H_3C-C equiv CH xrightarrow{NaNH_2} A xrighta | vedclass

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(D) $1$. The reaction starts with propyne $(H_3C-C \equiv CH)$.$2$. Treatment with $NaNH_2$ removes the acidic terminal proton to form the sodium acet

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cis-pent$-2-$ene

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(D) $1$. The reaction starts with propyne $(H_3C-C \equiv CH)$.$2$. Treatment with $NaNH_2$ removes the acidic terminal proton to form the sodium acetylide intermediate $(A)$: $H_3C-C \equiv C^- Na^+$.$3$. This intermediate undergoes an $S_N2$ reaction with bromoethane $(CH_3CH_2Br)$ to form pent$-2-$yne $(B)$: $H_3C-C \equiv C-CH_2CH_3$.$4$. Finally,hydrogenation with Lindlar's catalyst $(H_2/Pd, BaSO_4)$ reduces the internal alkyne to a $cis$-alkene.$5$. Therefore,the product $(C)$ is $cis$-pent$-2-$ene.

$H_3C-C \equiv CH$ $\xrightarrow{NaNH_2} A$ $\xrightarrow{CH_3CH_2Br} B$ $\xrightarrow{H_2/Pd, BaSO_4} C$
Identify product $(C)$?

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$H_3C-C \equiv CH$ $\xrightarrow{NaNH_2} A$ $\xrightarrow{CH_3CH_2Br} B$ $\xrightarrow{H_2/Pd, BaSO_4} C$Identify product $(C)$?