Question

A hydrocarbon A (C₄H₈) on reaction with HCl gives a compound B (C₄H₉Cl) which on reaction with 1 mol of NH₃ gives compound C (C₄H₁₀N). On reacting with NaNO₂ and HCl followed by treatment with water, compound C yields an optically active compound D. The D is

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is B

Let's break down the reaction step by step:

Starting compound A (C₄H₈): The given hydrocarbon is C₄H₈, which could be butene (C₄H₈), and when it reacts with HCl, it will form C₄H₉Cl (butyl chloride), compound B.

Reaction of B with NH₃ (ammonia): When butyl chloride (C₄H₉Cl) reacts with ammonia (NH₃), it undergoes nucleophilic substitution to replace the chloride with an amine group. This gives butylamine (C₄H₁₀N), which is compound C.

Reaction of C with NaNO₂ and HCl, followed by water treatment: Butylamine reacts with sodium nitrite (NaNO₂) and HCl to form a diazonium salt. When treated with water, the diazonium ion undergoes hydrolysis, resulting in the formation of an optically active product. The product formed is an alcohol with a chiral center, making it optically active.

Therefore, the correct compound B is butyl alcohol (C₄H₉OH), corresponding to Option B.
 

The correct answer is (B) : .