See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Key concept: Malonic acid derivatives (gem-dicarboxylic acids) undergo decarboxylation upon heating (loss of CO2), and the stereochemical outcome depends on the structure of the resulting product. (a) PhCHD-C(CH3)(CO2H)2 heated: Step 1: This is a malonic acid derivative (gem-diacid on one carbon). Upon heating, one CO2 is lost (decarboxylation), yielding PhCHD-CH(CH3)(CO2H) -- actually the product is PhCHD-C(CH3)H-CO2H, i.e., 2-methyl-3-deutero-3-phenylpropanoic acid. Step 2: The product PhCHD-CH(CH3)COOH has two stereocenters: C2 (bearing CH3, H, COOH, CHDPh) and C3 (bearing Ph, D, H, and C2 group). Step 3: Since decarboxylation can give both possible configurations at the newly formed sp3 carbon (C2, which was formerly the malonic acid carbon), and C3 (the CHDPh carbon) retains its configuration, two products are formed that are non-superimposable and non-mirror images of each other = diastereomers. Step 4: CO2 gas is also evolved during decarboxylation. Therefore (a) matches (p) diastereomers and (s) CO2 gas will evolve. (b) Et-C(CH3)(CO2H)2 heated: Step 1: This is also a malonic acid derivative. Upon heating, decarboxylation occurs giving Et-CH(CH3)-COOH = 2-methylbutanoic acid. Step 2: The product Et-CH(CH3)COOH has one stereocenter (C2 bearing Et, CH3, H, COOH). Step 3: Since the decarboxylation proceeds without stereospecificity (the carbanion/transition state is planar or both faces are equally accessible), both R and S configurations are formed equally, giving a racemic mixture. Step 4: CO2 gas is evolved. Therefore (b) matches (q) racemic mixture and (s) CO2 gas will evolve. (c) 1,1-di(CO2H)-4-methylcyclohexane (both COOH on wedge at C1, CH3 on wedge at C4) heated: Step 1: The molecule has both COOH groups on the same carbon (C1), making it a cyclic malonic acid derivative. Upon heating, decarboxylation occurs, losing one CO2 and generating a monocarboxylic acid: 4-methylcyclohexane-1-carboxylic acid. Step 2: The product has two stereocenters: C1 (now bearing H, CO2H, and ring) and C4 (bearing CH3 and ring). Step 3: The starting material has a defined geometry: both COOH on wedge at C1 and CH3 on wedge (same side) at C4, meaning C1 and C4 substituents are cis. After decarboxylation at C1, both cis and trans products could form, but because the molecule is symmetric and the resulting product has specific geometry -- the cis-4-methylcyclohexane-1-carboxylic acid is a diastereomer of the trans form. Two diastereomeric products (cis and trans) are formed. Step 4: CO2 gas evolves. Therefore (c) matches (p) diastereomers and (s) CO2 gas will evolve. (d) Bicyclic cis-dicarboxylic acid (both CO2H on wedge/same face at adjacent bridgehead carbons of a bicyclic system, e.g., cis-norbornane-2,3-dicarboxylic acid or similar): Step 1: This is NOT a gem-diacid (malonic acid type); the two COOH groups are on adjacent carbons of a bicyclic ring system with cis (same-face) geometry. Step 2: Cis-1,2-dicarboxylic acids on rigid ring systems can form cyclic anhydrides easily, but upon strong heating, they can undergo decarboxylation differently. However, in the context of this problem, the bicyclic cis-diacid upon heating gives a product. Step 3: The cis-bicyclic dicarboxylic acid, upon decarboxylation (losing CO2 from one COOH) or via anhydride formation and then decarboxylation, yields a monocarboxylic acid. Due to the rigid bicyclic framework and the cis relationship, the product has a defined stereochemistry. The product of decarboxylation of one COOH from the cis-bicyclic diacid gives a product where the remaining COOH and the ring junction create a meso-like or specific stereochemical situation. Step 4: For the specific structure shown (cis-bicyclo dicarboxylic acid where both COOH are on the same face), the decarboxylation product is a meso compound because the resulting molecule has an internal plane of symmetry. Step 5: Note: No CO2 gas option listed for (d), and the answer given is only (r), suggesting either no decarboxylation occurs giving CO2 freely, or the primary product is the meso compound via a different mechanism (e.g., anhydride formation then ring opening, or direct thermal rearrangement). The answer confirms (d) → (r) meso compound only. Therefore (d) matches (r) meso compound. Therefore, the correct answer is a-p,s; b-q,s; c-p,s; d-r.