JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Hydrocarbons

In free-radical bromination with $\mathrm{B}{\mathrm{r}}_2$ and light, bromine enters at the position that gives the most stable radical. For the methyl-substituted cyclohexene the allylic ring carbon already bearing the methyl group is tertiary, and hydrogen abstraction there gives a tertiary allylic radical far more stable than the primary radical on the methyl group, so bromination should occur on that ring carbon rather than on the methyl as drawn. The conversion of benzenediazonium chloride to bromobenzene with $\mathrm{C}{\mathrm{u}}_2\mathrm{B}{\mathrm{r}}_2/HBr$, the side-chain bromination of toluene to benzyl bromide with $\mathrm{B}{\mathrm{r}}_2/h\nu$, and the iron-catalysed ring bromination of toluene to o- and p-bromotoluene are all shown correctly. The reaction that is not correctly represented is the radical bromination yielding the methyl-brominated product.