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  3. Chemistry
  4. An optically active compound ' X ' has molecular formula C 4 H 8 O 3. It evolves CO 2 with aq NaHCO 3. ' X ' reacts with LiAlH 4 to give an achiral compound. ' X ' is

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Q. An optically active compound ' $X$ ' has molecular formula $C _4 H _8 O _3$. It evolves $CO _2$ with aq $NaHCO _3$. ' $X$ ' reacts with $LiAlH _4$ to give an achiral compound. ' $X$ ' is

NEETNEET 2022Alcohols Phenols and Ethers

Solution:

$X \left( C _4 H _8 O _3\right)$ - optically active, thus chiral carbon is present.
$-$ It evolves $CO _2$ with $NaHCO _3$, thus $- COOH$ present.
$- LiAlH { }_4$ converts $- COOH$ into $- CH _2 OH$
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