The hybrids that point outside the bonds

A double bond taken apart three ways shows that the two bent components and the σ–π description are one thing written twice, with the same hybrids underneath. There is a case where the word bent means something else entirely.

Cyclopropane’s carbons make three σ bonds each and its C–C–C angles are sixty degrees. Coulson’s relation says no set of equivalent s–p hybrids can produce that: at any s character between nothing and a half, two orthogonal hybrids meet at an angle between ninety and a hundred and eighty degrees, and sixty is not in the range.

So the ring’s bonds are bent in a second and different sense — the hybrids point outside the internuclear lines rather than along them — and the same relation that fixes the angle in a double bond says by how much.

The argument, which is a budget and one relation #

Each ring carbon carries four hybrids: two pointing at hydrogens and two pointing round the ring. They share one s orbital between them, and two equivalent hybrids are orthogonal only at the angle whose cosine is $-s/(1-s)$.

That is enough.

The measured H–C–H angle fixes the hydrogens’ share. Cyclopropane’s is 114°, so each C–H hybrid takes 0.2891 of the s orbital.

What is left goes round the ring. 1 − 2(0.2891) = 0.4217, so each ring hybrid takes 0.2109 — which is sp³·⁷⁴.

And the same relation gives the angle between the ring hybrids. At an s character of 0.2109 that is 105.50°.

The carbons are at 60°. So each ring hybrid points $(105.50 - 60)/2 = 22.75°$ outside the line joining the two atoms it bonds.

Nothing in that chain is fitted. Two measured angles go in — the H–C–H angle and the C–C–C angle — and the rest is orthogonality.

It is worth noticing what has not been assumed, because the usual account of cyclopropane assumes it. Nobody has said the carbons are sp³, or sp², or anything: the hybridisation is an output, and it comes out at sp³·⁷⁴ for the ring bonds and sp²·⁴⁶ for the C–H bonds. That is the point the angle does not fix the hybridisation makes, turned round — an angle does not fix a hybridisation, but two angles and a budget do.

The whole series #

Cyclohexane is the control. Its hybrids point along its bonds to better than a twentieth of a degree, which is far better than either measured angle is known to — so the relation is not manufacturing a bending where there is none, and the twenty-two degrees in cyclopropane is a real output rather than an artefact of the construction.

And the fall is steep. A factor of two from cyclopropane to cyclobutane and another factor of two to cyclopentane, then essentially nothing. The bending is a small-ring phenomenon in the sharp sense: it is confined to rings of three and four.

That the numbers come out where the literature puts them is worth saying once. Coulson and Moffitt’s classic treatment of cyclopropane gives ring hybrids of sp³·⁷⁴ and an interorbital angle a little over a hundred degrees, from the same relation and the same measured angle — so this is a reproduction rather than a discovery, and its value is that every step is computed here rather than quoted.

One detail is worth flagging because getting it wrong would produce a much less interesting table. The C–C–C angle used is the measured one, not the planar polygon’s. Cyclobutane puckers to 88° and cyclohexane’s chair sits at 111.5°, so using 90° and 120° would be measuring the puckering rather than the hybrids — and cyclohexane would come out with a bending of −4.2° that means nothing.

The floor, and why sixty degrees is different in kind #

The inequality behind all of this deserves separating out, because it is not a matter of degree.

Two equivalent s–p hybrids sharing one s orbital have $\cos\theta = -s/(1-s)$, and $s$ runs from 0 to $\tfrac12$. At $s = 0$ they are two pure p orbitals at 90°; at $s = \tfrac12$ they are two sp hybrids at 180°. There is nothing below ninety.

So:

A ring of five or more can be described with hybrids along its bonds. Its 108° needs an s character of 0.236 in each ring hybrid, which is available.

A ring of four is exactly at the boundary. Its 90° is reachable only with pure p ring hybrids, the whole s orbital going to the hydrogens — which is the Coulson–Moffitt description of cyclobutane and comes out here as an equality rather than an approximation.

And a ring of three is outside it. No s character produces 60°, so a description with hybrids pointing at the carbons does not exist, at any hybridisation, for any parameters. That is a statement of the same kind as the angle a ring cannot have — an impossibility rather than a strain — and the two are about different things: that one is about the nuclei and this one is about the orbitals.

Two senses of the word #

A double bond’s bent bonds and a small ring’s are different objects, and the word does for both because in each case a hybrid fails to lie along an internuclear line. The failures have nothing else in common.

In a double bond the two hybrids point away from the line because there are two of them and they have to go somewhere else. The molecule has a choice of descriptions — two bent bonds, or a σ and a π — and they are one thing written twice.

In a small ring there is one hybrid per bond and it still misses, because the geometry the nuclei have is outside the range orthogonality allows. There is no alternative description in which the hybrids point at the atoms, because no such set exists.

So the first is a choice of coordinates and the second is a constraint on them. Both are consequences of hybrids being a basis rather than a thing, and only the second says anything a measurement could contradict.

What is left of the bond #

If the hybrids do not point at each other, they overlap less than they would.

Two hybrids each misaligned by 22.75° retain $\cos 22.75° = 0.922$ of their alignment — an eight per cent loss, which is a real weakening and is much smaller than a naive reading of a forty-five-degree gap would suggest. Cyclobutane loses 1.8 per cent, cyclopentane 0.3, cyclohexane nothing.

That eight per cent is the whole of what the hybrid picture says about cyclopropane’s C–C bonds being unusual, and it is worth comparing against what is measured.

Cyclopropane’s C–C bond is shorter than an ordinary one, 1.510 Å against 1.535, which is the opposite of what an eight per cent overlap loss on its own would suggest. The model has an answer and it is not the bending: the ring hybrids carry less s character than sp³, so the C–H hybrids carry more — sp²·⁴⁶ rather than sp³ — and it is the C–H bonds that the extra s character shortens, while the C–C bond is short for a reason involving the ring’s own geometry rather than its hybrids.

And its C–H coupling constant is larger, which is the direct measurement of that extra s character and is the one number in the whole area that a hybrid description predicts cleanly. The s character computed here for cyclopropane’s C–H hybrids is 0.2891 against cyclohexane’s 0.2312, a ratio of 1.25; the measured one-bond carbon–hydrogen couplings are 161 and 123 Hz, a ratio of 1.31. Five per cent apart, from a construction with no fitted parameter in it — which is as close as a two-angle model has any right to come, and is stated as a ratio rather than as an agreement because the coupling is proportional to the s character only to leading order.

So the construction earns its keep on the C–H bonds rather than on the C–C ones, which is not where an essay about ring bonding would look.

What a chemist should take from it #

“Strained” and “bent” are two different failures. A strained bond has an angle away from its preference and pays for it; a bent bond has a hybrid away from its bond and loses overlap. Cyclopropane has both, they have different sizes, and the usual account merges them.

The hybridisation of a small ring is an output, not a label. Calling cyclopropane’s carbons sp³ and its angles strained gets the arithmetic backwards: with 114° between the hydrogens the ring hybrids are sp³·⁷⁴ and there is nothing left over to strain.

And the number to check a hybrid argument against is a coupling constant. It measures the s character of a C–H bond directly, it is routinely available, and it is the one place where an argument of this kind makes a prediction that a measurement can refuse.

What is quoted, and what is computed #

Twelve numbers are quoted and all twelve are measurements: four H–C–H angles, four C–C–C angles, two bond lengths and two coupling constants.

Everything else is computed: every s character, every hybridisation index, every interorbital angle, every bending and every overlap factor, from one relation and one budget. There is no fitting anywhere and no parameter that could have been chosen differently — which is unusual for a hybrid argument and is the reason this one can be checked against a control.

The control is cyclohexane, and it is worth saying why it is a real one. Its two measured angles are independent of each other and of everything in the construction, and the construction returns a bending of 0.035° from them. Nothing forced that: a relation with an error in it would have returned some other small number, and the odds of landing within a twentieth of a degree of zero by accident are not good.

What this cannot say #

Hybrids are a basis, not an observable. Everything about hybrids rests on that: a localised description is a unitary transformation of the canonical orbitals, and a statement about where a hybrid points is a statement about a chosen basis. What is not basis-dependent is the impossibility — no unitary transformation produces a set of equivalent s–p hybrids at 60°, and that is a fact about the space rather than about a choice inside it.

Two measured angles per molecule. The construction takes them and returns everything else, which makes it cheap and makes it inherit whatever those two numbers are worth. Cyclopentane’s H–C–H is an average over a puckered ring whose carbons are not equivalent, and cyclopentane is the ring that cannot hold still — so a single number for it is a convenience rather than a measurement of one structure.

Equivalent hybrids only. The relation assumes the two ring hybrids on a carbon are equivalent and the two C–H hybrids likewise, which symmetry supplies for cyclopropane and cyclobutane and does not for a puckered cyclopentane — the same equivalence three bent bonds and the same hybrid leaned on, and the same place it stops holding.

And there is no energy anywhere. The bending is a geometric consequence of orthogonality; what it costs, and whether a molecule would rather distort than pay it, is what strain calculations compute and is not here.

Nor is there a π contribution. Cyclopropane’s ring bonding is often described with an orbital picture that has nothing to do with hybrids at all — three p orbitals in the plane of the ring, combining like a small aromatic system — and that description is not compared against this one anywhere. Two descriptions of one wavefunction is the standing situation with hybrids, and comparing them properly needs the localisation transformation rather than an argument.

It is worth putting the whole description into numbers once, because the misalignment is easy to argue about in words and there is nothing to argue about in the arithmetic.

And it is worth showing that the description itself is free, so that the number above is a fact about the geometry rather than about a choice made when drawing it.

What was checked #

The four hybrids share exactly one s orbital, to twelve decimal places, in every ring — checked rather than imposed, because the ring hybrids’ share is computed from what the hydrogens leave.

And their interorbital angle is one two orthogonal hybrids can have, between 90° and 180°, which is the constraint the whole construction lives inside.

A small ring’s hybrids point well outside the lines joining its carbons, by more than five degrees, for the rings of three and four.

And a large ring’s do not, by less than a quarter of that — with a six-membered ring’s agreeing to better than half a degree, which is the control and is checked separately.

The bending falls monotonically as the ring grows, at every step.

A ring of three has an internuclear angle no pair of orthogonal s–p hybrids can reach, which is the impossibility the essay turns on.

And a ring of four is exactly at the limit: its ring hybrids would have to be pure p, computed to twelve decimal places rather than described.

While a ring of five can be described with hybrids along its bonds, at a real s character — so the boundary is a boundary and not the edge of the table.

The refusal is asking for the s character of a 60° pair, which is refused rather than answered with a number outside the range.

The bending is not the strain, and one comparison settles it #

The obvious reading of 22.75 degrees is that it is the strain — that a cyclopropane is high in energy because its bonds miss the atoms they join. The measured strain energies refuse that reading, and they refuse it with a single comparison that needs no calculation.

The bending falls off steeply with ring size and it falls off almost entirely between three and four. A three-membered ring’s hybrids miss by 22.75 degrees. A four-membered ring’s do not miss at all: its internuclear angle of ninety degrees is exactly the limit of Coulson’s relation, reachable with pure p hybrids pointing straight along the bonds.

The strain energies do not follow. Cyclopropane’s is about 115 kilojoules a mole and cyclobutane’s about 110 — a drop of five per cent across a step in which the misalignment goes from its largest value in the whole series to zero.

So whatever is straining a four-membered ring, it is not hybrid misalignment, because there is none; and whatever is straining a three-membered ring, at most about five kilojoules a mole of it can be the difference between having 22.75 degrees of misalignment and having none.

That is a clean negative result and it survives the crudeness of the numbers, because the two strain energies are known to a kilojoule or two and the gap between them is five.

What it leaves is a rearrangement of the accounting rather than a mystery. The strain shared by the three- and four-membered rings has to be something both have in common — angles far from tetrahedral, and eclipsed hydrogens on adjacent carbons — and the misalignment is a consequence of the small angle rather than the mechanism by which the small angle costs energy.

Which inverts the usual telling. The hybrids miss because the ring is strained; the ring is not strained because the hybrids miss.

Still open: the misalignment’s energy, and an unsymmetrical double bond #

The obvious open question is the energy the section above only bounds. The comparison between three- and four-membered rings caps the misalignment’s contribution at a few kilojoules a mole without saying what it actually is, and computing the overlap loss across the whole series would turn that bound into a number — which is the quantity any strain account needs in order to say what the rest of the strain is made of.

The nearer question is one not answered here: the bond with no symmetry to enforce equivalence. A carbonyl’s two bent components are inequivalent, so the mixing angle that maximises their localisation has to be searched for and their s characters are two different numbers. Whether they still average to the value the symmetric case gives is a question the same calculation answers with one search added, and it is the case every unsymmetrical double bond in chemistry is.