Identify 'A' From the given reaction:

The correct answer is Option (1) → 2-Butyne

A vicinal dihalide (two halogen atoms on adjacent carbons) loses two molecules of HX when treated with a strong base. Each step is a β-elimination (dehydrohalogenation): H is removed from one carbon and the halogen from the next, putting in one π bond. Two eliminations on the same pair of carbons give a triple bond. Sodamide (\( \ce{NaNH2} \)), a very strong base, is used because the second HX has to be removed from a vinylic halide, which alcoholic KOH does only with difficulty.

    \( \ce{CH3-CHBr-CHBr-CH3} \) + \( \ce{NaNH2} \) → \( \ce{CH3-CBr=CH-CH3} \) + NaBr + \( \ce{NH3} \)

    \( \ce{CH3-CBr=CH-CH3} \) + \( \ce{NaNH2} \) → \( \ce{CH3-C} \)≡\( \ce{C-CH3} \) + NaBr + \( \ce{NH3} \)

So A is but-2-yne (2-butyne).

Why the other options are incorrect

• Option (2) Buta-1,3-diene — Incorrect. It would need the two H atoms to be taken from the end carbons (C-1 and C-4); with \( \ce{NaNH2} \) the second HBr is lost from the vinylic bromide formed in the first step, which gives the triple bond between C-2 and C-3.

• Option (3) But-1-ene — Incorrect. It has only one double bond and that at the end of the chain; removing the two Br atoms alone (with zinc dust) would give but-2-ene, and \( \ce{NaNH2} \) removes HBr, not \( \ce{Br2} \).

• Option (4) 2-Bromobutane — Incorrect. Replacing one Br by H is a reduction; a base such as \( \ce{NaNH2} \) eliminates HBr and cannot do that.

NCERT (2026-27 edition):
Elimination Reactions – “… When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product. Since β-hydrogen atom is involved in elimination, it is often called β-elimination.”