Question

In the reaction,
${\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO + HCN }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{CH(OH)CN }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{CH(OH)COOH}}$
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

Answer 1

Hint: CN- ion attacks the molecule from behind which leads to the formation of a molecule with –OH group above and –CN group below. The other is when CN- attacks from above, this leads to the formation of molecules with –OH group below and –CN group above.

Complete step by step answer:
Let’s start with discussing the asymmetric centres and the type of mixtures that are formed. Asymmetric centres of an atom is the spatial arrangement of its ligands in such a way that it is not superimposable on its mirror image. Also the type of mixtures include those which contain symmetric centred molecules and those which contain asymmetric type of molecules.
In the above question, in reaction there are two possibilities with which the HCN can attack and the CH3CHO can be attacked. In first the CN- ion attacks the molecule from behind which leads to the formation of a molecule with –OH group above and –CN group below. The other is when CN- attacks from above, this leads to the formation of molecules with –OH group below and –CN group above. The possibility of both the attacks are equal. When these molecules further react with water they form the racemic mixture of acid which means 50% D + 50% L-isomer.

So, the correct answer is Option A .

Note:
A racemic mixture is the one which is optically inactive in nature. The reason behind its optical inactivity is the presence of equal amounts of D and L-isomers in the solution. Each individual enantiomer rotates the beam of light but in opposite directions which is canceled out due to the presence of an equal amount of two enantiomers. This leads to optical inactivity of a racemic mixture.