Doing a revision for a chemistry class test tomorrow, I stumbled upon thermodynamic and kinetic reaction control in the context of radical substitution.

Here, my chemistry teacher said, that thermodynamic reaction control would basically be about favoring stabilized (originally instable) transition states, e.g. by increasing temperatures. According to him, this would "thermodynamically prefer" the substitution of tertiary hydrogen atoms relative to primary ones.

However, he also told us that kinetic reaction control would, instead, prefer the substitution of primary hydrogen atoms. More precisely, the fastest reaction (which would result in the substitution of primary hydrogen atoms due to less steric hindrance) would be favored under certain conditions.

The question is, under which conditions a certain type of reaction control would be prefered.

  • $\begingroup$ I would certainly say that thermodynamic control is associated with higher temperatures, not lower. This is to allow reversibility of the kinetic pathway (abstraction of primary hydrogens) and supply sufficient energy to overcome the larger activation energy (since the TS is more unstable). $\endgroup$ Commented Sep 28, 2016 at 9:17
  • $\begingroup$ An example I can remember would be that a split (tertiary) hydrogen atom would lead to a charge at the carbon atom. In return, this could, be stabilized better by inductive effects (from neighbouring alkyl groups). $\endgroup$
    – Tacticus
    Commented Sep 28, 2016 at 9:20
  • $\begingroup$ 1) The carbon isn't positively charged. One could perhaps say it is electron-poor, but a radical is certainly not a carbocation. 2) I must confess I have never actually heard of any real examples of td vs kin control in radical halogenation. March 6ed and Clayden 2ed also have nothing on this. The classic example of td/kin control occurs with the addition of bromine to conjugated dienes, or addition of HX to α,β-unsaturated carbonyls. I am not really convinced that the concept is applicable here, although I may very well be wrong. $\endgroup$ Commented Sep 28, 2016 at 9:24
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    $\begingroup$ I realised, perhaps, a better explanation of why I am skeptical. Since a tertiary radical is most stable than a primary radical, the transition state leading to abstraction of a tertiary H is, by Hammond's postulate, more stable than that leading to abstraction of a primary H. This is solely based on electronic factors, and while steric factors certainly exist, I am not convinced that they are significant enough to overturn the relative energy of the transition states that we established earlier. Personally, I have not read anything that indicates so. $\endgroup$ Commented Sep 28, 2016 at 9:42

2 Answers 2


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.

  • $\begingroup$ 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)? I'm actually not sure. (It doesn't affect your argument in any way, since you explored both - just wondering.) $\endgroup$ Commented Sep 28, 2016 at 17:57
  • $\begingroup$ @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.) $\endgroup$
    – Jan
    Commented Sep 29, 2016 at 12:51

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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