Question

Which would undergo S_N1 reaction faster in the following pair and why?

Answer 1

Carbocations are the intermediates in the S_N1 reaction. Greater the stability of the carbocations, more easily will the product be formed and hence faster will be the rate of the reaction. Because the stability of the carbocations decreases in the order: 3° carbocation > 2° carbocation > 1° carbocation > CH₃+.

Therefore, the reactivity of alkyl halides towards S_N1 reactions decreases in the same order, i.e. 3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides.

The two structures are

Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a tertiary alkyl halide which forms a tertiary carbocation intermediate in the S_N1 reaction. Hence, 2-bromo-2-methylpropane undergoes an S_N1 reaction faster than bromoethane

Answer 2

A tertiary alkyl halide tends to undergo the S_N1 mechanism because it can form a tertiary carbocation, which is stabilised by the three alkyl groups attached to it. As alkyl groups are electron donating, they allow the positive charge in the carbocation to be delocalised by the induction effect. Hence, out of the given pairs, (CH₃)₃C-Br would undergo S_N1 reaction faster than CH₃-CH₂-Br.