Aromatic chemistry
The idea
Benzene's six π electrons are delocalized into a closed, especially stable ring — aromaticity, formalized by Hückel's 4n+2 rule. That stability changes everything about how the ring reacts: rather than adding across the double bonds like an alkene and destroying the aromatic system, benzene undergoes electrophilic aromatic substitution (EAS), where an electrophile replaces a hydrogen and the aromatic ring is preserved. Halogenation, nitration, sulfonation, and Friedel–Crafts reactions all share this two-step pattern: the ring attacks the electrophile to form a resonance-stabilized cation, then loses a proton to restore aromaticity.
When a substituent is already present, it controls both the rate and the position of the next substitution. Electron-donating groups (alkyl, OH, NH₂, OR) stabilize the intermediate cation, activate the ring, and direct the incoming electrophile to the ortho and para positions. Electron-withdrawing groups (NO₂, C=O, SO₃H, most directly bonded electronegative pulls) deactivate the ring and direct meta. Halogens are the instructive exception: they deactivate by induction yet direct ortho/para by resonance donation of a lone pair.
Guard against the assumption that aromatic rings behave like ordinary alkenes. Their reluctance to add, and their insistence on substitution, both trace to the energetic cost of breaking aromaticity in the intermediate — which is also why the directing effects can be predicted by drawing the resonance structures of that intermediate and asking which substituent best stabilizes the positive charge.
Worked example
Toluene (methylbenzene) is nitrated with a mixture of concentrated HNO₃ and H₂SO₄. Predict the major monosubstituted products and explain the directing and activating effect of the methyl group.
- Identify the electrophile: H₂SO₄ protonates and dehydrates nitric acid to generate the nitronium ion NO₂⁺, the active electrophile in nitration.
- Classify the existing substituent: the methyl group is electron-donating (through hyperconjugation and induction), so it activates the ring, making toluene nitrate faster than benzene itself.
- Determine orientation by stabilizing the intermediate: methyl is an ortho/para director, because attack at those positions produces a cationic intermediate with a resonance form bearing the positive charge on the methyl-substituted carbon, where the donating group stabilizes it best.
- Name the products: substitution occurs mainly at the carbons ortho and para to the methyl group, giving 2-nitrotoluene (ortho) and 4-nitrotoluene (para); the meta isomer is minor.
- Weigh statistics against sterics: there are two ortho positions for every one para, but attack beside the methyl group pays a steric price. Methyl is small enough that the statistical edge wins — toluene nitration gives roughly 3:2 ortho to para — while bulkier groups like tert-butyl push the split strongly toward para.
Answer. Nitration gives mainly 2-nitrotoluene and 4-nitrotoluene; the methyl group activates the ring and directs ortho/para.
Check your understanding
- Why does benzene undergo substitution rather than the addition that alkenes prefer, and what role does the intermediate play in that choice?
- How can a halogen substituent deactivate the ring yet still direct ortho/para, and why do those two effects not contradict each other?
- How would you decide the position of a second substitution when the ring already carries two groups with conflicting directing preferences?
- Why does drawing the resonance structures of the cationic intermediate let you predict directing effects without memorizing a list?
Build the foundations first
Aromatic chemistry builds on these concepts. If any feel shaky, start there.