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(C) In the presence of $HgSO_4$ and $H_2SO_4$,alkynes undergo hydration following Markovnikov's rule. Propyne $(CH_3-C \equiv CH)$ reacts with water t

Vedclass · Accepted Answer

$A: CH_3-C(OH)=CH_2, B: CH_3-C(=O)-CH_3$

Vedclass · Answer

(C) In the presence of $HgSO_4$ and $H_2SO_4$,alkynes undergo hydration following Markovnikov's rule. Propyne $(CH_3-C \equiv CH)$ reacts with water to form an unstable enol intermediate,$CH_3-C(OH)=CH_2$ $(A)$. This enol then undergoes tautomerization to form the more stable keto form,propanone $CH_3-C(=O)-CH_3$ $(B)$. $CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}/H^{+}} [CH_3-C(OH)=CH_2] 	o CH_3-C(=O)-CH_3$

Predict the correct intermediate and product in the following reaction: $CH_3-C \equiv CH \xrightarrow[HgSO_4]{H_2O, H_2SO_4} \text{Intermediate } (A) \to \text{Product } (B)$

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