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(C) The reaction is the hydration of an alkyne using $HgSO_4$ and $dil. H_2SO_4$ (Kucherov reaction). Step $-I$: The $Hg^{2+}$ ion acts as an electrop

Vedclass · Accepted Answer

$C_6H_5-CH_2-CO-(C_6H_4)-OCH_3$

Vedclass · Answer

(C) The reaction is the hydration of an alkyne using $HgSO_4$ and $dil. H_2SO_4$ (Kucherov reaction). Step $-I$: The $Hg^{2+}$ ion acts as an electrophile and forms a non-classical carbocation intermediate with the triple bond. Step $-II$: Water (nucleophile) attacks the more substituted carbon atom to form a stable carbocation intermediate,which eventually leads to the formation of an enol. Step $-III$: The unstable enol undergoes tautomerization to form the more stable ketone. In the given substrate $C_6H_5-C \equiv C-(C_6H_4)-OCH_3$,the attack of water occurs such that the carbonyl group is formed on the carbon closer to the electron-donating $-OCH_3$ group,resulting in $C_6H_5-CH_2-CO-(C_6H_4)-OCH_3$ as the major product.

The major product $(A)$ is:

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