Question

Consider the following sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

• A. 3
• B. 5
• C. 6
• D. 1

Hint:

Aniline is so strongly activating that bromination with bromine water results in substitution at all free ortho and para positions without needing a Lewis acid catalyst.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The sequence involves multiple steps: electrophilic aromatic substitution (bromination), reduction of a nitro group to an amine, poly-bromination of an activated ring, diazotization, and finally a Sandmeyer reaction to introduce another bromine atom.

Step 2: Detailed Explanation:
Step (1): Nitrobenzene undergoes bromination. Since the \( -NO_2 \) group is meta-directing, bromine is introduced at the meta position to form m-bromonitrobenzene.
Step (2) and (3): The nitro group is reduced using \( Sn/HCl \) and then neutralized to yield m-bromoaniline.
Step (4): Aniline is a highly activating group for electrophilic substitution. Reaction with bromine water (\( Br_2/H_2O \)) leads to bromination at all available ortho and para positions. For m-bromoaniline (with \( -NH_2 \) at C1 and \( -Br \) at C3), the activated positions are C2, C4, and C6. Thus, three more bromine atoms are added, forming 2,3,4,6-tetrabromoaniline.
Step (5) and (6): The amino group (\( -NH_2 \)) at C1 undergoes diazotization to form a diazonium salt, which is then replaced by a bromine atom via the Sandmeyer reaction.
The final product is 1,2,3,4,6-pentabromobenzene.
The total number of bromine atoms is \( 1 \text{ (initial meta)} + 3 \text{ (from bromine water)} + 1 \text{ (from Sandmeyer)} = 5 \).

Step 3: Final Answer:
The final product (P) contains 5 bromine atoms.