Schiemann reaction (fluoride): practice and mechanism
First NaNO₂/acid at 0 °C turns the amino group into a diazonium salt (Ar–N₂⁺). N₂ is an excellent leaving group. With HBF₄ the diazonium tetrafluoroborate precipitates; on heating fluoride replaces the N₂ group (aryl fluoride).
Example
The diazonium tetrafluoroborate can be isolated. On heating it loses N₂ and BF₃ and the aryl cation binds fluoride. This makes aryl fluorides that are not accessible by direct fluorination.
Mechanism step by step
- 1
NaNO₂ and acid form the nitrosyl cation (NO⁺) at 0 °C. The amine attacks it; after proton shifts and loss of water the aryl diazonium ion Ar–N≡N⁺ forms.
- 2
With HBF₄ the aryl diazonium tetrafluoroborate precipitates and is isolated.
- 3
On heating the salt decomposes: N₂ and BF₃ escape and fluoride bonds to the ring.
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: Sandmeyer reaction (chloride) · Phenol from a diazonium salt
Frequently asked questions
Why the detour via the diazonium salt?
It puts substituents where direct SEAr cannot: nitrate, reduce to the amine (a good director), then swap the amino group for Cl, Br, CN, OH, I, F or H.
Why not fluorinate directly with F₂?
F₂ reacts explosively and unselectively with arenes. The Schiemann reaction places the F at one defined position.