Ph - C equiv C - CH_3 xrightarrow{Hg^{2+}/H^+ | vedclass

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Question

(A) The reaction of an alkyne with $Hg^{2+}/H^+$ (Kuccherov reaction) involves the hydration of the triple bond to form an enol intermediate,which the

Vedclass · Accepted Answer

$1-$Phenylpropan$-1-$one

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(A) The reaction of an alkyne with $Hg^{2+}/H^+$ (Kuccherov reaction) involves the hydration of the triple bond to form an enol intermediate,which then tautomerizes to a more stable ketone.For the unsymmetrical alkyne $Ph-C \equiv C-CH_3$,the hydration follows Markovnikov's rule.The $OH$ group attaches to the carbon atom that can better stabilize the positive charge in the transition state (the carbon attached to the phenyl group).This forms the enol $Ph-C(OH)=CH-CH_3$,which tautomerizes to the ketone $Ph-CO-CH_2-CH_3$ ($1$-phenylpropan$-1-$one).

$Ph - C \equiv C - CH_3 \xrightarrow{Hg^{2+}/H^+} A$. What is $A$?

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