Which of the following reaction is NOT correctly represented?

Solution

Chemistry ma’am · 3 steps

Under light, bromine goes to the most stable radical, and in (a) that is not the CH₃ carbon.

1. Read the conditions. Light (hν) means a free-radical chain, so bromine replaces the H whose removal leaves the most stable radical. Fe in the dark means electrophilic attack on the ring. A diazonium salt with Cu₂Br₂/HBr is the Sandmeyer reaction.

   Option	What really happens	Drawn correctly?
   (a) 3-methylcyclohexene, Br₂, hν	H is removed from the ring carbon that carries CH₃. That radical is tertiary and allylic, so Br goes onto the ring, not onto CH₃	No
   (b) C₆H₅N₂⁺ Cl⁻, Cu₂Br₂/HBr	Sandmeyer reaction: N₂ leaves and Br takes its place, giving bromobenzene	Yes
   (c) Toluene, Br₂, hν	The benzylic radical is resonance-stabilised, so benzyl bromide forms	Yes
   (d) Toluene, Br₂, Fe, dark	FeBr₃ makes Br⁺. The CH₃ group directs it ortho and para	Yes

2. Why (a) fails. A radical on the CH₃ carbon would be primary and not next to the double bond, so it gets no allylic resonance. It is the least stable radical available, so the CH₂Br product shown does not form. Bromine ends up on an allylic ring carbon instead.
3. So. Options (b), (c) and (d) are shown correctly. The reaction that is not correctly represented is (a).

The idea to keep. In light-driven bromination, rank each C–H by the radical it would leave: allylic or benzylic first, tertiary over primary. Bromine is very selective and picks the best one.

Answer from the NTA final answer key. Solution by PrepWiser.