CH equiv CH xrightarrow[H_2SO_4]{H_2O/Hg^{2+} | vedclass

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Question

(C) The reaction sequence is as follows:$1$. $CH \equiv CH$ undergoes hydration in the presence of $Hg^{2+}/H_2SO_4$ (Kucherov reaction) to form aceta

Vedclass · Accepted Answer

Ethyl bromide

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(C) The reaction sequence is as follows:$1$. $CH \equiv CH$ undergoes hydration in the presence of $Hg^{2+}/H_2SO_4$ (Kucherov reaction) to form acetaldehyde $(X)$: $CH \equiv CH + H_2O \xrightarrow{Hg^{2+}/H_2SO_4} CH_3-CHO$.$2$. Acetaldehyde $(X)$ is reduced by $LiAlH_4$ to form ethanol $(Y)$: $CH_3-CHO \xrightarrow{LiAlH_4} CH_3-CH_2OH$.$3$. Ethanol $(Y)$ reacts with $P_4/Br_2$ (which generates $PBr_3$ in situ) to form ethyl bromide $(Z)$: $CH_3-CH_2OH \xrightarrow{P_4/Br_2} CH_3-CH_2Br$.

$CH \equiv CH$ $\xrightarrow[H_2SO_4]{H_2O/Hg^{2+}} X$ $\xrightarrow{LiAlH_4} Y$ $\xrightarrow{P_4/Br_2} Z$. Here $Z$ is

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