See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Friedel-Crafts Acylation (Benzene + Succinic Anhydride / AlCl3, 2.2 equiv): Succinic anhydride undergoes Friedel-Crafts acylation with benzene in the presence of AlCl3 (2.2 equivalents). The ring opens and one acyl group attacks the benzene ring via electrophilic aromatic substitution. The product (A) is 4-oxo-4-phenylbutanoic acid (also called beta-benzoylpropionic acid or 3-(phenylcarbonyl)propanoic acid): Ph-CO-CH2-CH2-COOH. The 2.2 equivalents of AlCl3 are needed because both the anhydride opening and coordination to the carboxylic acid product require AlCl3. Step 2 – Clemmensen Reduction (A + Zn(Hg)/HCl --> B): Zn(Hg)/HCl is the Clemmensen reduction, which reduces a ketone (C=O) to a methylene (CH2) without affecting carboxylic acids. The ketone carbonyl in Ph-CO-CH2-CH2-COOH is reduced to give Ph-CH2-CH2-CH2-COOH, i.e., 4-phenylbutanoic acid (gamma-phenylbutyric acid). This is compound (B). Step 3 – Intramolecular Friedel-Crafts Acylation (B + H3PO4 --> C): H3PO4 (polyphosphoric acid or phosphoric acid as a cyclization/dehydration agent) promotes intramolecular Friedel-Crafts-type acylation. The carboxylic acid end of 4-phenylbutanoic acid cyclizes onto the benzene ring to form a five-membered or six-membered ring. With a four-carbon chain (Ph-CH2-CH2-CH2-COOH), the acyl group attacks the ortho/para position of the attached phenyl to form a six-membered ring ketone. The product is alpha-tetralone (3,4-dihydronaphthalen-1(2H)-one), a bicyclic compound with a fused aromatic ring and a saturated six-membered ring bearing a ketone at C1 (adjacent to the aromatic ring). This corresponds to option (b). Why other options fail: - Option (a): Shows a ketone at C2 (beta position), which would require a different chain length or regiochemistry not consistent with this synthesis. - Option (c) and (d): Show hydroxyl groups, but H3PO4 cyclization gives a ketone, not an alcohol. Therefore, the correct answer is B.