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I came across aza[9]-annulene, which is given to be aromatic. I can clearly see it follows Hückel's rule with 8 π-electrons in the ring and the lone pair of nitrogen.

azonine

But the thing that concerns me is the planarity of structure. I'm not convinced that this structure is planar with so much angle strain (and thus aromatic).

Can anyone please explain this to me?

andselisk
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Nex
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    There is a relevant article on Wikipedia regarding the molecule, azonine. – Nicolau Saker Neto Jan 02 '21 at 05:37
  • @NicolauSakerNeto Thanks for the link. That surely answers my question to some extent. :) – Nex Jan 02 '21 at 05:40
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    related: https://chemistry.stackexchange.com/q/51484/102629 – cngzz1 Jan 02 '21 at 06:35
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    Please check the following link https://en.m.wikipedia.org/wiki/Azonine . It clearly states , this molecule is planar. – Chakravarthy Kalyan Jan 02 '21 at 07:08
  • There is no much of strain, even. It could be higher for closing a non-plantar ring! – Alchimista Jan 02 '21 at 08:35
  • @Alchimista Sir, So why do we consider angle strain (~24°) in [10]-annulene, which is merely 4° greater than in the given compound? – Nex Jan 02 '21 at 08:40
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    @Nex likely the non planar, 10 annulene counterpart is less strained. As far as a molecule do exist, there is not a clear border between stability or not. Also because strain is not the only parameter that affects the overall energy. Rather than compare different molecules you should compare the possible structures of the same molecule. If you would have molecular models at hand, you will certainly have to apply a force to close the azonine cycle while using localized double bonds, too. – Alchimista Jan 02 '21 at 08:46
  • @Nex above, where I say "there is no clear border between...." I mean you cannot set a fixed value for angle strain as a discriminant, and the same is true for every strain or factors contributing to the overall energy. When everything is balanced and results are compared, then you can conclude what is stable and what not. Still, nothing is better than experiments: most arguing is done at posteriori. – Alchimista Jan 02 '21 at 08:56
  • @Alchimista Got it. And yes, I completely agree with you on the posteriori part :P. Thanks for the input. – Nex Jan 02 '21 at 09:01

1 Answers1

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Comments cite Wikipedia as stating the molecule is planar. Here that is backed up with a technical reference from Somers et al 1.

Ref. 1 gives a drawing of the molecular structure, reproduced below from that source, showing bond angles in the ring whose average is $140°$ with good accuracy; this value is consistent with a planar enneagon.

enter image description here

The authors point out that this planar structure is easily distorted, but without necessarily losing aromaticity:

[I]nteraction with surrounding H2O molecules and alkali ions and substitution of the N−H hydrogen distorts the planarity of the ring. This distortion is such that the aromaticity remains. N-Methylazonine is characterized as nonplanar and the global minimum structures of the alkali salts have the metal residing on top of the distorted ring (cation−π interaction).

One reason for the methylated compound going away from planarity has to do with the $140°$ ring angles in the planar structure, which bring neighboring substituents close to each other. Substituents larger than hydrogen atoms are more prone to distort in such a situation. Hydrogen bonding of the protic hydrogen with water may also be regarded as a larger substituent.

Reference

K. R. F. Somers, E. S. Kryachko, and A. Ceulemans (2004). "Azonine, a 'Nearly' Forgotten Aromatic Molecule". J. Phys. Chem. A 108, 18, 4059–4068. https://doi.org/10.1021/jp037046+

Oscar Lanzi
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