See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Friedel-Crafts alkylation with AlCl3 and the given alkyl chloride (2-chloro-2,4,4-trimethylpentane, i.e., (CH3)3CCH2C(CH3)2Cl). With AlCl3, this tertiary alkyl chloride forms a carbocation. However, the initially formed tertiary carbocation at C2 [(CH3)3C-CH2-C+(CH3)2] can rearrange. The carbocation (CH3)3C-CH2-C+(CH3)2 is a tertiary carbocation; however, a hydride or methyl shift can occur. Actually, (CH3)3C-CH2-C+(CH3)2 is already tertiary but note the structure: it is 2,4,4-trimethylpent-2-yl cation. This cation can undergo a 1,2-hydride shift to give (CH3)3C+-CH(CH3)-CH(CH3)... wait, let us reconsider. The alkyl chloride is (CH3)3C-CH2-C(CH3)2Cl (2-chloro-2,4,4-trimethylpentane). The carbocation formed is (CH3)3C-CH2-C+(CH3)2. This tertiary carbocation can undergo a 1,2-hydride shift from the CH2 to give (CH3)3C+-CH(?)... Actually, a 1,2-hydride shift would move H from the CH2 adjacent to C+ to give: (CH3)3C+(the original CH2 becomes CH, the C+ shifts to the neopentyl-type carbon). More precisely: the cation (CH3)3C-CH2-C+(CH3)2 undergoes 1,2-hydride shift: H migrates from CH2 to the adjacent C+, giving (CH3)3C-CH+-CH(CH3)2... no. Let me be systematic. Numbering: C1=(CH3)3C, C2=CH2, C3=C+(CH3)2. H shifts from C2 to C3: gives C1=(CH3)3C, C2=CH+, C3=CH(CH3)2. That is (CH3)3C-C+H-CH(CH3)2, a secondary cation - that's worse. Alternatively, the carbocation can undergo intramolecular cyclization or the initial Friedel-Crafts gives a monoalkylated product, and then a second intramolecular cyclization occurs. The key insight: bromobenzene undergoes Friedel-Crafts alkylation. The alkyl group from (CH3)3CCH2C(CH3)2Cl with AlCl3 generates a carbocation. This carbocation attacks the bromobenzene ring. Given that the product shown in option (b) is a fused bicyclic compound (a tetralin/naphthalene-type skeleton with a saturated ring containing gem-dimethyl groups, and Br on the aromatic ring), this suggests a double alkylation or a cyclization reaction. The reaction proceeds as: first, Friedel-Crafts alkylation attaches the chain to the ring, then intramolecular Friedel-Crafts cyclization forms the second ring. The carbocation (CH3)3C-CH2-C+(CH3)2 first alkylates bromobenzene ortho to Br (since Br is ortho/para director), giving an intermediate with the chain attached. The chain then undergoes intramolecular cyclization: the terminal tert-butyl end (after rearrangement) attacks the ring again to form the fused 6-membered ring. The final product is a 1-bromo-5,5-dimethyl-8,8-dimethyl (or similar gem-dimethyl substituted) tetralin, which corresponds to option (b): a fused benzene-cyclohexane (tetralin) system with Br on the benzene ring and gem-dimethyl groups on the saturated ring. Option (a) has Br at bridgehead (not favored, violates Bredt's rule in a sense and steric issues). Option (c) lacks Br entirely - but starting material has Br on ring. Option (d) shows a non-aromatic product with Br on saturated carbon and indicated stereochemistry, which is not consistent with electrophilic aromatic substitution maintaining aromaticity. The bromobenzene ring retains its Br substituent (no removal of Br occurs under Friedel-Crafts conditions with AlCl3 alone), and the alkylation/cyclization gives the fused ring product with Br on the aromatic portion as in option (b). Therefore, the correct answer is B.