See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reaction conditions are HBr with R2O2/hv (peroxide and light), which indicates a free-radical (anti-Markovnikov) addition of HBr across a double bond. Step 1 - Identify the substrate: The starting material is a bicyclic compound consisting of a fused cyclohexane-cyclobutene ring system, where the cyclobutene double bond bears a methyl substituent. The double bond is between the two carbons shared by the cyclobutane ring (the bridgehead-adjacent position), with CH3 on one of those alkene carbons. Step 2 - Radical addition mechanism: Under radical conditions, HBr adds via anti-Markovnikov selectivity. The bromine radical adds to the carbon that gives the MORE stable (more substituted) radical intermediate, but in radical addition the Br• adds to the less hindered or less substituted carbon to give the more stable carbon radical. Step 3 - Determine regiochemistry: The double bond carbons are: C1 (the carbon bearing CH3, which is also at the ring junction/more substituted) and C6 (the other alkene carbon at the other ring junction). In radical addition, Br• adds to the terminal/less substituted end (anti-Markovnikov). The CH3-bearing carbon is more substituted, so Br• adds to the carbon without CH3 (the other vinyl carbon of the cyclobutene), generating a radical on the carbon bearing CH3. This radical is stabilized by the methyl group and the ring system. Step 4 - Product formation: H• then adds to the radical center (the carbon bearing CH3). Wait - re-examining: Br• adds to the less substituted carbon (the one without CH3), placing radical on the more substituted carbon (with CH3). Then H adds to that carbon. This gives Br on one cyclobutane ring carbon and CH3 on the adjacent ring carbon - which corresponds to option (c). Step 5 - Why other options fail: - Option (a): Shows Br and CH3 on the same carbon - this would be Markovnikov (ionic) addition product, not radical. - Option (b): Shows no Br at all - incomplete reaction product, not correct. - Option (d): Shows Br on the cyclohexane bridgehead - this would require addition in a different regiochemical sense not supported by radical mechanism here. Option (c) correctly shows anti-Markovnikov addition: Br on the cyclobutane carbon adjacent to the ring junction (the carbon that did not bear CH3 in the starting alkene), and CH3 on the neighboring cyclobutane carbon. Therefore, the correct answer is C.