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Structure of azete

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?

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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

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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  • $\begingroup$ Wouldn't a better explanation be that the nitrogen atom here with a pi bond can be either sp or sp2 hybridized, and in most cases, such as this one, an sp hybridization would cause too much ring strain, leading to an sp2 hybridization, which makes the nitrogen lone pair localized and thus unable to participate in conjugation? I don't see why theoretically every nitrogen with a pi bond has to have it's lone pair in a localized orbital if something like ring strain is absent, though I'm guessing that's rarely (or maybe even never) the case practically. $\endgroup$ – Ashish Ahuja Dec 27 '20 at 15:14
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    $\begingroup$ Even if sp hybridization were allowed it still would not work. Pi orbitals can be conjugated into the ring only if they are properly oriented, and when you have only s and p valence orbitals to work with there is only one properly orbital per atom. So if you have a pi orbital into the ring you're done. $\endgroup$ – Oscar Lanzi Dec 27 '20 at 15:36
  • $\begingroup$ Ahh, I get it now. $\endgroup$ – Ashish Ahuja Dec 27 '20 at 15:43

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