• # question_answer In this reaction $C{{H}_{3}}CHO+HCN\xrightarrow{{}}C{{H}_{3}}CH(OH)CN$ $\xrightarrow{\text{H}\text{.OH}}C{{H}_{3}}CH(OH)COOH$ an asymmetric centre is generated. The acid obtained would be A)  50% D + 50% L-isomer B)  20% D + 80% L-isomer C)  D-isomer D)  L-isomer

Lactic acid obtained in the given reaction is an optically active compound due to the presence of chiral C-atom. It exists as d and $l$ forms whose ratio is 1:1. $\underset{50%\,\text{L}-\text{isomer}}{\mathop{H-\underset{CN}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-OH}}\,\,\,\,\,\,+\,\,\,\,\,\,\underset{50%\,D\,-\text{isomer}}{\mathop{HO-\underset{CN}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H}}\,$