Q.1-Bromopropane is boiled with aqueous NaOH. Identify the mechanism of the reaction, the alcohol formed, and state why this route works cleanly for a primary halide.

Hydroxide ion attacks 1-bromopropane's electrophilic C-1 from the side directly opposite the leaving bromide, forming the new $\text{C--OH}$ bond as the $\text{C--Br}$ bond breaks in one concerted step ($S_N2$): $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{Br}^-$. This works cleanly for a primary substrate because the carbon under attack is sterically unhindered (only one alkyl group and two hydrogens around it), so hydroxide can approach the backside easily and there is no competing carbocation pathway to give side products.

Recognise the primary substrate as favouring $S_N2$; hydroxide directly displaces the halide with inversion at that one carbon.

Describing the mechanism as $S_N1$ (which would be expected for a tertiary, not primary, halide) or forgetting that the reaction proceeds in one single step rather than via a carbocation intermediate.