Question

$1,2$-Dibromopropane on treatment with $X$ moles of $Na{NH}_2$ followed by treatment with $C_2{H}_{5}Br$ gives a pentyne. The value of $X$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$NaN{H}_2$ is a strong base. It mediates formation of alkyne via elemination reaction. Treatment of $1,2$-dibromopropane for the formation of pentyne requires $3$ moles of $NaN{H}_2$. In which $2$ moles form a propyne by two consecutive elimination of $HBr$. The third mole of $NaN{H}_2$ produces propyne carbanion which reacts with $C_2{H}_{5}Br$ and forms Pentyne.