See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When toluene reacts with Br2 in the absence of any catalyst (no Lewis acid like FeBr3, no light specified but the product is side-chain bromination), the reaction involves bromination at the benzylic position (the CH3 group becomes CH2Br). Step 1: Identify the site of bromination. The product is benzyl bromide (PhCH2Br), meaning bromination occurred on the methyl side chain, not on the aromatic ring. Step 2: Identify the mechanism. Side-chain bromination of toluene with Br2 occurs via a free radical mechanism (typically initiated by heat or light, hv or Delta). Br2 undergoes homolytic cleavage to give bromine radicals, which abstract a benzylic hydrogen to form a benzylic radical (stabilized by resonance with the ring), which then reacts with another Br2 molecule to give the product and a new Br radical. Step 3: Rule out other options. (a) Nucleophilic addition: requires an electrophilic carbon (like a carbonyl), not applicable here. (b) Nucleophilic substitution: Br2 is not a nucleophile and toluene's methyl group is not an electrophilic carbon bearing a leaving group. (c) Electrophilic addition: applicable to alkenes, not to aromatic or benzylic C-H bonds. (d) Electrophilic substitution: this would occur on the aromatic ring (ring bromination) with a Lewis acid catalyst, giving bromobenzene derivatives, not side-chain bromination. (e) Free radical substitution: this correctly describes the mechanism where Br• abstracts a benzylic H (initiation/propagation), producing the benzylic radical intermediate, followed by reaction with Br2 to give PhCH2Br + HBr. This matches the observed transformation. Therefore, the correct answer is E.