When 1,2-dibromopropane is treated with X mol | vedclass

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(B) $1$. The reaction of $1,2-$dibromopropane $(CH_3-CHBr-CH_2Br)$ with $NaNH_2$ (a strong base) leads to dehydrohalogenation. $2$. Two moles of $NaNH

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$3$

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(B) $1$. The reaction of $1,2-$dibromopropane $(CH_3-CHBr-CH_2Br)$ with $NaNH_2$ (a strong base) leads to dehydrohalogenation. $2$. Two moles of $NaNH_2$ are required to remove two molecules of $HBr$ from the vicinal dihalide to form propyne $(CH_3-C \equiv CH)$. $3$. Propyne then reacts with another mole of $NaNH_2$ to form sodium propynide $(CH_3-C \equiv C^- Na^+)$. $4$. Sodium propynide reacts with ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ mechanism to form pentyne $(CH_3-C \equiv C-CH_2CH_3)$. $5$. Total moles of $NaNH_2$ used = $2$ (for dehydrohalogenation) + $1$ (for deprotonation) = $3$ moles. $6$. Thus,$X = 3$.

When $1,2-$dibromopropane is treated with $X$ moles of $NaNH_2$,the resulting compound reacts with ethyl bromide to form pentyne. What is the value of $X$?

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