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Standard XII

Chemistry

Stereoisomerism

An optically ...

Question

An optically active compound 'X' has molecular formula $C_{4} H_{8} O_{3}$ . It evolves $C O_{2}$ with aq. $N a H C O_{3}$ .
$X$ reacts with $L i A l H_{4}$ to give an achiral compound. $X$ is :

(1)

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(2)

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(3)

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(4)

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Solution

The correct option is B (2)
Since it liberates carbon dioxide with aq sodium bicarbonate solution, it is a carboxylic acid.
It is optically active. On reduction with lithium aluminum hydride, the COOH group is reduced to $- C H_{2} C O O H$ group.
Thus the chiral carbon will now have two $- C H_{2} C O O H$ groups.
Hence, it is no longer chiral.
Hence, compound X is represented by structure 2.

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An optically active compound 'X' has molecular formula C4H8O3. It evolves CO2 with aq. NaHCO3. X reacts with LiAlH4 to give an achiral compound. X is :

An optically active compound 'X' has molecular formula $C_{4} H_{8} O_{3}$ . It evolves $C O_{2}$ with aq. $N a H C O_{3}$ .
$X$ reacts with $L i A l H_{4}$ to give an achiral compound. $X$ is :