See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is phthalic anhydride, a cyclic anhydride where the two carbonyl groups are bridged by an oxygen atom in a five-membered ring fused to a benzene ring. Step 2 - Reaction with MeOH (A): When phthalic anhydride reacts with methanol (MeOH), the anhydride ring opens. Methanol acts as a nucleophile attacking one of the carbonyl carbons, opening the cyclic anhydride to give a half-ester/half-acid product: one carboxylic acid group (-COOH) and one methyl ester group (-COOMe) on the ortho positions of the benzene ring. This product A is mono-methyl hydrogen phthalate (2-(methoxycarbonyl)benzoic acid). Step 3 - Reaction with PCl3 (B): PCl3 converts a carboxylic acid (-COOH) into an acid chloride (-COCl). PCl3 selectively reacts with the free carboxylic acid group rather than the ester. Therefore, product B is methyl 2-(chlorocarbonyl)benzoate: one -COCl group and one -COOMe group at ortho positions on the benzene ring. Step 4 - Reaction with MeNH2 (C): Methylamine (MeNH2) is a nucleophile that reacts with the more reactive acid chloride (-COCl) preferentially over the ester (-COOMe). The amine attacks the carbonyl of the acid chloride, displacing chloride and forming an amide bond. This gives product C with one -C(=O)-NH-Me (methylamide) group and one -C(=O)-O-Me (methyl ester) group at ortho positions on the benzene ring. Step 5 - Match with options: Product C has -C(=O)-NH-Me and -C(=O)-O-Me groups ortho to each other on benzene. This matches option (b) exactly. Why option (a) fails: Option (a) shows a -C(=O)-Cl and -C(=O)-NH-Me, which would correspond to product B after only partial reaction, not after MeNH2 treatment. After MeNH2 reacts with the acid chloride, no chloride remains. Therefore, the correct answer is B.