JEE Main 2026 Apr 6 Shift 2, Chemistry Q15: Properties of Alkynes

JEE Main2026Apr 6, Shift 2Chemistry

An optically active alkyl bromide C $_{4}$ H $_{9}$ Br 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:

A
Haloform test
B
Lucas test
C
Silver mirror test
D
Benedict test

Solution

The optically active C $_{4}$ H $_{9}$ Br is 2-bromobutane (the only chiral isomer).

Step 1. Ethanolic KOH causes E2 elimination; major product (Saytzeff) is but-2-ene, $[A]$ = CH $_{3}$ –CH=CH–CH $_{3}$ .

Step 2. Br $_{2}$ addition gives 2,3-dibromobutane, $[B]$ .

Step 3. Double dehydrohalogenation (alc. KOH then NaNH $_{2}$ / $Δ$ ) yields but-2-yne, $[C]$ = CH $_{3}$ –C≡C–CH $_{3}$ .

Step 4. Kucherov hydration of an internal symmetric alkyne gives the ketone butan-2-one, $[D]$ = CH $_{3}$ –CO–CH $_{2}$ –CH $_{3}$ .

Butan-2-one is a methyl ketone, so it gives a positive haloform test (iodoform on treatment with NaOH/I $_{2}$ ).

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q15: Properties of Alkynes

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