In this reaction: $\begin{aligned} \mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCN} & \rightarrow \mathrm{CH}_3…

In this reaction: $\begin{aligned} \mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCN} & \rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CN} \stackrel{\mathrm{H} . \mathrm{OH}}{\longrightarrow} & \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} \end{aligned}$ an symmetric centre is generated. The acid obtained would be:
  1. D-isomer
  2. L-isomer
  3. $50 \% \mathrm{D}+50 \%$ L-isomer
  4. $20 \% \mathrm{D}+80 \% \mathrm{~L}$-isomer

Solution

Lactic acid is an optically active compound due to presence of assymmetric carbon. It exist in D and $\mathrm{L}$ form in the ratio $1: 1$, forming a racemic mixture. Related Theory Enantiomers are stereoisomers which are non superimposable, mirror images. A mixture of equal amounts of two stereoisomers of an optically active substance is called a racemic mixture or racemate.

Asked in: NEET 2003

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