According to my book, a compound is antiaromatic if it is cyclic, planar, and possesses a fully conjugated system of p-orbitals with $4n$ π-electrons.

However, I have also been told that the cyclopropenyl anion is neither aromatic nor antiaromatic.

Cyclopropenyl anion

How is this so? It has four π-electrons and therefore should satisfy the criteria listed above.


2 Answers 2


When you have a ring that has $4n$ π electrons it's anti-aromatic if you force it into a fully conjugated cycle.

Often such rings find a way to break the unfavorable (or at least, less favored) anti-aromatic conjugation. Large ones like cyclooctatetraene are likely to bend so all ring atoms are no longer in the same plane. Small ones like the cyclopropenyl anion shown here can't bend the ring out of planarity easily (or do it at all with only three atoms), but they can still break the conjugation by using unequal bond lengths or moving a ligand out of the plane giving that ring atom a pyramidal bonding geometry.

The pyramidal bonding geometry seems to be what cyclopropenyl anion actually does. A computational result is shown by Kass [1]; the top carbon atom has its bonds directed towards the observer indicating a pyramidal geometry (see below, from https://comporgchem.com/blog/archives/2987 and [1]):

enter image description here


1. Kass, S. R. "Cyclopropenyl Anion: An Energetically Nonaromatic Ion," J. Org. Chem. 2013, 78, 7370-7372, https://doi.org/10.1021/jo401350m.

  • 1
    $\begingroup$ I would not say that unequal bond lengths break the conjugation. See, the p-orbitals are still there, and still able to overlap and interact. Or look at butadiene: isn't it one conjugated structure? $\endgroup$ May 20, 2016 at 10:22
  • $\begingroup$ Cyclobutadiene (just butadiene" is a difgerent, non-cyclic molecule) has unequal carbon-carbon bond lengths and its ring is not considered fully conjugated. $\endgroup$ May 20, 2016 at 10:28
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    $\begingroup$ Yes, I intentionally asked about butadiene, which is different and non-cyclic. Though it applies to cyclobutadiene as well. Now you say they are not fully conjugated; this implies they are still conjugated a little, aren't they? Anyway, I see your point: the pattern of orbitals is no longer the same as predicted by the Frost circle, the degeneracy is broken, so cyclobutadiene is not a biradical, but just a diene, so it is not anti-aromatic, but just not aromatic. This makes sense. $\endgroup$ May 20, 2016 at 10:40
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    $\begingroup$ @OscarLanzi I couldn't get your point. What does cyclopropenyl do to break conjugation? $\endgroup$
    – Henry
    May 20, 2016 at 10:55
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    $\begingroup$ comporgchem.com/blog/?p=2987 has a good computational description and a 3D rendering which might help @Henry. Although mathematically, three points always form a plane, remember that the charged carbon has the option of adopting either an sp2 or sp3 hybridization. it can only be aromatic if it adopts sp2 so that the remaining p-orbital is planar and can interact effectively with the other p-orbitals. However, the bond angle puts a strain that is quite significant whereas adopting sp3 will reduce this significantly. But if that carbon is sp3, then there can be no continuous p-orbitals $\endgroup$
    – IT Tsoi
    May 21, 2016 at 3:11

Putting the lone pair in an sp3 hybrid orbital would break the conjugation and result in a more stable, non-aromatic rather than very unstable anti-aromatic.

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    $\begingroup$ Well, that looks kida like TL:DR for Oscar's answer - rather not enough to stand on its own. $\endgroup$
    – Mithoron
    Mar 7, 2018 at 20:29

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