One mole of 1, 2-dibromopropane is reacted wi | vedclass

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(C) The reaction of $1, 2-$dibromopropane $(CH_3-CHBr-CH_2Br)$ with $NaNH_2$ (a strong base) leads to dehydrohalogenation to form propyne $(CH_3-C \eq

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$3$

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(C) The reaction of $1, 2-$dibromopropane $(CH_3-CHBr-CH_2Br)$ with $NaNH_2$ (a strong base) leads to dehydrohalogenation to form propyne $(CH_3-C \equiv CH)$.This process requires $2$ moles of $NaNH_2$ to remove two molecules of $HBr$.The resulting propyne is then treated with a base to form the sodium propynide $(CH_3-C \equiv C^-Na^+)$,which requires an additional $1$ mole of $NaNH_2$.Finally,the sodium propynide reacts with ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ mechanism to form pentyne $(CH_3-C \equiv C-CH_2CH_3)$.Total moles of $NaNH_2$ required: $2$ (for dehydrohalogenation) $+ 1$ (for deprotonation) $= 3$ moles.

One mole of $1, 2-$dibromopropane is reacted with $x$ moles of $NaNH_2$ followed by treatment with ethyl bromide to yield a pentyne. What is the value of $x$?

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