6.1.2Organic Chemistry and Analysis
Electrophilic Substitution
5 min read 5 parts Lesson 3 of 33 Virtual lab
Section 1 of 5Why Substitution
The delocalised π system of benzene makes it unusually stable.
Delocalised ring
- Six π electrons spread over all six carbon atoms give benzene extra stability compared with a hypothetical ring with three separate double bonds.
Attack by electrophiles
- The high electron density of the ring attracts electron-deficient species, called electrophiles.
Substitution not addition
- Addition would destroy the delocalised system;
- substitution replaces one hydrogen and restores the stable ring.
Weaker than alkenes
- The π electron density is spread out,
- so benzene does not polarise bromine,
- and a halogen carrier or strong electrophile is needed.
Section 2 of 5General Mechanism
Every electrophilic substitution follows the same three stages.
1
An electrophile E⁺ is generated, often with the help of a catalyst.
2
Two electrons from the π system form a bond to E⁺, giving a positively charged intermediate in which the delocalisation is partly broken.
3
The C–H bond to the attacked carbon breaks and the electrons move back into the ring, restoring the delocalised system.
4
The H⁺ lost is picked up by a negative ion from the catalyst system, regenerating the catalyst.
5
The overall result is that one hydrogen on the ring is replaced by the group E.
Section 3 of 5Reactions and Conditions
Reactions and Conditions
Nitration
- Concentrated HNO₃ with concentrated H₂SO₄ at about 50 °C; H₂SO₄ is a catalyst: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.
- Productnitrobenzene, C₆H₅NO₂.
Halogenation
- Br₂ or Cl₂ with an anhydrous halogen carrier (FeBr₃, AlCl₃ or Fe).
- FeBr₃ + Br₂ → FeBr₄⁻ + Br⁺, giving C₆H₅Br + HBr.
Friedel–Crafts acylation
- An acyl chloride with anhydrous AlCl₃
- CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻.
- Product: phenylethanone, C₆H₅COCH₃.
Friedel–Crafts alkylation
- A haloalkane with anhydrous AlCl₃ gives an alkylbenzene,
- such as methylbenzene from chloromethane,
- but with a risk of further substitution.
Section 4 of 5Worked Example
Question
- 15.6 g of benzene (M = 78.0 g mol⁻¹) is nitrated and 18.5 g of nitrobenzene (M = 123.0 g mol⁻¹) is obtained.
- Calculate the percentage yield.
- n(C₆H₆) = 15.6 ÷ 78.0 = 0.200 mol, and the equation is 1 : 1.
- Theoretical mass of nitrobenzene = 0.200 × 123.0 = 24.6 g.
- Percentage yield = 18.5 ÷ 24.6 × 100 = 75.2%.
Answer The percentage yield is 75.2%.
Section 5 of 5Exam Checklist & Tips
6.1.2 – Key Points for the Exam
- Electrophile An electron-pair acceptor, attracted to regions of high electron density.
- Electrophilic substitution A reaction in which an electrophile replaces an atom, usually hydrogen, on a ring.
- Halogen carrier A catalyst such as FeBr₃ or AlCl₃ that helps to generate a strong electrophile.
- Acylation Introduction of an acyl group, RCO–, into a molecule.
Common Mistakes
- Drawing an addition mechanism for benzene, which would destroy the aromatic ring.
- Forgetting that sulfuric acid is regenerated in nitration and so acts as a catalyst.
- Using aqueous bromine or no catalyst when bromine is reacted with benzene.
- Using wet conditions for Friedel–Crafts reactions, where the catalyst AlCl₃ reacts with water.