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Organic Chemistry

Topic 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)

Aromatic substitution restores the delocalised ring

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

For monobromination, use Br2 with anhydrous AlBr3. The Lewis acid accepts electron density and strongly polarises Br2. A useful formal representation is Br2 + AlBr3 ⇌ Br+ + AlBr4-; this is electron bookkeeping for the activated electrophile rather than a claim that a bottle contains free isolated Br+.

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 H while restoring the delocalised system. Permanent addition would lose that aromatic stabilisation, explaining benzene's preference for substitution. Do not draw a full circle inside the sigma complex or leave the catalyst consumed in the final overall equation.