Question
In this reaction,
In this reaction,
$$C{H_3}CHO + HCN \to $$ \[C{{H}_{3}}CH\left( OH \right)CN\xrightarrow{H.OH}\] $$C{H_3}CH\left( {OH} \right)COOH$$
an asymmetric centre is generated. The acid obtained would be
A.
$$50\% \,D + 50\% \,L{\text{ - isomer}}$$
B.
$$20\% \,D + 80\% \,L{\text{ - isomer}}$$
C.
$$D{\text{ - isomer}}$$
D.
$$L{\text{ - isomer}}$$
Answer :
$$50\% \,D + 50\% \,L{\text{ - isomer}}$$
Solution :
Lactic 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.
Lactic 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.