10
$\begingroup$

Nucleophilic substitution at C-2 of 2,6-dichloro-3-nitropyridine

In the formation of ethyl 4-(6-chloro-3-nitropyridin-2-yl)piperazine-1-carboxylate G, why is the substitution favoured at the 2-position? I would have said that the 6-position substitution is preferred as:

  1. In pyridines, nucleophilic substitution occurs at the 2/6th or 4th position, and as chlorine is an effective leaving group compared to hydride this rules out position 4, leaving 2 and 6 as the most feasible...
  2. The electron withdrawing nature of the nitro group on the ring would favour nucleophilic substitution at the positions ortho or para to itself. The chlorine at the 2 -position is ortho, and the chlorine at the 6-position is para so both are feasible...
  3. If I didn't know what the product was, I would have said that position 6 is therefore the site of substitution as the amine reagent is quite bulky and the 6-position is less sterically hindered due to the absence of an adjacent nitro group

However, this evidently isn't the case! I found another similar reaction where the same thing occurs:

enter image description here

Is it something to do with the nitrogen in the pyridine ring hindering the electron density transfer to and from the para nitro group? I'm really stuck!

$\endgroup$
0

1 Answer 1

4
$\begingroup$

I'd agree that nucleophilic substitution at C-6, leaving the nitro and piperazine substituents para to each other, would lead to the thermodynamic product.

The observed regioselectivity is therefore probably kinetically controlled. Although the nitro group activates both ortho (C-2) and para (C-6) positions towards nucleophilic substitution via resonance, it's worth bearing in mind that the nitro group is also extremely electron-withdrawing via the inductive effect in its own right.

This would make C-2 more electron-deficient and hence prone to nucleophilic attack.

Recall that in an SNAr reaction the nucleophile must attack the π* orbitals of the arene, which are above and below the plane of the ring. In this case, the nitro substituent is also (for the most part) in the plane of the ring, and doesn't lie along the trajectory of nucleophilic attack. So, the nitro substituent doesn't present all that much steric hindrance to the nucleophile.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.