Electrophilic substitution: halogenation, nitration and sulfonation

The general mechanism

1. A curly arrow from the delocalised ring to the electrophile, E+. This forms a positively charged intermediate in which the delocalisation is partly broken (drawn as a horseshoe inside the ring, with + in it, and the H and E on the same carbon).

2. A curly arrow from the C-H bond into the ring: H+ is lost and the delocalised ring is restored.

Overall: C6H6 + E+ -> C6H5E + H+

Halogenation (spec 18.4)

Bromine with a halogen carrier, iron(III) bromide (or iron filings, which react with bromine to form it), at room temperature:

C6H6 + Br2 -> C6H5Br + HBr

Electrophile: Br2 + FeBr3 -> Br+ + FeBr4-. Afterwards H+ + FeBr4- -> FeBr3 + HBr, regenerating the catalyst. Chlorine works the same way with AlCl3 or FeCl3.

Nitration (spec 18.4)

Concentrated nitric acid and concentrated sulfuric acid, kept at about 50 °C (below 55 °C to limit further nitration):

C6H6 + HNO3 -> C6H5NO2 + H2O (nitrobenzene)

Electrophile, the nitronium ion: HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-. The sulfuric acid is a catalyst: H+ + HSO4- -> H2SO4. Nitration is the first step in making dyes and explosives.

Sulfonation and combustion (spec 18.4)

• Heating with fuming sulfuric acid (concentrated H2SO4 containing SO3) gives benzenesulfonic acid: C6H6 + SO3 -> C6H5SO3H. Sulfonic acids are used to make detergents and dyes.
• Combustion: 2C6H6 + 15O2 -> 12CO2 + 6H2O, with a smoky flame.

Worked example

Write equations to show how the electrophile forms in the nitration of benzene, and how the catalyst is regenerated.

1. HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-
2. After substitution: H+ + HSO4- -> H2SO4

Answer: NO2+ is the electrophile; sulfuric acid is regenerated

Check yourself

1. What are the conditions for the nitration of benzene?
   • Dilute nitric acid at room temperature
   • Nitrogen gas with a nickel catalyst
   • Concentrated nitric acid under reflux at 150 °C
   • Concentrated nitric and sulfuric acids at about 50 °C
   Show answer

   Concentrated nitric and sulfuric acids at about 50 °C. The mixed acids generate NO2+; a low temperature limits further substitution.

2. What is the electrophile in the nitration of benzene?
   • NO3-
   • NO2+
   • HNO3
   • NO2-
   Show answer

   NO2+. The nitronium ion is a strong electrophile.

3. What is the role of iron(III) bromide in the bromination of benzene?
   • It is the nucleophile
   • It removes water
   • It decolourises the bromine
   • It reacts with Br2 to produce the electrophile Br+
   Show answer

   It reacts with Br2 to produce the electrophile Br+. Br2 + FeBr3 -> Br+ + FeBr4-.

4. What are the products when benzene reacts with chlorine in the presence of AlCl3?
   • Chlorobenzene and hydrogen chloride
   • 1,2,3,4,5,6-hexachlorocyclohexane
   • Chlorobenzene and hydrogen
   • Chlorocyclohexane
   Show answer

   Chlorobenzene and hydrogen chloride. Substitution of one H by Cl, releasing HCl.

5. Which equation shows the formation of the electrophile for nitration?
   • HNO3 -> H+ + NO3-
   • H2SO4 + NO2 -> NO2+ + HSO4- + H
   • HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-
   • HNO3 + H2O -> NO3- + H3O+
   Show answer

   HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-. Sulfuric acid protonates nitric acid, which loses water.

6. Which reagent sulfonates benzene?
   • Fuming sulfuric acid
   • Dilute sulfuric acid
   • Sodium sulfate solution
   • Sulfur dioxide gas at room temperature
   Show answer

   Fuming sulfuric acid. Fuming H2SO4 contains SO3, the electrophile.

7. 7.80 g of benzene (Mr 78.0) gives 9.84 g of nitrobenzene (Mr 123.0). Calculate the percentage yield.
   Show answer

   80.0 %. Theoretical: 0.100 mol x 123.0 = 12.3 g; yield = 9.84 / 12.3 x 100 = 80.0%.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.