See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Under strongly acidic conditions (conc. H2SO4), tertiary or benzylic alcohols undergo dehydration via carbocation intermediates. The key here is an intramolecular Friedel-Crafts type alkylation (acid-catalyzed cyclization). Step 1 – Identify the starting material: The starting material is a compound where a cyclohexane ring carbon is connected to a carbon bearing two phenyl groups and one OH group. Looking more carefully, the structure shows a cyclohexane ring at the top whose bottom carbon is attached to C(Ph)2(OH). This is effectively 1,1-diphenyl-1-(cyclohex-1-yl)methanol, but the ring attachment suggests the OH-bearing carbon is directly bonded to the ring carbon — making it a tertiary benzylic alcohol. Step 2 – Protonation and carbocation formation: Conc. H2SO4 protonates the OH, giving water as a leaving group. This generates a tertiary carbocation at the carbon bearing two phenyl groups: C+(Ph)2 attached to the cyclohexane ring. Step 3 – Intramolecular cyclization: The carbocation is stabilized by the adjacent phenyl groups. One of the phenyl groups can undergo intramolecular electrophilic aromatic substitution (Friedel-Crafts cyclization). The carbocation acts as the electrophile attacking the ortho position of one of the phenyl rings, forming a new C–C bond and creating a five-membered ring fused between the two phenyl groups and the central carbon. Step 4 – Rearomatization: Loss of a proton from the intermediate arenium (Wheland) ion restores aromaticity. Step 5 – Product identification: The cyclization forms a fluorene-type skeleton. Since the starting cyclohexane ring is involved, and after ring contraction or direct cyclization of the diphenyl carbocation, the product is 9,9-diphenylfluorene — a fluorene (dibenzofuran-like bicyclic system with a five-membered central ring) bearing both phenyl groups on the sp3 carbon (C-9). This matches option (b). Why other options fail: - Option (a) places the two Ph groups on different carbons of the five-membered ring (1,3-positions), which would require a different mechanism and is not the product of simple intramolecular Friedel-Crafts on a gem-diphenyl carbocation. - Option (c) shows a methyl group, but no methyl group is present in the starting material. - Option (d) is eliminated because option (b) is correct. Therefore, the correct answer is B.