Question

in 2 methyl propane most stable radicle formed on hydrolysis is?

Answer 1

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

Br∙+H−RR∙+Br−Br⟶HBr+R∙⟶R−Br+Br∙(1)(2)

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 C−H bond;

formation of the H−Br bond.

Again, the second is the same no matter which radical is formed; therefore, it must be the cleavage of the 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 hyper conjugation to be a viable stabilisation mechanism for radicals; therefore, the energy lost by chromatically cleaving a tertiary C−H bond is slightly less (a less unstable product) than that lost cleaving a primary 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 Engage and carefully testing if a brick is move able or not. Chlorine on the other hand, is faster to react and thus has a less product-sided transition state.