N-alkylation of pyridine: practice and mechanism
Pyridine attacks CH₃I with its nitrogen (SN2): an N-methylpyridinium salt forms. No Friedel–Crafts-type C-methylation takes place.
Example
Pyridine is a base and a nucleophile at nitrogen (the lone pair is not part of the π system). It attacks CH₃I via SN2, giving an N-methylpyridinium salt. The ring is not methylated.
Mechanism step by step
- 1
The lone pair on N (in the ring plane, not part of the aromatic system) attacks the carbon of CH₃I.
- 2
In one SN2 step the N–C bond forms and iodide leaves.
- 3
N-methylpyridinium iodide forms; the ring stays aromatic but is now positively charged.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The ring is not attacked at carbon. The lone pair on nitrogen is the nucleophile, so the nitrogen reacts.
- The ring is not attacked at carbon. The lone pair on nitrogen is the nucleophile, so the nitrogen reacts.
Often confused with: Pyridine N-oxide · Friedel–Crafts alkylation
Frequently asked questions
Why is the N of pyridine nucleophilic but that of pyrrole not?
In pyrrole the N lone pair is part of the aromatic sextet; using it would destroy aromaticity. In pyridine it lies in the ring plane and is available.
Where do pyridinium ions occur?
In the coenzyme NAD⁺: the nicotinamide ring is an N-alkylated pyridinium ion that accepts hydride (NADH).