See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When HCN adds to an achiral carbonyl compound (acetaldehyde, CH3CHO), the nucleophile (CN-) can attack either face of the planar carbonyl carbon with equal probability. This is because the carbonyl carbon in CH3CHO is sp2 hybridized and has no pre-existing chirality, so both the re-face and si-face are attacked with equal likelihood (the transition states are mirror images of each other and have equal energy). Step 1: CH3CHO is achiral (no stereocenter). The carbonyl carbon is planar (sp2). Step 2: HCN adds via nucleophilic addition. CN- attacks the carbonyl carbon. Since both faces of the carbonyl are equally accessible (no chiral influence), CN- attacks from the top face and bottom face with exactly 50% probability each, producing equal amounts of (R)- and (S)-cyanohydrin. Step 3: Hydrolysis of the nitrile (CN) group with H2O (H-OH) converts CN to COOH without breaking or inverting the bond at the stereocenter. So the stereochemical ratio is preserved. Step 4: The product CH3CH(OH)COOH is lactic acid, which has one stereocenter. The reaction produces equal amounts of D- and L-lactic acid, i.e., a racemic mixture (50%D + 50%L). Why other options fail: (a) D-isomer only - would require selective attack on one face, which does not occur with an achiral substrate and achiral reagent. (b) L-isomer only - same reasoning as above. (c) 80%D + 20%L - an unequal mixture would require some chiral influence (asymmetric induction), which is absent here. Therefore, the correct answer is D.