# Is buckminsterfullerene aromatic?

According to Wikipedia,

The $\ce{C60}$ molecule is extremely stable,[26] withstanding high temperatures and high pressures. The exposed surface of the structure can selectively react with other species while maintaining the spherical geometry.[27] Atoms and small molecules can be trapped within the molecule without reacting.

Smaller fullerenes than $\ce{C60}$ have been distorted so heavily they're not stable, even though $\ce{M@C28}$ is stable where $\ce{M\,=\,Ti, Zr, U}$.

Some of us have heard and learned about the "rules" of aromaticity: The molecule needs to be cyclic, conjugated, planar and obey Huckel's rule (i.e. the number of the electrons in $\pi$-system must be $4n+2$ where $n$ is an integer).

However, I'm now very skeptical to these so-called rules:

• The cyclic rule is violated due to a proposed expansion of aromaticity. (See what is Y-aromaticity?)
• The must-obey-Huckel rule is known to fail in polycyclic compounds. Coronenefigure 1 and pyrene figure 2 are good examples with 24 and 16 $\pi$ electrons, respectively.
• Again, Huckel fails in sydnone. The rule tells you that it's aromatic, while it's not.

The planar rule is not a rule at all. We're talking about "2D" aromaticity when we're trying to figure out the $n$ in $4n+2$. The "3D" rule is as following:

In 2011, Jordi Poater and Miquel Solà, expended the rule to determine when a fullerene species would be aromatic. They found that if there were $2n^2+2n+1$ π-electrons, then the fullerene would display aromatic properties. - Wikipedia

This would mean $\ce{C60}$ is not aromatic, since there is no integer $n$ for which $2n^2+2n+1 = 60$.

On the other hand, $\ce{C60-}$ is ($n = 5$). But then this rule strikes me as peculiar because then no neutral or evenly-charged fullerene would be aromatic. Furthermore, outside the page for the rule, Wikipedia never explicitly states that fullerene is not aromatic, just that fullerene is not superaromatic. And any info on superaromaticity is unavailable or unhelpful to me; including the Wikipedia "article" on that topic.

So, is $\ce{C60}$ aromatic? Why, or why not?

• "The molecule needs to be cyclic, conjugated, planar and obey the Huckel's rule." This is redundant, because Hückel's Rule is "Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. [...]" see the goldbook. The definition acc. to IUPAC is much more fuzzy. – Martin - マーチン Sep 24 '15 at 11:33

Aromaticity is not binary, but rather there are degrees of aromaticity. The degree of aromaticity in benzene is large, whereas the spiro-aromaticity in [4.4]nonatetraene is relatively small. The aromaticity in naphthalene is not twice that of benzene.

Aromaticity has come to mean a stabilization resulting from p-orbital (although other orbitals can also be involved) overlap in a pi-type system. As the examples above indicate, the stabilization can be large or small.

Let's consider $\ce{C_{60}}$:

• Bond alternation is often taken as a sign of non-aromatic systems. In $\ce{C_{60}}$ there are different bond lengths, ~1.4 and 1.45 angstroms. However, this variation is on the same order as that found in polycyclic aromatic hydrocarbons, and less than that observed in linear polyenes.

Conclusion: aromatic, but less so than benzene.

• Magnetic properties are related to electron delocalization and are often used to assess aromaticity. Both experiment and calculations suggest the existence of ring currents (diamagnetic and paramagnetic) in $\ce{C_{60}}$.

Conclusion: Although analysis is complex, analysis is consistent with at least some degree of aromaticity.

• Reactivity - Substitution reactions are not possible as no hydrogens are present in $\ce{C_{60}}$. When an anion or radical is added to $\ce{C_{60}}$ the electron(s) are not delocalized over the entire fullerene structure. However, most addition reactions are reversible suggesting that there is some extra stability or aromaticity associated with $\ce{C_{60}}$.

Conclusion: Not as aromatic as benzene

• Resonance energy calculations have been performed and give conflicting results, although most suggest a small stabilization. Theoretical analysis of the following isodesmic reaction

$$\ce{C_{60} + 120 CH4 -> 30 C2H4 + 60 C2H6}$$

suggested that it only took half as much energy to break all of the bonds in $\ce{C60}$ compared to the same bond-breaking reaction with the appropriate number of benzenes.

Conclusion: Some aromatic stabilization, but significantly less than benzene.

This brief overview suggests that $\ce{C_{60}}$ does display properties that are consistent with some degree of aromatic stabilization, albeit less than that found with benzene.

• Yes The book MARCH'S ADVANCED ORGANIC CHEMISTRY talks sbout 3D aromaticity. And also a question was asked in the stack exchange previously chemistry.stackexchange.com/questions/88637/… – An enthusiast Apr 27 '20 at 15:28
• But although less stabilising than normal aromaticity – An enthusiast Apr 27 '20 at 15:29

There are quite a few conjugated $\pi$-systems out there; some of them are pretty stable, some are less so. The ultimate way to find out is to go and solve the eigenvalues problem for the corresponding matrix; in some cases this can be done manually, with pen and paper.

There are simple cases with analytical solutions. One of them is the case of cyclic polyenes. The result can be summarized in the Huckel's 4n+2 rule. Indeed, it applies exclusively to the cyclic (that is, not polycyclic) conjugated planar systems. Historically, these were the only systems called aromatic. Later, the concept began to stretch more and more.

With fullerenes, there is no simple parametric solution that would fit a more-or-less wide range of systems. The whole idea of $2n^2+2n+1$ electrons is downright wrong, exactly for the reasons you specify (odd number of electrons, to begin with), if not for any other. There is no simpler way than to solve the eigenvalues problem for a particular molecule and see. It turns out that $\ce{C60}$ is stable, and so is $\ce{C70}$ and many other fullerenes, including some isomers of $\ce{C80}$, but not the $I_h$-symmetric $\ce{C80}$, which may however be stabilized by adding six electrons. You see that the question of fullerene stability is highly complicated, and can't be resolved by considering the number of electrons alone.

Whether or not to call the stable fullerenes aromatic is but a question of convenience. I'd rather not, for their reactions are somewhat more like those of alkenes.

C$_{60}$ is antiaromatic as it does not follow Hirsch's rule for aromaticitity of spherically symmetric molecules, $2(n+1)^2$. If we add up all the $\pi$ electrons, we can see that there are 60, which is not twice a perfect square. Here is a great paper explaining spherical aromaticity.

Is C60 buckminsterfullerene aromatic?

In summary, if we look at the MOs of C60, we can see that for C$_{60}^{10+}$, the g-subshell is completely filled, and thus the species is aromatic. If we move to the neutral molecule, it has a partially filled subshell, and thus is not aromatic, which is in line with Hirsch's rule.

• While assigning the label 'aromatic' is itself rather meaningless mathematically, saying $\ce{C60}$ is antiaromatic is a bit too much. Yes, it's estimated that 80 percent of the enthalpy of formation is strain energy, but still, fullerene is a very stable molecule. – M.A.R. Sep 21 '15 at 16:02
• You need to define what you mean by stable. Fullerenes are not thermodynamically stable, they just don't have any decomposition routes with small activation barriers. From the conclusion of the C60 paper "C60 is ‘the most strained molecule ever isolated’ with the astronomically unstable experimental heat of formation of 610±30 kcal mol 1 (over 10 kcal/mol strain per carbon)!" – Jonathon Sep 21 '15 at 16:11
• @Jonathon I'm not sure I agree. Cook a bunch of carbon and take a mass spec, and you see a peak at 720 amu. There are plenty of examples before Smalley et al realized it was C60. If it was so unstable, other products would be likely. Empirically, there must be some stability. – Geoff Hutchison Sep 22 '15 at 12:52
• non aromatic or antiaromatic? – Archer Feb 24 '18 at 8:45