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Antiaromaticity is so destabilizing that it can cause compounds such as cyclobutadiene to elongate or manipulate their orbitals so that the pi system is no longer aromatic. I understand that aromatic compounds are stabilized because they have a full HOMO and that antiaromatic compounds have a half filled HOMO, which according to this question is the reason that they are unstable. I don't agree with this explanation because in other compounds, half filled orbitals cause compounds to be stabilized. For example, triplet oxygen is significantly more common than singlet oxygen in nature. Also, some transition metals will promote electrons from their S orbitals to D orbitals so that they can have these shells half filled. I know that full orbitals are more stable than half filled orbitals, but are half filled orbitals destabilizing in alkenes while they are stabilizing in oxygen and metals? If not, why are antiaromatic compounds so unstable?

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Niels Kornerup
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  • http://chemistry.stackexchange.com/questions/18557/is-cyclobutadiene-anti-aromatic – orthocresol May 08 '16 at 01:03
  • First of all, they are not unstable, just less stable and less symmetric, because the Jahn-Teller effect distorts their geometry to remove degeneracy. With $\ce{O2}$ it would not work that way, since you can't really distort a 2-atom molecule; no matter what you do, its symmetry is still the same. – Ivan Neretin May 08 '16 at 06:55
  • But, correct me if I am wrong, the Jahn-Teller effect distorts their geometry so that they are no longer antiaromatic. Compounds distort their orbitals to avoid the instability associeated with antiaromaticity. The Jahn-Teller effect is a symptom of antiaromatic instability, but I don't think it is it's cause. – Niels Kornerup May 08 '16 at 16:55
  • http://chemistry.stackexchange.com/questions/30927/what-is-the-justification-for-h%c3%bcckels-rule – Mithoron May 08 '16 at 17:28
  • What causes a diradical to be so unstable? I understand why free radicals such as superoxide are unstable, but triplet oxygen is a diradical and it is more stable than singlet oxygen, which has no radicals. I think there is something else that causes antiaromatic compounds to be unstable, but I do not know what it is. – Niels Kornerup May 08 '16 at 20:05
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    Re. your first comment, you are wrong about that. The antiaromatic instability is an instability with respect to a structure where the double bonds are localised, i.e. instability with respect to a Jahn-Teller distortion. The Jahn-Teller distortion leads to the adoption of exactly such a geometry: one with alternating bond lengths and that is not necessarily planar. So, the JT instability and the "antiaromatic instability" are one and the same. – orthocresol May 09 '16 at 18:46
  • For more information I would refer you to Orbital Interactions in Chemistry 2nd ed. by Albright, Burdett, & Whangbo, pp 282-285. I quote: "Strictly speaking, therefore, the distortion in cyclobutadiene is really of the second-order Jahn-Teller type." It is probably more common in inorganic chemistry to speak of it as a Jahn-Teller distortion, whereas organic chemists don't want to bother with all those wavefunction symmetry stuff and just call it antiaromaticity. – orthocresol May 09 '16 at 18:50
  • Thanks! I think I understand antiaromaticity now. Do you want to post what you were telling me as an answer, or should I write it up? – Niels Kornerup May 11 '16 at 01:22

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Antiaromatic compounds are not necessarily unstable - they are just less stable than a bunch of ethenes connected by sigma bonds. If the pi electrons in an antiaromatic compound were to delocalize across a large degenerate orbital, then the result would be a half filled shell. While a half filled shell is not the end of the world, it is a lot more stable to have a full shell of electrons. As a result, compounds such as cyclobutadiene undergo a Jahn Teller distortion (in this case, becoming a rectangle rather than a square) in order to form two separate (and non degenerate) orbitals. This creates a collection of separate, full, pi orbitals. The stability gained by having full rather than half full orbitals is more than enough to compensate for the energy required to break the symmetry in the first place. Cyclobutadiene is stabilized due to stretching not because it prevents antiaromaticity, but because it causes all the pi electrons to be part of complete shells. While it might seem weird that a physical distortion such as stretching can consume less energy than is released by going from a half to a full shell, let me remind you that this is not the case. Higher energy light (UV) is often required to stimulate electronic transitions while lower energy light (IR) can cause bond stretching in compounds.

I was able to figure this out thanks to the help of orthocresol.

Niels Kornerup
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