In this reaction : $CH_3CHO + HCN \rightarrow CH_3CH(OH)CN \xrightarrow{H.OH} CH_3CH(OH)COOH$, an asymmetric centre is generated. The acid obtained would be :
A
D-isomer
B
L-isomer
C
50% D + 50% L-isomer
D
20% D + 80% L-isomer
The answer is C.
Solution
The carbonyl group of acetaldehyde is planar ($sp^2$ carbon). The cyanide ion can attack this carbon equally from either face of the plane. Attack from one face gives one configuration of the new chiral carbon in $CH_3CH(OH)CN$, and attack from the other face gives the mirror image configuration. Both attacks are equally probable, since there is no chiral influence (no chiral reagent, catalyst or solvent). So the two enantiomers are formed in equal amounts. Hydrolysis of the cyanohydrin then gives lactic acid, $CH_3CH(OH)COOH$, also as a 1 : 1 mixture of the D and L forms, i.e. a racemic mixture which is optically inactive. So the acid obtained is 50% D + 50% L-isomer.
ConceptRacemic product from planar carbonyl
Aldehydes, Ketones and Carboxylic Acids in past papers
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