In the reactionHC equiv CH xrightarrow[(ii) C | vedclass

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(B) Step $1$: Acetylene $(HC \equiv CH)$ reacts with $NaNH_2$ in liquid $NH_3$ to form the acetylide ion $(HC \equiv C^-Na^+)$,which then undergoes $S

Vedclass · Accepted Answer

$X = 1-$butyne,$Y = 3-$hexyne

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(B) Step $1$: Acetylene $(HC \equiv CH)$ reacts with $NaNH_2$ in liquid $NH_3$ to form the acetylide ion $(HC \equiv C^-Na^+)$,which then undergoes $S_N2$ reaction with $CH_3CH_2Br$ to form $1-$butyne $(X = CH_3CH_2C \equiv CH)$.Step $2$: $1-$butyne $(CH_3CH_2C \equiv CH)$ reacts with $NaNH_2$ in liquid $NH_3$ to form the butynylide ion $(CH_3CH_2C \equiv C^-Na^+)$,which then undergoes $S_N2$ reaction with $CH_3CH_2Br$ to form $3-$hexyne $(Y = CH_3CH_2C \equiv CCH_2CH_3)$.

In the reaction
$HC \equiv CH \xrightarrow[(ii) CH_3CH_2Br]{(i) NaNH_2 / liq. NH_3} X$
$\xrightarrow[(ii) CH_3CH_2Br]{(i) NaNH_2 / liq. NH_3} Y$
$X$ and $Y$ are

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In the reaction$HC \equiv CH \xrightarrow[(ii) CH_3CH_2Br]{(i) NaNH_2 / liq. NH_3} X$$\xrightarrow[(ii) CH_3CH_2Br]{(i) NaNH_2 / liq. NH_3} Y$$X$ and $Y$ are