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Step-by-Step Solution
Step 1: Recognizing the Reaction
Acetaldehyde ($CH_3CHO$) reacts with hydrogen cyanide ($HCN$) to form a cyanohydrin ($CH_3CH(OH)CN$). On subsequent hydrolysis (treatment with water under acidic conditions), the nitrile ($-CN$) group of the cyanohydrin is converted into a carboxylic acid group ($-COOH$). Hence, the final product is an α-hydroxy carboxylic acid of the form $CH_3CH(OH)COOH$.
Step 2: Identifying the Newly Generated Chiral Centre
In the product $CH_3CH(OH)COOH$, the carbon bearing the $OH$ group (the former cyanohydrin carbon) is attached to four different groups:
$CH_3-$ (methyl group)
$-H$ (a hydrogen)
$-OH$ (hydroxyl group)
$-COOH$ (carboxylic acid group)
Because this carbon has four different substituents, it is a chiral (asymmetric) centre.
Step 3: Explaining the Formation of a Racemic Mixture
When $HCN$ attacks the planar carbonyl group of acetaldehyde ($CH_3CHO$), the nucleophile ($CN^-$) can approach from either face (above or below the plane) with equal probability. As a result, both possible configurations (R and S, or in biochemical notation D and L) of the newly generated chiral center are formed in equal proportions.
During the subsequent hydrolysis step, the configuration at that chiral center does not change, and the two enantiomers persist. Therefore, the final α-hydroxy carboxylic acid is obtained as a 1:1 mixture of its two enantiomeric forms.
Step 4: Conclusion
Because the incoming group can add with equal likelihood from either side, the product is a racemic mixture containing 50% D- and 50% L-isomer. Thus, the correct answer is:
50% D + 50% L-isomer
