CH_3-C equiv C-O-CH_3 xrightarrow{dil. H_2SO_ | vedclass

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Question

(A) The reaction of an alkyne with water in the presence of $dil. H_2SO_4$ and $HgSO_4$ is a hydration reaction. For the given substrate $CH_3-C \equi

Vedclass · Accepted Answer

Methyl propionate $(CH_3CH_2COOCH_3)$

Vedclass · Answer

(A) The reaction of an alkyne with water in the presence of $dil. H_2SO_4$ and $HgSO_4$ is a hydration reaction. For the given substrate $CH_3-C \equiv C-O-CH_3$,the hydration follows Markovnikov's rule. The $OH^-$ group attacks the more substituted carbon (or the carbon attached to the electron-donating group) to form an enol intermediate. However,in the presence of an alkoxy group $(-OCH_3)$,the reaction proceeds to form an ester. The hydration of the alkyne $CH_3-C \equiv C-O-CH_3$ leads to the formation of methyl propionate $(CH_3CH_2COOCH_3)$ as the major product.

$CH_3-C \equiv C-O-CH_3 \xrightarrow{dil. H_2SO_4, HgSO_4} \text{major product}$

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