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. $Zn/Zndust$

Answer 1

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

Vicinal dihalides on treatment with alcoholic KOH undergo dehydrohalogenation. One molecule of HX is eliminated to form alkenyl halide which on treating further with sodamide gives alkyne.

Alcoholic KOH alone will give dehydrohalogenation only from alkyl halides not from alkenyl halides.
Aqueous KOH is very weak, it act only as a nucleophile not as a base.
Zn/Zn dust will give only dehalogenation product.
Hence, the correct answer is option (b).