Question

Write the reaction mechanism of the unimolecular nucleophilic substitution (S_N1) reaction of 2-bromo-2-methylpropane.

Hint:

Two steps: slow heterolysis to a stabilised tertiary carbocation (rate-determining), then fast attack by OH^-; rate depends only on the halide.

Answer 1

Step 1: The substrate. 2-bromo-2-methylpropane is tert-butyl bromide, \( (CH_3)_3C\text{-}Br \). Take its reaction with aqueous alkali (OH^-) as the nucleophile. An S_N1 reaction goes in two steps through a carbocation.
Step 2: Step I – slow ionisation (rate-determining). The polar C–Br bond breaks heterolytically. Bromine leaves with the bonding electron pair as bromide ion, and a tertiary carbocation is formed:
\[ (CH_3)_3C\text{-}Br \rightarrow (CH_3)_3C^{+} + Br^{-} \]
This step is slow because breaking the bond needs energy; it decides the overall rate. The tertiary carbocation is stabilised by the +I (hyperconjugation and inductive) effect of the three methyl groups.
Step 3: Step II – fast attack of nucleophile. The hydroxide ion (or water) quickly attacks the planar carbocation from either face and forms the product:
\[ (CH_3)_3C^{+} + OH^{-} \rightarrow (CH_3)_3C\text{-}OH \]
giving 2-methylpropan-2-ol (tert-butyl alcohol).
Step 4: Rate law and conclusion. Since only the substrate is involved in the slow step, the rate depends on its concentration alone: \[ \boxed{\text{Rate} = k[(CH_3)_3C\text{-}Br]} \] The reaction is first order (unimolecular), hence S_N1.