JEE Main 2025 Apr 4 Shift 1, Chemistry Q17: Mechanism of Some Important Reactions of Alkenes

Step 1 (Br${}_2$/h$\nu$): Radical bromination selects the weakest C$-$H bond, which is the tertiary C$-$H at the ring carbon bearing the methyl group. This gives 1-bromo-1-methylcyclohexane.

Step 2 (alcoholic KOH, $\Delta$): An E2 elimination removes the Br and an adjacent ring H, giving the alkene 1-methylcyclohex-1-ene (the Saytzeff product, since the endocyclic double bond gives a more substituted alkene than methylenecyclohexane).

Step 3 (HBr / peroxide, h$\nu$): With peroxides, HBr adds via a radical mechanism showing anti-Markovnikov regioselectivity. The Br attaches to the less substituted vinyl carbon, i.e. C2 of the ring; the H goes to the more substituted carbon (C1) bearing the methyl.

Final product: 1-bromo-2-methylcyclohexane, option (2).