I am wondering about the electronic structure of pyridine.

If we look at nucleophilic substitution, which favored on electron poor centers, for 2-chloro and 4-chloro pyridine we see that the substitution in 4 position is about 5 times faster (Ethanol, 20°C, taken from The Chemistry of Heterocycles, page 277

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This SE question and answer explain why this might be the case and the explanation there is that the lone pair of the nitrogen is repelling the nucleophile. This might be the case to certain extend, but from geometry of the attack I doubt that it has much of an effect. In addition if we look at the electron distribution in the pi system we get this picture (also taken from The Chemistry of Heterocycles, page 271)

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So the 4 position seems to be electron poorer than the 2 position. Usually we explain electron poor and rich centers in heterocycles by using mesomeric forms:

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And the usual comment is that the 2 position is closer to the nitrogen so the electron withdrawing effect is stronger than in 4 position and it should be electron poorer, which isn't the case.

I also did some DFT calculations using 2- and 4-pyridyl substituents and 4-pyridyl is more electron withdrawing than 2-pyridyl in that case.

Is there an understandable explanation why the 4-position is electron poorer despite beeing furtehr away from the nitrogen?

  • $\begingroup$ I think when + charge is on the 4th position there is a type of cross conjugation which makes the cation lesser stable .. and when the + charge is on the second carbon there is extended conjugation which stabilizes the cation to more extent .. $\endgroup$ – Kushal Tripathi Jul 7 '17 at 9:48
  • $\begingroup$ @KushalTripathi besides mesomeric forms not being a real thing this explanation would lead to the exact opposite of what is observed. If the form with the positive charge on position 4 is less stable then it should contribute less to the overall electronic structure, which would mean it's not that electron poor in position 4. $\endgroup$ – DSVA Jul 7 '17 at 9:58
  • $\begingroup$ Theres a related issue which is why 2-pyridyl boronic acids are so unstable, I'll try and dig those references up, but the essence is that the bond is significantly lengthened when its at the 2 position (i.e. the sm is just more reactive, rather than any argument about nucleophiles being repelled) $\endgroup$ – NotEvans. Jul 7 '17 at 10:01
  • $\begingroup$ The relative energies of the transition state often are reflected in the relative rates (or vice-versa) so the electron distribution of the anionic intermediate is what should be calculated. $\endgroup$ – tim hockswender Jul 10 '17 at 11:45
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    $\begingroup$ You described the first scheme as $\mathrm{S_NAr}$ being 5 times faster at C-4, but the relative rates given indicate that it is 5 times faster at C-2. Typo, maybe? (I didn't want to assume which was correct.) $\endgroup$ – orthocresol Aug 6 '17 at 14:00

When you deal with an aromatic ring, pairwise covalent bond structures miss a lot of key features. You need to consider the entire conjugated ring as a unit and take account of symmetry.

Consider almost a benzene ring. We may describe its $\pi$ orbitals as linear combinations of the out-of-plane atomic $p$ orbital on the carbon atoms. Let $p_k$ represent the wave function of one such atomic, with the positive lobe on a specified side of the ring plane. The same orbital but with the positive lobe on the"wrong" side of the ring plane is rendered as $-p_k$. The subscript $k$ refers to atom number $k$ as determined in the standard IUPAC scheme.

We then describe the canonical, occupied $\pi$ orbital as the familiar linear combinations:

$\psi_1=(1/\sqrt{6})\times (p_1+p_2+p_3+p_4+p_5+p_6)$

$\psi_2=(1/\sqrt{12})\times (2p_1-p_2-p_3+2p_4-p_5-p_6)$


Now suppose that atom $1$ is changed from carbon to nitrogen to make pyridine. Now that atom, being more electronegative, acts an an electron sink on occupied orbitals having an interaction with that atom -- meaning $\psi_1$ and $\psi_2$ are vulberable to thus $\pi$-electron withdrawal but $\psi_3$ which has a node at atom $1$ is "protected" by its symmetry. Then positions $2$ and $3$, with some of their $\pi$- electron density on the protected $\psi_3$ orbital, are less susceptible than atom 4 whose $\pi$ electrons are all on the vulnerable $\psi_1$ and $\psi_2$ orbitals.

Now, going a little beyond the question, suppose we were to N-oxidize the pyridine, as with a peroxy acid. In pyridine N-oxide the oxygen acts as a $\pi$ donor so that the nitrogen (or more accurately, the N-oxide functional group) is now effectively an electron source instead of a sink. The "vulnerable" orbitals are now electron enriched and this characteristic is most strongly imparted, by conjugation and symmetry as argued above, to position 4. Then electrophilic substitution, enhanced (or enabled?) by the N-oxidation, is most strongly so at position 4. So, the primary electrophilic substitution product, after subsequently reducing the N-oxide, is the 4-substituted pyridine. Which is what we see empirically with a wide variety of pyridine N-oxide electrophilic substitutions.

  • $\begingroup$ Oh wow, thanks. I think I got it. Just one follow up question: In benzene the orbitals look like this: chemtube3d.com/benzenemos.html . I could now put the nitrogen in two different position, for example the most left one or one right next to it. You described the case where it's in the left position. In the other case the 4 position would be part of all three orbitals. Is it the described case "just" because there's more symmetry to it? $\endgroup$ – DSVA Jul 11 '17 at 8:13
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    $\begingroup$ @DVSA I might not be understanding your comment, but there shouldn't really be different positions to place the nitrogen in benzene. Whatever carbon you replace is effectively carbon $1$ due to the symmetry of benzene. Or, phrased another way, one of the linear combinations formed would not include the 4th carbon away from it. $\endgroup$ – Tyberius Aug 6 '17 at 15:44

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