CH_3 - CH_2 - C equiv CH xrightarrow{HgSO_4, | vedclass

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Question

(A) The reaction of an alkyne with water in the presence of $HgSO_4$ and $H_2SO_4$ is an electrophilic addition reaction that follows Markovnikov's ru

Vedclass · Accepted Answer

$CH_3 - CH_2 - C(=O) - CH_3$

Vedclass · Answer

(A) The reaction of an alkyne with water in the presence of $HgSO_4$ and $H_2SO_4$ is an electrophilic addition reaction that follows Markovnikov's rule.First,the alkyne $CH_3 - CH_2 - C \equiv CH$ undergoes hydration to form an unstable enol intermediate: $CH_3 - CH_2 - C(OH) = CH_2$.This enol intermediate then undergoes tautomerization to form a more stable keto form.The final product $A$ is butan$-2-$one,which is represented as $CH_3 - CH_2 - C(=O) - CH_3$.

$CH_3 - CH_2 - C \equiv CH \xrightarrow{HgSO_4, H_2SO_4} A$,the compound $A$ is

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