Is there a well defined way to discern the hybridization of a nitrogen atom in ring, like pyrrole? How can you know whether the nitrogen's lone pair are in the conjugated system or not?


If an $\alpha$-$\beta$ unsaturated heteroatom isn't already participating in a $\pi$-bond and it has a lone pair of electrons, these electrons will be delocalized and in the $\mathrm{p}$-orbital.

Let's compare pyridine and pyrrole:

$\hspace{5cm}$pyridine and pyrrole

The nitrogen in pyridine is already participating in a double bond, so it is clearly $\mathrm{sp^2}$ hybridized. The nitrogen in pyrrole is $\alpha$-$\beta$ unsaturated, so its lone pair is delocalized. Both of these molecules are heterocyclic aromatic rings, but these are not the only cases in which a nitrogen atom's lone pair are delocalized. They are also delocalized in amides (which resemble the $\alpha$-$\beta$ unsaturation of pyrrole):

$\hspace{4.6cm}$enter image description here

If you want to know more about criterion for aromaticity, read about Hückel's rule.

  • $\begingroup$ Hückel's rules actually cover only a tiny subset of aromatic compounds. It's a start, but it should not be used to attest aromaticity. The rules were derived from hydrocarbons and are strictly speaking only valid for those systems. Ergo pyridine and pyrrole are not really covered by them. Also, hybridisation is a model and not all the orbitals have to be hybridised in the same way - it is only depending on the geometry. In your last example I would doubt that the nitrogen is planar. $\endgroup$ – Martin - マーチン May 3 '16 at 12:07
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    $\begingroup$ I think it's bizarre to suggest that Hückel's rule wouldn't apply to heterocyclic rings. There are many, many examples on the web of them perfectly and accurately fitting the criterion. If anything I would caution against applying it to polycyclic species, where it is known to break down. Also, DMF is planar in the solid and liquid phases, and slightly non-planar in the gas phase according to this paper. $\endgroup$ – ringo May 3 '16 at 15:20
  • $\begingroup$ Strictly speaking, it does not apply. It is theoretically not justifiable and when you say that it fits perfectly and accurately, you are just over simplifying a very complex problem. Even trying to applying it to polycyclic compound is flat out wrong, it had nothing to do with cautioning. Just because it is easy, it does not mean it is correct. Just because many people make the same mistake, it does not make it correct. I find it bizarre, that you use the rules without questioning the validity. $\endgroup$ – Martin - マーチン May 3 '16 at 19:28

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