Topic 18 · page 6 of 10

Electrophilic substitution and bromination

Benzene is attacked by a strong electrophile in two steps. The ring donates two π electrons to E⁺, then loses H⁺ and recovers its delocalised ring. Bromination shows the pattern with a catalyst making Br⁺.

15 min read18.418.5

Read first: Why benzene resists bromination, Reactions of benzene at a glance

Why benzene reacts with electrophiles 18.5

The delocalised ring is a region of negative charge above and below the plane. So the ring attacks species that are short of electrons.

Benzene's electron density is low compared with an alkene's, so it needs a strong electrophile. Each reaction in this topic first makes one, using a catalyst or a strong acid. This is electrophilic substitution: an electrophile replaces a hydrogen atom on the ring.

The two-step mechanism 18.5

All five electrophilic substitutions of benzene in this topic follow one pattern. Here E⁺ stands for any electrophile.

General mechanism. Step 1: a curly arrow from the delocalised ring to E+. The intermediate has a positive horseshoe over five carbons, with H and E on the sixth. Step 2: a curly arrow from the C-H bond back into the ring. The product is the ring with E replacing H, plus H+.
The pattern for every substitution on benzene in this topic.

Step 1. Two electrons from the delocalised ring form a bond to E⁺. The curly arrow starts on the ring (the circle, or a double bond in the Kekulé form) and ends at E⁺. The carbon that bonds to E now has four σ bonds, so it can no longer take part in the π system. The intermediate is a cation: the π system covers only five carbons, with the positive charge spread over them. This is drawn as a horseshoe with + inside, open towards the carbon carrying H and E.

Step 2. The C–H bond breaks. Its electron pair moves back into the ring, which restores the six-carbon delocalised system. H⁺ is lost.

Bromination of benzene 18.418.5

Benzene reacts with liquid bromine in the presence of a catalyst, at room temperature. The catalyst is FeBr₃ or AlBr₃. Many labs add iron filings, which react with bromine to form the iron(III) bromide catalyst:

2 Fe(s)+3 BrX2(l)→2 FeBrX3(s) \ce{2Fe(s) + 3Br2(l) -> 2FeBr3(s)}

The apparatus must be dry, because water would destroy the catalyst.

CX6HX6(l)+BrX2(l)→CX6HX5Br(l)+HBr(g) \ce{C6H6(l) + Br2(l) -> C6H5Br(l) + HBr(g)}

The product is bromobenzene. Only one bromine atom is added.

Generating the electrophile

Br₂ is not a strong enough electrophile for benzene. The FeBr₃ catalyst has an electron-deficient iron atom that accepts a lone pair from one bromine atom of Br₂, so the Br–Br bond breaks and Br⁺ forms:

BrX2+FeBrX3→BrX++FeBrX4X− \ce{Br2 + FeBr3 -> Br^+ + FeBr4^-}

The mechanism

The electrophile is Br⁺. Follow the general pattern with E = Br.

Mechanism of bromination. A curly arrow from the ring to Br+. An intermediate with a positive horseshoe over five carbons and H and Br on the sixth. A curly arrow from the C-H bond back into the ring, giving bromobenzene and H+.
Bromination of benzene: the electrophile is Br+.

Regenerating the catalyst

The H⁺ lost in step 2 combines with FeBr₄⁻:

HX++FeBrX4X−→HBr+FeBrX3 \ce{H^+ + FeBr4^- -> HBr + FeBr3}

FeBr₃ is back where it started, which is why it is a catalyst. If a question asks you to show that it is, include this equation.

Chlorination works in the same way, with Cl₂ and a catalyst of AlCl₃ or FeCl₃. The electrophile is Cl⁺, formed by ClX2+AlClX3→ClX++AlClX4X−\ce{Cl2 + AlCl3 -> Cl^+ + AlCl4^-}, and the product is chlorobenzene.

Worked exampleMechanism for chlorination of benzene

Outline the mechanism for the reaction of benzene with chlorine and aluminium chloride.

  1. Electrophile: ClX2+AlClX3→ClX++AlClX4X−\ce{Cl2 + AlCl3 -> Cl^+ + AlCl4^-}
  2. Curly arrow from the delocalised ring to Cl⁺. The intermediate has a horseshoe over five carbons, with + inside, and H and Cl on the sixth carbon.
  3. Curly arrow from the C–H bond back into the ring. H⁺ is lost and chlorobenzene, C₆H₅Cl, forms.
  4. Catalyst regenerated: HX++AlClX4X−→HCl+AlClX3\ce{H^+ + AlCl4^- -> HCl + AlCl3}

Check yourself

  1. Define an electrophile.
  2. Write the equation for the formation of the electrophile when benzene reacts with bromine and FeBr₃, and the equation that regenerates the catalyst.
  3. Outline, with curly arrows, the mechanism of the bromination of benzene.
  4. Explain why cyclohexene reacts with bromine without a catalyst, but benzene needs one.
  5. Why does the intermediate lose H⁺ rather than bonding to Br⁻?
  6. Give the reagents and conditions for the chlorination of benzene, and name the product.

Answers cover these until you have tried the questions

  1. An electron pair acceptor.
  2. Electrophile: BrX2+FeBrX3→BrX++FeBrX4X−\ce{Br2 + FeBr3 -> Br^+ + FeBr4^-}. Regeneration: HX++FeBrX4X−→HBr+FeBrX3\ce{H^+ + FeBr4^- -> HBr + FeBr3}
  3. Curly arrow from the ring (the circle) to Br⁺. Intermediate: horseshoe over five carbons, + inside, with H and Br on the sixth carbon. Curly arrow from the C–H bond back into the ring. Products: bromobenzene and H⁺.
  4. Cyclohexene's localised π bond has high electron density, which polarises Br₂ so it reacts as it is. Benzene's delocalised π electrons give lower electron density, which cannot polarise Br₂ enough. So a catalyst is needed to form the much stronger electrophile Br⁺.
  5. Losing H⁺ restores the delocalised ring and its 152 kJ mol⁻¹ of extra stability. Bonding to Br⁻ would give an addition product with the delocalisation destroyed.
  6. Chlorine, with an AlCl₃ (or FeCl₃) catalyst, at room temperature in dry conditions. The product is chlorobenzene.