Chemistry Question from JEE Main 2025 · 2 April, Shift 1

An optically active alkyl halide $\mathrm{C_4H_9Br}$ [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$\mathrm{_2}$. During hydration $18 \, \text{gram}$ of water is added to $1 \, \text{mole}$ of gas [D] on warming with mercuric sulphate and dilute acid at $333 \, \text{K}$ to form compound [E]. The IUPAC name of compound [E] is :