See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: This is a Friedel-Crafts alkylation reaction. HF acts as a Lewis/Brønsted acid catalyst, and heat (Delta) is applied. The alkylating agent is methylenecyclohexane (an exocyclic alkene: cyclohexane with a =CH2 group). Step 1 - Protonation of the alkene: HF protonates the exocyclic double bond (C=CH2) of methylenecyclohexane. By Markovnikov's rule, the proton adds to the terminal =CH2 carbon (less substituted end), generating a carbocation at the ring carbon (C1 of cyclohexane), which is a tertiary carbocation. This tertiary carbocation at C1 of the cyclohexane ring is the electrophile. Step 2 - Electrophilic aromatic substitution: The tertiary carbocation (1-methylenecyclohexyl cation, i.e., a tertiary carbocation at C1 of cyclohexane which now bears a methyl group from the protonation... wait, let me reconsider. Methylenecyclohexane is cyclohexane with exocyclic =CH2. Protonation of CH2= end gives carbocation at the ring carbon C1, which is tertiary (connected to two ring CH2 groups and one CH3... no). Actually: the exocyclic alkene is C1(cyclohexane)=CH2. Protonation at CH2 gives +C1(cyclohexane), where C1 is connected to C2 and C6 of the ring and to =CH2 which becomes -CH3 after protonation. So the carbocation is at C1 of cyclohexane, tertiary, with a methyl group (from the former =CH2 now -CH3) - making it a tertiary carbocation (1-methylcyclohexyl cation). Step 3 - Attack on benzene: This tertiary 1-methylcyclohexyl carbocation attacks benzene in an electrophilic aromatic substitution, attaching the 1-methylcyclohexyl group to the benzene ring. Step 4 - Product: The product is (1-methylcyclohexyl)benzene, i.e., benzene with a 1-methylcyclohexyl substituent where both the phenyl and the methyl are on C1 of the cyclohexane ring. Why other options fail: - Option (a): Shows a methyl on the benzene ring (toluene derivative) with a cyclohexyl group - incorrect, no rearrangement leads to this. - Option (c): Appears similar to (b) but structural drawing differences - (b) correctly shows the methyl on the same carbon of cyclohexane that bears the phenyl. - Option (d): Shows a 4-methylcyclohexylbenzene - methyl at C4 of cyclohexane, which would require a different mechanism and is not the Markovnikov product. Therefore, the correct answer is B.