Identify ( Z ) in the following reaction: CH 3 COOH ⟶ LiAlH 4 ( X ) 573   K Cu ( Y ) ⟶…

The ethanal has the alpha hydrogen. So, the weak base will take the acidic hydrogen. It will form the carbanion. After that the carbanion will give the nucleophilic addition reaction with another mole of ethanal. The $\beta$-hydroxylcarbonyl compound is formed as a product.

Reaction :-

${\mathrm{CH}}_3\mathrm{COOH}\stackrel{{\mathrm{LiAlH}}_4}{⟶}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\underset{573\mathrm{K}}{\overset{\mathrm{Cu}}{\to }}{\mathrm{CH}}_3\mathrm{CHO}\stackrel{ \mathrm{dil}.\mathrm{NaOH} }{⟶}{\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_2-\mathrm{CHO}\left(\mathrm{Aldol}\right)$

Hence the $\mathrm{Z}$ in this reaction is aldol.