Question

In the reaction sequence, Y is: \[ CH_3CO_2H \xrightarrow{\text{(1) NH}_3, \text{(2)} \Delta} P \xrightarrow{\text{Br}_2/\text{NaOH}} Y \]

• A. \(\text{a primary amine with the same number of carbons as in P} \)
• B. \(\text{a primary amine with one carbon less than in P} \)
• C. \(\text{a secondary amine with the same number of carbons as in P} \)
• D. \(\text{a secondary amine with one carbon less than in P} \)

Hint:

The Hoffmann Bromamide Degradation reaction reduces the carbon count by one when converting an amide to an amine.

Answer 1

The Correct Option is B

Step 1: Understanding the Reaction Sequence 
1. First Reaction: Amide Formation 
- The given reactant is acetic acid (\( CH_3CO_2H \)).
- When reacted with ammonia (NH\(_3\)) and heated (\(\Delta\)), it forms an amide (P): \[ CH_3COOH + NH_3 \rightarrow CH_3CONH_2 \] - So, \( P \) is acetamide (\( CH_3CONH_2 \)).
2. Second Reaction: Hoffmann Bromamide Degradation - The Hoffmann degradation reaction involves treating an amide with Br\(_2\)/NaOH, which removes one carbon and forms a primary amine. \[ CH_3CONH_2 \xrightarrow{\text{Br}_2/\text{NaOH}} CH_3NH_2 \] - The product \( Y \) is methylamine (\( CH_3NH_2 \)), which has one carbon less than in P.
Step 2: Identifying the Correct Answer 
- \( P \) (acetamide) contains two carbon atoms.
- \( Y \) (methylamine) contains one carbon atom, indicating the loss of one carbon.
- The final product is a primary amine.
Thus, the correct answer is: Option (2): A primary amine with one carbon less than in P