See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The aldehyde shown is 3-methylpentanal: CH3CH2CH(CH3)CH2CHO. It has one pre-existing stereocenter at C3 (the carbon bearing the methyl group). However, as the starting material is drawn without specification of configuration, it is a racemic mixture (both R and S at C3). Step 2 - Reaction with NaCN then H2SO4: NaCN adds CN⁻ to the aldehyde carbonyl to give a cyanohydrin, creating a new stereocenter at the former carbonyl carbon (C1, now bearing OH, CN, H, and the rest of the chain). Subsequent H2SO4 hydrolysis converts CN to COOH, giving a 2-hydroxy acid (alpha-hydroxy acid). The new stereocenter is C1 (now C2 in the product numbering) bearing OH, COOH, H, and CH2CH(CH3)CH2CH3. Step 3 - Stereochemical analysis: The starting aldehyde has one stereocenter at C3. The CN addition creates a second stereocenter at C2 (the former carbonyl carbon). The CN⁻ can attack from either face of the carbonyl (re or si face) giving two possible configurations at C2, while the existing stereocenter at C3 is unchanged. Since the starting material is racemic (R and S at C3), and CN attack at each enantiomer gives both R and S at C2, we obtain four possible stereoisomers: (2R,3R), (2S,3R), (2R,3S), (2S,3S). These four products are two pairs. The (2R,3R) and (2S,3S) are enantiomers of each other; the (2R,3S) and (2S,3R) are enantiomers of each other. The pair (2R,3R)/(2S,3S) vs the pair (2R,3S)/(2S,3R) are diastereomers with respect to each other. The two stereocenters are not identical substituents (C2 has COOH, OH, H, CH2- chain vs C3 has CH3, Et, H, CH2-), so no meso compound is possible. The product is thus a mixture of two pairs of enantiomers, i.e., diastereomers. Step 4 - Why other options fail: (a) Racemic mixture - incorrect because two different stereocenters are created giving diastereomers, not just a single pair of enantiomers. (c) Meso - incorrect because the two stereocenters have different substituents, so no internal mirror plane is possible. (d) Mixture of meso and optically active - incorrect for the same reason as (c). Therefore, the correct answer is B.