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

Correct Answer: A

Solution :

 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. \[\underset{50%\,\text{L}-\text{isomer}}{\mathop{H-\underset{CN}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-OH}}\,\,\,\,\,\,+\,\,\,\,\,\,\underset{50%\,D\,-\text{isomer}}{\mathop{HO-\underset{CN}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H}}\,\]