2‑Iodo‑3‑methylpentane is refluxed with excess aqueous $\mathrm{NaOH}$. Name the major organic product and briefly explain the reaction mechanism.

Step-by-step solution

1. Realise that $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}(\mathrm{Br})(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$ is a tertiary halide, making $\mathrm{NaOH}$ favor an ${\mathrm{S}}_{\mathrm{N}}1$ process.
2. Predict formation of a tertiary carbocation ${\mathrm{C}}^{+}$ after loss of ${\mathrm{I}}^{-}$.
3. Water, present in excess, acts as a nucleophile and attacks the carbocation to give a protonated alcohol.
4. A second deprotonation step gives the neutral tertiary alcohol, 3‑methyl‑pentan‑2‑ol.
5. Conclude that the major product is 3‑methyl‑pentan‑2‑ol via an ${\mathrm{S}}_{\mathrm{N}}1$ mechanism.

Concept tested: S_N1 hydrolysis of tertiary haloalkane

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked