1, 2-Dibromopropane on treatment with X moles | vedclass

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

Question

(C) Step $1$: $1, 2$-Dibromopropane $(CH_3-CHBr-CH_2Br)$ reacts with $NaNH_2$ to undergo dehydrohalogenation. Two moles of $NaNH_2$ are required to re

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Step $1$: $1, 2$-Dibromopropane $(CH_3-CHBr-CH_2Br)$ reacts with $NaNH_2$ to undergo dehydrohalogenation. Two moles of $NaNH_2$ are required to remove two molecules of $HBr$ to form propyne $(CH_3-C \equiv CH)$. Step $2$: The terminal alkyne (propyne) reacts with another mole of $NaNH_2$ to form the sodium acetylide salt $(CH_3-C \equiv C^- Na^+)$. Step $3$: This salt then reacts with ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ reaction to form $2$-pentyne $(CH_3-C \equiv C-CH_2CH_3)$. Total moles of $NaNH_2$ used $(X)$ = $2$ (for dehydrohalogenation) + $1$ (for salt formation) = $3$.

$1, 2$-Dibromopropane on treatment with $X$ moles of $NaNH_2$ followed by reaction with ethyl bromide gives $2$-pentyne. The value of $X$ is:

Similar Questions

What is the major product obtained when propyne reacts with aqueous $H_2SO_4$ in the presence of $HgSO_4$?

Which of the following reactions will yield $2, 2-$dibromopropane?

Which of the following metal carbides gives propyne on hydrolysis?

For the alkyne with formula $C_6H_{10}$,the number of alkynes with acidic hydrogens is $x$ and the number of alkynes with no acidic hydrogens is $y$. $x$ and $y$ are respectively:

By heating a mixture of $CHCl_3$ with silver powder,the compound formed is

Vedclass Test Series

Exam Paper Generator

Online Exam Module