JEE MAIN - Chemistry (2025 - 2nd April Morning Shift - No. 12)

An optically active alkyl halide $\mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}[\mathrm{A}]$ 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 $\mathrm{NaNH}_2$. 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 :
Butan-2-ol
But-2-yne
Butan-2-one
Butan-1-al

Explanation

JEE Main 2025 (Online) 2nd April Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 12 English Explanation

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