Why is benzene different?

Here is what bench 1 does. Cyclohexene meets bromine and the orange goes at once, with no catalyst. Baeyer's test is cold, dilute, alkaline potassium manganate. A double bond takes it from purple to brown, and cyclohexene does. Benzene with bromine alone does nothing. The colour stays and the Baeyer test stays negative. There is a catalyst box on the bench. Tick it, with benzene selected, and read what the outcome line says then. Kekulé's picture of this ring has the bonds alternating, single, double, single, double. That picture predicts two different carbon-to-carbon lengths going round. The table below carries the measured lengths. It is often said Kekulé saw the ring in a dream about a snake. He told that story decades later. Historians doubt it, and it is not how the structure was found.

Start from the measured benzene length, 1.397 angstrom for all six carbon-to-carbon bonds. Ethane's single-bond stand-in is 1.536 angstrom, and 1.536 minus 1.397 gives 0.139 angstrom. Divide 0.139 by 1.536, and the benzene bond is 9.05 per cent shorter. Ethene's double-bond stand-in is 1.339 angstrom, and 1.397 minus 1.339 gives 0.058 angstrom. Divide 0.058 by 1.339, and the benzene bond is 4.33 per cent longer. One measured length sits between the two stand-ins, and there is only one of it.

An alternating ring should behave like three separate double bonds. Each one should take up bromine on its own. The measured ring has no alternating bonds to offer. All six of its bonds are one kind, at one length. Three double bonds means three extra pairs, so six electrons. Those six are not parked between fixed pairs of carbons. They are spread over the whole ring instead. That spreading has a name. Delocalisation is the sharing of bonding electrons over a whole group of atoms. So a delocalised ring has no plain double bond to attack. Whether that is the whole reason is a question for section 5. Either way, the spreading has to be worth something, and worth is measured in energy.

Bench 3 is a table of measured values, and nothing on it moves. Each row is one ring taken to cyclohexane with hydrogen. Cyclohexene releases 118.6 kilojoules a mole, measured in the gas phase. Benzene releases 205.3 kilojoules a mole, measured the same way. 1,3-cyclohexadiene's figure, 227.68 kilojoules a mole, is derived rather than measured directly. It comes from formation enthalpies for cyclohexane and for the diene. Every row carries its own source on the bench. All three enthalpies are negative, because every one of them releases heat.

Take every figure here as a size, because all three enthalpies are negative. Three separate double bonds should release three times cyclohexene's own figure. Multiply 118.6 kilojoules a mole by 3, and the expected release is 355.8. Benzene's own measured release is 205.3 kilojoules a mole, which is smaller. Subtract 205.3 from 355.8, and 150.5 kilojoules a mole never comes out as heat. That missing 150.5 is benzene's aromatic stabilisation, the energy its sharing is worth. Now run the same sum for two double bonds. Multiply 118.6 by 2, and the expected release is 237.2 kilojoules a mole. 1,3-cyclohexadiene's own release is 227.68, so subtracting leaves 9.52 kilojoules a mole. Divide 150.5 by 9.52, and benzene's figure is 15.8 times the diene's. Both gaps are energy that never appears as heat.

So benzene sits in a well, 150.5 kilojoules a mole deep. It starts lower than three separate double bonds would. The ladder did add hydrogen to it, so addition is not impossible. It is expensive, and the ladder has just priced it. Releasing a lot of heat is not the same as being unsteady. Here is the step that closes the argument. Section 1's catalysed reaction swapped one ring hydrogen for a bromine. Chemists call that swap substitution. Count the bonds in the product. Six equal bonds are still there, so the sharing survives whole. The 150.5 kilojoules a mole is never paid. Addition is different. Bromine across two carbons leaves one plain double bond behind. The sharing is broken, and the well has to be paid for. That is why the swap is the cheap reaction. Two double bonds side by side gain about 9.5 kilojoules a mole, which is ordinary. Benzene's gain is not more of the same.

The bench opens on benzene, a six-carbon ring with six pi electrons. Two letters label the levels. Here α is the energy of one pi electron left on a single carbon. And β is what the sideways touching between two neighbours is worth. Beta is negative, so a level below alpha is a bonding level. Six electrons fill the three lowest levels, two in each. The highest filled level is bonding, and it holds two of them. A shell filled like that is called closed. Now pick cyclobutadiene from the preset list. Watch where the last two electrons land as the levels refill.

Written in alpha and beta, benzene's lowest level is α + 2β. The next two levels are α + β each. Two electrons in the lowest give 2α + 4β, and four in the next pair give 4α + 4β. Add those and benzene's six electrons total 6α + 8β. One isolated double bond is two electrons at α + β, so 2α + 2β. Three of those give 6α + 6β in all. Subtract 6α + 6β from 6α + 8β, and the delocalisation energy is 2β. Divide 2β by the six electrons sharing it, and each electron gains 0.333β. The bench prints that per-electron figure itself. Cyclobutadiene's lowest level is α + 2β, and its middle two sit at α itself. Two electrons below and two at α total 4α + 4β. Two isolated double bonds give 4α + 4β as well. Subtract one total from the other, and the delocalisation energy is exactly zero beta.

Cyclobutadiene's last two electrons do not pair up. They land in two levels of exactly equal energy, one electron each. Hund's rule puts them that way. Those two levels sit at alpha itself, neither below it nor above. A level at alpha is non-bonding, so an electron there buys nothing. A shell left half empty like that is called open. The bench says as much on its own readout, in words. Cyclobutadiene is barely stable as a substance either. Two of its molecules join up within minutes above about 35 kelvin. So a ring shape on its own settles nothing. The count of electrons has to suit the ring as well.

Set the preset and read the bench. The reading is captioned planar and hypothetical, and that caption is the point. Real cyclooctatetraene is not flat. It folds into a shape called a tub. A flat eight-carbon ring is a geometry this molecule never adopts. The flat reading is a reference state, put there to be argued with.

Planar cyclooctatetraene's eight electrons total 8α + (4 + 4√2)β for the flat ring. That bracket works out at 9.657, so the total is 8α + 9.657β. Four separate double bonds would give 8α + 8β in all. Subtract 8α + 8β from 8α + 9.657β, and the delocalisation energy is 4(√2 − 1)β. That comes to 1.657 lots of β. Beta is negative, so that is an energy below the separate double bonds, which means a stabilisation. Now divide by the electrons sharing it. Divide 1.657β by 8, and each electron gains 0.207β. Benzene's own figure was 2β divided by 6, which is 0.333β an electron. So 0.207β against 0.333β, and the flat eight-ring gains less per electron than benzene.

The sum is not wrong. Its assumption is. It fills electrons into a flat ring, and this ring is not flat. Look at where the last two electrons go. They land in two levels of exactly equal energy, one electron each, by Hund's rule. Those levels sit at alpha, so they are non-bonding, and they buy nothing. That is cyclobutadiene's pattern, not benzene's closed shell. The bare 1.657β hides it. The real molecule leaves the plane instead, which is module 7's question about shape. Two separate things are wrong here. Eight is the wrong kind of count, and section 5 says which kind is right. The geometry is a second failure, not a rescue of the first. None of this makes the counting method wrong. The same method gives benzene 2β and cyclobutadiene zero, and both are right.

This ring still has to react as something. Section 1 worked two cases. This one has to match one of them.

Here is the rule in full, and it has four conditions. The molecule must be a ring. The ring must be flat, or close to flat. Every atom in the ring must carry a p orbital joining the next, with no break anywhere. And the shared pi electrons must number 4n + 2, where n is a whole number from zero up. With n as zero, one and two, 4n + 2 gives 2, 6 and 10. The table below applies all four conditions to three cases, and names them. The cyclopentadienyl anion is a five-carbon ring with a negative charge. The tropylium cation is a seven-carbon ring with a positive charge. Pyridine is a six-membered ring with one nitrogen in place of a CH. Three ring sizes, three charge states, and no two of them alike. All three carry exactly six shared pi electrons, and all three are aromatic. So the rule counts electrons. It does not ask about ring size, or charge, or whether every atom is carbon.

1,3-butadiene is a chain of four carbons rather than a ring, and the bench calls that linear. Its two lower levels come out at α + 1.618β and α + 0.618β. Its two upper levels are α − 0.618β and α − 1.618β. The four pi electrons fill the lower two levels, two in each. Two times α + 1.618β plus two times α + 0.618β gives 4α + 4.472β. Two separate double bonds would give 4α + 4β in all. Subtract 4α + 4β from 4α + 4.472β, and the delocalisation energy is 0.472β. That is a real, positive stabilisation, set against benzene's own 2β. Butadiene is still not aromatic, because it has no ring to close.

Butadiene is the sharpest case in this module. Its electrons really are delocalised. Its stabilisation is real, and it can be counted. And it is not aromatic at all. So section 1's reason was only half of it. Delocalisation alone does not protect butadiene. The sentence this page keeps says benzene's ring, and it says six. Butadiene fails twice over. It has no ring, and its count is four. Either failure is enough on its own.