An optically active alkyl halide C_4 H_9 Br[A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C] . The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH_2 . During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [ E ] is :

Aldehydes Ketones and Carboxylic Acids PYQ — JEE Main 2025 · 2 Apr · Shift 1 · Q12 with Solution

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Correct answer: C

Identify the optically active alkyl halide [A]

Among isomers of $\mathrm{C_4H_9Br}$ , the optically active one is: $\mathrm{CH_3-CH(Br)-CH_2-CH_3}$ i.e. 2-bromobutane, because the carbon bearing Br has four different groups.
Reaction with hot alcoholic KOH

Hot $\mathrm{KOH}$ in ethanol gives elimination ( $E2$ ), so 2-bromobutane forms alkene(s).

Major product by Saytzeff rule is: $\mathrm{CH_3-CH=CH-CH_3}$ i.e. but-2-ene = $[B]$ .
Addition of bromine to alkene

But-2-ene reacts with $\mathrm{Br_2}$ to give vicinal dibromide: $\mathrm{CH_3-CHBr-CHBr-CH_3}$ i.e. 2,3-dibromobutane = $[C]$ .
Reaction of dibromide with alcoholic $\mathrm{NaNH_2}$

A vicinal dibromide on treatment with alcoholic $\mathrm{NaNH_2}$ undergoes double dehydrohalogenation to form an alkyne: $\mathrm{CH_3-CHBr-CHBr-CH_3 \xrightarrow[alc.] {NaNH_2} CH_3-C\equiv C-CH_3}$

So gas $[D]$ is but-2-yne.
Hydration of alkyne

It is given that 18 g of water is added to 1 mole of gas $[D]$ .

Since $18\,$ g water $=1$ mole $\mathrm{H_2O}$ , this is normal hydration of one mole of alkyne: $\mathrm{CH_3-C\equiv C-CH_3 + H_2O \xrightarrow[333\,K]{HgSO_4/H^+} }$

Hydration of but-2-yne gives an enol, which tautomerizes to the ketone: $\mathrm{CH_3-CO-CH_2-CH_3}$

This compound is butan-2-one.
Check options
- A: Butan-2-ol Not formed by hydration of alkyne under these conditions
- B: But-2-yne This is $[D]$ , not $[E]$
- C: Butan-2-one Correct
- D: Butan-1-al Not obtained here

$\boxed{\text{Butan-2-one}}$ So, the correct option is C.

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Correct answer: C

Identify the optically active alkyl halide [A]

Among isomers of $\mathrm{C_4H_9Br}$ , the optically active one is: $\mathrm{CH_3-CH(Br)-CH_2-CH_3}$ i.e. 2-bromobutane, because the carbon bearing Br has four different groups.
Reaction with hot alcoholic KOH

Hot $\mathrm{KOH}$ in ethanol gives elimination ( $E2$ ), so 2-bromobutane forms alkene(s).

Major product by Saytzeff rule is: $\mathrm{CH_3-CH=CH-CH_3}$ i.e. but-2-ene = $[B]$ .
Addition of bromine to alkene

But-2-ene reacts with $\mathrm{Br_2}$ to give vicinal dibromide: $\mathrm{CH_3-CHBr-CHBr-CH_3}$ i.e. 2,3-dibromobutane = $[C]$ .
Reaction of dibromide with alcoholic $\mathrm{NaNH_2}$

A vicinal dibromide on treatment with alcoholic $\mathrm{NaNH_2}$ undergoes double dehydrohalogenation to form an alkyne: $\mathrm{CH_3-CHBr-CHBr-CH_3 \xrightarrow[alc.] {NaNH_2} CH_3-C\equiv C-CH_3}$

So gas $[D]$ is but-2-yne.
Hydration of alkyne

It is given that 18 g of water is added to 1 mole of gas $[D]$ .

Since $18\,$ g water $=1$ mole $\mathrm{H_2O}$ , this is normal hydration of one mole of alkyne: $\mathrm{CH_3-C\equiv C-CH_3 + H_2O \xrightarrow[333\,K]{HgSO_4/H^+} }$

Hydration of but-2-yne gives an enol, which tautomerizes to the ketone: $\mathrm{CH_3-CO-CH_2-CH_3}$

This compound is butan-2-one.
Check options
- A: Butan-2-ol Not formed by hydration of alkyne under these conditions
- B: But-2-yne This is $[D]$ , not $[E]$
- C: Butan-2-one Correct
- D: Butan-1-al Not obtained here

$\boxed{\text{Butan-2-one}}$ So, the correct option is C.