Question

The reagent (s) for the following conversion,

is/ are:

• A. Alcoholic $KOH$
• B. Alcoholic $KOH$ followed by $NaN{H}_2$
• C. Aqueous $KOH$ followed by $NaN{H}_2$
• D. $Zn/C{H}_{3}OH$

Answer 1

The correct option is B Alcoholic $KOH$ followed by $NaN{H}_2$

Here first the ellimination of $HBr$ occurs from the compound $BrC{H}_2-C{H}_{2}Br$ in presence of alcoholic $KOH$ to form $C{H}_2=CHBr$.
Then further elimination of $HBr$ from $C{H}_2=CHBr$ requires a stronger base because here, $Br$ accquries partial double bond character due to resonance and ethyne is formed.