}}{\text{expected} \; \Delta{T_f}} \tag{2}\] For the solute urea \(i\) = 1. For a strong electrolyte like NaCl that produces 2 moles of ions in a solution/ mole of solute dissolved, the effect on the freezing point depression would be expected to be twice as much as that for a nonelectrolyte. The expected \(i\) = 2. This leads the colligative properties to be rewritten as demonstrated in the table below. We can just substitute 1 for \(i\) for nonelectrolytes and for strong electrolytes, find the value of \(i\).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/12%3A_Thermodynamic_Processes_and_Thermochemistry/12.2%3A_The_First_Law_of_Thermodynamics_-_Internal_Energy_Work_and_Heat

The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. We can express this law mathematically as follows: \[ \Delta U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{12.4.1a} \] or \[\Delta{U_{sys}} =−ΔU_{surr} \label{12.4.1b} \] where the subscripts , , and refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings. The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy. An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the \(\ce{CO_2}/\ce{H_2O}\) product mixture is less than that of the isooctane \(\ce{O_2}\) reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (\(q\)) and the work done (\(w\)): \[\underbrace{ΔU_{sys} = q + w}_{\text{First Law of Thermodynamics}} \label{12.4.2}\] Although \(q\) and \(w\) are not state functions on their own, their sum (\(ΔU_{sys}\)) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings. Although \(q\) and \(w\) are not state functions, their sum (\(ΔU_{sys}\)) is independent of the path taken and therefore is a difference of a state function. A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (\(ΔU\)) of the gas in joules? : initial volume, final volume, external pressure, and quantity of energy transferred as heat Asked for: total change in internal energy A From Equation \(\ref{12.4.2}\), we know that \(ΔU = q + w\) ( ). We are given the magnitude of \(q\) (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, \(q\) is negative by convention. B Because the gas is being compressed, we know that work is being done on the system, so \(w\) must be positive. From Equation \(\ref{12.4.2}\), Thus \[\begin{align} ΔU &= q + w\nonumber \\[4pt] &= −140\; J + 284 \;J\nonumber \\[4pt] &= 144\; J\nonumber \end{align}\nonumber \] In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs. A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (\(ΔU\)) of the gas in joules? −216 J By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa. The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the \(PV\) work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for \(ΔU\).

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Kinetic_Molecular_Theory_of_Gases

To better understand the molecular origins of the , \[PV=nRT\] the basics of the ( ) should be understood. This model is used to describe the behavior of gases. More specifically, it is used to explain macroscopic properties of a gas, such as pressure and temperature, in terms of its microscopic components, such as atoms. Like the ideal gas law, this theory was developed in reference to ideal gases, although it can be applied reasonably well to real gases. In order to apply the kinetic model of gases, five assumptions are made: The last assumption can be written in equation form as: \[KE = \dfrac{1}{2}mv^2 = \dfrac{3}{2}k_BT\] where This equation says that the speed of gas particles is related to their absolute temperature. In other words, as their temperature increases, their speed increases, and finally their total energy increases as well. However, it is impossible to define the speed of any gas particle. As such, the speeds of gases are defined in terms of their root-mean-square speed. The macroscopic phenomena of pressure can be explained in terms of the kinetic molecular theory of gases. Assume the case in which a gas molecule (represented by a sphere) is in a box, length Figure 1). Through using the assumptions laid out above, and considering the sphere is only moving in the x-direction, we can examine the instance of the sphere colliding elastically with one of the walls of the box. The momentum of this collision is given by , in this case , since we are only considering the x dimension. The total momentum change for this collision is then given by \[mv_x - m(-v_x) = 2mv_x\] Given that the amount of time it takes between collisions of the molecule with the wall is we can give the frequency of collisions of the molecule against a given wall of the box per unit time as One can now solve for the change in momentum per unit of time: \[(2mv_x)(v_x/2L) = mv_x^2/L\] Solving for momentum per unit of time gives the force exerted by an object ( ). With the expression that = one can now solve for the pressure exerted by the molecular collision, where area is given as the area of one wall of the box, : \[P=\dfrac{F}{A}\] \[P=\dfrac{mv_x^2}{[L(L^2)}\] The expression can now be written in terms of the pressure associated with collisions from number of molecules: \[P=\dfrac{Nmv_x^2}{V}\] This expression can now be adjusted to account for movement in the x, y and z directions by using mean-square velocity for three dimensions and a large value of . The expression now is written as: \[P={\dfrac{Nm\overline{v}^2}{3V}}\] This expression now gives pressure, a macroscopic quality, in terms of atomic motion. The significance of the above relationship is that pressure is proportional to the mean-square velocity of molecules in a given container. Therefore, as molecular velocity increases so does the pressure exerted on the container.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/18%3A_Carboxylic_Acids_and_Their_Derivatives/18.10%3A_Reactions_of_Unsaturated_Carboxylic_Acids_and_Their_Derivatives

Unsaturated carboxylic acids of the type \(\ce{RCH=CH(CH_2)}_n \ce{COOH}\) usually exhibit the properties characteristic of isolated double bonds and isolated carboxyl groups when \(n\) is large and the functional groups are far apart. As expected, exceptional behavior is found most commonly when the groups are sufficiently close together to interact strongly, as in \(\alpha\),\(\beta\)-unsaturated acids, \(\ce{R} \overset{\beta}{\ce{C}} \ce{H=} \overset{\alpha}{\ce{C}} \ce{CO_2H}\). We shall emphasize those properties that are exceptional in the following discussion. In the presence of strong base, \(\alpha\),\(\beta\)- and \(\beta\),\(\gamma\)-unsaturated carboxylic acids tend to interconvert by migration of the double bond: Ester derivatives, \(\ce{RCH=CH-CH_2COOR'}\), and the corresponding unsaturated aldehydes and ketones, \(\ce{RCH=CH-CH_2COR'}\), are much more prone to this type of rearrangement than are the acids. Like alkenes, the double bonds of \(\alpha\),\(\beta\)-unsaturated acids can be brominated, hydroxylated, hydrated, and hydrobrominated, although the reactions often are relatively slow. In the addition of unsymmetrical reagents the direction of addition is to that observed for alkenes (anti-Markovnikov). Thus propenoic (acrylic) acid adds hydrogen bromide and water to form 3-bromo- and 3-hydroxypropanoic acids: These additions are analogous to the addition of halogens and halogen acids to 1,3-butadiene ( ). In the first step, a proton is transferred to the carbonyl oxygen. The resulting conjugate acid can be regarded as a resonance hybrid of structures \(16a\)-\(16d\): In the second step, a nucleophile (such as \(\ce{Br}^\ominus\) or a water molecule) attacks an electron-deficient carbon of the hybrid \(16\). Attack at the carboxyl carbon may occur but does not lead to a stable product. Attack of the nucleophile at the \(\beta\) carbon, however, produces the enol form of the \(\beta\)-substituted acid, which then is converted rapidly to the normal carboxylic acid: When the double bond of an unsaturated acid is farther down the carbon chain than between the alpha and beta positions, the so-called "conjugate addition" is not possible. Nonetheless, the double bond and carboxyl group frequently interact in the presence of acidic catalysts because the carbocation that results from addition of a proton to the double bond has a built-in nucleophile (the carboxyl group), which may attack the cationic center to form a cyclic ester called a . Lactone formation only occurs readily by this mechanism when a five- or six-membered ring can be formed: Five- or six-membered lactones also are formed by internal esterification when either \(\gamma\)- or \(\delta\)-hydroxy acids are heated. Under similar conditions, \(\beta\)-hydroxy acids are dehydrated to \(\alpha\),\(\beta\)-unsaturated acids, whereas \(\alpha\)-hydroxy acids undergo bimolecular esterification to substances with six-membered dilactone rings called : The foregoing examples of addition to the double bonds of unsaturated carboxylic acids all involve activation by an electrophilic species such as \(\ce{H}^\oplus\). Conjugate addition also may occur by nucleophilic attack on acid derivatives, the most important being the base-catalyzed Michael addition ( ) and 1,4-addition of organometallic compounds ( ). In all of these reactions a nucleophilic agent, usually a carbanion, attacks the double bond of an \(\alpha\),\(\beta\)-unsaturated compound in which the double bond is conjugated with, and activated by, a strongly electronegative unsaturated group (such as \(\ce{-CN}\), \(\ce{-NO_2}\), etc.). In the Michael addition, the carbanion usually is an enolate salt. The overall reaction is illustrated here by the specific example of the addition of diethyl propanedioate (diethyl malonate) to ethyl 3-phenylpropenoate (ethyl cinnamate): The mechanism of this kind of transformation, with diethyl propanedioate as the addend, is outlined in Equations 18-25 and 18-26. The basic catalyst required for the Michael addition (here symbolized as \(\ce{B:}\)) serves by forming the corresponding anion: A variety of nucleophilic agents can be used; propanedinitrile, 3-oxobutanoate esters, and cyanoethanoate esters all form relatively stable carbanions and function well in Michael addition reactions. Obviously, if the carbanion is stable, it will have little or no tendency to attack the double bond of the \(\alpha\),\(\beta\)-unsaturated acid derivative. Enamines ( and ) are excellent addends in many Michael-type reactions. An example is provided by the addition of \(\ce{N}\)-(1-cyclohexenyl)-azacyclopentane to methyl 2-methylpropanoate: and (1977)

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/Halide_Ions_as_Reducing_Agents

This page examines the redox reactions involving halide ions and concentrated sulfuric acid, using these reactions to discuss the trend in reducing ability of the ions from fluoride to iodide. Two types of reactions might occur when concentrated sulfuric acid is added to a solid ionic halide like sodium fluoride, chloride, bromide or iodide. The concentrated sulfuric acid can act as both an acid and an oxidizing agent. The concentrated sulfuric acid transfers a proton to the halide ion to produce a gaseous hydrogen halide, which immediately escapes from the system. If the hydrogen halide is exposed to moist air, steam fumes are formed. For example, concentrated sulfuric acid reacts with solid sodium chloride at low temperatures to produce hydrogen chloride and sodium bisulfate, as in the following equation: \[ NaCl + H_2SO_4 \rightarrow HCl + NaHSO_4\] All the halide ions behave similarly. Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize fluoride or chloride. In those cases, only the steamy fumes of the hydrogen halide—hydrogen fluoride or hydrogen chloride—are produced. In terms of the halide ions, fluoride and chloride are not strong enough reducing agents to reduce the sulfuric acid. This is not the case for bromides and iodides. Bromide is a strong enough reducing agent to reduce sulfuric acid. Bromide is oxidized to bromine in the process, as in the half-equation below: \[ 2Br^- \rightarrow Br_2 + 2e^-\] Bromide reduces sulfuric acid to sulfur dioxide gas, decreasing the oxidation state of sulfur from +6 to +4. The half-equation for this transition is as follows: \[ H_2SO_4 + 2H^+ + 2e^- \rightarrow SO_2 + 2H_2O\] These two half-equations can be combined into the overall ionic equation for the reaction: \[ H_2SO_4 + 2H^+ + 2Br^- \rightarrow Br_2 +SO_2 +2H_2O\] In practice, this reaction is confirmed by the steamy fumes of hydrogen bromide contaminated with the brown color of bromine vapor. The sulfur dioxide is a colorless gas, its presence cannot be directly observed. Iodide is a stronger reducing agent than bromide, and it is oxidized to iodine by the sulfuric acid: \[ 2I^- \rightarrow I_2 +2e^-\] The reduction of the sulfuric acid is more complicated than with bromide. Iodide is powerful enough to reduce it in three steps: The most abundant product is hydrogen sulfide. The half-equation for its formation is as follows: \[H_2SO_4 + 8H^+ + 8e^- \rightarrow H_2S + 4H2O\] Combining these two half-equations gives the following net ionic equation: \[ H_2SO_4 + 8H^+ + 8I^- \rightarrow 4I_2 + H_2S + 4H2O\] This is confirmed by a trace of steamy fumes of hydrogen iodide, and a large amount of iodine. The reaction is exothermic: purple iodine vapor is formed, with dark gray solid iodine condensing around the top of the reaction vessel. There is also a red color where the iodine comes into contact with solid iodide salts. The red color is due to the I ion formed by reaction between I molecules and I ions. Hydrogen sulfide gas can be detected by its "rotten egg" smell, but this gas is intensely poisonous. The following explanation is only (partially) accurate if fluoride is neglected works. When a halide ion acts as a reducing agent, it transfers electrons to something else. That means that the halide ion itself loses electrons. The larger the halide ion, the farther the outer electrons are from the nucleus, and the more they are shielded by inner electrons. It therefore gets easier for the halide ions to lose electrons down the group because there is less attraction between the outer electrons and the nucleus. This argument seems valid, but it is flawed. The energetics of the change must be examined. The amount of heat evolved or absorbed when a solid halide (like sodium chloride) is converted into an elemental halogen must be considered. Taking sodium chloride as an example, the following energetic quantities are important: Atomization energy is the energy needed to produce 1 mole of isolated gaseous atoms starting from an element in its standard state (gas for chlorine, and liquid for bromine, for example - both of them as X ). The figure below shows how this information fits together: The enthalpy change shown by the green arrow in the diagram for each of the halogens must be compared. The diagram shows that the overall change involving the halide ions is endothermic (the green arrow is pointing up toward a higher energy). This is not the total enthalpy change for the whole reaction. Heat is emitted when the changes involving the sulfuric acid occur. That is the same irrespective of the halogen in question. The total enthalpy change is the sum of the enthalpy changes for the halide ion half-reaction and the sulfuric acid half-reaction. The table below shows the energy changes that vary from halogen to halogen. The process is assumed to start from the solid sodium halide. The values for the lattice enthalpies for other solid halides would be different, but the pattern would be the same. The overall enthalpy change for the halide half-reaction: The sum of the enthalpy changes, in the final column, is decreasingly endothermic down the group. The total change in enthalpy (including the sulfuric acid) is also less positive. The amount of heat produced in the half-reaction involving the sulfuric acid must be great enough to make the reactions with the bromide or iodide feasible, but not enough to compensate for the more positive values produced by the fluoride and chloride half-reactions. In this section, the individual energy terms in the table that are most important in making the halogen half-reaction less endothermic down the group are determined. From chlorine to iodine, the lattice enthalpy changes most, decreasing by 87 kJ mol . By contrast, the energy required to remove the electron decreases by only 54 kJ mol . Both of these terms matter, but the decrease in lattice enthalpy is the more significant. This quantity decreases because the ions are getting larger. That means that they are farther away from each other, and so the attractions between positive and negative ions in the solid lattice are lessened. The simplified explanation mentioned earlier is misleading because it concentrates on the less-important decrease in the amount of energy needed to remove the electron from the ion. Fluoride ions are very difficult to oxidize to fluorine. The table above shows that this has nothing to do with the amount of energy required to remove an electron from a fluoride ion. It actually takes less energy to remove an electron from a fluoride ion than from a chloride ion. The generalization that an electron becomes easier to remove as the ion becomes larger does not apply here. Fluoride ions are so small that the electrons experience strong repulsion from each other. This outweighs the effect of their closeness to the nucleus and makes them easier to remove than the simplified argument predicts. There are two important reasons why fluoride ions are so difficult to oxidize. The first is the comparatively high lattice enthalpy of the solid fluoride. This is due to the small size of the fluoride ion, which means that the positive and negative ions are very close together and therefore strongly attracted to each other. The other factor is the small amount of heat that is released when the fluorine atoms combine to make fluorine molecules (see the table above). This is due to the low bond enthalpy of the F-F bond. The reason for this low bond enthalpy is discussed on a separate page. This discussion has focused on the energetics of the process starting from solid halide ions because that is the standard procedure when using concentrated sulfuric acid. Halides could also be oxidized in solution with another oxidizing agent. The trend is exactly the same. Fluoride is difficult to oxidize and it becomes easier down the group toward iodide; in other words, fluoride ions are not good reducing agents, but iodide ions are. The explanation starts from the hydrated ions in solution rather than solid ions. In a sense, this has already been done on another page. Fluorine is a very powerful oxidizing agent because it very readily forms its negative ion in solution. It is therefore energetically difficult to reverse the process. By contrast, for energetic reasons, iodine is relatively reluctant to form its negative ion in solution. Therefore, it is relatively easy for it to revert back to iodine. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Structure_and_Nomenclature_of_Coordination_Compounds/Isomers/Structural_Isomers_-_Ionization_Isomerism_in_Transition_Metal_Complexes

Coordination isomerism is a form of structural isomerism in which the composition of the complex ion varies. In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change. Ionization isomers can be thought of as occurring because of the formation of different ions in solution. Ionization isomers are identical except for a ligand has exchanged places with an anion or neutral molecule that was originally outside the coordination complex. The central ion and the other ligands are identical. For example, an octahedral isomer will have five ligands that are identical, but the sixth will differ. The non-matching ligand in one compound will be outside of the coordination sphere of the other compound. Because the anion or molecule outside the coordination sphere is different, the chemical properties of these isomers is different. The difference between the ionization isomers can be viewed within the context of the ions generated when each are dissolved in solution. For example, when pentaaquabromocobaltate(II)chloride is dissolved in water, \(Cl^-\) ions are generated: \[CoBr(H_2O)_5Cl {(s)} \rightarrow CoBr(H_2O)^+_{5} (aq) + Cl^+ (aq)\] whereas when pentaaquachlorocobaltate(II)bromide is dissolved, \(Br^-\) ions are generated: \[CoCl(H_2O)_5Br {(s)} \rightarrow CoCl(H_2O)^+_{5} (aq) + Br^+ (aq).\] If one dissolved \([PtBr(NH_3)_3]NO_2\) and \([Pt(NO_2)(NH_3)_3]Br\) into solution, then two different set of ions will be general. \[[PtBr(NH_3)_3]NO_2 (s) \rightarrow [PtBr(NH_3)_3]^+ (aq) + NO_2^- (aq) \label{R1}\] \[[Pt(NO_2)(NH_3)_3]Br (s) \rightarrow [Pt(NO_2)(NH_3)_3]^+ (aq) + Br^- (aq) \label{R2}\] Notice that these two ionization isomers differ in that one ion is directly attached to the central metal, but the other is not. Equations \(\ref{R1}\) and \(\ref{R2}\) are valid under the assumption that the platinum-ligand bonds of the complexes are stable (i.e., not labile). Otherwise, they may break and other ligands (e.g., water) may bind. Are \(\ce{[Cr(NH3)5(OSO3)]Br}\) and \(\ce{[Cr(NH3)5Br]SO4}\) coordination isomers? First, we need confirm that each compound has the same number of atoms of the respective elements (this requires viewing both cations and anions of each compound). Now, let's look at what these two compounds look like (Figure \(\Page {2}\)). The sulfate group is a ligand with a dative bond to the chromium atom and the bromide counter ion (\(\ce{[Cr(NH3)5(OSO3)]Br}\)). For \(\ce{[Cr(NH3)5Br]SO4}\), this is the the reverse. Yes, \(\ce{[Cr(NH3)5(OSO3)]Br}\) and \(\ce{[Cr(NH3)5Br]SO4}\) are coordination isomers. Are [Co(NH ) (SO )]Br and [Co(NH ) Br]SO ionization isomers? In the first isomer, SO is attached to the Cobalt and is part of the complex ion (the cation), with Br as the anion. In the second isomer, Br is attached to the cobalt as part of the complex and SO is acting as the anion. A hydrate isomer is a specific kind of ionization isomer where a water molecule is one of the molecules that exchanges places. A very similar type of isomerism results from replacement of a coordinated group by a solvent molecule ( ). In the case of water, this is called . The best known example of this occurs for chromium chloride "CrCl .6H O" which may contain 4, 5, or 6 coordinated water molecules. These isomers have very different chemical properties and on reaction with \(AgNO_3\) to test for \(Cl^-\) ions, would find 1, 2, and 3 \(Cl^-\) ions in solution respectively.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Basics_of_Kinetic_Molecular_Theory

The ideal gas equation \[PV = nRT\] describes gases behave, e.g.: However, the ideal gas law (nor any of the constituent gas laws) does not explain gases behave this way? What happens to gas particles when conditions such as pressure and temperature change? This is addressed via Kinetic Molecule Theory. The molecules of a gas are in a state of perpetual motion in which the velocity (that is, the speed and direction) of each molecule is completely random and independent of that of the other molecules. This fundamental assumption of the kinetic-molecular model helps us understand a wide range of commonly-observed phenomena. The five basic tenets of the kinetic-molecular theory are as follows: The Kinetic-Molecular Theory is "the theory of moving molecules." -Rudolf Clausius, 1857 If gases do in fact consist of widely-separated particles, then the observable properties of gases must be explainable in terms of the simple mechanics that govern the motions of the individual molecules. The kinetic molecular theory makes it easy to see why a gas should exert a pressure on the walls of a container. Any surface in contact with the gas is constantly bombarded by the molecules. At each collision, a molecule moving with momentum strikes the surface. Since the collisions are elastic, the molecule bounces back with the same velocity in the opposite direction. This change in velocity Δ is equivalent to a on \(a\); according to Newton's second law, \[F = ma\] with a force, \(F\), that is exerted on the surface of area \(A\) exerting a pressure \[P = \dfrac{f}{A}\] According to the kinetic molecular theory, the average kinetic energy of an ideal gas is directly proportional to the absolute temperature. Kinetic energy is the energy a body has by virtue of its motion: \[ KE = \dfrac{1}{2}m v^2\] with As the temperature of a gas rises, the average velocity of the molecules will increase; a doubling of the temperature will increase this velocity by a factor of four. Collisions with the walls of the container will transfer more momentum, and thus more kinetic energy, to the walls. If the walls are cooler than the gas, they will get warmer, returning less kinetic energy to the gas, and causing it to cool until thermal equilibrium is reached. Because temperature depends on the kinetic energy, the concept of temperature only applies to a statistically meaningful sample of molecules. We will have more to say about molecular velocities and kinetic energies farther on. Although the molecules in a sample of gas have an average kinetic energy (and therefore an average speed) the individual molecules move at various speeds. Some are moving fast, others relatively slowly At higher temperatures at greater fraction of the molecules are moving at higher speeds (Figure 3). What is the speed (velocity) of a molecule possessing average kinetic energy? KMT theory shows the the average kinetic energy (KE) is related to the (rms) speed \(u\) \[KE = \dfrac{1}{2} m u^2\] This is different from the typical definition of an average speed \( \langle v \rangle\) as demonstrated in Example 5.6.1. Suppose a gas consists of four molecules with speeds of 3.0, 4.5, 5.2 and 8.3 m/s. What is the difference between the average speed and root mean square speed of this gas? The speed is: \[ \langle v \rangle = \dfrac{3.0 + 4.5 + 5.2 + 8.3}{4}=5.25\; m/s\] The s: \[u= \sqrt{\dfrac{3.0^2+ 4.5^2+5.2^2+8.3^2}{4}}= 5.59 \; m/s\] ( ) )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/Halogens_as_Oxidizing_Agents

This page examines the trend in oxidizing ability of the Group 17 elements (the halogens): fluorine, chlorine, bromine and iodine. It considers the ability of one halogen to oxidize the ions of another, and how this changes down the group. Consider a situation in which one halogen (chlorine, for example) is reacted with the ions of another (iodide, perhaps) from a salt solution. In the chlorine and iodide ion case, the reaction is as follows: \[\ce{Cl_2 + 2I^- \rightarrow 2Cl^- + I_2}\] This is therefore a redox reaction in which chlorine acts as an oxidizing agent. Fluorine must be excluded from this discussion because its oxidizing abilities are too strong. Fluorine oxidizes water to oxygen, as in the equation below, and so it is impossible to carry out reactions with it in aqueous solution. \[\ce{2F_2 + H_2O \rightarrow 4HF + O_2}\] In each case, a halogen higher in the group can oxidize the ions of one lower down. For example, chlorine can oxidize bromide ions to bromine: \[\ce{Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2}\] The bromine forms an orange solution. As shown below, chlorine can also oxidize iodide ions to iodine: \[\ce{Cl_2 +2I^- \rightarrow 2Cl^- + I_2}\] The iodine appears either as a red solution if little chlorine is used, or as a dark gray precipitate if the chlorine is in excess. Bromine can only oxidize iodide ions, and is not a strong enough oxidizing agent to convert chloride ions into chlorine. A red solution of iodine is formed (see the note above) until the bromine is in excess. Then a dark gray precipitate is formed. \[\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2}\] Iodine won't oxidize any of the other halide ions, except possibly the extremely radioactive and rare astatide ions. In short, oxidizing ability decreases down the group. Whenever one of the halogens is involved in oxidizing a species in solution, the halogen ends is reduced to a halide ion associated with water molecules The following figure illustrates this process: Down the group, the ease with which these hydrated ions are formed decreases; the halogens become less effective as oxidizing agents, taking electrons from something else less readily. The reason that the hydrated ions form less readily down the group is due to several complicated factors. Unfortunately, this explanation is often over-simplified, giving a faulty and misleading explanation. The wrong explanation is dealt with here before a proper explanation is given. The following explanation is normally given for the trend in oxidizing ability of chlorine, bromine and iodine. The ease of ionization depends on how strongly the new electrons are attracted. As the atoms get larger, the new electrons are further from the nucleus and increasingly shielded by the inner electrons (offsetting the effect of the greater nuclear charge). The larger atoms are therefore less effective at attracting new electrons and forming ions. This is equivalent to saying electron affinity decreases down the group. Electron affinity is described in detail on another page. The problem with this argument is that it does not include fluorine. Fluorine's tendency to form a hydrated ion is much higher than that of chlorine. However, fluorine's electron affinity is less than that of chlorine. This contradicts the above argument.This problem stems from examining a single part of a very complicated process. The argument about atoms accepting electrons applies only to isolated atoms in the gas state picking up electrons to form isolated ions, also in the gas state. The argument must be generalized. In reality: The table below shows the energy involved in each of these changes for atomization energy, , and (hydration energy): Consider first the fifth column, which shows the overall heat evolved, the sum of the energies in the previous three columns. The amount of heat evolved decreases quite dramatically from the top to the bottom of the group, with the biggest decrease between fluorine to chlorine. Fluorine generates a large amount of heat when it forms its hydrated ion, chlorine a lesser amount, and so on down the group. The first electron affinity is defined as the energy released when 1 mole of gaseous atoms each acquire an electron to form 1 mole of gaseous 1- ions, as in the following equation: In symbol terms: \[ X(g) + e^- \rightarrow X^-(g)\] The fifth column measures the the energy released when 1 mole of gaseous ions dissolves in water to produce hydrated ions, as in the following equation, which is not equivalent to that above: \[ X^-(g) \rightarrow X^- (aq)\] There are two main factors. First, the atomization energy of fluorine is abnormally low. This reflects the low bond enthalpy of fluorine. The main reason, however, is the very high hydration enthalpy of the fluoride ion. That is because fluoride is very small. There is a very strong attraction between fluoride ions and water molecules. The stronger the attraction, the more heat is evolved when the hydrated ions are formed. The decrease in atomization energy between these three elements is relatively small, and would tend to make the overall change more negative down the group. It is helpful to look at the changes in electron affinity and hydration enthalpy down the group. Using the figures from the previous table: Both of these effects contribute, but that the more important factor—the one that changes the most—is the change in the hydration enthalpy. Down the group, the ions become less attractive to water molecules as they get larger. Although the ease with which an atom attracts an electron matters, it is not as important as the hydration enthalpy of the negative ion formed. The faulty explanation is incorrect even if restricted to chlorine, bromine and iodine: Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/Z017_Chemistry_of_Chlorine_(Z17)/The_Manufacture_of_Chlorine

This page describes the manufacture of chlorine by the electrolysis of sodium chloride solution using a diaphragm cell and a membrane cell. Both cells rely on the same underlying chemistry, but differ in detail. Chlorine is manufactured by electrolyzing sodium chloride solution. This process generates three useful substances: chlorine, sodium hydroxide and hydrogen. Sodium chloride solution contains the following: The protons and hydroxide ions come from the equilibrium: \[ H_2O (l) \rightleftharpoons H^+(aq) + OH^- (aq) \label{1}\] At any time, the concentration of protons or hydroxide ions is very small; the position of equilibrium lies well to the left. The negative ions, chloride and hydroxide, are attracted to the positively charged anode. It is easier to oxidize hydroxide ions to oxygen than to oxidize chloride ions to chlorine, but there are far more chloride ions arriving at the anode than hydroxide ions. The major reaction at the anode is therefore: \[ 2Cl^-_{(aq)}\rightarrow Cl_{2(g)}+2e^- \label{2}\] Two chloride ions each give up an electron to the anode, and the atoms produced combine into chlorine gas. The chlorine is, however, contaminated with small amounts of oxygen because of a reaction involving hydroxide ions, which also give up electrons: \[ 4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^- \label{3}\] The chlorine must be purified by removing this oxygen. Sodium ions and protons (from the water) are attracted to the negative cathode. It is much easier for a proton to pick up an electron than for a sodium ion to do so. Therefore, the following reaction occurs: \[ 2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)} \label{4}\] As protons convert into hydrogen gas, the equilibrium below shifts to the right to replace them: The net effect of this process is a buildup of sodium ions and newly-produced hydroxide ions at the cathode. In other words, sodium hydroxide solution is formed. If chlorine comes into contact with hydrogen, it produces a mixture that explodes violently on exposure to sunlight or heat, producing hydrogen chloride gas. Clearly these gases must remain separated. However, chlorine also reacts with sodium hydroxide solution to produce a mixture of sodium chloride and sodium chlorate(I), also known as sodium hypochlorite; this mixture is commonly sold as bleach. In addition, when the desired products are chlorine and sodium hydroxide rather than bleach, chlorine and sodium hydroxide must also be kept apart. The diaphragm and membrane cells are designed to keep all the products separate. The diaphragm is made of a porous mixture of asbestos and polymers. The solution can seep through it from the anode compartment into the cathode compartment. Notice that there is a higher level of liquid on the anode side. This ensures that the liquid always flows from left to right, preventing any of the sodium hydroxide solution from coming into contact with chlorine products. Chlorine is produced at the titanium anode according to the following equation: \[2Cl^- (aq) - 2e^- \rightarrow Cl_2(g)\] The product is contaminated with some oxygen because of the reaction below: \[ 4OH^-(aq) - 4e^- \rightarrow 2H_2O(l) + O_2(g)\] The chlorine is purified by liquefaction under pressure. The oxygen remains a gas when compressed at ordinary temperatures. Hydrogen is produced at the steel cathode by the following process: \[ 2H^+ (aq) + 2e^- \rightarrow H_2 (g)\] A dilute solution of sodium hydroxide solution is also produced at the cathode (see above for the explanation of what happens at the cathode). It is highly contaminated with unreacted sodium chloride. The sodium hydroxide solution leaving the cell is concentrated by evaporation. During this process, most of the sodium chloride crystallizes out as solid salt. The salt can be separated, dissolved in water, and passed through the cell again. Even after concentration, sodium hydroxide still contains a small percentage of sodium chloride. The membrane is made from a polymer that only allows the diffusion of positive ions. That means that the only sodium ions can pass through the membrane; the chloride ions are blocked. The advantage of this is that the sodium hydroxide formed in the right-hand compartment is never contaminated with sodium chloride. The sodium chloride solution must be pure. If it contains any other metal ions, these can also pass through the membrane and so contaminate the sodium hydroxide solution. Chlorine is produced at the titanium anode according to the following equation: \[2Cl^- (aq) + 2e^- \rightarrow Cl_2(g)\] It is contaminated with some oxygen because of the parallel reaction below: \[4OH^-(aq) + 4e^- \rightarrow 2H_2O(l) + O_2\] The chlorine is purified by liquefaction under pressure. The oxygen remains in the gas phase when compressed at normal temperatures. Hydrogen is produced at the nickel cathode as follows: \[ 2H^+ (aq) +2e^- \rightarrow H_@ (g)\] An approximately 30% solution of sodium hydroxide solution is also produced at the cathode (see the background chemistry section for an explanation of what happens at the cathode). Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/05%3A_Stereoisomerism_of_Organic_Molecules/5.06%3A_Molecules_with_More_Than_One_Chiral_Center._Diastereomers

We have seen examples of molecules with one chiral center that exist in two mirror-image configurations, which we call enantiomers. What happens when there is more than one chiral center? How many stereoisomers should we expect? Consider the stereoisomers of the important amino acid, threonine, (2-amino-3-hydroxybutanoic acid). For this substance, if we write all of the possible configurations of its chiral carbons, we have different projection formulas, \(19\)-\(22\), corresponding to four different stereoisomers: Because each chiral center added to a chain doubles the number of possible configurations, we expect eight different stereoisomers with three chiral carbons, sixteen with four, and so on, the simple rule the is \(2^n\) possible different stereoisomers for \(n\) chiral centers. As we shall see later, this rule has to be modified in some special cases. What is the relationship between stereoisomers \(19\)-\(22\)? This will be clearer if we translate each of the projection formulas into a three-dimensional representation, as shown in Figure 5-13. You will be helped greatly if you work through the sequence yourself with a ball-and-stick model. Drawn as , \(19\)-\(22\) come out as shown in \(19a\)-\(22a\): It should be clear (and, if it isn't, ball-and-stick models will be invaluable) that \(19a\) and (20a\) are mirror images of one another and that \(21a\) and \(22a\) are similarly mirror images.\(^5\) What about other combinations such as \(19a\) and \(21a\) or \(20a\) and \(22a\)? If you look at the pairs closely you will find that they are not mirror images and are not identical. Such substances, related to each other in this way and which can be converted one into the other only by changing the configurations at one or more chiral centers, are called . The difference between enantiomers and diastereomers is more than just geometry. Diastereomers have substantially different chemical and physical properties, whereas enantiomers have identical physical properties (apart from their optical rotations). This is illustrated in Table 5-1 for the threonine stereoisomers. The reason for the difference in physical properties between diastereomers can be seen very simply for a substance with two chiral centers by noting that a right shoe on a right foot (\(D,D\)) is a mirror image, or has the same physical properties, as a left shoe on a left foot (\(L,L\)), but is not a mirror image, nor does it have the same physical properties, as a left show on a right foot (\(L,D\)), or a right shoe on a left foot (\(D,L\)). All of the threonine stereoisomers \(19\)-\(22\) are chiral substances; that is, they are not identical with their mirror images. However, it is important to recognize that not all diastereomers are chiral. To illustrate this point, we return to the tartaric acids mentioned previously in connection is Pasteur's discoveries ( ). Proceeding as we did for threonine, we can write four projection formulas for tartaric acid, 2,3-dihydroxybutanedioic acid, as shown by \(23\)-\(26\): There are two pairs of mirror images \(23a\) and \(24a\), as well as \(25a\) and \(26a\). However, what will not be so immediately clear, but what you verify for yourself is that \(25a\) and \(26a\) are, in fact, identical. This means that \(25a\) and \(26a\) are representations of a achiral substance, identical with its mirror image. . The condition that makes possible the existence of meso compounds is an appropriate degree of molecular symmetry. There are several kinds of such molecular symmetry. In the case of projection formulas \(25\) (or \(26\)) there is a , which means that a plane can be placed through the molecule such that . The mirror plane for -tartaric acid can be seen easily from its projection formulas \(25b\) and \(26b\). These two formulas are superimposable if one is rotated \(180^\text{o}\) in the plane of the paper. The Newman representations \(25a\) and \(26a\) of -tartaric acid does have a mirror plane. Why is it different from the Fischer projections in this respect? The reason is that the projection formulas represent a particular conformation \(27\) of -tartaric acid that does have a mirror plane: Therefore, if you are confronted with a particular sawhorse or Newman formula and you have to decide whether it represents a compound, the best procedure is to make a ball-and-stick model of the conformation and then rotate around the bonds to see if it can be brought into a conformation (staggered or eclipsed) that has a plane of symmetry (such as \(27\)) or is identical with its mirror image. As expected from our previous discussions diastereomers of tartaric acid have different physical properties (Table 5-2). If you find yourself confused about the \(D,L\) and forms of tartaric acid, a simple analogy may help keep matters straight. Consider three sets of shoes. A right shoe beside a left shoe is a meso combination with a plane of symmetry. A left shoe next to a left shoe is not identical with, but is the mirror image of, a right shoe next to a right shoe. None of the three combinations are identical. Each right or left shoe corresponds to a right or left configuration of a tartaric acid carbon so the three sets correspond to -, \(L\)-, and \(D\)-tartaric acid, respectively. There is another symmetry test for meso configurations that is applicable to staggered conformations and can be illustrated with the tartaric acids. If you make models of \(25a\) and \(26a\) you will find that they are mirror images identical but, as we have said, they have no plane of symmetry. In this conformation, the molecules do have a . Thus a line drawn at any angle through the midpoint of the central \(C-C\) bond of \(25a\) (or \(26a\)) has an identical environment on each side of the midpoint. Another way of putting it is that each half of the molecule is the (i.e., reverse) of the other half. For a molecule with chiral centers, if its projection formula has a plane of symmetry or if we can find a rotational conformation with either a plane or center of symmetry, then it will be meso and achiral. The idea that for every \(n\) chiral centers there can be \(2^n\) different configurations will be true only if none of the configurations has sufficient symmetry to be identical with its mirror image. For every meso form there will be one less pair of enantiomers and one less total number of possible configurations than is theoretically possible according to the number of chiral centers. At most, one meso compounds is possible for structures with four chiral centers. An example is offered by the meso forms of tetrahydroxyhexanedioic acid which, with four chiral atoms, have configurations \(28\) and \(29\): \(^5\)The same information can be obtained from projection formulas. You can see that projections \(19\) and \(20\) are mirror images and that \(20\), \(21\), or \(22\) can not be superimposed on \(19\). However, in some situations confusion can result in making such comparisons and it is important to be able to translate the projection formulas into ball-and-stick models or perspective drawings. and (1977)

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Interpreting_Infrared_Spectra/IR4._Carbon_Carbon_Multiple_Bonds

Unsaturated hydrocarbons contain only carbon and hydrogen, but also have some multiple bonds between carbons. One type of unsaturated hydrocarbon is an olefin, also known as an alkene. Alkenes contain double bonds between carbons. One example of an alkene is 1-heptene. It looks similar to hexane, except for the double bond from the first carbon to the second. Look at the IR spectrum of 1-heptene. You should see: So far, these peaks are the same as the ones seen for hexane. We can assign them as the C-H stretching and bending frequencies, respectively. Looking further, you will also see: The peak at 1650 cm can be identified via computational methods as arising from a carbon-carbon double bond stretch. It is a weak stretch because this bond is not very polar. Sometimes it is obscured by other, larger peaks. ,