An optically active compound X has the molecu | vedclass

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(B) $1.$ The evolution of $CO_2$ with $NaHCO_3$ indicates the presence of a carboxylic acid group $(-COOH)$.$2.$ The molecular formula $C_4H_8O_3$ and

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$CH_3-CH(CH_2OH)-COOH$

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(B) $1.$ The evolution of $CO_2$ with $NaHCO_3$ indicates the presence of a carboxylic acid group $(-COOH)$.$2.$ The molecular formula $C_4H_8O_3$ and optical activity suggest the presence of a chiral center.$3.$ Reaction with $LiAlH_4$ reduces the $-COOH$ group to a $-CH_2OH$ group.$4.$ For Option $B$: $CH_3-CH(CH_2OH)-COOH \xrightarrow{LiAlH_4} CH_3-CH(CH_2OH)-CH_2OH$. The product,$2$-methylpropane-$1,3$-diol,is achiral because it has a plane of symmetry (the central carbon is attached to two identical $-CH_2OH$ groups).$5.$ Options $A$ and $D$ yield chiral products upon reduction,and Option $C$ has the incorrect molecular formula $(C_3H_6O_3)$.

An optically active compound $X$ has the 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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