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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:
Options:
$\begin{aligned}
\mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCN} & \rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CN} \\
\stackrel{\mathrm{H} . \mathrm{OH}}{\longrightarrow} & \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}
\end{aligned}$
an symmetric centre is generated. The acid obtained would be:
Solution:
2317 Upvotes
Verified Answer
The correct answer is:
$50 \% \mathrm{D}+50 \%$ L-isomer
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.
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.
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