A and D respectively are:

The correct answer is Option (3) → But-2-ene; Butan-2-one

Each step of the sequence can be read from the reagent and checked against the given structures of B and C:

• A: 2-bromobutane with alcoholic KOH undergoes dehydrohalogenation. By Saytzeff’s rule the major alkene is the one with more alkyl groups on the double-bonded carbons, so A is but-2-ene:

      \( \ce{CH3CH(Br)CH2CH3} \) + KOH(alc) → \( \ce{CH3CH=CHCH3} \) + KBr + \( \ce{H2O} \)

• B: \( \ce{Br2} \) adds across the double bond: \( \ce{CH3CH(Br)CH(Br)CH3} \), 2,3-dibromobutane, as shown. (But-1-ene would have given 1,2-dibromobutane instead.)

      \( \ce{CH3CH=CHCH3} \) + \( \ce{Br2} \) → \( \ce{CH3CH(Br)CH(Br)CH3} \)

• C: double dehydrohalogenation with alc. KOH and then \( \ce{NaNH2} \) gives \( \ce{CH3C} \)≡\( \ce{CCH3} \), but-2-yne, as shown.

      \( \ce{CH3CH(Br)CH(Br)CH3} \) + KOH(alc) → \( \ce{CH3C(Br)=CHCH3} \) + KBr + \( \ce{H2O} \)
      \( \ce{CH3C(Br)=CHCH3} \) + \( \ce{NaNH2} \) → \( \ce{CH3C} \)≡\( \ce{CCH3} \) + NaBr + \( \ce{NH3} \)

• D: hydration of the alkyne with dilute \( \ce{H2SO4} \)/\( \ce{Hg^{2+}} \) adds water to the triple bond; the enol formed changes to the ketone:

      \( \ce{CH3C} \)≡\( \ce{CCH3} \) + \( \ce{H2O} \) → \( \ce{[CH3C(OH)=CHCH3]} \) → \( \ce{CH3COCH2CH3} \) (butan-2-one)

Why the other options are incorrect

• Option (1) But-2-ene; Butanoic acid — Incorrect. A is right, but hydration of an alkyne gives a ketone (or ethanal from ethyne), never a carboxylic acid.

• Option (2) But-1-ene; Butan-2-one — Incorrect. but-1-ene is only the minor product of the elimination, and it would not give the 2,3-dibromo compound B shown.

• Option (4) But-1-ene; Butanoic acid — Incorrect. Both parts are wrong, for the reasons given for Options (1) and (2).

NCERT (2026-27 edition):
Elimination Reactions – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: