Reagents for Dehydrohalogenation of Dibromoethane to Ethyne

To convert 1,2-dibromoethane ($\text{Br}-\text{CH}_2-\text{CH}_2-\text{Br}$) into ethyne ($\text{H}-\text{C}\equiv\text{C}-\text{H}$), two successive dehydrohalogenation ($-\text{HBr}$) steps are required:

1. First Dehydrohalogenation:
   Treatment of 1,2-dibromoethane with a moderately strong base, such as alcoholic $\text{KOH}$, results in an E2 elimination to yield vinyl bromide:
   $\text{Br}-\text{CH}_2-\text{CH}_2-\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}-\text{Br} + \text{KBr} + \text{H}_2\text{O}$

2. Second Dehydrohalogenation:
   In vinyl bromide ($\text{CH}_2=\text{CH}-\text{Br}$), the lone pair of electrons on the bromine atom is in resonance with the $\pi$-bond:
   $\text{CH}_2=\text{CH}-\overset{\cdot\cdot}{\underset{\cdot\cdot}{\text{Br}}}\text{:} \longleftrightarrow \overset{-}{\text{C}}\text{H}_2-\text{CH}=\overset{+}{\text{Br}}\text{:}$
   This resonance confers partial double-bond character to the $\text{C}-\text{Br}$ bond, making it significantly stronger and less reactive toward elimination. Additionally, the vinylic hydrogen is attached to an $sp^2$-hybridized carbon. Consequently, alcoholic $\text{KOH}$ is not a sufficiently strong base to eliminate the second equivalent of $\text{HBr}$ efficiently.

   A much stronger base, such as sodamide ($\text{NaNH}_2$ in liquid $\text{NH}_3$), is required to abstract the vinylic proton and drive the second elimination to yield ethyne:
   $\text{CH}_2=\text{CH}-\text{Br} \xrightarrow{\text{NaNH}_2} \text{H}-\text{C}\equiv\text{C}-\text{H} + \text{NaBr} + \text{NH}_3$

Analysis of other options:

• (A) Alcoholic $\text{KOH}$: It only carries out the first elimination efficiently, primarily forming vinyl bromide.
• (C) Aqueous $\text{KOH}$ followed by $\text{NaNH}_2$: Aqueous $\text{KOH}$ acts as a nucleophile rather than a base, favoring substitution over elimination to form ethylene glycol ($\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}$).
• (D) $\text{Zn}/\text{CH}_3\text{OH}$: This reagent causes debromination (removal of $\text{Br}_2$) to form ethene ($\text{CH}_2=\text{CH}_2$), not ethyne.

Therefore, the appropriate reagents are alcoholic $\text{KOH}$ followed by $\text{NaNH}_2$.

Correct Answer: B