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Thermodynamic vs kinetic reaction control with radical substitution
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Martin David McCoy
Thermodynamic vs kinetic reaction control with radical substitution
Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).
$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$
On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.
$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$
One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.
To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.
A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.
Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.
Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.
Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).
$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$
On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.
$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$
One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.
To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.
A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.
Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.
Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.
As you said, kinetic control compares the TS leading to the alkyl radical. Does TD control here compare the intermediate (tertiary vs primary radical), or the final product (tertiary vs primary alkyl halide)? Im actually not sure. (It doesnt affect your argument in any way, since you explored both - just wondering.)More
@orthocresol I’m going to say the final product, although I can’t back it with examples. (The stereotypical Diels-Alder example doesn’t work since we’re comparing transition states there.)More
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Basically any reaction will proceed according to the conditions available. While predicting the product of a chemical reaction we generally tend to guess the most stable product we can think of. But the important thing we forget are the condition needed for it. If a reaction occurs in an environment where the temperature is very low, in some cases the a certain product may be stable but the chemical factors required for the proper collision at that position to give the desired product would be difficult. Thus the reaction will form some other product as a major product by making the reactant molecules collide at a position where there may be less steric crowding or something else on the same lines. This may alter the transition states, thus altering the product. But at high temperatures due to increase in the velocity of molecules more collisions happen and we may get our desired stable product. Thus the product on the first case is kinetically favoured and the product in the second case is thermodynamically favoured.
And to be case specific as I said before kinetically favoured product can be one which is less steric ally hindered so in your case primary carbons are less sterically hinderedd,thus when high temperature isnt available , it will collide with the primary carbon.
Hope it helps
Basically any reaction will proceed according to the conditions available. While predicting the product of a chemical reaction we generally tend to guess the most stable product we can think of. But the important thing we forget are the condition needed for it. If a reaction occurs in an environment where the temperature is very low, in some cases the a certain product may be stable but the chemical factors required for the proper collision at that position to give the desired product would be difficult. Thus the reaction will form some other product as a major product by making the reactant molecules collide at a position where there may be less steric crowding or something else on the same lines. This may alter the transition states, thus altering the product. But at high temperatures due to increase in the velocity of molecules more collisions happen and we may get our desired stable product. Thus the product on the first case is kinetically favoured and the product in the second case is thermodynamically favoured. And to be case specific as I said before kinetically favoured product can be one which is less steric ally hindered so in your case primary carbons are less sterically hinderedd,thus when high temperature isnt available , it will collide with the primary carbon.Hope it helps
Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).
$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$
On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.
$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = \frac{k_1}{k_2}\tag{4}$$
One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.
To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.
A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.
Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.
Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.
Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).
$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$
On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.
$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = \frac{k_1}{k_2}\tag{4}$$
One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.
To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.
A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.
Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.
Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.
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Basically any reaction will proceed according to the conditions available. While predicting the product of a chemical reaction we generally tend to guess the most stable product we can think of. But the important thing we forget are the condition needed for it. If a reaction occurs in an environment where the temperature is very low, in some cases the a certain product may be stable but the chemical factors required for the proper collision at that position to give the desired product would be difficult. Thus the reaction will form some other product as a major product by making the reactant molecules collide at a position where there may be less steric crowding or something else on the same lines. This may alter the transition states, thus altering the product. But at high temperatures due to increase in the velocity of molecules more collisions happen and we may get our desired stable product. Thus the product on the first case is kinetically favoured and the product in the second case is thermodynamically favoured. And to be case specific as I said before kinetically favoured product can be one which is less steric ally hindered so in your case primary carbons are less sterically hinderedd,thus when high temperature isnt available , it will collide with the primary carbon. Hope it helps
Basically any reaction will proceed according to the conditions available. While predicting the product of a chemical reaction we generally tend to guess the most stable product we can think of. But the important thing we forget are the condition needed for it. If a reaction occurs in an environment where the temperature is very low, in some cases the a certain product may be stable but the chemical factors required for the proper collision at that position to give the desired product would be difficult. Thus the reaction will form some other product as a major product by making the reactant molecules collide at a position where there may be less steric crowding or something else on the same lines. This may alter the transition states, thus altering the product. But at high temperatures due to increase in the velocity of molecules more collisions happen and we may get our desired stable product. Thus the product on the first case is kinetically favoured and the product in the second case is thermodynamically favoured. And to be case specific as I said before kinetically favoured product can be one which is less steric ally hindered so in your case primary carbons are less sterically hinderedd,thus when high temperature isnt available , it will collide with the primary carbon.Hope it helps
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