JEE Main 2026 · 6 Apr · Shift 2 · Chemistry Q65

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ChemistrySingle correct+4 / -1Question 65 of 75

An optically active alkyl bromide $\ce{C4H9Br}$ , reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by :

Haloform test

Lucas test

Silver mirror test

Benedict test

Answer and worked solution

Official answer: A

Sequence.

The optically active $\ce{C4H9Br}$ is 2-bromobutane. Ethanolic KOH gives but-2-ene (Saytzeff).
$\ce{Br2}$ gives 2,3-dibromobutane.
Ethanolic KOH and then $\ce{NaNH2}$ give but-2-yne.
Hydration ( $\ce{HgSO4}$ , dilute $\ce{H2SO4}$ ) gives butan-2-one.

Test. Butan-2-one is a methyl ketone, so the haloform test confirms it.

(A)

Topics:Alkynes Elimination reactions Named reactions

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JEE Main 2026 · 6 Apr · Shift 2, Chemistry question 65