See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: The reaction of 1,3-indandione with Br2/KOH is a haloform reaction combined with base-mediated hydrolysis. Step 1 - Identify the reactive site: 1,3-Indandione has a CH2 group flanked by two carbonyl groups (position 2). This activated methylene is highly acidic and reactive. Step 2 - Haloform reaction on the CH2: Under Br2/KOH conditions, the CH2 (which is equivalent to a methyl ketone-like system with respect to halogenation) undergoes successive bromination. The carbon bearing two carbonyls with the CH2 group reacts similarly to how a methyl ketone undergoes haloform reaction. The two C=O groups adjacent to CH2 make this carbon very electon-poor and susceptible. Step 3 - CHBr3 formation: The haloform reaction cleaves off CHBr3 (bromoform), as indicated by the CHBr3 precipitate in the equation. This cleavage breaks the five-membered ring at the C2 position. Step 4 - Product identification: After loss of CHBr3 from the C2 position, the remaining fragment is the benzene ring bearing two carboxylate groups (from the two C=O groups that were at C1 and C3 of the indandione). Under KOH (basic) conditions, the carboxylic acids are deprotonated to give carboxylate anions (CO2^-). Therefore the product (A) is potassium phthalate / phthalate dianion: a benzene ring with two CO2^- groups in ortho positions, which corresponds to option (c). Why other options fail: (a) shows the free acid form (CO2H), but reaction is done in KOH so product is the salt/anion form CO2^-. (b) shows one carboxylic acid and one acetyl group, which is not consistent with the cleavage pattern. (d) shows a methyl group, which is incorrect as no methyl group is generated. Therefore, the correct answer is C.