Resonance Structures

Resonance Structures

A resonance structure is one of two or more valid Lewis structures for the same molecule that differ only in where the electrons are drawn. The atoms stay exactly where they are. Only π electrons and lone pairs move. When a molecule has more than one such drawing, none of them is the whole story: the real molecule is a resonance hybrid, a weighted average of all of its contributors. Its electrons are delocalized, meaning spread over several atoms, and that makes it more stable than any single drawing suggests.

This lesson covers how to draw resonance structures by hand, the rules every structure has to obey, how to tell which contributor matters most, and why resonance decides so much of organic reactivity. To check your own answers, draw the molecule in the Resonance Solver and it will generate the resonance structures for you.

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What is resonance?

Some molecules cannot be described by one Lewis structure. The formate ion, HCOO⁻, can be drawn with its C=O double bond on either oxygen, and the two drawings are equally good. Experiment settles the question: both C–O bonds in formate are the same length, in between a single and a double bond, and the negative charge is shared equally by the two oxygens. Neither drawing is right on its own. The real ion is the hybrid of the two.

Two resonance contributors of the formate ion joined by a double-headed arrow, with curved arrows moving an oxygen lone pair and the C=O pi bond, beside the resonance hybrid drawn with two partial carbon-oxygen bonds
The two contributors of formate, and the single hybrid they describe.

Three ideas do most of the work:

  • Contributors are drawings, not molecules. Formate does not flip back and forth between two structures. It is one species, all the time, and it looks like the hybrid.
  • The double-headed arrow (↔) means "same molecule, different drawing." It is not the equilibrium arrow (⇌), which connects two different species that really do interconvert.
  • Delocalization is stabilizing. Spreading charge or π electrons over more atoms lowers the energy. That is why resonance comes up whenever you compare the stability of ions, acids, and reaction intermediates.

How to draw resonance structures, step by step

The worked example is hex-3-en-2-one, a ketone with a C=C double bond next to its C=O. The carbons are numbered so the electrons are easy to follow.

Bond-line structure of hex-3-en-2-one with carbons numbered 1 to 6: a C=O double bond at carbon 2 with two lone pairs on oxygen, and a C=C double bond between carbons 3 and 4

Step 1: Find the electrons that can move. π bonds (the second bond of a double bond, or the second and third of a triple bond) and lone pairs can move. σ bonds, the single bonds that hold the skeleton together, stay put. Here the mobile electrons are the C2=O π bond, the C3=C4 π bond, and the lone pairs on oxygen.

Step 2: Push one pair toward an atom that can take it. Electrons move onto a neighboring atom or into a neighboring bond. Start with the C2=O π bond and move it onto oxygen, the more electronegative atom. A curved arrow always starts at the electrons and points to where they end up.

Curved arrow moving the pi electrons of the C=O bond in hex-3-en-2-one onto the oxygen atom

Step 3: Recount the formal charges. Formal charge is the atom's valence electrons, minus its lone-pair electrons, minus its number of bonds. Oxygen now has three lone pairs and one bond, so 6 − 6 − 1 = −1. C2 has three bonds and no lone pair, so 4 − 0 − 3 = +1. The molecule is still neutral overall, because moving electrons never changes the net charge.

Charge-separated resonance structure of hex-3-en-2-one with three lone pairs and a negative charge on oxygen and a positive charge on carbon 2

Step 4: Look for the next move. C2 is now electron-poor, and it sits right next to the C3=C4 π bond. Those π electrons can shift over to make a new C2=C3 double bond, which moves the positive charge out to C4.

Curved arrow moving the C3=C4 pi electrons toward the bond between carbon 2 and carbon 3
Shift the C3=C4 π electrons toward C2.
Resonance structure with a new double bond between carbon 2 and carbon 3, a negative charge on oxygen, and a positive charge on carbon 4
The positive charge is now on C4.

Step 5: Check each drawing against the rules, then connect them with ↔. Hex-3-en-2-one has three contributors: the neutral structure you started with, and two charge-separated structures with the positive charge on C2 or on C4.

The rules every resonance structure must follow

  1. Only electrons move. Every atom keeps the same neighbors in every structure. If you had to move an atom, even a hydrogen, you have drawn a different compound, not a resonance structure.
  2. Every drawing is a valid Lewis structure. All contributors have the same total number of electrons and the same net charge. Second-row atoms (C, N, O, F) never exceed an octet, so a carbon with five bonds is always wrong.
  3. Curved arrows start at electrons. The tail sits on a lone pair or a bond, and the head points at the atom or bond where that pair ends up. An arrow never starts at a positive charge, because there are no electrons there to move.
  4. Contributors are linked with ↔. Never use equilibrium arrows between resonance structures.

Five patterns that signal resonance

Nearly every resonance problem in a first organic chemistry course is one of five patterns. Spot the pattern and you know which arrow to draw.

1. A lone pair next to a π bond. The lone pair comes down to form a new π bond, and the old π bond becomes a lone pair on the far atom. The same pattern explains enolates, carboxylates, and amides.

Two resonance structures of an allylic carbanion, with the lone pair and negative charge on either end of the three-carbon pi system

2. A positive charge next to a π bond. The π electrons shift toward the electron-poor carbon, so the double bond and the positive charge trade places. This is an allylic carbocation.

Two resonance structures of an allylic carbocation, with the positive charge on either end of the three-carbon pi system

3. A lone pair next to a positive charge. The lone pair forms a π bond to the electron-poor atom. An oxygen or nitrogen lone pair stabilizing a neighboring carbocation is the classic case.

4. A π bond between atoms of different electronegativity. In C=O, C=N, and C≡N, the π electrons can be pushed onto the more electronegative atom. This is exactly Step 2 of the worked example.

5. Alternating double bonds around a ring. The three π bonds of benzene can be drawn in two ways. The two drawings are equivalent, so they contribute equally, all six C–C bonds are identical, and the hybrid is often drawn with a circle.

The two Kekulé resonance structures of benzene, labelled Form A and Form B, beside the resonance hybrid drawn as a hexagon with a circle inside

Major and minor contributors

Contributors are rarely equal. The hybrid looks most like its most stable contributor, so ranking them tells you where the electrons really are. Apply these tests in order:

  1. Full octets win. A structure in which every second-row atom has an octet beats one with an electron-deficient atom.
  2. Fewer formal charges win. A neutral structure beats a charge-separated one.
  3. Charges belong where they are most comfortable. Negative charge is best on the more electronegative atom, and positive charge on the less electronegative one.
  4. Equivalent structures contribute equally, as in formate and benzene.
Two resonance structures of an enolate ion. The structure with the negative charge on oxygen is marked with a green check, and the structure with the negative charge on carbon is marked with a red X
Both are valid. The check marks the major contributor and the X marks the minor one.

Both drawings of this enolate are valid resonance structures. Oxygen is more electronegative than carbon, so the structure with the negative charge on oxygen is the major contributor and the one with the charge on carbon is the minor contributor. Minor does not mean unimportant: enolates usually form new bonds at that carbon.

The same ranking applies to the worked example. The neutral drawing of hex-3-en-2-one is the major contributor, since every atom has an octet and there are no formal charges. The two charge-separated drawings are minor, but they are the ones that explain how the molecule reacts.

Why resonance matters for reactivity

The minor contributors of hex-3-en-2-one put positive charge on C2 and on C4, so both carbons are electron-poor. A nucleophile can attack the carbonyl carbon (1,2-addition) or the carbon at the far end of the C=C bond (1,4-addition, also called conjugate addition). The oxygen carries the negative charge, so that is where a proton or a Lewis acid binds.

Charge-separated resonance structure of hex-3-en-2-one with one water molecule pointing a hydrogen at the negatively charged oxygen and a second water molecule pointing its oxygen at the positively charged carbon 4
Water's hydrogen is drawn to the electron-rich oxygen, and water's oxygen is drawn to the electron-poor C4.

The same reasoning runs through the rest of the course:

  • Stabilized intermediates. Allylic and benzylic carbocations, carbanions, and radicals form more easily than their localized counterparts, because the charge or the unpaired electron is shared.
  • Acidity. A carboxylic acid (pKa about 5) is far more acidic than an alcohol (pKa about 16), in large part because the carboxylate's negative charge is shared by two oxygens.
  • Where a molecule reacts. Resonance structures show which atoms are electron-rich and which are electron-poor, so they predict where nucleophiles and electrophiles attack.
  • Aromatic substitution. Benzene undergoes substitution rather than addition, because substitution keeps its delocalized ring intact.

Check your work with the Resonance Solver

Drawing contributors by hand is the skill an exam tests, so try each problem on paper first. Then draw the same molecule in the Resonance Solver to generate its resonance structures, see how the charges and π bonds move from one to the next, and catch any contributor you missed.

Summary

  • Resonance structures are different Lewis drawings of one molecule: the same atoms in the same places, with the electrons drawn differently.
  • The real molecule is the resonance hybrid. It is more stable than any single contributor because its electrons are delocalized.
  • To draw them, move only lone pairs and π electrons with curved arrows, then recount the formal charges.
  • Look for the five patterns: a lone pair or a positive charge next to a π bond, a lone pair next to a positive charge, a polar π bond, and alternating double bonds in a ring.
  • Rank contributors by octets first, then by the number of charges, then by where the charges sit.
  • Minor contributors often explain reactivity, because they show which atoms are electron-rich or electron-poor.

For structured practice, work through the Resonance section of the Learn course or take the resonance structures quiz.