In the reaction sequence $CH_{3}COOH \xrightarrow{LiAlH_{4}} X \xrightarrow{Cu, 300^{\circ}C} Y \xrightarrow{dil.\ NaOH} Z$, the final product Z is
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- butanol
- aldol
- ketol
- acetal
The Correct Option is B
Solution and Explanation
$CH_3COOH$ is reduced by $LiAlH_4$ to form ethyl alcohol ($X = CH_3CH_2OH$).
Step 2: Second Reaction
Dehydrogenation of ethanol over hot Copper produces acetaldehyde ($Y = CH_3CHO$).
Step 3: Final Reaction
Acetaldehyde reacts with dilute $NaOH$ to undergo an Aldol Condensation reaction, forming aldol ($Z = CH_3CH(OH)CH_2CHO$).
Final Answer: (b)
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