Describe the electrophilic substitution mechanism for bromination
Explain catalyst-assisted bromination through ring attack and proton loss that restores aromaticity.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 30. Follow the listed reagents, catalysts, temperatures and reaction scope for benzene and methylbenzene. Directing effects are limited to –NH₂, –OH, –R, –NO₂, –COOH and –COR. Written practice is self-checked
Before you begin
- Use the Topic 29 model of benzene as a planar aromatic ring with a delocalised π system, and recognise electrophiles, nucleophiles and organic functional groups.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the role of AlBr₃ during benzene bromination?
What you will learn
- Explain how AlBr₃ activates bromine
- Describe ring attack and the positively charged intermediate
- Show deprotonation restoring aromaticity
The Lewis-acid catalyst activates Br₂
AlBr₃ interacts with and polarises Br₂, making one bromine sufficiently electrophilic for the aromatic π system to attack. The exact syllabus conditions use Br₂ and AlBr₃ for bromination of an arene.
The ring π electrons form a bond to electrophilic bromine. A positively charged intermediate forms and the ring's delocalisation is temporarily interrupted.
Deprotonation completes substitution
A base removes H⁺ from the carbon where bromine has bonded. The C–H electron pair restores delocalisation in the ring, giving bromobenzene and HBr while the catalyst is regenerated.
The key logic is the same as nitration: electrophile attack followed by proton loss. Name the bromine catalyst condition and show that the aromatic system is restored in the product.
Pause and explain
Which step restores the ring after bromine attaches?
Worked example
Outline the steps in bromination of benzene using AlBr₃.
Show the worked solution
- AlBr₃ polarises and activates Br₂.
- The ring π system attacks electrophilic Br and forms an intermediate.
- Loss of H⁺ restores aromaticity and produces bromobenzene.
Answer AlBr₃ activates Br₂; the ring attacks electrophilic bromine, then deprotonation restores the aromatic system to give bromobenzene.
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What change does the aromatic π system cause when it attacks activated bromine?
Give me a hint
Track the new bond and charge in the intermediate.
Compare my reasoning
- A ring electron pair forms a C–Br bond.
- The ring intermediate becomes positively charged.
- Aromatic delocalisation is temporarily interrupted.
The π system donates an electron pair to electrophilic Br, forming C–Br and a positively charged intermediate.
Look for these in your work
- Electron donation stated
- Bond formed
- Intermediate charge described
Why is HBr formed in the overall bromination of benzene?
Give me a hint
Account for the ring hydrogen and the other bromine atom.
Compare my reasoning
- One Br atom enters the ring.
- A ring H is removed to restore aromaticity.
- The removed H combines with the remaining Br to form HBr.
The ring loses H⁺ during restoration, and it combines with the remaining bromide/proton equivalent to give HBr overall.
Look for these in your work
- Ring Br traced
- Hydrogen loss traced
- By-product accounted
A student draws bromine addition across a benzene bond and stops before proton loss. What two key features are missing?
Give me a hint
Compare a substitution product with the required stable aromatic product.
Compare my reasoning
- The intermediate must lose H⁺.
- That step restores delocalisation.
- The final product is bromobenzene, not a permanently added dibromide.
The mechanism must show deprotonation and restoration of the aromatic π system; the product is bromobenzene rather than a permanent addition product.
Look for these in your work
- Deprotonation included
- Aromaticity restoration included
- Product class corrected
Common mistakes
- Saying Br₂ reacts without activation under the stated conditions
- Keeping the intermediate neutral after ring attack
- Leaving H attached after bromine substitution
- Forgetting aromaticity is restored by proton loss
What can you explain now?
What is the role of AlBr₃ during benzene bromination?
Compare with the explanation
It activates/polarises Br₂
This helps create an electrophilic bromine species.
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Make it stick
What do AlBr₃ activation and proton loss each contribute to bromination of benzene?
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