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JEE CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

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Analyzing +4 | -1
An optically active compound 'X{X}' has molecular formula C4H8O3{C}_4 {H}_8{O}_3. It evolves CO2{CO}_2 with NaHCO3{NaHCO}_3.
'X{X}' reacts with LiAlH4{LiAlH}_4 to give an achiral compound 'X{X}' is
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Correct option: C
CH3−CH∣CH2OH−COOHC{H_3} - \mathop {CH}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_2}OH} \end{array}} - COOH
Option C is correct because the compound reacts with NaHCO3 to release CO2, indicating the presence of a carboxylic acid group, and reduction with LiAlH4 yields an achiral compound.

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