Arenes and Electrophilic Substitution
Explain electrophilic substitution in arenes and predict its products.
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Arenes combine a delocalised pi system with condition-sensitive chemistry. This lesson covers the complete 9476 sequence: structure and reactivity, electrophilic substitution, exact benzene/methylbenzene conditions, side-chain reactions and substitution-position reasoning.
Review Mechanisms and Curly Arrows first, then use the Organic Chemistry hub to connect arenes to halogenoarenes, phenols and phenylamine.
Definitions (Must Know)
A. Arene and delocalisation
An arene is an aromatic hydrocarbon containing one or more benzene rings. In benzene, each carbon is trigonal planar and contributes a p orbital to a continuous, delocalised pi system above and below the ring.
B. Electrophilic substitution
An electrophile replaces a ring hydrogen. The ring temporarily loses aromaticity while the new bond forms, then loses H⁺ and restores the delocalised system.
C. Ring and side chain
In methylbenzene, ring substitution replaces a ring hydrogen. Side-chain substitution replaces a hydrogen in the methyl group. Reagents and conditions decide which site reacts.
D. Position notation
Relative to an existing substituent, positions 2 and 6 are ortho, 3 and 5 are meta, and 4 is para.
Key Ideas (What Earns Marks)
| Transformation | Reagents and essential conditions | Main outcome |
|---|---|---|
| chlorination of ring | Cl₂, AlCl₃ catalyst | chloroarene + HCl |
| bromination of ring | Br₂, AlBr₃ catalyst | bromoarene + HBr |
| nitration of benzene | conc. HNO₃/H₂SO₄, 50 °C | nitrobenzene |
| nitration of methylbenzene | conc. HNO₃/H₂SO₄, 30 °C | mainly 2- and 4-nitromethylbenzene |
| Friedel–Crafts alkylation | halogenoalkane, AlCl₃ or AlBr₃ | alkylbenzene |
| methylbenzene side-chain halogenation | Cl₂ or Br₂, UV, room temperature | radical substitution in side chain |
| methylbenzene side-chain oxidation | hot alkaline KMnO₄ then dilute acid, or hot acidified KMnO₄ | benzoic acid |
Halogen plus a Lewis-acid halogen carrier targets the aromatic ring by electrophilic substitution. Halogen plus UV light targets the alkyl side chain by free-radical substitution.
Detailed Explanations
A. Why benzene differs from an alkene
The six pi electrons are delocalised over the ring rather than confined to three isolated double bonds. This stabilises benzene, makes every carbon–carbon bond equivalent and reduces its readiness to attack electrophiles. Addition would permanently destroy the delocalised system; substitution restores it, so benzene prefers electrophilic substitution.
B. Mono-bromination mechanism
- AlBr₃ acts as a Lewis acid and polarises Br₂ strongly, generating the effective electrophile.
- A pair from the benzene pi system moves to electrophilic bromine as the Br-Br bond breaks.
- The sigma-complex intermediate has a new C-Br bond and a positive charge delocalised around the ring.
- A base removes H⁺; the C-H bond pair returns to the ring, restoring aromaticity and regenerating the catalyst.
The mechanism must show loss and restoration of the delocalised pi system, not alkene-style addition of two bromine atoms.
C. The prescribed ring reactions
- Halogenation: use Cl₂/AlCl₃ or Br₂/AlBr₃ and recognise the aluminium halide as a Lewis-acid catalyst.
- Nitration: concentrated nitric and sulfuric acids generate NO₂ +. Maintain about 50 °C for benzene but 30 °C for more reactive methylbenzene.
- Friedel–Crafts alkylation: a halogenoalkane with AlCl₃ or AlBr₃ supplies an alkyl electrophile and forms an alkylbenzene.
D. Methylbenzene: ring versus side chain
The methyl group releases electron density and activates the ring, especially at positions 2 and 4. With a halogen carrier, substitution therefore occurs mainly on the ring. Under UV light, homolytic fission starts a radical chain at the side chain. Strong hot permanganate oxidises an alkyl side chain that has a benzylic hydrogen all the way to -CO₂H.
E. Predicting substitution position
Electron-donating groups such as alkyl, -OH and -NH₂ generally favour positions 2 and 4. Strong electron-withdrawing groups such as -NO₂ and -CO₂H generally favour position 3. Halogens are deactivating but direct further substitution mainly to positions 2 and 4. Justify the prediction using the substituent’s effect on the stability of the substitution intermediate.
Worked Examples
Modelled example 1
Nitration of Benzene
Problem
Study the worked solution
Choose the nitrating mixture
Method
Use concentrated nitric acid and concentrated sulfuric acid.Reason
The acid mixture generates the electrophile needed for aromatic substitution.Working
Concentrated HNO₃ + concentrated H₂SO₄.Control the temperature
Method
Warm the mixture to about 50–60 °C.Reason
Controlled warming gives a useful reaction rate while limiting further nitration.Working
C₆H₆ → [conc. HNO₃/conc. H₂SO₄, 50-60 ^\circ C] C₆H₅NO₂.
Guided practice 2
Electrophile in Nitration
Problem
Identify the attacking species
Hints
Hint 1: charge
Hint 2: formula pattern
View solution step by step
Apply the electrophile definition
Method
Select the electron-pair-accepting species generated by the acid mixture.Reason
The benzene π system donates an electron pair to this species.Working
Electrophile: NO₂ +.Name the species
Method
Name NO₂ + as the nitronium ion.Reason
The formula and name together identify the nitrating electrophile unambiguously.Working
NO₂ +, nitronium ion.
Common misconception 3
Why Benzene Needs a Halogen Carrier
Learner claim
Connect delocalisation to electrophile strength
View solution step by step
Correct the bonding model
Method
Describe benzene as an aromatic ring with delocalised π electrons.Reason
Its delocalisation stabilises the ring, so benzene is less reactive toward bromine than an alkene.Working
Benzene is not modelled as three independent alkene bonds.Explain the catalyst
Method
Use a halogen carrier to generate a sufficiently strong electrophile from bromine.Reason
The stronger electrophile can attack the stabilised aromatic π system.Working
Br₂ + halogen carrier → electrophilic substitution conditions.
Challenge 4
Chlorination of Benzene
Halogenation transfer
Predict products and catalyst
Hints
Hint 1: reaction family
Hint 2: mass balance
View solution step by step
Write the substitution equation
Method
Replace one ring hydrogen by chlorine and form hydrogen chloride.Reason
Electrophilic substitution preserves the aromatic ring and releases the displaced hydrogen as HCl.Working
C₆H₆ + Cl₂ → C₆H₅Cl + HCl.State a halogen carrier
Method
Name aluminium chloride or iron(III) chloride.Reason
Either Lewis-acid halogen carrier activates chlorine for aromatic substitution.Working
Catalyst: AlCl₃ or FeCl₃.
Common Mistakes
- Saying benzene contains three ordinary isolated C = C bonds.
- Explaining substitution preference only as “benzene is stable” without linking it to delocalised pi electrons and restoration of aromaticity.
- Using bromine water as a benzene test or omitting the halogen carrier.
- Quoting one nitration temperature for benzene and methylbenzene.
- Omitting Friedel–Crafts alkylation from the arene reaction map.
- Predicting ring or side-chain halogenation without using the catalyst-versus-UV condition.
- Stopping side-chain oxidation at an alcohol or aldehyde instead of benzoic acid under the prescribed vigorous conditions.
- Treating every substituent as a 2,4-director or claiming one exclusive product where a mixture is expected.
Exam Tips
- For “benzene versus alkene”, compare electron localisation, reactivity towards electrophiles and substitution versus addition.
- Give exact paired conditions: Cl₂/AlCl₃ and Br₂/AlBr₃ for the prescribed ring examples.
- Write 50 °C for benzene nitration and 30 °C for methylbenzene nitration.
- In the bromination mechanism, show formation of the electrophile, the positively charged intermediate, deprotonation and catalyst regeneration.
- Use “major” or “mainly” for directed substitution unless the question establishes a single product.
- For side-chain oxidation, check for at least one hydrogen on the carbon directly attached to the ring.
Mind Stretchers
Mind stretcher 1: One reagent, two reaction sitesExtension
A student proposes Br₂ for two separate conversions of methylbenzene: making 4-bromomethylbenzene and making phenylmethyl bromide, C₆H₅CH₂Br. Specify how the conditions must differ and explain why each selects its site.
Show Hint
Decide whether the reactive species should be an electrophile or a radical.
Show Answer
For ring bromination, use Br₂/AlBr₃ without UV. The Lewis acid generates a stronger electrophile, and the methyl group directs substitution mainly to positions 2 and 4. For side-chain bromination, use Br₂ under UV light at room temperature without the halogen carrier. Homolytic fission generates radicals, and abstraction at the benzylic position gives a resonance-stabilised radical before C₆H₅CH₂Br forms.
Mind stretcher 2: Testing an arene oxidation claimExtension
Hot acidified permanganate converts ethylbenzene to benzoic acid, but does not convert tert-butylbenzene under the same side-chain rule. Explain the structural distinction and account for the fate of the extra side-chain carbon atoms in ethylbenzene.
Show Hint
Inspect the carbon directly bonded to the ring and ask whether it bears a hydrogen.
Show Answer
Ethylbenzene has a benzylic hydrogen on the carbon directly attached to the ring, so vigorous oxidation can convert that carbon to the carboxyl carbon; the remaining side-chain carbon is oxidised further, ultimately to carbon dioxide under complete conditions. The benzylic carbon of tert-butylbenzene has no hydrogen, so it does not satisfy the usual prescribed side-chain oxidation requirement.
Syllabus and review details
- GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus
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