Resonance Structures: Worked Examples
Resonance structures examples trip up organic chemistry students not because the electron-pushing is mysterious, but because two separate skills compound: drawing the arrows correctly, then ranking what you drew. A carboxylate ion has two equivalent contributors and no ranking decision. An enolate has two inequivalent contributors and the ranking determines which atom attacks in the next reaction. Getting the mechanics right on simpler molecules first is the only way to build the fluency the harder examples demand.
The five examples below follow a consistent method drawn from the resonance sections of OpenStax Organic Chemistry and the approach laid out in MIT OpenCourseWare 5.12. Each example shows the starting structure, every curved arrow, the resulting structure, formal charges where they appear, and the ranking rationale. The most common error for each example is flagged at the end of its section.
What Are Resonance Structures?
Resonance structures are alternative Lewis dot representations of the same molecule that differ only in the placement of electrons, never in the connectivity of atoms. You draw them by moving lone pairs and pi bonds according to strict arrow conventions. The molecule itself does not oscillate between them; it exists at all times as a single species whose electron density is the weighted average of all contributors.
The Molecule Is the Hybrid, Not Either Structure
Acetate (CH3COO-) is drawn with one C=O double bond and one C-O single bond, then an arrow shifts the lone pair on the single-bond oxygen to produce a second structure with the bonds reversed. Neither drawing is "correct" by itself. The real acetate ion has two C-O bonds of identical length (about 125 pm, between a single bond at 143 pm and a double bond at 120 pm) because bond order is 1.5 across both. X-ray crystallography confirms this intermediate bond length, as described in the OpenStax discussion of resonance.
That intermediate geometry cannot be shown in a single Lewis structure. Resonance notation solves the problem by drawing multiple structures connected by double-headed arrows (not equilibrium arrows), acknowledging that the truth lies in between.
Five Rules Before You Draw a Single Arrow
Violating any of these rules produces a structure that is not a valid resonance contributor, which costs marks on an exam and produces wrong predictions in mechanisms.
| Rule | What it means | Common violation |
|---|---|---|
| 1. Atoms stay fixed | Only electrons move; connectivity cannot change | Drawing a structure where a hydrogen has moved to a new atom |
| 2. Arrows start on electrons | Tail on a lone pair or the midpoint of a pi bond | Placing the tail on an atom symbol rather than a bond or lone pair |
| 3. Respect electron count | Total charge and electron count must match across all structures | Losing or gaining an electron during the arrow-pushing |
| 4. Second-period octet rule | C, N, O, F cannot exceed 8 electrons (no expanded octets) | Drawing a five-bond carbon after an arrow push |
| 5. Conjugation required | Lone pair or pi bond must be adjacent to the accepting system | Pushing a lone pair across a saturated (sp3) carbon with no adjacent pi bond |
Violating rules 2 or 4 is the most common exam error. Check every new structure for octet compliance and correct arrow placement.
The Formal Charge Formula You Must Know
Formal charge determines which contributor ranks higher. The formula is:
FC = (valence electrons) - (non-bonding electrons) - (1/2 × bonding electrons)
For a neutral oxygen with one double bond and two lone pairs: FC = 6 - 4 - (1/2 × 4) = 0. For that same oxygen after accepting a negative charge with two single bonds and three lone pairs: FC = 6 - 6 - (1/2 × 2) = -1. Calculating this after every arrow push, rather than guessing from the drawing, prevents the most common ranking errors.
Curved-Arrow Conventions
Full curved arrows (two-headed) show movement of an electron pair. The tail sits on the source: a lone pair drawn as two dots or the bond line of a pi bond. The head points to the destination: an adjacent pi bond being broken or an empty orbital. Half-headed arrows (fishhook arrows) belong to radical mechanisms, not resonance. Using a half-arrow in a resonance drawing is always wrong.
You cannot push electrons through a sigma bond in a resonance structure. Delocalization requires a continuous system of p orbitals. If a saturated carbon (sp3, no pi bond, no adjacent empty orbital) sits between the source and destination, the electrons cannot flow. Students who try to draw resonance across a sp3 carbon produce invalid contributors that do not represent any real delocalization.
Example 1: The Carboxylate Ion (Acetate)
The carboxylate group (RCO2-) appears in amino acids, fatty acids, and countless organic reaction intermediates. It produces two equivalent resonance structures and requires no ranking decision, making it the standard starting point for resonance practice problems.
Step-by-Step Solution
Start with acetate drawn as CH3-C(=O)-O(-). Label the atoms: C1 is the methyl carbon, C2 is the carbonyl carbon, O1 carries the double bond, and O2 carries the negative charge with three lone pairs.
Arrow 1: Draw a curved arrow from one of O2's lone pairs toward the C2-O2 bond position. This will convert the C2-O2 single bond into a double bond.
Arrow 2: Draw a curved arrow from the C2=O1 pi bond toward O1. This breaks the double bond and places a new lone pair on O1, giving it the negative charge.
Result: The new structure has a C2=O2 double bond and a C2-O1 single bond with three lone pairs on O1 (charge -1). The two structures are mirror images in terms of electron distribution.
Formal charges (structure 1): O1 with double bond: FC = 6 - 4 - 2 = 0. O2 with single bond and three lone pairs: FC = 6 - 6 - 1 = -1. Both structures carry the same -1 overall charge, and neither has formal charges on carbon. The two contributors are equivalent, so no ranking is needed.
When two resonance structures are equivalent (same connectivity, same atom types carrying the same formal charges, just mirrored), the resonance delocalization is maximal and the molecule gains the most stabilization. The carboxylate is one of the most stabilized anions in organic chemistry precisely because its two contributors are perfectly equivalent. This stabilization is why carboxylic acids are far more acidic than alcohols.
Example 2: Benzene Delocalization
Benzene (C6H6) has two Kekule resonance structures, each showing alternating single and double bonds around the six-membered ring. Neither structure alone predicts the equal C-C bond lengths (all 140 pm) or the aromatic stabilization energy of around 150 kJ/mol reported by OpenStax Organic Chemistry, Chapter 15.
Drawing and Interpreting the Hybrid
Structure 1 places double bonds between C1-C2, C3-C4, and C5-C6. Structure 2 places them between C2-C3, C4-C5, and C6-C1. Both are valid; neither violates the octet rule; formal charges are zero on every carbon in both.
The hybrid shows a circle inside the ring rather than alternating bonds, representing the six electrons spread equally across all six carbon atoms. In an exam context, drawing both Kekule structures and labeling them as resonance contributors with a double-headed arrow earns full marks for the resonance portion of a question, even when the shorthand circle notation is used for subsequent steps.
The stabilization from this delocalization explains why benzene undergoes electrophilic aromatic substitution rather than addition. Adding across one double bond would produce a non-aromatic intermediate, losing the delocalization energy. Substitution preserves it.
Example 3: Nitrate Ion
Nitrate (NO3-) produces three equivalent resonance structures. The nitrogen atom carries a formal charge of +1 in each, and one oxygen carries -1 while two oxygens are neutral, cycling through all three positions across the three contributors.
Tracking Formal Charges Across All Three Structures
Starting from N at the center with one N=O double bond and two N-O single bonds, and the overall charge distributed as one oxygen carrying -1:
| Structure | N formal charge | O with double bond FC | Each single-bond O- FC |
|---|---|---|---|
| 1 | +1 | 0 | -1 (one of the two single-bond O atoms) |
| 2 (push lone pair from O2) | +1 | 0 (now O2) | -1 (now O1 or O3) |
| 3 (push lone pair from O3) | +1 | 0 (now O3) | -1 (now O1 or O2) |
Nitrogen at +1 in all three contributors. Each oxygen cycles through carrying -1, confirming the three structures are equivalent. Overall charge remains -1 across all.
The nitrogen formal charge of +1 can surprise students who expect a neutral nitrogen. It is correct: nitrogen in nitrate has four bonds (one double, two single), giving FC = 5 - 0 - 4 = +1. This does not mean nitrate is unstable; the overall charge of the ion is still -1, and the three equivalent contributors provide strong delocalization. The+1 on nitrogen is an artifact of the formal charge convention, not a prediction of instability.
Example 4: The Allyl Cation
The allyl cation (CH2=CH-CH2+) demonstrates resonance involving an empty p orbital rather than a lone pair. The empty orbital on the terminal carbon accepts electrons from the adjacent pi bond in the same way a pi bond accepts electrons from an adjacent lone pair.
Why All Three Contributors Are Equal Weight
Starting structure: CH2=CH-CH2+ (double bond between C1 and C2; positive charge as empty orbital on C3).
Arrow: Draw a curved arrow from the C1=C2 pi bond (midpoint of the bond) toward C3. This breaks the pi bond and moves electron density onto C3.
Result: C1 now bears the positive charge (empty orbital), the C2-C3 bond becomes a double bond, and C3 is neutral.
Formal charges: Structure 1: C1 = 0 (in double bond), C2 = 0, C3 = +1 (empty p orbital, FC = 4 - 0 - 3 = +1). Structure 2: C1 = +1, C2 = 0, C3 = 0. The two structures are mirror images through C2, so they are equivalent contributors.
The allyl cation is significantly more stable than a simple secondary carbocation because the positive charge is delocalized over two carbons. This delocalization is why allylic substrates react faster in SN1 pathways. For a deeper look at carbocation stability and substitution mechanism selection, the SN1 and SN2 worked examples post covers how resonance stabilization shifts the mechanism decision.
The allyl anion (CH2=CH-CH2-) produces two equivalent contributors by the same logic: the lone pair on C3 delocalizes into the C1=C2 pi bond, placing the negative charge on C1. Both the cation and anion gain stability through delocalization. Allylic anions are important intermediates in enolate chemistry and in reactions involving alpha-carbon deprotonation adjacent to pi systems.
Example 5: The Enolate (Exam Level)
The enolate ion (formed by deprotonating the alpha carbon of a carbonyl compound) produces two resonance structures that are not equivalent in energy. This is where ranking major and minor contributors becomes the exam skill. OpenStax Organic Chemistry Chapter 22 covers enolate reactivity in detail; the resonance treatment here focuses on the structure-ranking step.
Ranking Major vs Minor Contributors
Start with the enolate drawn as CH3-C(=O)-CH2- (acetaldehyde deprotonated at the alpha carbon, overall charge -1). The negative charge sits on C2 (the alpha carbon) in the starting structure.
Arrow: Draw a curved arrow from the C-C2 pi bond (which will form after the lone pair on C2 delocalizes) toward the carbonyl oxygen. More precisely: the lone pair on C2 becomes a pi bond C1=C2, and simultaneously the C1=O pi bond breaks to place a lone pair on O, giving O the negative charge.
Structure A (starting): C2 bears the negative charge, C1=O double bond intact. Formal charges: C2 = -1, O = 0, C1 = 0.
Structure B (after arrow): O bears the negative charge, C1=C2 double bond formed (enol system), C1-O single bond. Formal charges: O = -1, C1 = 0, C2 = 0.
Ranking: Structure B is the major contributor. Oxygen (electronegativity 3.44) stabilizes a negative charge far better than carbon (electronegativity 2.55). Structure A, with the carbanion on carbon, is the minor contributor.
| Criterion | Structure A (C- negative) | Structure B (O- negative) | Winner |
|---|---|---|---|
| Formal charges | C at -1 | O at -1 | Tie (same count) |
| Electronegative atom bears charge | No: carbon at -1 | Yes: oxygen at -1 | Structure B |
| Octet satisfaction | All atoms have octets | All atoms have octets | Tie |
| Overall ranking | Minor contributor | Major contributor | Structure B |
Both structures have one formal charge of -1. The tiebreaker is the atom type: oxygen stabilizes negative charge better than carbon, so Structure B is the major contributor.
The practical consequence: enolate anions react at oxygen with hard electrophiles (O-alkylation) and at carbon with soft electrophiles (C-alkylation). Drawing both structures before predicting the site of reaction is the standard method taught in advanced organic chemistry courses. A student who draws only Structure A misses the oxygen nucleophile entirely.
The Four Errors That Cost the Most Marks
These four errors appear in organic chemistry resonance structures practice problems across universities. All four can be caught in under two minutes if you run the check list below after drawing each structure.
| Error | What happens | How to catch it |
|---|---|---|
| Moving an atom | A hydrogen migrates to a new position between two structures | Verify atom count and connectivity match the original in every new structure |
| Exceeding the octet on carbon | An arrow push gives carbon five bonds | Count bonds on every carbon after drawing; five bonds on C = invalid |
| Arrow tail on atom symbol rather than lone pair | The arrow has no clear electron source | Lone pairs must be drawn explicitly before placing an arrow tail on them |
| Forgetting to recalculate formal charges | Wrong ranking of major vs minor contributor | Recalculate FC on every atom that changed bonding after the arrow push |
The first two errors produce invalid structures; the last two produce valid but misranked structures. Both cost marks on an exam.
The check list after every arrow: count total electrons (must match original), check carbon bond count (max 4), check octet on all second-period atoms, then calculate formal charge on any atom that changed. Running this as a habit on resonance structures practice problems in your first few weeks removes most errors before they appear on an assessment.
For subject-specific tools and calculators that support chemistry and other science modules, the subject calculators hub has tools across chemistry, physics, mathematics, and statistics. The university resources hub covers the full range of academic tools available.
For more worked examples in adjacent subject-mastery topics, see the step-by-step walkthroughs in equilibrium and Le Chatelier worked examples and conservation of energy worked examples. Both follow the same show-every-step method used here.
If you want to work through resonance problems interactively with immediate feedback on your arrow pushing and formal charge calculations:
Key Takeaways
- Resonance structures examples show that only electrons move between contributors; atoms hold their positions. The real molecule is the single hybrid species, not a mixture of the drawn structures.
- Every curved arrow has its tail on a lone pair or pi bond (the electron source) and its head on an adjacent empty orbital or pi bond (the acceptor). Arrows cannot cross sp3 carbons that lack adjacent pi systems.
- Formal charge (FC = valence electrons minus non-bonding electrons minus half of bonding electrons) must be recalculated on every atom that changes bonding. Skipping this step produces wrong rankings.
- When all contributors are equivalent (carboxylate, benzene, nitrate), no ranking is needed. When contributors differ (enolate, amide), rank by: fewer formal charges beats more; negative charges on O beats N beats C; complete octets beat incomplete octets.
- The enolate has two inequivalent contributors. The O-negative structure (major) and the C-negative structure (minor) both represent real nucleophilic sites, explaining why enolates react at both oxygen and alpha carbon depending on the electrophile hardness.
- The four errors that cost the most marks are: moving an atom, giving carbon five bonds, misplacing the arrow tail, and failing to update formal charges. Running a four-point check after each arrow push catches all four before they reach an exam paper.
- Resonance delocalization explains measurable physical properties: equal C-O bond lengths in carboxylate, aromatic stabilization in benzene, and the acidity difference between carboxylic acids and alcohols.
