(A) xrightarrow[dil. H_2SO_4]{HgSO_4} (B) xri | vedclass

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(C) The reaction sequence is as follows:$1$. The hydration of an alkyne $(A)$ in the presence of $HgSO_4$ and $dil. H_2SO_4$ yields a ketone $(B)$.$2$

Vedclass · Accepted Answer

$CH_3-C \equiv C-CH_3$

Vedclass · Answer

(C) The reaction sequence is as follows:$1$. The hydration of an alkyne $(A)$ in the presence of $HgSO_4$ and $dil. H_2SO_4$ yields a ketone $(B)$.$2$. The reduction of ketone $(B)$ with $LiAlH_4$ yields a secondary alcohol $(C)$.$3$. For the product $(C)$ to be a racemic mixture,it must contain a chiral center. This implies that the ketone $(B)$ must be an unsymmetrical ketone such that its reduction creates a chiral carbon.$4$. Looking at the options,$CH_3-C \equiv C-CH_3$ (but$-2-$yne) is symmetrical and would yield butan$-2-$one,which upon reduction gives a racemic mixture of butan$-2-$ol.$5$. However,the provided image shows the reaction of but$-2-$yne $(CH_3-C \equiv C-CH_3)$ to form butan$-2-$one,which is then reduced to a racemic mixture of butan$-2-$ol.$6$. Therefore,reactant $(A)$ is $CH_3-C \equiv C-CH_3$.

$(A)$ $\xrightarrow[dil. H_2SO_4]{HgSO_4} (B)$ $\xrightarrow{LiAlH_4} \underset{\text{racemic mixture}}{(C)}$
$\therefore$ reactant $(A)$ is

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$(A)$ $\xrightarrow[dil. H_2SO_4]{HgSO_4} (B)$ $\xrightarrow{LiAlH_4} \underset{\text{racemic mixture}}{(C)}$$\therefore$ reactant $(A)$ is