Electrophilic Substitution On Aniline
When aniline is treated with bromine water at room temperature, a white precipitate forms rapidly; what is the structure of this insoluble product?
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Solution
2,4,6-tribromoaniline
Aniline is extremely reactive toward electrophilic aromatic substitution because the nitrogen lone pair strongly activates the ring through resonance, raising electron density at the ortho and para positions. Bromine water is a mild brominating agent, yet aniline is so activated that bromination is not limited to a single position; all three highly activated sites react, giving 2,4,6-tribromoaniline, which precipitates as a white solid. The option of only para-bromoaniline is wrong because the strong activation prevents stopping at monosubstitution under these conditions; selective monobromination instead requires first reducing reactivity by acetylation. The ortho-only product is similarly incorrect, both because ortho is more hindered and because substitution does not stop at one site. The meta product is impossible since –NH2 is an ortho/para director, never a meta director. This rapid trisubstitution is a hallmark NCERT and JEE demonstration of how powerfully an amino group activates a ring, and it contrasts sharply with benzene, which needs a Lewis acid catalyst even for monobromination. As a plausibility check, the immediate formation of an insoluble solid without any catalyst confirms multiple substitutions on a strongly activated ring rather than a controlled single substitution, and acetylation of the amine first is required if only one bromine is wanted.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing nitrogen
- Topic
- electrophilic substitution on aniline
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2,4,6-tribromoaniline
Aniline is extremely reactive toward electrophilic aromatic substitution because the nitrogen lone pair strongly activates the ring through resonance, raising electron density at the ortho and para positions. Bromine water is a mild brominating agent, yet aniline is so activated that bromination is not limited to a single position; all three highly activated sites react, giving 2,4,6-tribromoaniline, which precipitates as a white solid. The option of only para-bromoaniline is wrong because the strong activation prevents stopping at monosubstitution under these conditions; selective monobromination instead requires first reducing reactivity by acetylation. The ortho-only product is similarly incorrect, both because ortho is more hindered and because substitution does not stop at one site. The meta product is impossible since –NH2 is an ortho/para director, never a meta director. This rapid trisubstitution is a hallmark NCERT and JEE demonstration of how powerfully an amino group activates a ring, and it contrasts sharply with benzene, which needs a Lewis acid catalyst even for monobromination. As a plausibility check, the immediate formation of an insoluble solid without any catalyst confirms multiple substitutions on a strongly activated ring rather than a controlled single substitution, and acetylation of the amine first is required if only one bromine is wanted.
This medium difficulty chemistry question is from the chapter organic compounds containing nitrogen, covering the topic of electrophilic substitution on aniline. It appeared in the 2025 exam.
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