CH_3 - C equiv CH xrightarrow{Na} X xrightarr | vedclass

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(B) Step $1$: Propyne $(CH_3 - C \equiv CH)$ reacts with sodium $(Na)$ to form sodium propynide $(CH_3 - C \equiv C^- Na^+)$ and releases hydrogen gas

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$CH_3 - C \equiv C - CH_3$

Vedclass · Answer

(B) Step $1$: Propyne $(CH_3 - C \equiv CH)$ reacts with sodium $(Na)$ to form sodium propynide $(CH_3 - C \equiv C^- Na^+)$ and releases hydrogen gas $(H_2)$. Here,$X$ is $CH_3 - C \equiv C^- Na^+$.Step $2$: Sodium propynide $(CH_3 - C \equiv C^- Na^+)$ reacts with methyl chloride $(CH_3 - Cl)$ via an $S_N2$ mechanism to form but$-2-$yne $(CH_3 - C \equiv C - CH_3)$ and sodium chloride $(NaCl)$. Here,$Y$ is $CH_3 - C \equiv C - CH_3$.

$CH_3 - C \equiv CH$ $\xrightarrow{Na} X$ $\xrightarrow{CH_3 - Cl} Y$,
$Y$ is

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$CH_3 - C \equiv CH$ $\xrightarrow{Na} X$ $\xrightarrow{CH_3 - Cl} Y$,$Y$ is