See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Reaction of benzene with phthalic anhydride under Friedel-Crafts conditions (AlCl3, then H3O+): Phthalic anhydride acts as an acylating agent. One carbonyl of the anhydride acylates benzene via Friedel-Crafts acylation. After aqueous workup (H3O+), the ring-opened product is 2-(benzoyl)benzoic acid (o-carboxybenzophenone). This is compound A. Step 2 – Clemmensen reduction of A (Zn-Hg, HCl, heat): The ketone carbonyl (C=O) in the benzoyl group is reduced to a CH2 group by Clemmensen reduction. The carboxylic acid is not reduced under these conditions. The product B is 2-(benzyl)benzoic acid (o-benzylbenzoic acid), i.e., a benzene ring bearing a -CH2C6H5 group and a -COOH group at ortho positions. Step 3 – Conversion of B to C (SOCl2, then AlCl3, then H3O+): SOCl2 converts the carboxylic acid (-COOH) of B to an acid chloride (-COCl), giving o-benzylbenzoyl chloride. AlCl3 then catalyzes an intramolecular Friedel-Crafts acylation: the acid chloride attacks the pendant phenyl ring of the benzyl group, forming a new six-membered ring with a ketone. H3O+ workup gives the cyclized product. The product is a tricyclic compound: two aromatic rings bridged by a -CH2- and a -C(=O)- group forming the central ring, which is 1-oxo-1,2,3,4-tetrahydroanthracene (anthrone / 10H-anthracen-9-one analogue). This corresponds to option (b): a structure with two benzene rings fused to a central ring containing one C=O and one CH2 (i.e., one position is a ketone, the other is a methylene, making it an anthrone-type compound). Why other options fail: - (a) Anthraquinone would require two ketone groups; Clemmensen reduction removed one carbonyl, so only one remains after cyclization. - (c) o-Dibenzoylbenzene would require two intact benzoyl groups; Clemmensen reduced one carbonyl. - (d) Has both benzoyl and COOH groups, which corresponds to compound A before reduction, not the final product C. Therefore, the correct answer is B.