The ring with a twist in it

The Hückel rule is usually presented as an observation with a rationalisation attached: rings with six π electrons are stable, rings with four are not, and the level diagram of a ring happens to have one orbital at the bottom and degenerate pairs above it, so the counts that fill a shell are $2, 6, 10, \ldots$

That level pattern is not an accident and it is not inevitable either. It follows from one assumption that is never stated because it is never violated in an ordinary molecule: that every neighbouring pair of p orbitals around the ring is in phase.

Take the ribbon of p orbitals and give it a half turn before closing it. One pair — whichever pair the closure falls on — now meets head to tail, and the interaction between them has the same magnitude and the opposite sign. Nothing else about the ring changes.

The consequence is total. The eigenvalues become

$$x_k = 2\cos\frac{(2k+1)\pi}{n}$$

instead of $2\cos(2\pi k/n)$ — the same cosine sampled half a step along. The single nodeless level at the bottom disappears, the levels come in degenerate pairs from the bottom up, and the count that closes a shell becomes $4n$.

What actually changes in the matrix #

The whole of the construction is one sign, and it is worth being precise about which one and why it is allowed.

A Hückel matrix has zeroes on the diagonal for a hydrocarbon and ones between bonded atoms, in units of the resonance integral $\beta$. The entry between two atoms is proportional to the overlap between their p orbitals, and $\beta$ is negative — which is why a larger positive eigenvalue means a more bonding orbital in the usual convention.

Turning one p orbital through half a turn about the ring axis changes its sign relative to its neighbour. The overlap integral has the same magnitude and the opposite sign, so the matrix entry becomes $-1$ rather than $+1$.

In ordinary Hückel theory that is forbidden, and for good reason. A negative resonance integral between neighbours in an untwisted molecule means a mistake — a heteroatom’s resonance parameter is a strength, and a negative strength is nonsense. The twist is the one case where the sign is physical: the rule is not that a negative entry is impossible but that it needs a reason, and a half-turn of the p-orbital basis around the loop is one.

Where the twist is put does not matter. Any odd number of reversed bonds gives the same spectrum, because moving a reversal from one bond to the next is a change of sign of one coefficient, which is a similarity transformation by a diagonal matrix of $\pm 1$ and leaves every eigenvalue where it was. One reversal is the simplest representative and is the one used throughout.

The closed form, and the check that goes with it #

The eigenvalues of a twisted ring are $2\cos((2k+1)\pi/n)$, and this can be confirmed rather than assumed: diagonalising the matrix at every size from three to twelve reproduces the formula to better than $10^{-15}$.

The shape of the spectrum is what the rule turns on. For the untwisted ring, $k = 0$ gives $x = 2$ — a single level, nodeless, the totally symmetric combination — and then the levels pair up. For the twisted ring there is no $k$ giving a single level, because $(2k+1)\pi/n$ never equals zero. The pairing starts immediately.

So the filling arithmetic changes from “two electrons in the bottom orbital plus four in each pair above” to “four in each pair, starting at the bottom”:

$$2 + 4n \qquad\longrightarrow\qquad 4n.$$

That is the whole rule, and it can be produced rather than recalled: fill each ring and ask whether the highest occupied shell comes out full, exactly as aromaticity as a shell closure does for the flat case.

What the two counting rules are about #

Suppose the two verdicts are required to disagree at every ring size and every electron count. That is a stronger demand than testing each rule separately, since a flat ring mistaken for a twisted one would pass both rules at once.

The demand fails immediately, at three atoms and six electrons.

Six electrons in a three-ring fills every level twice over. The highest occupied shell is full whatever the spectrum looks like, so both rules call it closed, and the two verdicts agree. The same happens for a four-ring at eight, a five-ring at ten, and so on.

That is not a bug in either rule; it is the boundary of what they are about. $4n+2$ and $4n$ are statements about an incomplete shell — the first count at which the filling stops inside the level diagram rather than past the top of it — and writing them as statements about any count at all is the commonest way they are misquoted. Stated properly, the rules apply up to $2n - 2$ electrons, and a completely filled ring closes both ways — an exclusion worth stating rather than hiding.

Cyclobutadiene both ways #

The sharpest single case is the ring the flat rule condemns.

Flat. Four electrons in levels $2, 0, 0, -2$ give a π energy of $4.000\beta$. Two isolated double bonds give $4\beta$. The delocalisation energy is exactly zero — not small, zero — and the two levels at zero are the half-filled degenerate shell that makes the molecule a triplet in this model and drives the Jahn–Teller distortion in reality.

Twisted. The same four electrons in levels $\sqrt2, \sqrt2, -\sqrt2, -\sqrt2$ give $4\sqrt2 = 5.657\beta$. The delocalisation energy is $1.657\beta$, the shell is closed, and there are no unpaired electrons.

Benzene runs the other way. Flat it gives $8.000\beta$ against $6\beta$ localised, a delocalisation energy of $2\beta$ — the number aromaticity as a shell closure computes against a reference state written down as a count of isolated double bonds. Twisted it gives $6.928\beta$, which is below the localised reference, and its six electrons leave a degenerate shell half filled.

So the twist does not merely relabel the two molecules. It exchanges them.

Heilbronner predicted this in 1964, and it took thirty-nine years #

The construction is Edgar Heilbronner’s, and the paper is a page and a half of exactly the arithmetic above. He noticed that the assumption of a uniform sign was an assumption, worked out what dropping it does, and predicted that a sufficiently large annulene with a half twist would be aromatic at $4n$ electrons.

He also said why nobody would see it soon. A twist has to be distributed around the ring, and a small ring cannot distribute it — the p orbitals have to rotate relative to one another between neighbours, which means the π system is misaligned everywhere, and the resonance integrals fall by the cosine of the misalignment. In a ring of $n$ atoms with one half turn shared out, each neighbouring pair is off by $180°/n$, so the interactions are multiplied by $\cos(180°/n)$: at $n = 4$ that is $0.707$ and at $n = 16$ it is $0.981$.

The energy gained by the twisted closure is therefore competing against a cost that grows sharply as the ring shrinks, and the ring is also being asked to bend in a way the angle a ring cannot have says small rings cannot afford.

The first molecule anybody made with a genuine Möbius π system was Rainer Herges’s C₁₆ annulene, in 2003. Sixteen carbons, which is $4n$ with $n = 4$, and about as small as the geometry allows.

What the model does not include, said plainly #

The comparison above holds every resonance integral at the same magnitude, and that is the assumption doing the most work.

It is right for the topological question — what does reversing a sign do to the level pattern — and it is what makes the two spectra comparable at all. It is wrong as a prediction about a molecule, because a real twisted ring has integrals reduced by the misalignment factor above and bond angles distorted by the bending, and neither cost is in a Hückel matrix with equal integrals.

So the honest form of the result is: the twist changes the shell-closure rule, exactly, and whether any particular ring can afford the twist is a question about strain and misalignment energies that Hückel theory does not contain. That is the same boundary one coordinate, three point groups drew around hydrogen peroxide’s dihedral angle — the geometry decides a symmetry exactly, and which geometry occurs is a barrier that is quoted rather than computed.

The multiplication by $\cos(180°/n)$ is easy enough to put in, and the reason for leaving it out was that it would decide a competition whose other half — the bending — is still missing. The section below puts it in anyway, because what it produces is not a plausible number but an identity, and an identity settles the objection rather than falling to it.

Paying for the twist costs exactly what it buys #

The misalignment factor was set aside above as a one-sided correction. It is worth putting in after all, because what comes out is not a plausible number but an exact cancellation, and an exact cancellation says something the omission was hiding.

Distribute the half turn evenly and every neighbouring pair is off by $180°/n$, so every resonance integral is multiplied by $\cos(180°/n)$ and the whole twisted spectrum scales by that one factor. Apply it to the $4n$ rings, where the twist is supposed to pay:

The last column is the first, at every size, to every digit computed. The twisted closure of a $4n$ ring gains exactly as much π energy as the even distribution of the twist costs, and the two cancel identically rather than approximately.

So the correction is not one-sided after all; it is total. At this level of theory a Möbius $4n$ ring and a planar $4n$ ring have the same π energy, and the entire question of which one a molecule adopts falls to the σ frame — the bending, the strain, the loss of overlap where the ribbon has to curve — none of which is computed here.

That is a stronger version of the refusal above rather than a retraction of it. The reason not to publish the corrected π number was that it would decide a competition whose other half was missing. It turns out it decides nothing at all: the π term is a draw by an identity, and everything that separates the two structures lives in the σ frame, which a π-only model does not have.

The same table run at $4n+2$ shows the asymmetry. Benzene’s twisted energy is 6.928β against 8.000β flat, and the misalignment takes it to 6.000β — so an aromatic ring loses twice, once to the level pattern and once to the geometry. Only the $4n$ case is a tie, which is exactly the case Heilbronner’s argument was about.

What the twist does to the bond orders #

The level pattern is not the only thing the sign change moves, and the bond orders are worth a look because they answer a question the energies leave open.

A flat benzene has every π bond order at $0.6667$, identical around the ring, which is the equal-row-sum result what one pair can hold together traces to a property of the matrix rather than of the graph. Reversing one entry does not break that property — the row sums are still equal in magnitude — so a twisted ring at a closed shell also has uniform bond orders, and the reversed bond has one of the same size and the opposite sign.

That sign is the interesting part. A bond order computed from the eigenvectors is a statement about how much the two atoms’ coefficients reinforce, and at the twisted bond they reinforce in the opposite sense — which is precisely what a head-to-tail overlap means. The magnitude being unchanged is the arithmetic saying that the twisted bond is as strong as the others, and the sign is saying that a sign convention has been carried around the loop and come back inverted.

So a twisted ring is uniformly bonded, with one bond whose bookkeeping sign differs, and nothing local distinguishes that bond from any other. The twist is genuinely a property of the whole loop and not of the bond where the reversal happened to be placed — which is the same statement the similarity transformation made algebraically, arrived at from the eigenvectors instead.

Where else a sign appears in a loop #

The mathematics is not confined to annulenes and the pattern is worth recognising, because this site meets it in two other places already.

A ring in a magnetic field picks up a phase factor around the loop, and the levels shift continuously between the two limits above as the flux changes. The Möbius spectrum is the half-flux case. No field enters any calculation here, so the connection is noted rather than used.

A transition state can be Möbius. The Woodward–Hoffmann rules distinguish reactions whose cyclic array of interacting orbitals has one sign inversion from those that have none, and the counting for the two cases is $4n$ and $4n+2$ respectively — the same pair of rules from the same argument. Reaction chemistry is outside this site’s target and it is worth naming the connection precisely once so that a reader who has met the rule elsewhere can see it is the same object.

A twisted chain is not the same as a twisted ring. Reversing a sign in an open chain does nothing at all: the diagonal matrix of $\pm 1$ that moves the reversal along can be used to remove it entirely, because a chain has no loop for the sign to be trapped in. That is the cleanest statement of what the twist is — a property of the loop rather than of any bond in it — and it is why the construction has nothing to say about polyenes.

The odd rings, where the twist does something else again #

Rings with an odd number of atoms are the case where the two spectra differ in kind rather than in offset, and they are worth separating out.

A flat odd ring is not bipartite, so it has no pairing symmetry: its levels are not arranged symmetrically about zero, and the pairing theorem must fail for cyclopropenyl, tropylium and cyclopentadienyl. That failure is the model working, and it is why those three ions have level diagrams with a lopsided look.

A twisted odd ring has the opposite property. Its eigenvalues $2\cos((2k+1)\pi/n)$ include $k = (n-1)/2$, which gives $2\cos\pi = -2$ exactly — a single level at the bottom of the antibonding end rather than at the top of the bonding one — and the rest pair up above it. So a twisted odd ring is the mirror image of a flat one: where the flat ring has a lone nodeless level at its most bonding end, the twisted ring has a lone fully-noded level at its most antibonding end.

The consequence for the counting is that the two rules swap for odd rings too, and in the direction that makes the tropylium cation the interesting case. Flat, its six electrons close a shell and it is the standard aromatic cation. Twisted, its six electrons leave a degenerate pair half filled and its π energy falls from $8.988\beta$ to $8.098\beta$.

That is a smaller change than benzene’s, and the reason is the same one that makes large rings insensitive: an odd ring’s levels are already less symmetric, so shifting the sampling by half a step disturbs less.

What the twist adds to the counting rule #

The familiar refinements of the aromaticity rule — the closure rule itself, the delocalisation energy against a stated reference, the ring carrying a charge, the ring with one atom that is not carbon — all change what is in the ring. The twist changes its topology instead.

The reason it belongs with the rule rather than beside it is that it converts the rule from an observation into a consequence. Before it, $4n+2$ is a fact about ring level diagrams. After it, $4n+2$ is a fact about ring level diagrams when the orbitals are in phase around the loop, and the qualification has a companion rule attached showing what happens without it.

A rule with a stated scope and a known counterexample is a different object from a rule with neither, and the difference is exactly the amount of confidence it deserves.