See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Decarboxylation (loss of CO2) occurs most readily when a carboxylic acid has a carbonyl group (C=O) at the beta position (two carbons away, i.e., the beta-keto acid arrangement). This allows a cyclic six-membered (or in simple cases four-membered) transition state where the beta-carbonyl assists in removing CO2 via a concerted mechanism. Step 1 - Identify beta-keto acid structures: For easy decarboxylation, we need the pattern: -C(=O)-CH2-COOH (or -C(=O)-CHR-COOH), where the carbonyl is directly beta to the carboxyl group. Step 2 - Analyze option (a): Ph-C(=O)-CH2-CH2-COOH. The ketone is gamma to the carboxyl (there are two CH2 groups between C=O and COOH). This is a gamma-keto acid, not a beta-keto acid. Decarboxylation is not as facile. Step 3 - Analyze option (b): HO-C(=O)-CH(Ph)-CH2-C(=O)-Et. The carboxyl carbon is C1, the alpha carbon bears Ph, then CH2, then C=O. The carbonyl is at the gamma position relative to the carboxyl. Not a true beta-keto acid. Step 4 - Analyze option (c): This is a diacid-like compound with a benzene ring linker; meta-(HOOC)-C6H4-(CH2)3-COOH. No beta-keto arrangement; decarboxylation would require very harsh conditions. Step 5 - Analyze option (d): HO-C(=O)-CH(Ph)-CH2-C(=O)-CH2CH3. Here C1=COOH, C2=CH(Ph) (alpha), C3=CH2, C4=C=O (ketone). Wait - re-examining the structure: the carboxyl is at one end, and looking at the drawing, the carbonyl (C=O) appears to be directly at the beta carbon (C3 position from COOH). Actually from the image, option (d) shows Ph-CH(COOH)-CH2-CO-Et, meaning the keto group is beta to the carboxyl: HOOC-CHPh-CH2-CO-Et. The beta carbon to COOH is CHPh (alpha), and the next carbon bears the ketone - this is a beta-keto acid (the carbonyl is on the beta carbon counting from the carboxyl). This arrangement -CO-CHR-COOH most readily loses CO2. Step 6 - Comparing (b) and (d): Both appear similar, but in option (d) the carbonyl is at the beta position directly (C=O is on C3, which is beta to the carboxyl carbon C1, with C2 being alpha carbon bearing Ph). In option (b), the arrangement places the ketone further away. Option (d) has the classic beta-keto acid pattern: R-CO-CHPh-COOH, which forms a stable enol after decarboxylation and proceeds through a favorable transition state. Step 7 - Why (d) is best: Option (d) is a beta-keto acid (ketone at beta position relative to carboxylic acid: Et-CO-CH2-CHPh-COOH where the keto is beta), enabling the well-known facile thermal decarboxylation via a cyclic transition state, releasing CO2 and forming a ketone. The phenyl group at alpha provides additional stabilization. This makes (d) the most readily decarboxylated compound. Therefore, the correct answer is D.