Toluene, when heated with Br2 in the presence of light, mainly gives:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Benzyl bromide
In the presence of light, bromine reacts by a free radical path. The radicals attack the C–H bonds of the side chain (–CH3), not the benzene ring, so one H of the methyl group is replaced by Br:
\( \displaystyle \ce{C6H5CH3 + Br2} \xrightarrow{h\nu} \ce{C6H5CH2Br + HBr} \)
The main product is benzyl bromide (bromomethylbenzene). Chlorine reacts with toluene in the same way in light (Cl2/hν), giving side chain chlorination.
Why the other options are incorrect
Option (1) p-bromotoluene — Incorrect. Bromine enters the ring by electrophilic substitution, which needs a Lewis acid catalyst such as iron, not light.
Option (3) o-bromotoluene — Incorrect. It is also a ring substitution product, formed with a Lewis acid catalyst, not in light.
Option (4) A mixture of o-bromotoluene and p-bromotoluene — Incorrect. This mixture is what toluene gives with bromine and a Lewis acid catalyst; in light the side chain reacts.
NCERT (2026-27 edition):
8.2.2 Preparation of Aldehydes – “Side chain chlorination of toluene gives benzal chloride, which on hydrolysis gives benzaldehyde.”