Brominate phenylamine with aqueous bromine
Phenylamine reacts with Br₂(aq) at room temperature to substitute bromine at positions 2, 4 and 6 of the ring.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 34. Use the stated reagents, conditions and product scope. Written practice is self-checked This online lesson cannot demonstrate supervised laboratory skills.
Before you begin
- Recall nucleophilic substitution, reduction, acyl-chloride addition–elimination, acid–base equilibria and covalent bonding.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which conditions brominate phenylamine in this syllabus outcome?
What you will learn
- Recall aqueous bromine at room temperature
- Predict 2,4,6-substitution on phenylamine
- Explain why the –NH₂ group activates the ring
Phenylamine activates the ring
The nitrogen lone pair in –NH₂ donates electron density into the aromatic ring. Phenylamine therefore reacts readily with Br₂(aq) at room temperature.
Bromination gives 2,4,6-tribromophenylamine: the two ortho positions and the para position relative to –NH₂ are substituted.
Use the specified conditions and positions
Unlike benzene bromination, this reaction does not need an AlBr₃ catalyst. Aqueous bromine at room temperature is the stated condition.
The aromatic ring remains intact; substitution replaces three ring hydrogens with bromine. Do not mistake the product for bromination of the nitrogen atom.
Pause and explain
Which positions are brominated in the named phenylamine product?
Worked example
State the product and conditions when phenylamine reacts with bromine water.
Show the worked solution
- Use Br₂(aq) at room temperature.
- The –NH₂ group directs to ortho and para positions.
- Substitute at 2, 4 and 6.
Answer 2,4,6-tribromophenylamine forms with Br₂(aq) at room temperature.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Name the major bromination product of phenylamine and state the reagent condition.
Give me a hint
Mark both ortho positions and the para position.
Compare my reasoning
- Use bromine water.
- Keep the reaction at room temperature.
- Substitute ring positions 2, 4 and 6.
2,4,6-tribromophenylamine from phenylamine + Br₂(aq) at room temperature.
Look for these in your work
- Aqueous bromine
- Room temperature
- Three positions
Explain why no Lewis-acid catalyst is needed for phenylamine bromination under these conditions.
Give me a hint
Compare electron density with benzene.
Compare my reasoning
- Nitrogen donates a lone pair into the ring.
- This activates the ring toward electrophilic substitution.
- Aqueous bromine at room temperature is sufficient in the syllabus reaction.
The –NH₂ group activates the ring by lone-pair donation, so phenylamine reacts with Br₂(aq) at room temperature without the benzene catalyst conditions.
Look for these in your work
- Lone-pair donation
- Ring activation
- Condition stated
A learner predicts 3,5-dibromophenylamine. Correct the directing pattern.
Give me a hint
Treat –NH₂ as an ortho/para director.
Compare my reasoning
- Position 1 carries –NH₂.
- Ortho positions are 2 and 6; para is 4.
- The listed product is substituted at all three sites.
The product is 2,4,6-tribromophenylamine, not a 3,5-dibromo product.
Look for these in your work
- Reference position
- Ortho/para identified
- Product corrected
Common mistakes
- Using Br₂/AlBr₃ and heat
- Predicting only monobromination
- Putting bromine at positions 3 and 5
- Changing –NH₂ into a bromo group
What can you explain now?
Which conditions brominate phenylamine in this syllabus outcome?
Compare with the explanation
Br₂(aq) at room temperature
Phenylamine's ring is strongly activated.
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Make it stick
What does the –NH₂ group cause when phenylamine reacts with bromine water?
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