An optically active compound 'X' having molec | vedclass

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Question

(A) The compound $X$ is $CH_3-CH(OH)-CH_2-COOH$ ($3-$hydroxybutanoic acid).It is optically active due to the presence of a chiral center at the $C-3$

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$CH_3-CH(OH)-CH_2-COOH$

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(A) The compound $X$ is $CH_3-CH(OH)-CH_2-COOH$ ($3-$hydroxybutanoic acid).It is optically active due to the presence of a chiral center at the $C-3$ position.It evolves $CO_2$ gas when treated with $NaHCO_3$ because it contains a carboxylic acid group $(-COOH)$.Upon reduction with $LiAlH_4$,the carboxylic acid group is reduced to a primary alcohol,yielding $1,3-$butanediol $(CH_3-CH(OH)-CH_2-CH_2OH)$.Since $1,3-$butanediol is achiral (it does not possess a chiral center as the $C-3$ carbon is no longer chiral in the product),it satisfies all the given conditions.

An optically active compound $'X'$ having molecular formula $C_4H_8O_3$ evolves $CO_2$ with $NaHCO_3$. $'X'$ on treatment with $LiAlH_4$ gives an achiral compound. Then $'X'$ is:

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