In my organic chemistry class, we were asked what determines the relatively high degree of selectivity in the reaction of a bromine radical with primary, secondary, and tertiary C–H bonds.

I said it is the overall free energy of the reaction because according to Hammond's postulate, endothermic (endergonic) reactions have transition states closely resembling the product side, making the difference between primary, secondary, and tertiary radicals more important in the transition state.

Would this logic be wrong?


1 Answer 1


First, let’s take a look at the radical chain reaction that leads to the substitution in question:

$$\begin{align}\ce{Br^. + H-R &-> HBr + R^.}\tag{1}\\[0.4em] \ce{R^. + Br-Br &-> R-Br + Br^.}\tag{2}\end{align}$$

Obviously, the selectivity is determined in the first step as long as no radical recombinations take place. If we break down the process in the first step into sub-processes, we get:

  • cleavage of the $\ce{C-H}$ bond;
  • formation of the $\ce{H-Br}$ bond.

Again, the second is the same no matter which radical is formed; therefore, it must be the cleavage of the $\ce{C-H}$ bond that leads to selectivity. Indeed, you can imagine radicals as being electron-deficient compounds and therefore behave similar to cations. Being similar to cations means that you can assume hyperconjugation to be a viable stabilisation mechanism for radicals; therefore, the energy lost by homolytically cleaving a tertiary $\ce{C-H}$ bond is slightly less (a less unstable product) than that lost cleaving a primary $\ce{C-H}$ bond. Exploiting this energy difference is the basis of selectivity.

This selectivity is higher if bromine is used than if chlorine is used. The reason is precisely the Hammond postulate you cited: the reaction with bromine is predicted to be more endothermic and thus the transition state more product-sided. A layman’s explanation would be that bromine can approach slowly and try out whether it feels happy about reacting with this hydrogen — think of playing Jenga and carefully testing if a brick is moveable or not. Chlorine on the other hand, is faster to react and thus has a less product-sided transition state.

There is a second factor that reduces selectivity. Assume 2-methylbutane; this compound conveniently has primary, secondary and tertiary hydrogens. Probabilities of bromine radicals to react with each of these hydrogens have been calculated; since I don’t have actual numbers on me right now, I am going to ballpark the following: assume, that a secondary hydrogen has a five times as high probability of reacting with bromine when compared to a primary one and that the tertiary hydrogen has a 15 times as high probability. Naïvely, we could assume that the reaction yields products with a ratio of $1:5:15$ — but we need to remember how many hydrogens of a given type there is.

The possible products are 1-bromo-3-methylbutane, 1-bromo-2-methylbutane (both primary attacks), 2-bromo-3-methylbutane (secondary attack) and 2-bromo-2-methylbutane (tertiary attack). For each of these, the following number of hydrogens can react to form the product in question:

  • 1-bromo-3-methylbutane: 3
  • 1-bromo-2-methylbutane: 6
  • 2-bromo-3-methylbutane: 2
  • 2-bromo-2-methylbutane: 1

Thus, to accurately calculate the ratios, we need to multiply the preference factor with the number of hydrogens, resulting in a product ratio of $3:6:10:15$ in the same order as presented — a lower fraction of tertiary bromide than presumed.

  • 1
    $\begingroup$ By way of comparison, see my answer to a similar question for the numbers for the chlorination of 2-methylbutane. $\endgroup$
    – user7951
    Commented Oct 21, 2017 at 7:53

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