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According to wikipedia and the references given therein, $\pi\cdots\pi$ stacking interactions are the result of interaction between the quadrupole moments of two aromatic rings, rationalising the stabilisation of perpendicular and offset-parallel association modes of the benzene dimer.

Conventional wisdom is that the supramolecular complexes of buckminsterfullerene and functionalised corannulenes (the so-called buckycatchers) are stabilised by $\pi\cdots\pi$ stacking. How can the quadrupole interaction explanation be reconciled with the pseudo-spherical symmetry of buckminsterfullerene? Surely the interaction must be entirely Van der Waals in nature?

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In short, I think the answer is that although the overall symmetry of the buckminsterfullerene is almost spherical, the quadrupole–quadrupole interaction with other molecules with aromatic cycles are local (in addition to the ever-present van der Waals interactions).

I tried to think of a way to exemplify that, and drew the 2D picture below:

enter image description here

In this 2D model, the “double bonds” of the circular structure represent the individual aromatic rings of your buckminsterfullerene (which isn't superaromatic, i.e. electrons are not delocalized over its entire structure; think may be a key point), bearing each a local quadrupole moment (symbolized in the red/blue thingy). The other particle with a quadrupole is close to one of the ball's quadrupoles, and this local interaction dominates. In fact, other interactions maybe be stabilizing or not, depending on the orientation of the individual quadrupoles, but they are weaker that the local quadrupole–quadrupole interaction.

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  • $\begingroup$ This sounds reasonable, and I didn't appreciate the lack of superaromaticity in $\mathrm{C}_{60}$. Your suggestion is similar (but with crucial differences) to that of a researcher (focused on graphite intercalates) I got the chance to speak with, who speculated that the dihedral symmetry elements of the $I_{h}$ symmetry may permit a molecular quadrupole. I won't accept the answer just yet to give others a chance to chime in. $\endgroup$ – Richard Terrett May 1 '12 at 3:17

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