question_answer

                In this reaction:                                                                                                                                                    \[\begin{align}   & C{{H}_{3}}CHO+HCN\xrightarrow[{}]{{}}C{{H}_{3}}CH(OH)CN \\  & \xrightarrow{H.OH}C{{H}_{3}}CH(OH)COOH \\ \end{align}\]                 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 :

                Lactice acid obtained in the given reaction is an optically active compound due to the presence of chiral C-atom. It exits as d and \[l\] forms whose ratio is 1 : 1.