Shake cyclohexene with bromine, and the red-brown color of the bromine vanishes at once as Br₂ adds across the C=C. Shake benzene with bromine, and nothing happens. On paper benzene has three C=C bonds, so it ought to be even more reactive than cyclohexene. It is far less.
Benzene resists hydrogenation too. As Catalytic hydrogenation showed, an ordinary alkene takes up H₂ under mild conditions, while a benzene ring needs high pressure and high temperature. Rings that behave this way are called aromatic. This page explains why, and the same explanation reaches well beyond benzene, into the rings found throughout biology and medicine.
Benzene's real structure
Start with benzene's bond lengths. Its six carbon–carbon bonds all have the same length, 139 pm. That lies between a typical C–C single bond (154 pm) and a C=C double bond (134 pm). If benzene really had three fixed double bonds, it would have three short bonds and three long ones.
The explanation comes from Resonance. Benzene has two Kekulé structures, which differ only in where the double bonds are drawn. They are two drawings of one molecule, not two compounds that switch back and forth. The real molecule is their hybrid, with the six π electrons spread evenly around the whole ring.
Measuring the extra stability
Heats of hydrogenation, from Alkene structure, put a number on benzene's stability. Adding H₂ to cyclohexene releases 28.6 kcal/mol. A ring with three separate C=C bonds should therefore release about three times as much, 85.8 kcal/mol, on the way to cyclohexane. Benzene releases only 49.8.
The difference, 85.8 − 49.8 = 36 kcal/mol, is how much more stable benzene is than an imaginary "cyclohexatriene" with three fixed double bonds. This extra stability is called the resonance energy, or aromatic stabilization energy.
Ordinary conjugation cannot account for it. Cyclohexa-1,3-diene, with two conjugated C=C bonds, releases 55.4 kcal/mol, only 1.8 less than the 57.2 expected for two separate ones. Something about a closed ring of six π electrons is special, and the rest of this page is about what that is.
The four requirements
Look at benzene's p orbitals. Each of its six carbons is sp² (see Hybridization), so each has one p orbital left over, standing straight up from the ring. Because the ring is flat, all six p orbitals are parallel, and each one overlaps with its neighbors on both sides. The overlap runs all the way around, so the six π electrons belong to one continuous ring of orbitals.
Benzene's features give the general rule: a ring is aromatic only if it meets all four of these requirements.
- Cyclic. The atoms form a closed ring, so the overlap can go all the way around.
- Planar. The ring is flat, so the p orbitals are parallel and can overlap.
- Fully conjugated. Every ring atom has a p orbital that joins the loop. In practice every ring atom is sp² (or sp). A single sp³ atom has no p orbital, and it breaks the loop.
- 4n + 2 π electrons. The loop holds 2, 6, 10, 14 … π electrons, where n is a whole number (0, 1, 2, 3 …). This is Hückel's rule.
Failing any one test disqualifies a ring, and which test it fails often explains its chemistry.
Why the count must be 4n + 2
The number looks arbitrary, but it follows from how the π electrons fill their orbitals. Take it in three steps.
First, as Conjugated systems showed, n overlapping p orbitals combine into n π molecular orbitals. Benzene's six p orbitals give six. In a ring, these orbitals always come in the same pattern: one orbital alone at the lowest energy, then pairs of orbitals above it. The two orbitals in a pair have exactly the same energy: they are degenerate, as in Orbitals.
Second, a degenerate pair that holds electrons is stable only when it is completely full. Suppose a pair holds just two. By Hund's rule, also from Orbitals, they go one in each orbital, unpaired. The ring is left with two unpaired electrons, a reactive arrangement much like two radicals. With four electrons the pair is full, and the problem disappears.
Third, count. The lowest orbital holds 2 electrons, and each pair above it holds 4 more. So the totals that fill every level completely are 2, then 6, then 10, then 14. That sequence is exactly 4n + 2.
A drawing called a Frost circle gives the orbital pattern for any ring. Draw a circle, and fit the ring inside it as a regular polygon with one corner pointing straight down. The height of each corner is the energy of one π molecular orbital. Corners below the circle's center are bonding orbitals, corners on the center line are nonbonding, and corners above it are antibonding. Always point a corner down; the other way round gives the wrong pattern.
Filling the levels from the bottom shows the difference. Benzene's six electrons fill the lowest orbital and the pair above it, so every bonding orbital is full. Cyclobutadiene's four electrons fill the lowest orbital, and the last two have to split across the nonbonding pair, one electron in each with parallel spins. They add no bonding, and they are unpaired: the half-filled arrangement the rule forbids.
This clean filling is where the 36 kcal/mol comes from. Resonance alone is not enough; the count has to fill the orbitals completely.
Counting π electrons
Most mistakes happen here, so it helps to state the rules one at a time.
- Each C=C (or C=N) in the ring contributes 2.
- A carbon with a negative charge has a lone pair in its p orbital, and it contributes 2.
- A carbon with a positive charge has an empty p orbital, and it contributes 0. Its empty orbital still completes the loop.
- A lone pair on a ring heteroatom (an atom other than carbon, such as N, O or S) contributes 2 only if it sits in the p orbital that joins the loop. A lone pair in an sp² orbital, lying in the plane of the ring, contributes 0.
To apply the last rule, ask one question of each heteroatom: does it already have a double bond in the ring? If it does not, its lone pair must sit in the p orbital, because that is the only way the atom can join the loop. If it does, the double bond already puts the atom in the loop, and its lone pair lies in the ring plane instead.
Pyrrole and pyridine show both cases. Pyrrole is a five-membered ring whose nitrogen carries an H and has no ring double bond. Its lone pair therefore sits in the p orbital: two C=C (4 electrons) plus the lone pair (2) make 6, and pyrrole is aromatic.
Pyridine is a six-membered ring, like benzene with one CH replaced by N. Its nitrogen already has a C=N, so its lone pair lies in an sp² orbital in the ring plane, pointing outward. Two C=C and one C=N make 6, and pyridine is aromatic too, without any help from the lone pair.
This one difference changes how the two nitrogens behave. Pyridine's lone pair is outside the π system, so a proton can take it without disturbing the ring, and pyridine is a base. Pyrrole's lone pair is part of the six, so taking it would destroy the aromatic ring, and pyrrole's nitrogen is not basic. Later, in Amines, Structure & basicity puts numbers on both.
Furan and thiophene are five-membered rings like pyrrole, with an O or an S in place of the N–H. Each of these atoms has two lone pairs. One lone pair sits in the p orbital and joins the two C=C to make six. The other lies in an sp² orbital in the ring plane. So both rings are aromatic, and each keeps one lone pair outside the π system.
Imidazole has two nitrogens that do different jobs. N1 carries an H and has no ring double bond, so it behaves like pyrrole's nitrogen and puts its lone pair into the ring. N3 has a C=N, so it behaves like pyridine's nitrogen and keeps its lone pair in the plane. The ring holds 6 π electrons and is aromatic, and N3 is the basic nitrogen.
Aromatic ions
A charge can complete an aromatic ring or ruin one. Cyclopentadiene and the two ions made from it show the three possible outcomes.
Cyclopentadiene has two C=C bonds and one sp³ CH₂. The CH₂ carbon has no p orbital, so the loop is broken. It is nonaromatic.
The cyclopentadienyl anion forms when the CH₂ loses a proton. That carbon becomes sp², and its lone pair sits in a p orbital. Two C=C (4 electrons) plus the lone pair (2) make 6, in a flat ring of five sp² carbons. It is aromatic.
The cyclopentadienyl cation forms when the CH₂ loses a hydride ion (H⁻, a hydrogen with two electrons) instead. That carbon is left with an empty p orbital. The loop is complete, but it holds only the 4 electrons of the two C=C bonds, the same bad count as cyclobutadiene. The cation is antiaromatic (defined in the next section) and very hard to make.
The anion's aromaticity shows up in the pKa. As Factors affecting acidity previewed, cyclopentadiene's CH₂ hydrogens have a pKa of about 16, close to water's. An ordinary alkane C–H has a pKa of about 50, so losing the proton is about 34 pKa units easier here, largely because it creates an aromatic ring. For a carbon acid that is an enormous shift: the sp effect in Factors affecting acidity, the biggest orbital effect, is worth only about 25.
The same count works for other ring sizes. Cycloheptatriene is a seven-membered ring with three C=C bonds and one CH₂. Remove a hydride from the CH₂, and that carbon is left with an empty p orbital. All seven carbons are now sp², the ring is flat, and the three C=C supply 6 π electrons, since 4(1) + 2 = 6.
This tropylium ion (the cycloheptatrienyl cation) is aromatic, and unusually stable for a carbocation. Tropylium bromide is an ionic salt that dissolves in water, which no ordinary carbocation salt can do.
The cyclopropenyl cation is the smallest case. It has three sp² carbons, one C=C and one empty p orbital, so it holds 2 π electrons. With n = 0, 4(0) + 2 = 2, so it is aromatic. Its 60° bond angles are badly strained, yet the ion is stable enough to make, because aromatic stabilization outweighs the strain.
Compare the two cations with the cyclopentadienyl cation. A positive charge decides nothing by itself: count the electrons in the loop, and let the count decide.
Antiaromatic: the right shape, the wrong count
A ring that is cyclic, planar and fully conjugated but holds exactly 4n π electrons (4, 8, 12 …) is antiaromatic. It is not merely missing the extra stability; it is less stable than a similar open-chain compound with the same number of π electrons. Its top two electrons sit one in each orbital of a degenerate pair, unpaired: a reactive arrangement much like two radicals.
Cyclobutadiene, with 4 π electrons, is the clearest example. It is so unstable that it was first observed trapped in frozen argon a few degrees above absolute zero. The cyclopentadienyl cation's reluctance to form points the same way. Together they show that antiaromaticity is a real destabilization, not just a missing bonus.
Cyclooctatetraene (COT), with 8 π electrons, avoids the problem instead. It is not flat. It folds into a tub shape, in which neighboring C=C bonds tilt away from each other. The p orbitals on either side of each single bond then point in different directions and barely overlap, so the eight electrons never form one loop.
COT therefore behaves like four ordinary, separate alkenes, and it is nonaromatic. The tub also relieves angle strain, since a flat octagon would force 135° angles on sp² carbons that prefer 120°. So COT is not the cleanest evidence for antiaromaticity; cyclobutadiene and the cyclopentadienyl cation are.
Nonaromatic: a requirement missing
Cyclohexa-1,3-diene has two C=C bonds, but its two sp³ CH₂ carbons break the ring's conjugation. It is nonaromatic: neither specially stabilized nor specially destabilized.
The general rule: a ring that fails any of the first three requirements (cyclic, planar, fully conjugated) is nonaromatic. It behaves like an ordinary alkene or diene, and the addition chemistry of Alkenes & Alkynes applies to it as usual. Cyclopentadiene, cycloheptatriene and tub-shaped COT all belong here.
| Category | Cyclic, planar, fully conjugated? | π electron count | Examples |
|---|---|---|---|
| aromatic | yes | 4n + 2 (2, 6, 10 …) | benzene, pyridine, pyrrole, cyclopentadienyl anion, tropylium |
| antiaromatic | yes | 4n (4, 8 …) | cyclobutadiene, cyclopentadienyl cation |
| nonaromatic | no, at least one fails | does not matter | cyclopentadiene, cyclohexa-1,3-diene, tub-shaped COT |
Fused rings
Naphthalene is two benzene rings that share one edge. The two carbons on that edge belong to both rings. Naphthalene has 10 π electrons spread over 10 carbons, and it is aromatic.
Hückel's rule was worked out for a single ring, so apply it to fused rings with care. The other fused systems in this course are also flat, fully conjugated and hold a 4n + 2 count, and all are aromatic: anthracene (three benzene rings in a row, 14 π), indole (a benzene ring fused to pyrrole, 10 π) and purine (the double ring of adenine and guanine, 10 π). Learn them as known examples rather than deriving them from the rule.
The count still matters. Pentalene, two fused five-membered rings, is flat and fully conjugated, but its 8 π electrons are a 4n count. It is not aromatic; like cyclobutadiene, it is too unstable to keep at room temperature. Heptalene, two fused seven-membered rings with 12 π electrons, escapes the way COT does: it twists out of plane and is nonaromatic.
Large single rings
Large single rings with alternating single and double bonds are called annulenes, and each is named by its ring size in brackets. [18]Annulene has eighteen carbons and eighteen π electrons, which fits 4n + 2 with n = 4. It is flat enough for its p orbitals to overlap, and it is aromatic.
[10]Annulene also has the right count, 10, but it is not aromatic. A flat ten-membered ring with every double bond cis would need 144° bond angles. Making two of the double bonds trans relieves that strain, but then two hydrogens point into the middle of the ring and collide, so the ring twists out of plane. Planarity is a real requirement.
Why it matters beyond the exam
Aromatic rings are everywhere in biology. They include the bases of DNA and RNA, the amino acids phenylalanine, tyrosine, tryptophan and histidine (whose side chain is an imidazole), and the porphyrin ring at the center of heme and chlorophyll. Most drugs contain at least one. Part of the reason is stability, since an aromatic ring survives conditions in the body. Part is shape: flat rings with delocalized electrons can stack against each other and bind in ways that saturated rings cannot.
What carries forward
Aromatic stabilization is why benzene reacts by substitution rather than addition, which is the subject of Electrophilic aromatic substitution. Losing aromaticity and getting it back is the energy story of every mechanism in this chapter.
The general rule is simple. Any equilibrium that creates an aromatic ring is pulled strongly forward, and any equilibrium that destroys one is pushed strongly back. Cyclopentadiene's acidity and pyrrole's non-basic nitrogen both show it. Whenever a step in a later chapter looks surprisingly easy or surprisingly hard, check whether an aromatic ring is being made or broken.