Aryl iodide from a diazonium salt: 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. Iodide replaces the N₂ group even without copper: an aryl iodide forms.
Example
Iodide is so easily oxidised that it reacts with the diazonium salt without a copper catalyst: N₂ escapes and the aryl iodide forms. Direct iodination of the ring with I₂ hardly works.
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
Iodide transfers an electron to the diazonium ion (possible without copper because I⁻ is easily oxidised); N₂ escapes.
- 3
The aryl radical combines with iodine: aryl iodide.
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 (bromide) · Phenol from a diazonium salt
Frequently asked questions
Why work at 0 °C?
Aryl diazonium salts decompose on warming (N₂ escapes) and can even explode when dry. In the cold they stay stable in solution long enough for the second step.
Why no copper as for Cl and Br?
Iodide is a good reducing agent itself and starts the radical reaction alone. Cl⁻ and Br⁻ cannot.