I recently answered the question of whether twisted polyene chains were known by pointing towards Möbius aromatic system. While in conventional Hückel aromatic systems the lowest-lying π-orbital obviously has a nodal plane containing all the nuclei, this is not possible in a Möbius aromat since topologically these only have one side.

A number of Möbius aromatic systems have been both synthesised[1] and calculated;[2] for an X-ray determined structure of one, see the second structure in the figure below.[3]

Hückel and Möbius aromatic systems of an octaporphyrin

One can easily see that the π chain would have green and white or plus phase and minus phase overlapping. This got me thinking and wondering: what would this system’s molecular orbitals look like? Specifically: is there a set of MOs that would correspond approximately to the π system of such a Möbius aromat? If not, how are the π bonds seen in an MO analysis? The only calculations I could find searching the literature were both very old and without nice pictures.


[1]: M. Stępień, N. Sprutta, L. Latos-Grażyński, Angew. Chem. Int. Ed. 2011, 50, 4288–4340. DOI: 10.1002/anie.201003353.

[2]: H. S. Rzepa, Chem. Rev. 2005, 105, 3697−3715. DOI: 10.1021/cr030092l.

[3]: Y. Tanaka, S. Saito, S. Mori, N. Aratani, H. Shinokubo, N. Shibata, Y. Higuchi, Z. S. Yoon, K. S. Kim, S. B. Noh, J. K. Park, D. Kim, A. Osuka, Angew. Chem. Int. Ed. 2008, 47, 681–684. DOI: 10.1002/anie.200704407.

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    $\begingroup$ Isn't what the colours in the picture represent? $\endgroup$
    – Alchimista
    Commented Dec 25, 2017 at 21:18

2 Answers 2


I suppose you know better than many that aromaticity is not about a single MO, but a whole set thereof. We may just solve the system by the simple Huckel method and see what they are like. Sure, it is terribly old-fashioned and won't give you nice pictures (unless you make some yourself), but I don't think we need them all that much. What we need is the qualitative description, and it will be there all right.

From a cursory glance, it seems to me that the eigenvalues of a Möbius system are given by the same Frost circles, only the polygon is pointing down with its side, rather than a vertex. This, BTW, explains the reversed stability rules ($4n$ instead of $4n+2$). The molecular orbitals will look kinda like those in usual aromatic systems, except that the lowest orbital will be doubly degenerate and indeed will have a node plane across the chain.

So it goes.

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    $\begingroup$ I actually do want a (set of) pictures though. And in Hückel cases there is a picture of the lowest pi symmetry which is the nice one showing the system. $\endgroup$
    – Jan
    Commented Dec 26, 2017 at 14:49

The best way to understand the Moebius aromatic orbital structure is to render it as a modification of the familiar Hueckel structure.

Look at benzene. As you know, the single lowest occupied pi orbital has no nodes orthogonal to the ring plane. One step above is a pair of degenerate orbitals that are commonly regarded as each having one ring-orthogonal nodal plane. But note that the nodal plane, in each case, passes through the rotational symmetry axis and so can be divided into two half-planes, 180 degrees apart. Likewise, at the second, unoccupied degenerate level each orbital may be considered to have four nodal half-planes orthogonal to the ring plane, at 90 degrees from each other. In each case we have, in all, an even number of nodal half-planes orthogonal to the ring plane. At the highest level we would expect another degenerate pair, but when the number of nodal half-planes ( in thus case six) matches the number of ring atoms we lose one degenerate orbital. With an even number of nodal half-planes this drop-off of one orbital from the highest "pair" occurs for even size rings.

Now in a Moebius aromatic system, the twist leads to an odd number of nodal half-planes emerging from the center of the ring. At the lowest bonding level there is one such nodal half-plane per orbital. To match the molecular symmetry at this energy level there must then be a second orbital also having one nodal half-plane, orthogonal to the first. To meet the orthogonality reqirement the second orbital must be turned 180 degrees from the first one, and match the molecular geometry th second orbital must have its nodal half-plane twisted by 90 degrees from the first. The pair of nodal half-planes corresponding to the two degenerate orbitals look like somebody tried to assemble two half-planes into a full plane but did not get them aligned right.

That is the lowest degenerate pair. The next pair will have three nodal half-planes per orbital, again with the nodes twisted to match the molecular twist. To meet symmetry and orthogonality requirements we again need a second, degenerate orbital. As in the first degenerate pair, the second orbital has its nodal half-planes rotated 180 degrees and twisted 90 degrees from those of the first orbital. The same applies for the the third pair with five nodal half-planes per orbital, etc up to the size of the ring. At the highest level in an odd-size ring such as $\ce{C9H9+}$ (https://en.wikipedia.org/wiki/Möbius_aromaticity), one orbital drops out in the same manner as we saw for even-size rings like benzene in the Hueckel aromatic case.


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