JEE Advanced 2026 · Paper 1, Chemistry question 11

X. $\ce{HO2C-CH(CH3)-CO-CH(CH3)-CO2H}$ is a β-keto diacid. On heating, both carboxyl groups are lost as $\ce{CO2}$, giving pentan-3-one, $\ce{CH3CH2COCH2CH3}$: 1 C=O.

Y. In 4-oxocyclopentane-1,2-dicarboxylic acid the carboxyl groups sit on the β-carbons of the ketone, not the α-carbons, so neither is a β-keto acid and neither decarboxylates. The adjacent cis carboxyl groups lose water to form a five-membered cyclic anhydride. Y keeps the ketone and adds two anhydride C=O groups: 3 C=O.

Total $1+3=4$.