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: What S_N1 means. 'Unimolecular' tells us only one species (the haloalkane) appears in the slow, rate-determining step. The mechanism therefore splits into ionisation followed by capture.
Step 2: Ionisation. In \((CH_3)_3CBr\) the bond to bromine snaps heterolytically; the electrons stay with bromine, producing \(Br^-\) and a flat, \(sp^2\) tertiary carbocation \((CH_3)_3C^+\). Three electron-donating methyl groups spread the positive charge, so this cation is comparatively stable, which is why tertiary halides favour S_N1. This is the slow step.
Step 3: Nucleophilic capture. A hydroxide ion, being electron rich, is drawn to the electron-poor carbon and bonds to it rapidly. Because the cation is planar, attack from both faces is possible, giving \((CH_3)_3C\text{-}OH\), tert-butyl alcohol.
Step 4: Kinetics. With only the halide in the slow step, doubling \([OH^-]\) does not change the rate, so \[ \boxed{\text{Rate} = k[(CH_3)_3CBr]} \] confirming a first-order, unimolecular pathway.