Aniline reacts with bromine water at room temperature and forms a compound X. Identify X.

2,4,6-tribromoaniline

The correct answer is Option (3) → 2,4,6-tribromoaniline

Aniline contains an –NH₂ (amino group) attached to a benzene ring.

The –NH₂ group is a strongly activating group due to its +M (electron donating) effect. This increases the electron density in the benzene ring and makes it highly reactive toward electrophilic substitution reactions.

The amino group directs incoming electrophiles to the ortho and para positions.

Reaction with bromine water

When aniline reacts with bromine water at room temperature, the strong activating effect of the amino group causes multiple substitutions to occur easily.

As a result, bromine substitutes at all the activated positions:

• 2-position (ortho)
• 4-position (para)
• 6-position (ortho)

Thus the product formed is:

2,4,6-tribromoaniline

During this reaction, a white precipitate of 2,4,6-tribromoaniline is formed.

Option-wise Explanation

Option 1: 2-Bromoaniline

This would be formed only if controlled monobromination occurred, but in bromine water the reaction proceeds further to multiple substitution. Hence this option is incorrect.

Option 2: 4-Bromoaniline

Similarly, para substitution alone does not occur under these conditions because the ring is highly activated. Therefore this option is incorrect.

Option 3: 2,4,6-tribromoaniline

Due to the strong activating effect of the amino group, bromination occurs at the two ortho and one para positions, producing 2,4,6-tribromoaniline. Hence this option is correct.

Option 4: 2-Bromoaniline and 4-Bromoaniline

These products may form under controlled conditions, but with bromine water at room temperature the reaction proceeds to the tribromo product. Hence this option is incorrect.