I read that the molecular orbital diagram of cyclobutadiene predicts that it should have a diradical nature as it has two unpaired electrons in two degenerate non bonding orbitals (one in each).

Why does this diradical state exist only in cyclobutadiene? Shouldn't other anti aromatic compounds, like cyclooctatetraene, also exist in a similar diradical state?

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    $\begingroup$ It predicts that if cyclobutadiene was square it would be diradical, but it's not. $\endgroup$ – Mithoron Dec 1 '20 at 15:40
  • $\begingroup$ This isn't a duplicate of Cyclobutadiene - Jahn–Teller effect or not?, although the answer there may be helpful. $\endgroup$ – orthocresol Dec 1 '20 at 18:46
  • $\begingroup$ And C8H8 isn't even nearly flat. There's just nothing antiaromatic about it. $\endgroup$ – Karl Dec 1 '20 at 21:12

Anti-aromaticity is often not taught very clearly. Let me start, then, by emphasising that this anti-aromatic diradical state should not be taken as a real thing. It is a purely hypothetical state that may arise if the molecule adopted the shape of a planar regular polygon (i.e. square for $\ce{C4H4}$, octagon for $\ce{C8H8}$).

Because of various reasons, this state never truly arises in real-life $\ce{C4H4}$ or $\ce{C8H8}$.

So cyclobutadiene is not really special in this regard. Both $\ce{C4H4}$ and $\ce{C8H8}$ have the same status: if they were to have a regular polygon structure, they would possess a diradical electronic state.* However, neither of them have such a structure, so neither of them have such an electronic state.

* For advanced readers: even square $\ce{C4H4}$ does not really possess a diradical state either, as is explained in Cyclobutadiene - Jahn–Teller effect or not?. This is because the MO theory used to predict the diradical state is not sufficiently advanced. (Technically, the diradical state arises from Hückel theory, and the proof of its non-existence for square $\ce{C4H4}$ arises from post-HF theories, such as configuration interaction.)


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