2019 H2 Chem P1 Q20 — Halogenoalkane Solution

The answer is D.

A racemic mixture (a 1:1 mixture of enantiomers) forms only when the nucleophile attacks from either side with equal chance. This happens in \(\text{S}_\text{N}1\) hydrolysis through a planar carbocation. \(\text{S}_\text{N}2\) at a chiral carbon gives inversion, i.e. a single enantiomer, not a racemic mixture.
So the carbon bonded to Cl in X must be:
• tertiary, so that it reacts by \(\text{S}_\text{N}1\), and
• chiral, so that the alcohol formed is chiral.
In D, \((\text{CH}_3)_2\text{CHC(CH}_3\text{)(CH}_2\text{CH}_3\text{)Cl}\), the C-Cl carbon carries Cl, \(\text{CH}_3\), \(\text{C}_2\text{H}_5\) and \(\text{CH(CH}_3\text{)}_2\): four different groups, and it is tertiary. The planar carbocation is attacked by \(\text{OH}^-\) equally from both faces, giving a racemic mixture of \((\text{CH}_3)_2\text{CHC(CH}_3\text{)(C}_2\text{H}_5\text{)OH}\).

• A: a primary chloroalkane, hydrolysed by \(\text{S}_\text{N}2\), and its C-Cl carbon carries two H atoms, so it is not chiral and the product is achiral.
• B: tertiary, but the C-Cl carbon carries three identical ethyl groups, so it is not chiral and the product is achiral.
• C: tertiary, but the C-Cl carbon carries two identical methyl groups, so it is not chiral and the product is achiral.