Electrophilic Aromatic Substitution Reactions: Bromination

Core Concepts

Why substitution and not addition

Alkenes react with Br₂ by addition: the π bond breaks and two new C–Br bonds form. If benzene did the same, the product would lose its aromatic sextet and the reaction would be uphill in resonance energy. Substitution avoids this: the C–H bond electrons that replace the leaving proton reform the aromatic π system; the driving force is the re-aromatization in step 2. Experimentally: cyclohexene decolorizes bromine instantly at room temperature; benzene needs a catalyst and mild heating — and even then it substitutes rather than adds.

The Lewis-acid trick: activating bromine

Molecular Br₂ is not electrophilic enough to attack benzene. FeBr₃ acts as a Lewis acid: it accepts electron density from one bromine, leaving the other electron-poor. The complex can be pictured as Br–Br···FeBr₃, the distal bromine carrying substantial positive character — the "Br⁺" equivalent. FeBr₃ is usually generated in situ by adding iron filings (Fe + Br₂ → FeBr₃). FeCl₃ plays the same role in chlorination.

Step 1: Formation of the arenium ion (rate-determining)

The electrophilic bromine approaches the π cloud and bonds to one carbon, which rehybridizes to sp³ (it now bears Br and H and has lost its p orbital from the aromatic loop). The positive charge is not localized — resonance delocalizes it over the ortho and para carbons (describe the curved arrows in words: from the π bond to the electrophile, then shifting adjacent π bonds to place the positive charge at the ortho positions and the para position). Because the intermediate has an sp³ carbon and no continuous p-orbital loop, aromaticity is lost — hence the large activation energy and rate-determining status. The arenium ion is a true (if short-lived) intermediate, not a transition state.

Step 2: Deprotonation restores aromaticity

A base — in practice FeBr₄⁻ or Br⁻ — removes the proton from the sp³ carbon; the C–H bond electrons collapse into the ring, reforming the p-orbital loop and the aromatic sextet. This step is fast, strongly exothermic, and regenerates the catalyst: the proton combines with the bromide fragment to give HBr, freeing FeBr₃ to activate another Br₂. Energy profile: reactants → high-energy arenium-ion valley (rate-determining barrier) → low-energy product.

Regiochemistry and scope

All six benzene positions are equivalent, so monosubstitution gives a single product, bromobenzene, with no positional isomers and no carbocation rearrangements (a complication of Friedel–Crafts alkylation, Topic 3). The product ring is less reactive than benzene — bromine is weakly deactivating, ortho/para-directing — so disubstitution needs more forcing conditions. Chlorination is analogous (Cl₂/FeCl₃); fluorination with F₂ is dangerously vigorous, while iodination with I₂ is too slow and needs an oxidant — both treated next topic.

Common Confusions