See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Under H2SO4 (acid catalyst), cyclohexene undergoes protonation to form a carbocation (cyclohexyl carbocation at C1). This carbocation acts as an electrophile toward benzene in a Friedel-Crafts type alkylation (acid-catalyzed electrophilic aromatic substitution). Step 1: H2SO4 protonates cyclohexene to give the cyclohexyl carbocation (secondary carbocation, C1 of cyclohexane ring). Step 2: The cyclohexyl carbocation attacks benzene via electrophilic aromatic substitution (EAS) to give cyclohexylbenzene (monoalkylation product). Step 3: Since 2 moles of cyclohexene are used per mole of benzene, a second alkylation occurs. The cyclohexylbenzene undergoes a second EAS with the second equivalent of cyclohexyl carbocation. Step 4: The second cyclohexyl group adds to the para position of the ring (para product is favored over ortho due to steric reasons - the bulky cyclohexyl group already present directs incoming electrophile to para position preferentially). Result: 1,4-dicyclohexylbenzene (para-dicyclohexylbenzene) is formed - a benzene ring with cyclohexyl substituents at C1 and C4 positions. Why other options fail: - Option (a): Shows a structure with a cycloheptane ring and two phenyl groups - cycloheptane is not formed from cyclohexene under these conditions; also requires two benzene molecules. - Option (c): Shows phenyl groups on a cyclohexane-cyclohexane framework - this would require C-C bond formation between two cyclohexane rings and phenyl groups in a different connectivity. - Option (d): Shows Ph attached to cyclohexyl-cyclohexyl chain - this implies two cyclohexene molecules added in sequence to one end without going onto benzene ring for the second alkylation, and only one phenyl group is present. Option (b) correctly shows 1,4-dicyclohexylbenzene: benzene with two cyclohexyl groups at para positions, consistent with two moles of cyclohexene reacting with one mole of benzene under H2SO4. Therefore, the correct answer is B.