See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed ring expansion (cationic rearrangement) of methylenecycloalkanes under thermodynamic control. Step 1 – Identify the starting material: The reactant is (cyclopentylidene) with an isopropylidene unit — more precisely, it is 1-(propan-2-ylidene)cyclopentane, i.e., cyclopentane bearing an exocyclic =C(CH3)2 double bond. This is a cyclopentane ring with an exocyclic double bond to C(CH3)2. Step 2 – Protonation of the exocyclic double bond: Under H+/heat conditions, the exocyclic alkene is protonated according to Markovnikov's rule. Protonation at the terminal carbon of =C(CH3)2 gives a tertiary carbocation at the carbon bearing two methyl groups attached to the ring (i.e., a tertiary cyclopentyl carbocation at C1 of the ring), OR protonation at C1 of the ring gives a tertiary carbocation on the =CMe2 carbon. The more stable carbocation is the tertiary one on the exocyclic carbon: +C(CH3)2 adjacent to the ring — actually both positions give tertiary carbocations, but the one that can ring-expand is the cyclopentylcarbocation at C1. Step 3 – Ring expansion via 1,2-alkyl shift: The tertiary carbocation at C1 of the cyclopentane ring undergoes a 1,2-carbon shift (ring expansion): one of the C–C bonds of the cyclopentane ring migrates to give a cyclohexyl carbocation. This expands the five-membered ring to a six-membered ring (cyclohexane), placing the positive charge at an endocyclic carbon, with a methyl substituent. Step 4 – Deprotonation under thermodynamic control: Loss of a proton from the cyclohexyl carbocation gives the most stable (most substituted, Zaitsev) alkene. The endocyclic trisubstituted double bond in a methylcyclohexene ring is more stable than an exocyclic methylenecyclohexane. This yields a trisubstituted endocyclic alkene: 1-methylcyclohex-2-ene or equivalently a cyclohexene with a methyl group, corresponding to option (d). Why other options fail: - (a) is an aromatic/conjugated product that would require additional dehydrogenation steps not possible under these mild conditions. - (b) shows an exocyclic double bond on cyclohexane — this would be the kinetic product but not the thermodynamic one; endocyclic alkenes are more stable. - (c) shows a terminal exocyclic =CH2 on methylcyclohexane — less substituted, thermodynamically disfavored. - (d) shows an endocyclic trisubstituted double bond in a methylcyclohexene framework — most thermodynamically stable product, consistent with thermodynamic (heat) conditions. Therefore, the correct answer is D.