C_6H_5 - C equiv C - CH_3 xrightarrow{HgSO_4, | vedclass

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Question

(A) The reaction of an alkyne with $HgSO_4$ and $H_2SO_4$ is a hydration reaction (Kucherov reaction).For an unsymmetrical alkyne like $C_6H_5 - C \eq

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$1-$phenylpropan$-1-$one

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(A) The reaction of an alkyne with $HgSO_4$ and $H_2SO_4$ is a hydration reaction (Kucherov reaction).For an unsymmetrical alkyne like $C_6H_5 - C \equiv C - CH_3$,the addition of water follows Markovnikov's rule.The $OH^-$ group attaches to the more substituted carbon atom.Thus,the intermediate enol formed is $C_6H_5 - C(OH) = CHCH_3$.This enol tautomerizes to form the ketone,$C_6H_5 - CO - CH_2CH_3$ ($1$-phenylpropan$-1-$one).

$C_6H_5 - C \equiv C - CH_3 \xrightarrow{HgSO_4, H_2SO_4} A$

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