JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

Reaction 1: The reactant is 2,4-dimethyl-3-oxopentanedioic acid, which is a -keto acid (both - CO _2 H groups are at the -position with respect to the central ketone group). -keto acids readily undergo decarboxylation upon heating. Since there are two such groups, it will undergo double decarboxylation. HO _2 C - CH ( CH _3)- C (= O )- CH ( CH _3)- CO _2 H CH _3- CH _2- C (= O )- CH _2- CH _3 + 2 CO _2 The major product X is 3-pentanone. Number of carbonyl groups (> C=O ) in X = 1. Reaction 2: The reactant is *cis*-3,4-dicarboxycyclopentanone. The two carboxylic acid groups are on adjacent carbons (a 1,2-dicarboxylic acid system). Heating a 1,2-dicarboxylic acid leads to dehydration, forming a stable 5-membered cyclic anhydride. The major product Y is the cyclic anhydride of the reactant, which is a fused bicyclic system containing the original cyclopentanone ring and a newly formed succinic anhydride ring. The structure of Y contains one ketone carbonyl group and two anhydride carbonyl groups. Number of carbonyl groups (> C=O ) in Y = 1 + 2 = 3. Sum of total number of carbonyl groups in X and Y = 1 + 3 = 4. Answer: 4