An optically active alkyl halide $$C_4H_9Br$$ [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:

The alkyl halide [A] has formula $$C_4H_9Br$$ and is given to be optically active.
For a compound with four carbon atoms to show optical activity it must possess a chiral centre. The only $$C_4H_9Br$$ isomer that contains such a centre is $$CH_3-CHBr-CH_2-CH_3$$, i.e. $$2$$-bromobutane.
Therefore, $$[A] =$$ 2-bromobutane.

Hot ethanolic $$KOH$$ carries out $$\beta$$-elimination (dehydrohalogenation). Eliminating $$HBr$$ from 2-bromobutane gives the more substituted alkene as the major product:

$$[A] \xrightarrow[\text{EtOH}]{\text{hot }KOH} \; CH_3-CH=CH-CH_3$$

Thus $$[B] =$$ but-2-ene (mixture of $$E$$ and $$Z$$ forms).

Adding bromine across the double bond gives a vicinal dibromide:

$$[B] + Br_2 \rightarrow CH_3-CHBr-CHBr-CH_3$$

Hence $$[C] =$$ 2,3-dibromobutane (meso + racemic mixture).

Alcoholic $$NaNH_2$$ is a very strong base; with a vic-dibromide it removes two molecules of $$HBr$$ to form an alkyne:

$$[C] \xrightarrow[\text{alc.}]{NaNH_2} CH_3-C \equiv C-CH_3$$

Therefore $$[D] =$$ but-2-yne, a gaseous alkyne (molar mass $$=54$$ g mol$$^{-1}$$). During this step nothing else is added or lost other than $$2HBr$$, so the formula of $$[D]$$ is $$C_4H_6$$.

The statement “18 g of water is added to 1 mol of gas $$[D]$$” confirms that exactly one mole of $$H_2O$$ (18 g) adds to each mole of alkyne, matching hydration of a single triple bond.

Markovnikov hydration of an internal alkyne is carried out with dilute acid and mercuric sulphate catalyst at 333 K. First an enol forms, which then tautomerises to a ketone:

$$CH_3-C \equiv C-CH_3 \xrightarrow[\text{HgSO}_4]{H_2O/H^+, \;333\,\text{K}} CH_3-CO-CH_2-CH_3$$

The product $$[E]$$ is $$CH_3-CO-CH_2-CH_3$$, whose IUPAC name is butan-2-one.

Thus the correct option is Option C: Butan-2-one.