See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When a dicarboxylic acid (or a compound with two carboxylic acid groups in close proximity) is treated with acetic anhydride (Ac2O) and heat, intramolecular dehydration can occur to form a cyclic anhydride. Step 1: Identify the starting material. The starting material is 2-(carboxymethyl)benzoic acid (homophthalic acid), which has a -CH2-CO2H group and a -CO2H group on adjacent (ortho) carbons of a benzene ring. Step 2: Determine the reaction. Acetic anhydride (Ac2O) with heat is a classic reagent for dehydrating carboxylic acids to form anhydrides. When two -COOH groups are in proximity, they undergo intramolecular condensation to form a cyclic anhydride with loss of water. Step 3: Count the ring size of the anhydride. The two acid groups are: ArCO2H (directly on ring) and ArCH2CO2H (one carbon away from ring). When these two form an anhydride linkage (-C(=O)-O-C(=O)-), the resulting ring includes: Ar-C(=O)-O-C(=O)-CH2- closing back to the Ar carbon, forming a six-membered ring (counting: C(aryl)-C(=O)-O-C(=O)-CH2-C(aryl), which is a 6-membered ring). This is homophthalic anhydride (2H-isochromen-1,3-dione), a bicyclic compound with benzene fused to a six-membered anhydride ring containing a CH2 group. Step 4: Match to options. Option (c) shows a benzene ring fused to a six-membered ring with two C=O groups and one oxygen (the anhydride linkage) and a CH2 group — this is homophthalic anhydride, consistent with our product. Step 5: Eliminate other options. - Option (a): Shows CCl2Ac group, which would require chlorine atoms not present in the reagents. Incorrect. - Option (b): Shows a mixed product with one acid and one acetate ester. While Ac2O can form mixed anhydrides, the intramolecular cyclic anhydride is strongly favored under heating conditions. Incorrect. - Option (d): Shows a five-membered anhydride ring (like phthalic anhydride), which would form from two directly ring-attached COOH groups (ortho-dicarboxylic acid). Here one acid is via a CH2 spacer, so the ring is six-membered, not five-membered. Incorrect. Therefore, the correct answer is C.