One mole of 1,2-dibromopropane on treatment w | vedclass

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

Question

(C) The reaction proceeds in two main stages:$1$. Dehydrohalogenation: $1,2$-dibromopropane $(CH_3-CH(Br)-CH_2Br)$ reacts with $2$ moles of $NaNH_2$ t

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The reaction proceeds in two main stages:$1$. Dehydrohalogenation: $1,2$-dibromopropane $(CH_3-CH(Br)-CH_2Br)$ reacts with $2$ moles of $NaNH_2$ to undergo double dehydrohalogenation,forming propyne $(CH_3-C \equiv CH)$.$2$. Alkylation: The terminal alkyne (propyne) reacts with another $1$ mole of $NaNH_2$ to form the sodium acetylide intermediate $(CH_3-C \equiv C^-Na^+)$,which then reacts with ethyl bromide $(CH_3-CH_2Br)$ via an $S_N2$ mechanism to form pentyne $(CH_3-C \equiv C-CH_2-CH_3)$.Thus,the total number of moles of $NaNH_2$ required is $2 + 1 = 3$ moles. Therefore,$X = 3$.

One mole of $1,2$-dibromopropane on treatment with $X$ moles of $NaNH_2$ followed by treatment with ethyl bromide gave a pentyne. The value of $X$ is

Similar Questions

Which reaction is correct for the conversion of chloroform to acetylene?

The general formula of an alkyne is:

In the given reactions,identify $A$ and $B$.

$Mg_2C_3 + H_2O \xrightarrow{X}$ (organic compound).
Compound $X$ is

The number of structural isomers of alkyne possible for the molecular formula $C_6H_{10}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module