JEE Chemistry 2025 Question with Solution

Given: An optically active alkyl halide $\mathrm{C_4H_9Br}$ undergoes elimination, bromine addition, double dehydrohalogenation, and finally hydration with HgSO$_4$ and dilute acid.

Find: The IUPAC name of compound [E].

The optically active isomer of $\mathrm{C_4H_9Br}$ is 2-bromobutane. Therefore,

With hot alcoholic KOH, [A] undergoes E2 elimination. By Zaitsev's rule, the major alkene is but-2-ene:

So, [B] is but-2-ene.

Now [B] reacts with bromine to form the vicinal dibromide:

Hence, [C] is 2,3-dibromobutane.

Treatment of [C] with alcoholic NaNH$_2$ causes double dehydrohalogenation and forms an alkyne:

Therefore, [D] is but-2-yne.

Hydration of the alkyne [D] with mercuric sulphate and dilute acid gives an enol first, which tautomerizes to the ketone:

Thus [E] is $\mathrm{CH_3COCH_2CH_3}$, whose IUPAC name is Butan-2-one.

Therefore, the correct option is C.

Given: The sequence is alkyl halide $\rightarrow$ alkene $\rightarrow$ dibromide $\rightarrow$ alkyne $\rightarrow$ hydrated product.

Find: The final compound [E].

The key clue is that [A] is optically active. Among isomers of $\mathrm{C_4H_9Br}$, only 2-bromobutane is chiral.

Its elimination gives two alkenes, but the major product is the more substituted one:

So [B] must be but-2-ene.

Addition of bromine across the double bond gives 2,3-dibromobutane, and removal of two molecules of HBr using NaNH$_2$ gives but-2-yne as [D].

A symmetrical internal alkyne such as but-2-yne on hydration under HgSO$_4$ and dilute acid conditions gives a ketone after enol-keto tautomerism. The product is butan-2-one.

Hence, the correct answer is C.