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) : .

Answer 2

To solve this problem, let's carefully analyze the steps described in the reaction and deduce the correct product D.

1. Reaction of Hydrocarbon A with HCl:
The hydrocarbon A is a compound with the molecular formula C₄H₈. When it reacts with HCl, the product B (C₄H₉Cl) is formed. This suggests that hydrocarbon A is an alkene, and its reaction with HCl results in the addition of HCl across the double bond, leading to the formation of 1-chlorobutane (C₄H₉Cl).

2. Reaction of Compound B with Ammonia (NH₃):
Compound B, 1-chlorobutane, reacts with 1 mol of ammonia (NH₃) to give compound C (C₄H₉N). This indicates that the chlorine atom in 1-chlorobutane is replaced by an amino group (-NH₂), forming butan-1-amine (C₄H₉N).

3. Reaction of Compound C with NaNO₂ and HCl:
When butan-1-amine (C₄H₉N) reacts with sodium nitrite (NaNO₂) and hydrochloric acid (HCl), a diazotization reaction occurs. The amino group (-NH₂) is converted to a diazonium ion (C₄H₈N₂⁺). The reaction is then followed by the addition of water, leading to the formation of an optically active compound D. The conversion of the diazonium ion to a hydroxyl group (-OH) results in the formation of an optically active alcohol. This process is known as the Sandmeyer reaction, which introduces a hydroxyl group (OH) to form butan-2-ol (C₄H₉OH), a compound with a chiral center, making it optically active.

4. Identifying the Correct Product (D):
Based on the reaction described, the optically active compound D is butan-2-ol (C₄H₉OH), which corresponds to option (B), "CH₃-CH₂-CH(OH)-CH₃." This is the product formed after diazotization and hydrolysis, leading to the formation of a chiral alcohol.

Final Answer:
The correct answer is (B) "CH₃-CH₂-CH(OH)-CH₃."