See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: NaH^14CO3 (sodium bicarbonate with 14C-labeled carbon) reacts with acids to produce ^14CO2 gas as the labeled carbon source. The ^14CO2 gas then acts as the electrophilic carbon in a Grignard reaction with PhMgBr. Step 2 - Reaction pathway for acetic acid (CH3COOH): - CH3-C(=O)-O-H reacts with NaH^14CO3. The acid-base reaction produces ^14CO2 gas (labeled CO2) as compound (A). The acetate ion (CH3COO-) is the byproduct but (A) is the gas = ^14CO2. - ^14CO2 + PhMgBr (Grignard reagent) undergoes nucleophilic addition: Ph attacks the ^14C of CO2, forming Ph-^14C(=O)-OMgBr after addition. - Upon workup with H3O+, this gives Ph-^14C(=O)-OH, i.e., benzoic acid with the carbonyl carbon labeled as 14C. This is product (B). Step 3 - Reaction pathway for methanesulfonic acid (CH3SO3H): - CH3-S(=O)(=O)-O-H reacts with NaH^14CO3. Similarly, the acidic proton reacts with bicarbonate to release ^14CO2 gas as compound (C). - ^14CO2 + PhMgBr: same Grignard addition gives Ph-^14C(=O)-OMgBr, then after H3O+ workup gives Ph-^14C(=O)-OH, benzoic acid with 14C label. This is product (D). Step 4 - Key insight: Both reactions produce the same ^14CO2 gas intermediate. Both react with PhMgBr followed by H3O+ to give Ph-^14COOH (benzoic acid with 14C at carbonyl carbon). Therefore both B and D are Ph-C(=O)-OH with 14 on the carbonyl carbon. Step 5 - Why other options fail: - Option (a): Shows unlabeled products, ignoring the 14C label entirely. - Option (b): Shows 14C label only on B but not correctly on D, and shows a sulfinic acid for D which is wrong. - Option (d): Shows unlabeled benzoic acid and a sulfinic acid derivative, both incorrect. - Option (c): Correctly shows both B and D as Ph-^14C(=O)-OH (benzoic acid with 14C label on the carboxyl carbon), which matches the mechanism where ^14CO2 is the electrophile in both Grignard reactions. Therefore, the correct answer is C.