It's well established that angular twisting can decrease the degree of delocalization across a $\pi$ system (e.g. biphenyl).

Moreover, there's considerable debate about aromaticity in spherical species like $\ce{C_{60}}$.

So let's ask a naive question for discussion. Does aromatic stabilization require a planar delocalized $\pi$ system?

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    $\begingroup$ Good prompt points would include buckybowls and Mobius aromatic systems. $\endgroup$ May 20, 2016 at 20:34
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    $\begingroup$ Then helicenes would also fit I guess. $\endgroup$
    – Mithoron
    May 20, 2016 at 21:44

1 Answer 1


The $\pi$ system is not obliged to be planar at all; the only limitation is to keep the angular twist between the adjacent atoms low enough so they still can be conjugated. (Biphenyl violates that, so it is out of luck.) The whole structure can be pretty much anything. It can be slightly out-of-plane, like corannulene or circulene; it may have cylindrical shape (see the example below), in which case the Hückel's rule still applies, or it may be all twisted to the topology of a Möbius strip, in which case the said rule is reversed.

Fullerenes are not aromatic, unless you stretch the definition of aromaticity to utter pointlessness. Some of their derivatives are, though. $\ce{C60Cl30}$ comes to mind: its 30 remaining $sp^2$ carbons are grouped in two 6-membered benzene-like rings and one 18-membered belt, which (judging by its equalized bond lengths) is an example of a cylindrical aromatic system.


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