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

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(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:

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