CH equiv CH xrightarrow{HCl text{ (Excess)}} | vedclass

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(C) Step $1$: Reaction of ethyne with excess $HCl$ gives $1,1-$dichloroethane $(A)$.$CH \equiv CH + 2HCl \rightarrow CH_3-CHCl_2 (A)$Step $2$: Hydroly

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ethanol

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(C) Step $1$: Reaction of ethyne with excess $HCl$ gives $1,1-$dichloroethane $(A)$.$CH \equiv CH + 2HCl \rightarrow CH_3-CHCl_2 (A)$Step $2$: Hydrolysis of $1,1-$dichloroethane $(A)$ followed by heating gives ethanal $(B)$.$CH_3-CHCl_2 + H_2O \xrightarrow{\Delta} CH_3CHO (B) + 2HCl$Step $3$: Reduction of ethanal $(B)$ with $Na-Hg / H_2O$ gives ethanol $(C)$.$CH_3CHO + 2[H] \xrightarrow{Na-Hg / H_2O} CH_3CH_2OH (C)$Thus,compound $C$ is ethanol.

$CH \equiv CH$ $\xrightarrow{HCl \text{ (Excess)}} A$ $\xrightarrow{\text{Hydrolysis}, \Delta} B$ $\xrightarrow{\text{Reduction}, Na-Hg / H_2O} C$
Compound $C$ is

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$CH \equiv CH$ $\xrightarrow{HCl \text{ (Excess)}} A$ $\xrightarrow{\text{Hydrolysis}, \Delta} B$ $\xrightarrow{\text{Reduction}, Na-Hg / H_2O} C$Compound $C$ is