JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Hydrocarbons

Phthalic anhydride undergoes Friedel–Crafts acylation with benzene ($AlC{l}_3$ catalyst); the anhydride ring opens, one carbonyl acylates the arene, and the other remains as a free carboxylic acid. This gives 2-benzoylbenzoic acid (A) — a biphenyl-type skeleton carrying a ketone linkage on one ring and a free $-COOH$ on the other.
Clemmensen reduction ($Zn$–$Hg/HCl$) then reduces only the ketone carbonyl to a methylene ($-C{H}_2-$) group; it does not touch the carboxylic acid. This converts A into 2-benzylbenzoic acid (B). So both A and B retain the $-COOH$ group, with A carrying the $C=O$ and B the corresponding $C{H}_2$ — matching the option in which both structures are shown bearing $-COOH$.