During the hydrolysis of 1‑iodopropane ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{I}$) with excess aqueous $\mathrm{NaOH}$, which rate law correctly describes the reaction?

Step-by-step solution

1. 1. Recognise that 1‑iodopropane is a primary alkyl halide.
2. 2. $\mathrm{NaOH}$ acts as a strong nucleophile, so the reaction proceeds via an “S${}_{\mathrm{N}}$(2)” mechanism.
3. 3. “S${}_{\mathrm{N}}$(2)” has a second‑order rate law: one factor for the substrate and one for the nucleophile.
4. 4. Therefore the rate is proportional to the product of the concentrations of 1‑iodopropane and $\mathrm{NaOH}$.

Concept tested: Second‑order rate law for SN2

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked