Chemistry · 2025
JEE Main · 2 April 2025, Shift 1 · Q66
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…
An optically active alkyl halide $\displaystyle \mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}[\mathrm{A}]$ 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 $\displaystyle \mathrm{NaNH}_2$. During hydration $\displaystyle 18$ gram of water is added to $\displaystyle 1$ mole of gas [D] on warming with mercuric sulphate and dilute acid at $\displaystyle 333$ K to form compound [E]. The IUPAC name of compound $\displaystyle [\mathrm{E}]$ is :
Official answer
From NTA’s final answer key for this paper.
(1)
Butan-$\displaystyle 2$-one
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JEE Main 2025 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.