I was reading some organometallics and I came upon the preparation of uranocene, $\ce{U(C8H8)2}$. It has two cyclooctatetraenide anions, $\ce{C8H8^2-}$, as ligands, with a uranium atom between them.

It was mentioned that the neutral cyclooctatetraene molecule $\ce{C8H8}$ is non aromatic and has a tub-shaped conformation, but the dianion $\ce{C8H8^2-}$ is planar but aromatic. Why is this so? Can fulfilling Hückel's rule force a cyclic molecule to become planar?


2 Answers 2


Fulfilling Hückel's rule means that you get to be aromatic.

The reason cyclooctotetraene is non-planar is because its planar arrangement would be antiaromatic. It twists to avoid it, because antiaromatic compounds are unstable with respect to localisation of the π-electrons (see here: What is the justification for Hückel's Rule?)

The veritable nanosecond that a molecule has the potential to become aromatic though, it will leap at the opportunity. Any angle strain from being planar is so quickly overtaken by the huge jump in stability that aromaticity brings that the molecule will twist to form an aromatic structure in a blink.

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    $\begingroup$ Fulfilling Huckel's rule is just one of the conditions to become aromatic. Planarity is another one. I think there are compounds that fulfill the rule but aren't planar and hence not aromatic. $\endgroup$
    – Papul
    Commented Jun 25, 2015 at 15:20
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    $\begingroup$ In Cyclodecapentaene angle strain overcomes stabilisation $\endgroup$
    – Mithoron
    Commented Jun 25, 2015 at 15:36
  • $\begingroup$ Under most circumstances, aromaticity > angle strain. $\endgroup$ Commented Jun 25, 2015 at 15:45
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    $\begingroup$ Yeah, in smaller, more common rings rings $\endgroup$
    – Mithoron
    Commented Jun 25, 2015 at 16:22
  • $\begingroup$ No, the jump in stability is not huge, except for small aromatic rings. The energetic effect of aromaticity decreases with larger ring size. This explains why [10]annulene notoriously fails to find an aromatic conformation and larger [4n+2]annulenes, when planar, tend not to show the peculiar reaction characteristics we see in benzenoid systems. $\endgroup$ Commented Jan 28, 2018 at 16:30

If cyclooctatetraene were planar, it would be an antiaromatic compound according to Hückel's rule, because it has 8 π-electrons. However, since its lowest energy conformation is non-planar, cyclooctatetraene can also be considered a non-aromatic compound.

Conformation of cyclooctatetraene

(source: Wikimedia Commons)

The dianion has 10 π-electrons. Since this fulfills the $4n+2$ criterion, it adopts a planar conformation to enjoy aromatic stabilisation.


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