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(C) $1$. The molecular formula $C_4H_8O_3$ and the evolution of $CO_2$ with $NaHCO_3$ indicate that $X$ is a carboxylic acid.$2$. $X$ is optically act

Vedclass · Accepted Answer

$CH_3CH(CH_2OH)COOH$

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(C) $1$. The molecular formula $C_4H_8O_3$ and the evolution of $CO_2$ with $NaHCO_3$ indicate that $X$ is a carboxylic acid.$2$. $X$ is optically active,meaning it must contain a chiral center.$3$. Reduction of $X$ with $LiAlH_4$ yields an achiral compound. Let us analyze option $C$: $CH_3CH(CH_2OH)COOH$.$4$. $CH_3CH(CH_2OH)COOH$ has a chiral center at the $C-2$ position.$5$. Upon reduction with $LiAlH_4$,the carboxylic acid group $-COOH$ is reduced to $-CH_2OH$,resulting in $CH_3CH(CH_2OH)CH_2OH$.$6$. In $CH_3CH(CH_2OH)CH_2OH$,the central carbon is attached to two identical $-CH_2OH$ groups,making the molecule achiral.$7$. Thus,$X$ is $CH_3CH(CH_2OH)COOH$.

An optically active compound $X$ has molecular formula $C_4H_8O_3$. It evolves $CO_2$ with $NaHCO_3$. $X$ reacts with $LiAlH_4$ to give an achiral compound. $X$ is

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