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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

cis-pent$-2-$ene

Vedclass · Answer

(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 $(C)$?

Similar Questions

The product formed in the above reaction is

Which of the following compounds do not react with $NaNH_2$? Why?
$CH_3-C \equiv C-CH_3, CH_3-CH_2-C \equiv CH, C_2H_5-C \equiv C-C_2H_5, HC \equiv CH, CH_3-CH=CH_2, CH_2=CH-C \equiv CH$

In the reaction
$R - C \equiv C - R \xrightarrow{Pd + CaCO_3} B$
$R - C \equiv C - R \xrightarrow{Na + liq. NH_3} A$
$A$ and $B$ are geometrical isomers. Then:

The most suitable reagent for the following conversion is:
$H_3C-C \equiv C-CH_3 \rightarrow \text{cis-2-butene}$

$KMnO_4$ will oxidise acetylene to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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