Why isn’t azete aromatic?

[![Structure of azete](https://i.stack.imgur.com/3pyEO.png)](https://i.stack.imgur.com/3pyEO.png)


I’m confused, because it seems like Hückel’s rule is satisfied (two electrons from the nitrogen lone pair and four from the double bonds) and that the molecule should be aromatic. Where am I going wrong here?




The compound is anti-aromatic. While counting the number of π-electrons, you count the electrons which are delocalized over the ring. In this case the nitrogen lone pair is localised and does not participate in resonance. The nitrogen lone pair is in an $\mathrm{sp^2}$ orbital (red) which is orthogonal to the π system (blue):


[![Orbitals in azete](https://i.stack.imgur.com/3dQvw.png)](https://i.stack.imgur.com/3dQvw.png)


So, the total number of π-electrons is only four: two from each double bond.


Please read this for a better understanding : <http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf>


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The rule for lone pairs in this situation goes like this: you put one lone pair into the conjugated circuit if the atom otherwise has no pi bond into the ring. Any other lone pairs after you've done that are localized in outward pointing orbitals. Here the nitrogen already has a pi bond into the ring, so the lone pair is outward pointing and you can't get to the six conjugated ring electrons you wanted.


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