Question

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 :

• A. But-2-yne
• B. Butan-2-ol
• C. Butan-2-one
• D. Butan-1-al

Hint:

For alkene to alkyne conversions, always check for the necessary reagents such as NaNH\(_2\) and the correct conditions for hydration (HgSO\(_4\)).

Answer 1

The Correct Option is C

The reaction sequence is as follows:

(A) Reacts with KOH to form alkene [B] (Elimination of HBr).
(B) Reacts with Br\(_2\) to give dibromide [C].
(C) Dibromide reacts with NaNH\(_2\) to form [D] (Alkyne formation).
(D) The alkyne [D] undergoes hydration with mercuric sulfate to give [E].