See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Thermal decarboxylation of malonic acid derivatives (beta-keto acids or malonic acid-type diacids) proceeds via a cyclic six-membered transition state. When a compound bearing two carboxylic acid groups (a malonic acid-type or 1,1-dicarboxylic acid arrangement) is heated, one CO2 is lost. The key question is which CO2 is lost and what stereochemistry results. Step 1: Identify the starting material. The starting material is a bicyclic compound (norbornane-type or bicyclo framework) bearing two COOH groups on the same carbon (geminal dicarboxylic acid, i.e., malonic acid derivative at a ring carbon). One CO2H appears to be on a bridgehead-adjacent carbon and another on the same carbon (gem-diacid), making this a substituted malonic acid. Step 2: Thermal decarboxylation mechanism. A gem-dicarboxylic acid (malonic acid type) loses CO2 upon heating via a cyclic transition state to give a monocarboxylic acid. The product retains one COOH group. Step 3: Stereochemical outcome. In the bicyclic system, the carbon bearing both COOH groups loses one CO2. The remaining COOH can be in two orientations (the proton delivered in the cyclic TS can come from either COOH group acting as donor), giving two possible stereochemical products at that carbon - one with COOH in the configuration shown in (a) and one shown in (b). Since decarboxylation via the cyclic 6-membered TS can proceed through either COOH group donating the proton, both epimers at the resulting carboxylic acid carbon can form. Step 4: Why both (a) and (b) are correct. The thermal decarboxylation of a malonic acid derivative embedded in a rigid bicyclic system can produce both possible stereoisomers of the monocarboxylic acid product, because the cyclic transition state geometry allows loss of CO2 with the proton delivered to either face. Both option (a) and option (b) represent the two stereoisomeric monocarboxylic acid products differing in the configuration of the COOH-bearing carbon. Step 5: Why (d) fails. Both (a) and (b) are legitimate products, so 'none of these' is incorrect. Therefore, the correct answer is C.