Hückel's rule says that any planar compound with a ring of conjugated p orbitals with $4n+2$ electrons is aromatic. Here, can $n$ be any integer, or does $n$ have to be related to the number of p orbitals in the conjugated π system (as shown in the Frost circle)? For example, in benzene, there are 6 electrons in the pi system ($n=1$), which fill all 3 bonding orbitals. Thus, benzene is very stable. But say we remove 4 electrons so that there's only 2 electrons in the π system. Here, $n=0$, so is this cation considered aromatic?

On the other hand, if we add 4 electrons to benzene's π system, it has 10 electrons ($n=2$). Is this anion aromatic? It doesn't seem very stable to me, since it has now filled 2 antibonding orbitals as well.

In general, it is a requirement that the $4n+2$ p electrons exactly fill all the bonding orbitals of the pi system for the molecule to be aromatic?

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    $\begingroup$ Indeed too much charge disallows aromaticity, that's a very good question. $\endgroup$
    – Mithoron
    Commented Jun 23, 2016 at 22:47
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    $\begingroup$ Agreed. Even $\ce{C_8H_8^{2-}}$ is pushing it. Some calculations suggest it's unstable as a free ion. The thing we make when we combine potassium and COT in liquid ammonia is surely solvated by the ammonia and ion-paired with the $\ce{K^+}$. $\endgroup$ Commented Jun 23, 2016 at 23:47
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    $\begingroup$ If the benzene tetraanion and tetracation are more stable than say the hexatriene tetraanion and tetracation, and I suspect they are, then the benzene ions could be said to be "aromatic". $\endgroup$
    – ron
    Commented Jun 24, 2016 at 1:03
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    $\begingroup$ This question is pointless, as the tetraionic compounds are completely hypothetic. Phantastic chemistry, that. ;-) Or putting it another way: It's Hückels rule, and you're bending it far beyond the point of breakage. $\endgroup$
    – Karl
    Commented Jun 24, 2016 at 19:46
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    $\begingroup$ Or maybe not ... onlinelibrary.wiley.com/doi/10.1002/anie.197902291/abstract $\endgroup$ Commented Jun 26, 2016 at 0:37

1 Answer 1


No, you do not have to have as many π electrons as p-orbitals or any type of direct relationship. As long as you have a continuous, circular arrangement of p-orbitals and an electron count $4n+2$ (for monocycles), you can expect your molecule to be aromatic.

  • Pyrrole, furan and related molecules are aromatic. They have 6 π electrons like benzene, but only five p-orbitals.
  • The cyclobutadienyl dianion has been synthesised as a dilithium salt. The authors claim evidence for aromaticity.[1]
  • The dianion of cyclooctatetraene has also been synthesised and examined by ESR as early as the 1960’s. The authors state:

    The assumption that both the dianion and the radical-anion are planar […] is consistend with [their equation] within experimental error.

    Thus it can be assumed that $\ce{cot^2-}$ is also an aromatic system.[2]

  • Disubstituted dications of cyclooctatetraene have been synthesised a decade later by Olah et al. The published proton NMR data implies that the protons are more deshielded than a simple triene moiety would imply. The summary the authors give states the following:

    The $\ce{^{1}H}$ and $\ce{^{13}C}$ NMR spectroscopic characterization of the reported new cyclooctatetraene dications indicates these ions possess those characteristics expected of a $\ce{C8}$ 6 π aromatic system and hence further extends the evidence for the great predictive utility of the $4n+2$ Hückel rule.

    $\ce{cot^2+}$ is thus also an aromatic system.[3]

These last two bullet points must be the ones you are looking for. The cyclooctatetraene dianion, a cyclic $\ce{C8}$ species with 10 π electrons ($n=2$) is aromatic, as is also the cyclooctatetraene dication, the same $\ce{C8}$ species albeit with 6 π electrons ($n=1$).

For most systems, aromaticity is either already given (as they have e.g. six π electrons in their neutral state; compare benzene etc.) or they are only one electron away (cyclopentadienyl, cycloheptatrienyl or tropylium). The way these molecules obtain aromaticity is relatively fixed since greater charges destabilise molecules.

Only for those systems starting with $4n$ π electrons such as cyclooctatetraene, which are antiaromatic by the Hückel definition, can there be a choice. Both the dication and the dianion are generally unfavourable, but they are similarly unfavourable thus the system can be pushed into either direction.

It is highly unlikely that a tetraanion or tetracation of benzene exist outside of a $4~\mathrm{K}$ matrix, much like it is very unlikely for a cyclopentadienyl-trication or a cycloheptatrienyl-trianion to exist outside of similar conditions due to their great charge density.


[1]: A. Sekiguchi, T. Matsuo, H. Watanabe, J. Am. Chem. Soc. 2000, 122, 5652. DOI: 10.1021/ja0004175.

[2]: H. L. Strauss, T. J. Katz, G. K. Fraenkel, J. Am. Chem. Soc. 1963, 85, 2360. DOI: 10.1021/ja00899a004.

[3]: G. A. Olah, J. S. Staral, G. Liang, L. A. Paquette, W. P. Melega, M. J. Carmody J. Am. Chem. Soc. 1977, 99, 3349. DOI: 10.1021/ja00452a027.

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    $\begingroup$ But all those systems have one specific value of $n$ for the given ring system. The OP seems to ask whether a given ring system can show aromaticity for multiple values of $n$. I would guess probably not unless you have a very large ring. $\endgroup$ Commented Jun 25, 2016 at 9:38
  • $\begingroup$ @OscarLanzi Agreed, I was trying to ask if a given ring system is aromatic for every value of $n$. The examples of pyrrole and furan have all the bonding MOs filled, so of course they're stable. Cyclobutadienyl and cyclooctatetraenyl dianions have all bonding and nonbonding MOs filled, so that would still be pretty stable. For more exotic values of $n$, I suppose it would still be aromatic because as ron mentioned in a comment, it's more stable than the isolated double bonds? $\endgroup$
    – carbenoid
    Commented Jun 25, 2016 at 13:41
  • $\begingroup$ @swenger Okay, it seems I slightly misunderstood the question. But not to worry, I dug up the research paper claiming the cyclooctatetraenyl dication, too. Thus, $\ce{cot^2+}$ and $\ce{cot^2-}$ are both aromatic with $6$ and $10$ π electrons, respectively ($n = 1$ and $n=2$ in Hückel’s $4n+2$ rule). $\endgroup$
    – Jan
    Commented Jun 25, 2016 at 16:18
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    $\begingroup$ Pyrrole, furan, etc. don't strictly obey Hückel's rule, since it was only derived for hydrocarbons, and they do not have the same symmetry restrictions. $\endgroup$ Commented Jun 25, 2016 at 21:08

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