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9476 / 2027
Organic Chemistry

Lesson 10 of 34

Aromatic substitution restores the delocalised ring

Benzene first forms a non-aromatic intermediate, then loses a proton to recover aromatic stabilisation.

A-Level 9476 (2026-2027)

  • I can compare benzene and alkene reactivity through delocalisation.Syllabus

    Syllabus 9476, 11.3(e). Compare benzene and alkene reactivity through delocalisation.

  • I can describe aromatic electrophilic substitution.Syllabus

    Syllabus 9476, 11.3(m). Describe aromatic electrophilic substitution.

  • I can describe the specified aromatic-ring reactions.Syllabus

    Syllabus 9476, 11.4(d). Describe the specified aromatic-ring reactions.

Aromatic substitution restores the delocalised ring

Benzene first forms a non-aromatic intermediate, then loses a proton to recover aromatic stabilisation.

Make a guess. You are not marked.

Bromine is mixed with benzene in the dark at room temperature, with no catalyst. What happens?

  1. Bromobenzene forms quickly without any help
  2. The colour fades as bromine adds across the ring
  3. Nothing visible; the bromine colour stays
Show the answer

Nothing visible; the bromine colour stays

The delocalised ring is stable, and Br2 alone is not a strong enough electrophile. A catalyst such as FeBr3 is needed to make Br+.

To add one bromine to benzene, use Br2 with anhydrous AlBr3. AlBr3 is a Lewis acid. It accepts electron density from Br2 and makes it strongly polar. You can write this as Br2 + AlBr3 ⇌ Br+ + AlBr4-. This equation shows where the electrons go. It does not mean free Br+ ions sit in the bottle.

The aromatic pi system attacks the activated bromine

A curly arrow starts at the aromatic pi system and points to electrophilic bromine. Forming a carbon-bromine sigma bond temporarily interrupts the cyclic delocalisation.

This step forms a positively charged sigma complex. One ring carbon now bears both H and Br and is temporarily tetrahedral; the whole ring must not retain a full aromatic circle in that intermediate.

Loss of H+ restores the aromatic pi system

One resonance form of the sigma complex has a ring carbon bearing H and Br, two remaining double bonds and a positive charge elsewhere. A base accepts H; the carbon-hydrogen bond pair moves into the adjacent ring bond, restoring the aromatic system.

The base is a bromide ligand in AlBr4-. Removing the proton forms HBr and regenerates AlBr3. The displayed positive-charge position is one resonance contributor, not a fixed isolated charge.

Overall, C6H6 + Br2 → C6H5Br + HBr. Substitution replaces an H and keeps the delocalised ring. Addition would destroy the ring and its extra stability, which is why benzene prefers substitution. Do not draw a full circle inside the intermediate. The catalyst is made again, so it must not appear as used up in the overall equation.

Exam-style question

Benzene reacts with bromine in the presence of anhydrous AlBr3. Name and describe the mechanism. Show how the catalyst is regenerated. [3]

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Check your understanding

In alkene bromination the carbocation gains Br-. In benzene bromination the intermediate loses H+ instead. Why?

  1. Br- is too large to approach the ring, so the intermediate has to lose H+ to become neutral.
  2. Losing H+ restores the delocalised ring of six pi electrons, which is especially stable.
  3. The AlBr3 catalyst holds on to every Br- ion, so no bromide is free to add to the ring.
Show the answer

Losing H+ restores the delocalised ring of six pi electrons, which is especially stable.

The C-H bond pair returns to the ring, so the pi electrons are delocalised over all six carbons again. Adding Br- would leave two separate C=C bonds and lose that stability, so substitution wins.