The angle a ring cannot have

Every account of ring strain starts from the observation that a small ring cannot have tetrahedral bond angles, and almost none of them says how much of that is a fact about geometry and how much is a fact about energy. The two are separable, and the geometric half turns out to be a theorem with no chemistry in it at all.

Write a ring as its sequence of bond vectors rather than its atom positions. The ring closes when the vectors sum to zero. Consecutive bonds meet at the bond angle $\theta$, which means the direction turns by $180° - \theta$ at each atom. So a ring of $n$ atoms turns through $n(180° - \theta)$ in total.

A closed curve in space must turn through at least a full circle. That is Fenchel’s theorem, it is about curves and not about molecules, and it gives

$$n(180° - \theta) \ge 360° \qquad\Longrightarrow\qquad \theta \le \frac{180°(n-2)}{n}$$

which is exactly the interior angle of the regular planar polygon. Equality holds only when the ring is planar and convex — that is Fenchel’s equality case — so puckering always costs angle and never buys it.

For three, four, five, six and seven atoms the ceilings are $60°$, $90°$, $108°$, $120°$ and $128.6°$. The tetrahedral angle is $109.47°$, and it falls between the fifth and the sixth.

The bound, checked by trying to break it #

A bound stated is worth less than a bound tested, and the test is a search.

The ring is written as $n$ unit vectors with two conditions: consecutive pairs have a fixed dot product, which is the bond angle, and the whole set sums to zero, which is the closure. Everything else — where the ring sits, how it is oriented, which atom is first, how long the bonds are — has been divided out before the search begins. What is left is a small minimisation, run from two dozen starts, and the number that matters is the residual it returns.

At the ceiling, every ring size closes with a residual near $10^{-16}$, and the ring that does it is planar to the same precision. Three degrees above the ceiling, every ring size refuses: the residual sits between $0.08$ and $0.11$ and does not fall when the search is given five times as many starts and five times as many steps.

Two orders of magnitude separate a ring that exists from one that does not, which is what makes the refusal readable. A search that fails is weaker evidence than a search that succeeds, and quoting the residual is how the weaker evidence is made to say something rather than nothing.

Three and five have no other geometry at all #

The search was expected to find a ring at every angle below the ceiling and does not. Three and five refuse below their ceilings as well as above.

The reason is a count. Writing the ring as $n$ unit vectors gives $2n$ freedoms, of which three are the orientation of the whole thing, so $2n - 3$ are real. The conditions are $n$ angles plus three components of closure, which is $n + 3$. Freedoms outnumber conditions only from six atoms up:

At six they balance, and a six-ring is rigid at each bond angle but exists — that is the chair, one shape for each angle rather than a family. At seven and above there is a family. At three and five the conditions win and there is nothing.

For three that is obvious once said: three equal bonds meeting at equal angles are an equilateral triangle, and an equilateral triangle has $60°$ angles and no other option. Cyclopropane’s geometry is not a compromise. It is the only arrangement its bonds admit, and everything else about the molecule has to accommodate it.

For five it is not obvious, and it is the one place the arithmetic above needs a caveat. Four is over-determined by the same count and escapes anyway, because its symmetry makes two of the conditions the same condition — a folded square with all four bonds equal has all four angles equal automatically. Five has no such rescue, and the search refuses it at $106°$, at $104°$ and at $100°$ with residuals of $0.025$, $0.050$ and $0.100$.

That does not mean cyclopentane is planar. It means cyclopentane’s five bond angles are not all equal, which is exactly what the envelope and half-chair conformations have: four atoms in a plane and one out of it, with angles that differ from each other by a degree or two. The equal-angle condition is what fails, not the closure.

Six atoms, and the ten degrees the chair spends #

The chair is the case where the ceiling has room to spare and the geometry uses it.

A planar hexagon has $120°$ angles. The tetrahedral angle is $109.47°$, ten and a half degrees lower. Fenchel says a ring can go below its ceiling only by leaving the plane, and the search confirms it in the sharpest possible form: asked for a six-ring at exactly $109.4712°$, it returns a closed ring with a residual of $8\times10^{-12}$ and an out-of-plane spread of 0.222 bond lengths.

That is the chair, produced by a minimisation that was told the bond angle and nothing else. Not assumed, not built from a template, not read off a model kit — a closure condition and an angle, and the shape that satisfies both is the one every organic chemistry course draws.

The margin is why cyclohexane is the ring that behaves. Its ceiling is above the angle it wants, so it can reach the angle exactly, and it does. Cycloheptane’s ceiling is higher still and it reaches the angle too, with more freedom than it can use — which is why medium rings have many conformations of similar energy and small ones have one or two.

Cyclopropane’s orbitals, which cannot follow its bonds #

Sixty degrees is thirty-nine and a half below the tetrahedral angle, and hybrids that were never orthogonal computes what that does to the usual description: two quarter-s hybrids drawn at $60°$ overlap by $0.625$, and there is no s fraction at all for which two hybrids are orthogonal below a right angle.

So an sp³ description of cyclopropane fails as arithmetic, not as an approximation. The standard repair is bent bonds — let the orbitals point outside the triangle so the interorbital angle is near $104°$ while the internuclear angle stays at $60°$ — and it works, at the cost of separating the direction of a bond from the direction of an orbital.

The measured consequences are all in the same direction and are worth quoting because they are the evidence that the picture is right rather than convenient. Cyclopropane’s C–H bonds have unusually high s character, which shows in a C–H coupling constant of 161 Hz against ethane’s 125. Its C–C bonds behave like double bonds in several reactions. Its electron density has a maximum outside the internuclear lines rather than on them, which diffraction sees directly.

Every one of those is what a bent bond predicts, and none of them is predicted by a description that puts the orbitals along the bonds.

What this does not compute, and the word that hides it #

Everything above is geometry. Not one number in it is an energy, and the energies the subject is usually about are not computed here.

Ring strain energies are measured, from heats of combustion, and they are quoted here as quoted: about $115$ kJ/mol for cyclopropane, $110$ for cyclobutane, $26$ for cyclopentane and essentially nothing for cyclohexane. Those numbers are what “strain” means in practice and geometry does not produce them.

The interesting thing is how badly the angles predict them. Cyclopropane and cyclobutane have almost the same strain and their angle deficits are $49.5°$ and $19.5°$ — a factor of two and a half apart. Cyclopentane’s deficit is $1.5°$, the smallest of the four, and it carries a quarter of cyclopropane’s strain rather than a fiftieth.

So the angles explain three and four in the right direction and nothing about the size, and they do not explain five at all.

What explains five is a different quantity entirely. A planar cyclopentane has all its hydrogens eclipsed with their neighbours’, and the torsional cost of that is comparable with the angle cost of puckering — which is why cyclopentane puckers, trading an angle it could afford to keep for a torsion it could not. The word “strain” carries two unrelated quantities, and the one this essay bounds is the one that turns out not to be the dominant term for the ring where the bound is closest to being satisfied.

Fenchel, and why the theorem is about curves rather than molecules #

The bound is unusual in chemistry in resting on a result from differential geometry rather than from quantum mechanics, and it is worth saying what the result is and how much of it is needed.

Fenchel’s theorem, from 1929, says the total curvature of a closed curve in space is at least $2\pi$, with equality exactly for a plane convex curve. The intuition is that the tangent direction traces a closed path on the unit sphere, and a closed path on a sphere that does not enclose an entire hemisphere can be shrunk to a point — which would mean the tangent never reverses, and a curve whose tangent never reverses does not come back.

The version needed here is the discrete one, for a polygon, where “total curvature” is the sum of the exterior angles. That version is older and easier: it follows from the same spherical argument applied to the finite set of tangent directions, and for a plane polygon it is the elementary fact that the exterior angles of a convex polygon sum to $360°$.

What the theorem does not require is anything about the bond lengths being equal. It bounds the sum of the turns for any closed polygon, so the general statement is about the mean bond angle rather than about a common one: $\overline{\theta} \le 180°(n-2)/n$ for any closed ring whatever. The equal-bond, equal-angle case searched for above is a special case chosen because it is the one where a single number is the answer.

That generality is what makes the bound useful for the rings where the equal-angle search refuses. Cyclopentane’s five angles are not equal and the search finds no equiangular ring below $108°$ — but the mean of its five angles is still bounded by $108°$, and the measured mean in the envelope conformation is $104.5°$, comfortably underneath.

Why VSEPR has nothing to say here #

It is worth being explicit about which familiar calculation does not apply, since a bond angle is exactly what VSEPR, computed computes.

VSEPR minimises repulsion between electron domains around one centre and returns the angles that fall out. For a carbon with four domains it returns $109.47°$, and it returns that for a carbon in cyclopropane too, because the model has no way of knowing that two of the four directions are joined to each other somewhere else.

The ring closure is a constraint that acts between centres. It is not in the model, it cannot be added without changing what the model is, and the result is that VSEPR’s prediction for a small ring is simply wrong by up to fifty degrees — not because the repulsion argument is bad, but because it is answering a question about one atom while the ring is asking one about several.

Which angles are symmetry and which are the model sorts molecular angles into those decided by symmetry and those decided by the repulsion exponent. A ring’s ceiling belongs to neither category and is a third thing: an angle decided by a topological constraint, holding for any interaction whatever, and therefore harder than either.

The other ring sizes, briefly #

Above six, the ceiling stops binding and other things take over.

Seven, eight, nine and ten atoms all have ceilings above the tetrahedral angle with room to spare, and all of them can close at $109.47°$ — the search finds rings for each with residuals near $10^{-11}$. Their strain energies are nonetheless not zero: about $26$, $41$, $53$ and $50$ kJ/mol, rising to a maximum around nine and falling away after.

None of that is angle strain. It is transannular — hydrogens on opposite sides of the ring pushed into each other by a shape that has to close — and it is a fact about distances rather than angles. Medium rings are the case where the geometry this essay computes has stopped saying anything and the chemistry has not.

Below three there is nothing, which is worth one sentence: a two-membered ring is a bond, and the closure condition with $n = 2$ requires $\theta \le 0°$, so the closure condition says a two-ring does not exist rather than that it is strained.

What one carbon wants, on its own, is four directions at 109.47° from a repulsion minimisation with nothing else in it — and a ring of three cannot supply two of them at that angle however it is drawn. That is the collision the whole essay is about, and it is a statement about geometry rather than about strain.

Unequal bonds move the angles and not the bound #

The assumption of equal bonds is wrong for every heterocycle, and dropping it changes less than it looks as though it should.

Fenchel’s theorem is about the total turning of a closed curve, and a curve’s total turning does not know how long its segments are. The turning at each atom is still $180° - \theta_i$, the sum of the turnings is still at least a full circle, and so

$$\sum_{i=1}^{n} \theta_i \le 180°(n-2)$$

holds with the bond lengths anywhere. The bound on the mean angle is therefore unchanged: an oxygen in a ring cannot raise the average angle by a single degree, however short its bonds are.

What unequal bonds change is how the slack is shared out. The sum is fixed and the individual angles are not, so a short bond buys an angle somewhere by spending one somewhere else — and for a three-membered ring the sharing is completely determined, because a triangle’s angles follow from its sides with no energy in the argument at all.

Running that on the measured bond lengths: cyclopropane’s three equal 1.510 Å bonds give 60.00° everywhere. Oxirane, with a C–C of 1.470 Å and two C–O of 1.436 Å, gives 61.57° at the oxygen and 59.21° at each carbon. Aziridine, at 1.481 and 1.475 Å, gives 60.27° and 59.87° — barely distinguishable from cyclopropane, which is what nearly equal bonds must give. Thiirane, whose C–S bonds are 1.815 Å against a C–C of 1.484 Å, gives 48.26° at the sulfur and 65.87° at each carbon.

Every one of those is within a few hundredths of a degree of the measured angle, and none of it is a calculation about bonding. The three-ring’s shape is its three lengths, read through the cosine rule.

So the heteroatom’s effect on the angles is real, large in the case of sulfur, and entirely geometric. What it is not is an escape: the three angles still sum to 180°, thiirane’s carbons pay 65.87° for the sulfur’s 48.26°, and the average is 60° in all four.

Still open: the energies, and larger heterocycles #

This is the geometric part of ring strain, and it is deliberately taken first.

Everything beyond it is energy. What a torsional barrier is, why an eclipsed conformation costs what it does, how the two contributions to “strain” separate quantitatively, why the strain maximum across ring sizes is at nine rather than at three — every one of those needs a quantity not computed here, and several need quantities nobody computes without a fitted force field.

What the geometry supplies is the part of the answer that no calculation can change. Cyclopropane’s angle is $60°$ in every method, at every level of theory, for every substituent, because it follows from three equal bonds closing. That is a different kind of statement from “cyclopropane’s angle is computed to be $60°$”, and separating the two is the reason to start with the geometry rather than the energy.

The extension that stays inside the geometry is unequal bonds, and the section above takes it: the bound is untouched, the individual angles are not, and for a three-ring the sharing is fixed by the cosine rule alone. What is left undone is the same question for larger heterocycles, where the lengths no longer determine the shape and the closure condition admits a family of them.