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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.

orthocresol
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Henry
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2 Answers2

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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

Reference

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

Oscar Lanzi
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    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? – Ivan Neretin May 20 '16 at 10:22
  • Cyclobutadiene (just butadiene" is a difgerent, non-cyclic molecule) has unequal carbon-carbon bond lengths and its ring is not considered fully conjugated. – Oscar Lanzi May 20 '16 at 10:28
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    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. – Ivan Neretin May 20 '16 at 10:40
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    @OscarLanzi I couldn't get your point. What does cyclopropenyl do to break conjugation? – Henry May 20 '16 at 10:55
  • I checked Wikipedia, they say cyclobutadiene forms an equilibrium between square and rectangular structures, as if the square structure were a transition state between two petpendiculatly oriented rectangles. We get teo stereoisomers of 1,2-dideutetocyclobutadiene depending on whether the deuterium atoms are attached ot a long of short side of the rectangle. Whether any conjugation remains in the rectangular forms is ultimately immaterial: the molecule is destabiluzed by stetic strain, and only aromatic conjugation could overcome that. – Oscar Lanzi May 20 '16 at 11:02
  • Henry please read my answer. Cyclopropenyl anion breaks the planar structure by pyramidalizing a bond to a ligand atom. – Oscar Lanzi May 20 '16 at 11:04
  • @OscarLanzi "Cyclopropenyl anion breaks the planar structure by pyramidalizing a bond to a ligand atom." Can you please provide a diagram for reference? Thanks. – Henry May 20 '16 at 13:13
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    http://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 – IT Tsoi May 21 '16 at 03:11
  • @IvanNeretin Altering bond lengths does not necessarily break the conjugation, you're right about that. It does lower the symmetry of the molecule, which means it splits degeneracies of orbitals. The main reason for the destabilisation is that you have either too many or to few electrons to fill those degenerate orbitals completely or avoid it. In this sense, it breaks the ideal conjugation. – Martin - マーチン May 18 '17 at 14:35
  • Wouldn't the carbanion keep its lone pair in a hybrid orbital instead of a p orbital to avoid anti-aromaticity, hence making it non-aromatic due to absence of complete conjugation? – User Jul 25 '17 at 04:08
  • That is what pyramidalization effectively does. – Oscar Lanzi Jul 26 '17 at 19:21
  • I don't understand why cyclopropenyl anion is planar(necessary condition for anti-aromaticity). I learnt that a carbanion is always sp3 hybridised. – user600016 Oct 31 '18 at 13:50
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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.

Ian Hunt
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