Compare the rate of $\mathrm{S_N1}$ reaction in the following compounds:

isobutyl chloride; ethyl chloride

We see that both compounds form primary carbocations. However, there is more hyperconjugation in the second compound as compared to the first one.

Also as the step in which the leaving group leaves is rate determining step (slowest step), which is before the rearrangement of the carbocation, I concluded that the second compound must have a greater rate of $\mathrm{S_N1}$ as compared to the first one.

But the answer says that the first has a greater reactivity than the second. Is the answer correct? If yes, then why?

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    $\begingroup$ I don't know the answer without either looking into the literature or (maybe) running some calculations (unlikely unless I'm v bored), but what I want to point out is that the ionisation (i.e. loss of chloride) and the rearrangement (alkyl / hydride migration) are not necessarily separate: they can occur in a concerted fashion. $\endgroup$ Apr 14, 2019 at 18:47

1 Answer 1


The inductive effect of the methyl groups should help stabilize the carbocation formed from 1. The answer may be a bit clearer if you add equilibrium to your thinking. enter image description here

The initial dissociation of the 1 to form carbocation A is the rate determining step. However the rearrangement of A to B should be relatively quick, and will drive the initial dissociation forward by Le Châtelier's Principle.

In contrast, 2 has less stabilization by the inductive effect, and there is no possibility of a rearrangement.


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