Question

The product B is

• A. N, N-Dimethyl phenyl methanamine
• B. N, N-Dimethyl benzenamine
• C. N-Benzyl-N-methyl methanamine
• D. phynyl-N-N-dimethyl methanamine

Answer 1

The Correct Option is A

To solve this problem, we need to analyze the reactions and determine the product B formed from the given starting compound.

1. Understanding the Reactions:
The starting compound is C₆H₅CH₂Cl (benzyl chloride). In the first step, benzyl chloride reacts with alcoholic ammonia (alc. NH₃), which is a nucleophilic substitution reaction. The ammonia (NH₃) substitutes the chlorine atom (Cl⁻), forming benzylamine (C₆H₅CH₂NH₂). This is compound A.

2. Further Reaction with Methyl Chloride (CH₃Cl):
The second step involves the reaction of benzylamine (A) with excess methyl chloride (CH₃Cl). The amino group (-NH₂) of benzylamine reacts with the methyl chloride, leading to the formation of a N-methylated product. The final product will be N, N-dimethylphenylmethanamine (C₆H₅CH₂N(CH₃)₂), which is compound B.

3. Identifying the Correct Product (B):
The product formed, as described above, is N, N-dimethylphenylmethanamine. This corresponds to option (A), "N, N-dimethylphenylmethanamine," which correctly describes the final product after the methylation of benzylamine.

Final Answer:
The correct answer is (A) "N, N-dimethylphenylmethanamine."

Answer 2

Let's analyze the reaction step by step:

The starting compound is benzyl chloride (C₆H₅CH₂Cl), which contains a benzyl group (C₆H₅CH₂-) attached to a chlorine atom.

The reaction with alcoholic NH₃ (ammonia) leads to a nucleophilic substitution where ammonia replaces the chlorine atom to form benzylamine (C₆H₅CH₂NH₂). This is the product A.

In the second step, the reaction with excess CH₃Cl (methyl chloride) results in the methylation of the amine group, leading to N, N-dimethylphenylmethanamine (C₆H₅CH₂NH(CH₃)₂). This is the product B.
 

The correct answer is (A) : N, N-Dimethyl phenyl methanamine.