See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Determine X - moles of CO2 released from the hexacarboxylic cyclohexanone compound upon heating. The starting material is a cyclohexanone with six carboxylic acid groups: two at C2 (geminal), two at C6 (geminal), and two at C4 (geminal). When heated, malonic acid-type (geminal dicarboxylic acid) groups undergo decarboxylation. Each pair of geminal diacid groups on a carbon can lose one CO2 via malonic ester-type decarboxylation, but we must also consider beta-keto acid decarboxylation. Let's analyze each position: - C2 (alpha to ketone): has two COOH groups. The carbon is a malonic acid derivative AND a beta-keto acid. Beta-keto acid decarboxylation removes one CO2. Then the remaining gem-COOH (now a single COOH on alpha carbon, or it could further decarboxylate as malonic type). Actually, with two COOH groups at C2 (alpha to C=O), both can decarboxylate: first as beta-keto acid (1 CO2), then as another beta-keto acid after the first loss (1 CO2). So 2 CO2 from C2. - C6 (alpha to ketone on other side): same situation, 2 CO2. - C4 (not alpha to ketone): has two COOH groups (malonic acid type). Malonic decarboxylation gives 1 CO2. Total X = 2 + 2 + 1 = 5 moles of CO2. Wait, let me reconsider. The structure shows C1 has the ketone, C2 has HO2C and CO2H (two COOH), C3 is a CH2, C4 has HO2C and CO2H, C5 is CH2, C6 has HO2C and CO2H. Actually re-reading: the ring is cyclohexanone with gem-diCOOH at three positions. Positions alpha to ketone (C2 and C6): each is a beta-keto malonic acid - can lose 2 CO2 each (both COOH groups decarboxylate because after first loss, the remaining COOH is still beta to keto). So 2+2=4 CO2 from alpha positions. Position C4 (beta to ketone, not alpha): malonic-type, loses 1 CO2. Total X = 4 + 1 = 5. Step 2: Determine Y - number of beta-keto acids (including stereoisomers) that give phenylacetone (PhCH2COCH3) upon heating. The product is PhCH2-CO-CH3 (1-phenylpropan-2-one). Beta-keto acids lose CO2 upon heating: R-CO-CH(R')-COOH → R-CO-CH2R' + CO2. Possible beta-keto acids that give PhCH2COCH3: 1. Ph-CH2-CO-CH2-COOH (4-phenyl-3-oxobutanoic acid): decarboxylation gives PhCH2COCH3. No stereocenters. 1 compound. 2. Ph-CH(decarboxylation from other side)-CO-CH3... i.e., Ph-CH(COOH)-CO-CH3 (2-methyl-3-oxo-3-phenylpropanoic acid? No): Ph-C(COOH)H-CO-CH3. This has a stereocenter at the alpha carbon. Gives 2 stereoisomers (R and S). So 2 compounds. Total Y = 1 + 2 = 3. Step 3: X + Y = 5 + 3 = 8. Therefore, the correct answer is 8.