Question

The major products A and B formed in the following reaction sequence are : 
 

• A. A
• B. B
• C. C
• D. D

Hint:

Protection-deprotection strategy is vital for controlling regioselectivity and reactivity of highly active aromatic systems like amines and phenols. Remember: \(-\text{NHCOCH}_3\) is milder than \(-\text{NH}_2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This reaction sequence demonstrates the protection of an amine group followed by electrophilic aromatic substitution. Direct bromination of aniline often leads to polysubstitution (2,4,6-tribromoaniline) because the \(-\text{NH}_2\) group is highly activating. To get a monosubstituted product, the amino group must be protected.
Step 2: Detailed Explanation:
Stage 1: Acetylation
Aniline reacts with acetic anhydride (\((\text{CH}_3\text{CO})_2\text{O}\)) to undergo nucleophilic acyl substitution. The lone pair on Nitrogen attacks the carbonyl group of the anhydride, resulting in the formation of Acetanilide (A). The \(-\text{NHCOCH}_3\) group is less activating than \(-\text{NH}_2\) because the lone pair on Nitrogen is involved in resonance with the carbonyl oxygen.
\[ \text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 \text{ (A)} + \text{CH}_3\text{COOH} \]
Stage 2: Bromination
Acetanilide (A) is then treated with bromine in acetic acid. The acetamido group (\(-\text{NHCOCH}_3\)) is an \(ortho, para\)-directing group. Due to the steric bulk of the acetyl group, substitution at the \(ortho\) position is hindered. Therefore, the \(para\) position is favored, yielding \(p\)-bromoacetanilide (B) as the major product.
\[ \text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{Br}_2, \text{CH}_3\text{COOH}} p\text{-Br-C}_6\text{H}_4\text{NHCOCH}_3 \text{ (B)} \]
Step 3: Final Answer:
Product A is Acetanilide and Product B is \(p\)-bromoacetanilide.