Some molecules cannot be shown correctly by any single Lewis structure. You met one in Lewis structures: in nitrate, the double bond could go to any of the three oxygens, and all three N–O bonds turn out to be identical. This page explains what such a molecule really looks like and how to draw it. It uses the curved arrows from Curved arrows throughout.
One ion, two drawings: acetate
Take acetate, CH₃COO⁻, the ion left when acetic acid (the acid in vinegar) loses the H of its O–H group. Its central carbon is bonded to a CH₃ group and to two oxygens. When you draw its Lewis structure, you reach a choice. One oxygen gets a double bond to carbon. The other gets a single bond, three lone pairs and the negative charge. Nothing in the ion says which oxygen should be which, so drawings A and B below are equally good.
Two curved arrows turn A into B. The first starts on a lone pair of the charged oxygen and ends between that oxygen and the carbon, where it makes a new π bond. The second starts on the old C=O π bond and ends on the other oxygen as a new lone pair. That oxygen now carries the negative charge.
Measurements agree with neither drawing on its own. Both C–O bonds in acetate are 126 pm long, between a C–O single bond (143 pm) and a C=O double bond (123 pm). The two bonds are identical, and the negative charge is shared equally, half on each oxygen.
Drawings like A and B are called resonance structures, or contributors. They are Lewis structures of the same species with every atom in the same place; only the electrons are placed differently. A double-headed arrow, ↔, links them. The real ion is the resonance hybrid, a single structure that blends all of its contributors. A and B are equivalent: they are the same drawing with the two oxygens swapped, so they contribute equally to acetate's hybrid.
Electrons that spread over more atoms than one Lewis structure can show are delocalized. Delocalization lowers a molecule's energy, and it is one of the most important stabilizing effects in organic chemistry. Resonance structures are how chemists draw it.
Where resonance can happen: conjugation
Compare two nitrogen compounds. In ethylamine, CH₃CH₂–NH₂, the nitrogen's neighbors are a CH₂ carbon and two hydrogens, and none of them has a π bond. An arrow from the nitrogen lone pair has nowhere legal to go, so ethylamine has only one structure. Its lone pair stays on nitrogen and is fully available. That is why an amine nitrogen bonds readily to H⁺, and why it acts as a nucleophile: an electron-pair donor that bonds to an electron-poor atom.
In acetamide, CH₃–C(=O)–NH₂, the same NH₂ group is bonded to the carbon of a C=O. Its lone pair now sits right beside the C=O π bond. The two arrows that worked for acetate work here too, so the pair spreads toward oxygen and much less of it is left on nitrogen. The element, the lone pair and the charge are all the same as in ethylamine. Only the neighbor has changed.
What the lone pair needs is a p orbital next door. A π bond is two p orbitals overlapping side by side, as Bonding showed. Electrons can spread along a row of p orbitals on neighboring atoms, all standing parallel so that each overlaps the next. A lone pair in a p orbital at the end of such a row joins it. Hybridization showed that the amide nitrogen puts its lone pair in exactly that kind of p orbital.
A single sp³ atom breaks the row. Compare CH₂=CH–OH with CH₂=CH–CH₂–OH. In the first, the oxygen is bonded directly to a carbon of the C=C, and one of its lone pairs can spread onto the far carbon. In the second, an sp³ CH₂ sits between the oxygen and the double bond. That carbon used all three of its p orbitals to make its four hybrids, so it has none left to pass electrons along. The oxygen's lone pairs are cut off, and there is no second structure to draw.
An unbroken row of p orbitals like this is a conjugated system. Delocalization runs the length of the row and stops where the row stops. So when a question asks whether something is conjugated, it is asking one thing: is there an unbroken row of p orbitals? Check the hybridization of each atom along the way. Any sp³ atom ends the row.
A row can also be built from two π bonds side by side, as in CH₂=CH–CH=CH₂. Each of the four carbons has a p orbital, and each one overlaps the next, so the two π bonds form one conjugated system.
The three moves
Nearly every pair of resonance structures in this course is linked by one of three moves. Each starts from a different arrangement of p orbitals, and the figure below shows one example of each.
1. A lone pair beside a π bond. Take CH₂=CH–OH again. One arrow takes an oxygen lone pair into the C–O bond, where it becomes a C=O π bond. A second arrow moves the C=C π pair onto the far carbon as a lone pair. The new structure, ⁻CH₂–CH=OH⁺, has a negative charge on that carbon and a positive charge on oxygen. Acetate uses this move, and so do amides and esters.
2. A π bond beside an empty p orbital. A carbocation is a carbon with three bonds, a positive charge and an empty p orbital (see Hybridization). In the allyl cation, CH₂=CH–CH₂⁺, a carbocation sits next to a C=C. One arrow moves the π pair over into the C–C bond on the other side, where it forms a new π bond. The empty p orbital, and with it the positive charge, ends up on the carbon at the far end. "Allyl" is the name of the three-carbon unit CH₂=CH–CH₂–. This shift of the charge from one end to the other lies behind the allylic rearrangements in Substitution & Elimination.
3. A π bond to an electronegative atom. In formaldehyde, H₂C=O, the simplest aldehyde, one arrow moves the C=O π pair onto oxygen. That gives H₂C⁺–O⁻, with a positive charge on carbon and a negative charge on oxygen. A C=N bond can do the same thing, with nitrogen taking the pair. The pair always goes to the more electronegative atom. Sending it the other way would leave oxygen positive and with only six electrons, so that structure is not drawn.
Legal first, then ranked
Before asking how much a structure contributes, check that it is a structure at all. Suppose you draw only the first arrow of the acetate move. The lone pair makes a new C–O π bond, but the old C=O π bond stays where it was. The carbon now has five bonds, which means ten electrons around it, and no such structure exists.
Three tests catch almost every drawing like that. Run each one on every structure you draw until it becomes a habit.
- No second-row atom may exceed an octet. Carbon, nitrogen, oxygen and fluorine have only four valence orbitals, so eight electrons is a hard ceiling. For carbon that means four bonds at most, and a nitrogen or oxygen that keeps lone pairs has room for fewer. A carbon with five bonds, or a nitrogen with four bonds and a lone pair, is not a poor contributor: it does not exist. This is the most common way a drawn structure goes wrong.
- The electron count stays the same. Add up the bonding pairs and lone pairs in one structure, then in the next. If the totals differ, an arrow created or lost electrons. The atoms do not change either, so the total charge must match as well.
- The number of unpaired electrons stays the same. Resonance moves electrons in pairs and never splits one. A species with no unpaired electrons cannot have a structure with two. A radical, a species with one unpaired electron, has exactly one in every structure. The allyl radical, CH₂=CH–CH₂•, has its unpaired electron on one end carbon in one structure and on the other end carbon in the other, and never gains a second.
How many structures?
You may be asked how many resonance structures a molecule has. The answer is usually small. Benzene, C₆H₆, has five. Only two of them matter: the Kekulé structures, which place the three C=C bonds in the two possible alternating patterns around the ring. The other three are Dewar structures. Each has only two C=C bonds and pairs its last two electrons in a long, weak bond drawn straight across the ring. They are high in energy, and chemists normally ignore them.
Naphthalene, C₁₀H₈, is two benzene rings sharing one edge, and it has three Kekulé structures. The allyl cation from move 2 has two, and so do the allyl anion, CH₂=CH–CH₂⁻, and the allyl radical.
The number itself is never the point. What matters is that when a molecule has several equivalent contributors, its charge or π electrons really are spread over all the atoms those structures place them on.
Ranking contributors
Acetate's two structures are equivalent, so they contribute equally. Most pairs are not like that. When two structures differ, the more stable one contributes more, and the hybrid looks more like it. That structure is the major contributor, and the other is a minor contributor. Two examples show how to decide which is which.
An enolate is the anion formed when an H⁺ is removed from a carbon next to a C=O. Removing an H⁺ from the CH₃ of acetaldehyde, CH₃–CH=O, gives the simplest one, which is usually drawn CH₂=CH–O⁻. Rank its two structures.
Identify the move. A lone pair on the charged oxygen sits beside the C=C π bond. That is move 1.
Draw the second structure. The oxygen lone pair becomes a C=O π bond, and the C=C π pair moves onto the end carbon as a lone pair. That gives ⁻CH₂–CH=O.
Work out the charges. In the first structure, oxygen has one bond and three lone pairs: 6 − 6 − 1 = −1. Every carbon is neutral. In the second, oxygen has two bonds and two lone pairs: 6 − 4 − 2 = 0. The end carbon has three bonds and one lone pair: 4 − 2 − 3 = −1. Each structure carries one negative charge, as it must.
Compare. Every atom has an octet in both structures, and each structure has one charge. They differ only in where that charge sits: on oxygen in the first, on carbon in the second. Oxygen is much more electronegative than carbon, so it holds a negative charge far better. The O⁻ structure is the major contributor.
The minor structure still matters. The hybrid carries some real negative charge on the end carbon, and that is why an enolate often forms its new bond through carbon even though most of its charge sits on oxygen. Enolate Chemistry puts this to work. Ranking tells you where the electrons mostly are. On its own, it does not tell you which atom reacts.
Acetone is (CH₃)₂C=O. When an H⁺ bonds to its oxygen, the result is (CH₃)₂C=OH⁺. Rank its two structures.
Identify the move. The C=O π bond can move onto oxygen. That is move 3.
Draw the second structure. The π pair becomes a lone pair on oxygen, which keeps its bonds to carbon and hydrogen and now has two lone pairs. Carbon is left with three bonds and only six electrons. The positive charge has moved from oxygen to carbon: (CH₃)₂C⁺–OH.
Work out the charges. In the first structure, oxygen has three bonds and one lone pair: 6 − 2 − 3 = +1. Carbon is neutral. In the second, oxygen is 6 − 4 − 2 = 0 and carbon is 4 − 0 − 3 = +1. There is one positive charge each time.
Compare. Charge placement favors the second structure, because a positive charge sits better on carbon, the less electronegative atom. But octets are checked first, and the second structure leaves carbon with six electrons. The first structure, with every octet full, is the major contributor.
Here too the minor structure counts. It fails only one check, on one atom, so it still makes up a real share of the hybrid. That share puts part of the positive charge on carbon. This is why adding H⁺ to a carbonyl oxygen makes the carbon much more attractive to a nucleophile.
A neutral carbonyl follows the same rules. Formaldehyde's second structure from move 3, H₂C⁺–O⁻, fails two checks at once. It has two charges where the first structure has none, and it leaves carbon with six electrons, two short of an octet. That makes it a very minor contributor. It is still part of the hybrid, though, and it adds to the δ+ on the carbonyl carbon that Bond polarity traced to electronegativity.
These examples applied the checks in a fixed order. Here is the full list, most important first.
- Full octets win. A structure in which every second-row atom has an octet beats one that leaves an atom short. This outranks every rule below it.
- Fewer formal charges win. No charges beats a +1/−1 pair, which beats +2/−2.
- Negative charge belongs on the more electronegative atom, and positive charge on the less electronegative one.
- Like charges should not sit on neighboring atoms. Two negative charges, or two positive ones, side by side repel each other, and that costs energy.
- More bonds and less separation of charge win when everything else is close.
The second, third and fourth rules are the ones Formal charge used to choose between Lewis structures. Every legal structure is worth drawing. The rules only say which one contributes more.
What resonance buys: stability
The energy drop from delocalization is large, and acidity shows it clearly. Acid strength is measured as pKa: a lower pKa means a stronger acid, and each unit is a factor of ten. Acetic acid (pKa ≈ 4.8) is about 10¹¹ times more acidic than ethanol (pKa ≈ 16), even though both give up the H⁺ of an O–H group.
The difference lies in the anion each acid leaves behind. Acetate shares its negative charge between two oxygens. Ethoxide, CH₃CH₂O⁻, has to keep the whole charge on one. The more stable the anion left behind, the more readily the acid gives up its proton. Acids & Bases names that anion, the conjugate base, and makes the comparison exact.
Cations are stabilized too. The allyl cation from move 2 spreads its positive charge over two carbons. A similar cation with no double bond beside it has to keep the charge on one carbon. Gas-phase measurements put the allyl cation roughly 15 kcal/mol more stable than such a cation. How Reactions Happen compares cations more fully.
Benzene is about 36 kcal/mol more stable than an imaginary "cyclohexatriene" with three fixed C=C bonds. That extra stability is the reason benzene and its relatives, the aromatic compounds, behave as a separate class.
In an amide, the second structure (minor, because it carries two charges) puts a double bond between carbon and nitrogen, so the real C–N bond is partly double. Twisting about it costs roughly 18 kcal/mol. That is enough to hold six atoms in one flat plane: the carbonyl C, O and N, the carbon attached to the carbonyl carbon, and the two atoms attached to N. The peptide bonds that join the amino acids of a protein are amides, so they are flat too, and the shapes that protein chains fold into depend on it (Biomolecules).
What carries forward
Resonance is the most reused idea in the rest of the course. It sets the acidity of carboxylic acids, phenols and the hydrogens next to a C=O in Acids & Bases and Enolate Chemistry. It stabilizes the carbocations of Substitution & Elimination and Alkenes & Alkynes. It makes an amide unreactive and an acid chloride very reactive in Carboxylic Acids & Derivatives. In Aromatic Chemistry it helps explain why aromatic rings are so stable, though resonance alone is not the whole story. The directing effects there are resonance arguments almost from start to finish. Each of those arguments uses the three moves and the ranking rules on this page.