JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Haloalkanes and Haloarenes

An ${\mathrm{S}}_{\mathrm{N}}1$ reaction proceeds through a carbocation, so within each molecule the halogen that leaves to give the more stable cation is the more reactive one. In A, loss of $\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$ gives a benzylic cation stabilised by the ring, far more stable than the secondary cation from $\mathrm{B}{\mathrm{r}}_{(\mathrm{b})}$, so $\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$ is favoured. In B, loss of ${\mathrm{I}}_{(\mathrm{a})}$ gives an allylic cation delocalised over the double bond, whereas ${\mathrm{I}}_{(\mathrm{b})}$ would give an unstabilised primary cation, so ${\mathrm{I}}_{(\mathrm{a})}$ is favoured. In C, $\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$ sits at a bridgehead where a planar cation cannot form, so the non-bridgehead tertiary $\mathrm{B}{\mathrm{r}}_{(\mathrm{b})}$ is favoured. In D, $\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$ is on a tertiary carbon while $\mathrm{B}{\mathrm{r}}_{(\mathrm{b})}$ is primary, so $\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$ is favoured. The most reactive halogens are therefore $\mathrm{A}-\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$, $\mathrm{B}-{\mathrm{I}}_{(\mathrm{a})}$, $\mathrm{C}-\mathrm{B}{\mathrm{r}}_{(\mathrm{b})}$ and $\mathrm{D}-\mathrm{B}{\mathrm{r}}_{(\mathrm{a})}$.