CH_3COOH xrightarrow{LiAlH_4} X xrightarrow{C | vedclass

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(B) $Step \ 1$: Reduction of acetic acid $(CH_3COOH)$ with $LiAlH_4$ gives ethanol $(CH_3CH_2OH)$ as $X$.$Step \ 2$: Dehydrogenation of ethanol $(X)$

Vedclass · Accepted Answer

aldol

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(B) $Step \ 1$: Reduction of acetic acid $(CH_3COOH)$ with $LiAlH_4$ gives ethanol $(CH_3CH_2OH)$ as $X$.$Step \ 2$: Dehydrogenation of ethanol $(X)$ over $Cu$ at $300^{\circ}C$ gives acetaldehyde $(CH_3CHO)$ as $Y$.$Step \ 3$: Acetaldehyde $(Y)$ undergoes aldol condensation in the presence of dilute $NaOH$ to form $3-hydroxybutanal$ $(CH_3CH(OH)CH_2CHO)$,which is commonly known as aldol,as $Z$.Therefore,the correct option is $B$.

$CH_3COOH$ $\xrightarrow{LiAlH_4} X$ $\xrightarrow{Cu, 300^{\circ}C} Y$ $\xrightarrow{\text{Dilute } NaOH} Z$. In the above reaction,$Z$ is:

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