2017 H2 Chem P1 Q27 — Halogenoalkane Solution

The answer is A (\(\text{H}_2\text{O}\)).
The \(\text{S}_\text{N}1\) rate-determining step is ionisation of \((\text{CH}_3\text{CH}_2)_3\text{CBr}\) into a carbocation and \(\text{Br}^-\). A polar solvent stabilises these ions and speeds up this step. However, the carbocation is highly electrophilic, so any solvent molecule with a lone pair can attack it and form an adduct.
Water is polar. When it attacks the carbocation (and loses \(\text{H}^+\)) it gives \((\text{CH}_3\text{CH}_2)_3\text{COH}\), the same product as from \(\text{OH}^-\), so no by-product forms.
• B is wrong: hexane, \(\text{C}_6\text{H}_{14}\), is non-polar. It does not stabilise the carbocation and \(\text{Br}^-\), so it does not increase the rate.
• C is wrong: ethylamine is polar, but its N lone pair makes it a nucleophile. It attacks the carbocation to give the by-product \((\text{CH}_3\text{CH}_2)_3\text{C-NHCH}_2\text{CH}_3\).
• D is wrong: ethanol is polar, but its O lone pair attacks the carbocation to give the ether by-product \((\text{CH}_3\text{CH}_2)_3\text{C-OCH}_2\text{CH}_3\).