In the following reaction,CH_3-C equiv CH xri | vedclass

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(B) In the first step,propyne $(CH_3-C \equiv CH)$ undergoes hydration in the presence of $Hg^{2+}$ and $H_3O^+$ to form acetone $(CH_3COCH_3)$ as $X$

Vedclass · Accepted Answer

$X = CH_3COCH_3$,$Y = CH_3COCH=CHPh$

Vedclass · Answer

(B) In the first step,propyne $(CH_3-C \equiv CH)$ undergoes hydration in the presence of $Hg^{2+}$ and $H_3O^+$ to form acetone $(CH_3COCH_3)$ as $X$.In the second step,acetone reacts with benzaldehyde $(PhCHO)$ in the presence of dilute $NaOH$. This is a Claisen-Schmidt condensation (a type of cross-aldol condensation) where the $\alpha$-hydrogen of acetone reacts with the carbonyl group of benzaldehyde to form $4$-phenylbut-$3$-en-$2$-one $(CH_3COCH=CHPh)$ as $Y$.

In the following reaction,
$CH_3-C \equiv CH$ $\xrightarrow[H_3O^+]{Hg^{2+}} X$ $\xrightarrow[PhCHO]{dil. NaOH} Y$
$X$ and $Y$,respectively,are

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In the following reaction,$CH_3-C \equiv CH$ $\xrightarrow[H_3O^+]{Hg^{2+}} X$ $\xrightarrow[PhCHO]{dil. NaOH} Y$$X$ and $Y$,respectively,are