In the given reaction x is: CH_3 - C equiv CH | vedclass

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(C) In the first step,propyne reacts with excess $HCl$ to form $2,2$-dichloropropane $(A)$ via Markovnikov addition.In the second step,alkaline hydrol

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$CH_3-CO-CH_3$

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(C) In the first step,propyne reacts with excess $HCl$ to form $2,2$-dichloropropane $(A)$ via Markovnikov addition.In the second step,alkaline hydrolysis of $2,2$-dichloropropane replaces the two chlorine atoms with hydroxyl groups,forming an unstable gem-diol which loses water to give propanone $(x)$.$CH_3-C \equiv CH + 2HCl \rightarrow CH_3-CCl_2-CH_3 (A)$$CH_3-CCl_2-CH_3$ $\xrightarrow{H_2O/OH^{-}} [CH_3-C(OH)_2-CH_3]$ $\rightarrow CH_3-CO-CH_3 (x) + H_2O$

In the given reaction $x$ is: $CH_3 - C \equiv CH$ $\xrightarrow{\text{Excess } HCl} (A)$ $\xrightarrow{H_2O/OH^{-}} (x)$

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