Nucleophilic aromatic substitution (SNAr): practice and mechanism
Strong electron-withdrawing groups (NO₂) ortho or para to the halogen make the ring electron-poor. Methoxide adds, the anion (Meisenheimer complex) is stabilised by NO₂, then the halide leaves.
Example
A nitro group in the ortho or para position withdraws electrons so strongly that a nucleophile can attack the carbon bearing the halogen. The negative charge of the intermediate (Meisenheimer complex) is taken up by the nitro group, then the halide leaves.
Mechanism step by step
- 1
Methoxide attacks the carbon bearing the halogen. The ring takes up the negative charge.
- 2
The Meisenheimer complex forms: the charge is spread onto the NO₂ groups in ortho/para positions. This is the slow step.
- 3
The halide leaves and the ring becomes aromatic again: an aryl methyl ether.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Williamson ether synthesis
Frequently asked questions
Why does fluorine react fastest?
The slow step is the attack, not the leaving. The very electronegative F makes the carbon especially positive and stabilises the anion. Order: F > Cl > Br > I.
And an NO₂ group in the meta position?
Barely helps, because the negative charge of the Meisenheimer complex can only spread onto the NO₂ group from ortho/para.