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In nucleophilic aromatic substitution (SNAr) reactions on pyridines, displacement of leaving groups at C-2 and C-4 is favoured because the negative charge in the intermediate can be delocalised onto the nitrogen atom.

However, is there any intrinsic difference in the rates of substitution at C-2 and C-4 (assuming the leaving groups at both positions are the same)? If so, why?

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True, the para (C4) and ortho (C2) positions give similar resonance structures but consider the fact that:

  1. For an $\ce{S_{N}Ar}$ reaction the electron rich nucleophile must attack. Nitrogen, being highly electronegative, does draw electrons away from the ortho and para positions but because the electron clouds would be larger around nitrogen, it is harder for the nucleophile to approach the ortho than the para. You could think of it as the lone pair on Nitrogen being a steric shield of some sort.

  2. An attack on the para position gives a symmetric intermediate. This would be lower in energy compared to the intermediate produced by an attack on the ortho position.

As such, the para product would be the kinetic product of the reaction whereas the ortho product would be the thermodynamic product. Statistically ortho should be attacked more often as well, as there are two positions whereas there is only one para position.

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    $\begingroup$ I really don't think it's that simple. If you look in the literature (pyridine SNAr is very common in med chem for example), there are examples of both selective C-2 and selective C-4 substitution, and I think it really depends on conditions and exact substrate. See e.g. Baran group meeting "Pyridines readily undergo SNAr, usually faster at C-4 than C-2/6, but highly dependent on nucleophile and conditions" $\endgroup$ – orthocresol Oct 17 '18 at 17:15

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