A-Level Amines, amides and amino acids: worked solution

Question

Both 2-chloropyridine and 4-chloropyridine take part in nucleophilic aromatic substitution with relative ease, whereas 3-chloropyridine does not. This mechanism proceeds first by addition of a nucleophile and then by an elimination step. (2-chloropyridine + R2NH -> a 2-aminopyridine product + HCl; 4-chloropyridine + R2NH -> a 4-aminopyridine product + HCl; 3-chloropyridine + R2NH -> no reaction.) Using an appropriate diagram, propose an explanation for why this difference is seen.

The nucleophile (R2NH) first adds to the ring carbon bearing the chlorine, giving a negatively charged (carbanion/Meisenheimer-type) intermediate; the negative charge is delocalised around the ring. Diagram: draw the addition intermediate showing the negative charge spread by resonance. For 2-chloro- and 4-chloropyridine the carbon attacked is directly conjugated with the ring nitrogen, so one resonance form places the negative charge ON the electronegative nitrogen, which stabilises the intermediate - reaction proceeds, then Cl- is eliminated. For 3-chloropyridine the position of attack is NOT conjugated to nitrogen, so the negative charge cannot be delocalised onto N; the intermediate is much less stable and the reaction does not occur.