Nucleophilic substitution of halopyridines: practice and mechanism
A halogen at C2 or C4 of pyridine is easily replaced by nucleophiles because the ring N takes the negative charge of the intermediate. A methoxypyridine forms.
Example
Halogen at C2 or C4 of pyridine is easily replaced, without any nitro group: the ring nitrogen takes the negative charge of the addition complex (like an NO₂ group in SNAr). At C3 this is impossible, so 3-chloropyridine hardly reacts.
Mechanism step by step
- 1
Methoxide attacks the carbon bearing the chlorine (C2 or C4).
- 2
In the addition complex (Meisenheimer-like) the negative charge sits on the electronegative ring N.
- 3
Chloride leaves and the ring becomes aromatic again: 2- or 4-methoxypyridine.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Methoxide is not a methylating agent. As a nucleophile it attacks the carbon bearing the halogen; the ring nitrogen is not methylated.
- The nucleophile is methoxide, not hydroxide. The methyl ether forms, not a hydroxypyridine (pyridone).
Often confused with: Nucleophilic aromatic substitution (SNAr) · Chichibabin reaction
Frequently asked questions
How does pyridine differ from benzene?
N is more electronegative than C and pulls electron density out of the ring: electrophiles react poorly (as with nitrobenzene), nucleophiles readily at C2 and C4. The lone pair on N lies in the ring plane and makes pyridine basic (pKa of pyridinium ≈ 5).
Why does 3-chloropyridine hardly react?
With attack at C3 the negative charge cannot be placed on N, only on carbons. It is like chlorobenzene without a nitro group.