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Organic Chemistry I · Structure and Bonding

Resonance structures practice, with arrow pushing and answers

Resonance structures are different ways of drawing the same molecule, where only electrons have moved. You push lone pairs and π bonds with curved arrows. Atoms and σ bonds stay exactly where they are. Get that rule right, add formal charges, and you can tell which contributor matters most.

Already know the basics? Jump to the 8 practice questions or the worked examples.

The rules: only electrons move

Resonance structures are different Lewis drawings of the same molecule. The real molecule is a blend of all of them, called the resonance hybrid. It doesn't flip back and forth between them. You connect contributors with a double-headed arrow (↔), never with equilibrium arrows.

  1. Atoms never move. Every atom keeps the same position and the same neighbors. If a hydrogen has moved, you've drawn a different compound (a tautomer or an isomer), not a resonance structure.
  2. Only π electrons and lone pairs move. A curved arrow starts at a lone pair or a π bond. It ends either between two atoms (making a new π bond) or on an atom (making a lone pair). Single (σ) bonds are never broken in a resonance arrow.
  3. Don't break the octet rule. C, N, and O can have eight valence electrons at most. A carbon with five bonds or a nitrogen with five bonds is never a valid contributor. Fewer than eight is allowed (a carbocation has six), but it costs stability.
  4. The overall charge stays the same. Add up the formal charges in every contributor. The total must match the starting structure.

Formal charge is the bookkeeping that keeps you honest: formal charge = valence electrons − nonbonding electrons − ½(bonding electrons). Check every atom whose bonds or lone pairs changed.

Formal charges you'll see constantly
AtomBondsLone pairsFormal charge
C400
C30+1 (carbocation)
C31−1 (carbanion)
N310
N40+1
N22−1
O220
O31+1
O13−1

Count a double bond as two bonds and a triple bond as three.

Four patterns that generate almost every resonance structure

You don't have to guess where to push arrows. Look for these setups. Each one is a group of three atoms (or a ring) with p orbitals lined up next to each other.

Resonance patterns and what the arrows do
PatternArrowsExample
Lone pair next to a π bond (allylic lone pair)Lone pair → new π bond; old π bond → lone pair on the far atomCarboxylate, enolate, amide
Positive charge next to a π bond (allylic carbocation)π bond → toward the positive carbon. The charge moves to the far end.CH₂=CH–CH₂⁺ ↔ ⁺CH₂–CH=CH₂
π bond between atoms of different electronegativityπ bond → lone pair on the more electronegative atom, leaving a positive charge behindC=O ↔ C⁺–O⁻ (a minor contributor)
Alternating π bonds in a ringEvery π bond shifts one position around the ring at the same timeBenzene's two Kekulé structures

Notice the allylic carbocation has no lone pair to push. The arrow starts at the π bond, because arrows always start at electrons, never at a positive charge. In the allyl cation the positive charge is shared between the two end carbons. The middle carbon never carries it.

Which resonance contributor matters most?

Contributors don't all count equally. The hybrid looks most like the most stable one. Rank them with these checks, in this order:

  1. Full octets win. A contributor where every C, N, and O has an octet beats one with an electron-poor atom, even if that means putting a positive charge on oxygen or nitrogen.
  2. Fewer formal charges win. A neutral contributor beats one with separated + and − charges.
  3. Charges should sit on the right atoms. Negative charge belongs on the more electronegative atom (O over C). Positive charge belongs on the less electronegative one.
  4. Equivalent contributors count equally. When two structures look the same, like the two halves of a carboxylate, the hybrid is an exact average, and the extra stability is largest.

Why it matters in your course: resonance explains why carboxylic acids are more acidic than alcohols, why amides are planar and poor bases, why allylic and benzylic carbocations form easily in SN1 reactions, and why benzene is so stable.

Worked examples

Example 1: acetate (carboxylate)

Draw the second resonance structure of acetate, CH₃–C(=O)–O⁻, and describe the hybrid.

  1. Find the pattern: the O⁻ has a lone pair right next to the C=O π bond (lone pair next to a π bond).
  2. Arrow 1: a lone pair on O⁻ moves in to form a C=O π bond.
  3. Arrow 2: the original C=O π bond moves onto its oxygen as a lone pair. Carbon never has more than eight electrons.
  4. Result: CH₃–C(–O⁻)=O. Same atoms, same total charge (−1), and the two contributors are equivalent.

Answer: CH₃–C(=O)–O⁻ ↔ CH₃–C(–O⁻)=O. Both C–O bonds are identical, between single and double, and each oxygen carries about half the negative charge.

Example 2: the acetone enolate (which contributor is major?)

The enolate of acetone can be drawn as CH₂=C(CH₃)–O⁻. Draw the other contributor and decide which one matters more.

  1. Pattern: the O⁻ lone pair sits next to the C=C π bond.
  2. Arrow 1: an O lone pair forms a C=O π bond. Arrow 2: the C=C π bond moves onto the CH₂ carbon as a lone pair.
  3. Result: ⁻CH₂–C(CH₃)=O, with the negative charge on carbon.
  4. Compare: both contributors have full octets and one charge. Oxygen is more electronegative than carbon, so it handles the negative charge better.

Answer: CH₂=C(CH₃)–O⁻ ↔ ⁻CH₂–C(CH₃)=O. The contributor with the charge on oxygen is major. The carbon contributor still matters, because it explains why enolates react at carbon.

Example 3: an amide

Draw the second resonance structure of acetamide, CH₃–C(=O)–NH₂, and explain what it says about the C–N bond.

  1. Pattern: nitrogen's lone pair is next to the C=O π bond.
  2. Arrow 1: N's lone pair forms a C=N π bond. Arrow 2: the C=O π bond moves onto oxygen.
  3. Formal charges: N now has 4 bonds and no lone pair (+1). O has 1 bond and 3 lone pairs (−1).
  4. Rank: the first structure has no charges, so it's major. The charge-separated structure is minor, but every atom still has an octet, so it contributes noticeably.

Answer: CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺. The C–N bond has partial double-bond character, so the amide group is planar, rotation around C–N is restricted, and the nitrogen is a very weak base.

Example 4: nitromethane formal charges

Nitromethane is CH₃NO₂, with nitrogen bonded to the carbon and both oxygens. Draw its two contributors and assign formal charges.

  1. Try a neutral structure with two N=O double bonds. Nitrogen would have five bonds (10 electrons), so that structure is invalid.
  2. Valid structure: CH₃–N(=O)–O. N has 4 bonds, no lone pair: 5 − 0 − 4 = +1. The double-bonded O: 6 − 4 − 2 = 0. The single-bonded O has 3 lone pairs: 6 − 6 − 1 = −1.
  3. Arrows: a lone pair on O⁻ forms an N=O π bond, while the other N=O π bond moves onto its oxygen.
  4. Check: total charge is +1 − 1 = 0 in both contributors.

Answer: CH₃–N⁺(=O)–O⁻ ↔ CH₃–N⁺(–O⁻)=O. The two are equivalent, so both N–O bonds are the same length. Nitrogen is +1 in both.

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Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 1Which pair is not a pair of resonance structures?

Show solution

Resonance structures differ only in where electrons are placed. In the keto and enol forms, a hydrogen atom has moved and a σ bond has changed. They're two real compounds in equilibrium (tautomers), which is why they're connected by equilibrium arrows, not ↔.

Question 2In the ozone Lewis structure where the central oxygen has one double bond, one single bond, and one lone pair, what is the central oxygen's formal charge?

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Central O: 6 − 2 − ½(6) = +1. The end oxygen with a double bond is 6 − 4 − 2 = 0, and the end oxygen with a single bond is 6 − 6 − 1 = −1. Ozone is O=O⁺–O⁻ ↔ ⁻O–O⁺=O, with a total charge of 0.

Question 3Which contributor contributes most to the acetone enolate hybrid?

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Same number of charges, octets everywhere, so compare where the negative charge sits. Oxygen is more electronegative than carbon, so the oxygen-centered contributor is major. The carbon-centered one is minor but explains why enolates react at carbon.

Question 4Which statement about the amide group in acetamide, CH₃C(=O)NH₂, is true?

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CH₃C(=O)NH₂ ↔ CH₃C(–O⁻)=NH₂⁺. The neutral contributor is major, but the second one matters. It gives the C–N bond partial double-bond character, so the amide group is planar with restricted rotation, and nitrogen's lone pair is much less available.

Question 5Which curved-arrow description correctly converts CH₂=CH–CH₂⁺ into its other resonance structure?

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One arrow: tail on the C1=C2 π bond, head between C2 and C3. The new structure is ⁺CH₂–CH=CH₂. The positive charge is shared by C1 and C3, never C2.

Question 6Which of these is not a valid resonance contributor of nitromethane, CH₃NO₂?

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A second-row atom can't exceed eight valence electrons. The neutral drawing gives nitrogen five bonds, so it isn't allowed. The two valid, equivalent contributors put +1 on N and −1 on one oxygen.

Question 7How many equivalent resonance contributors does the carbonate ion, CO₃²⁻, have?

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Each contributor has one C=O and two C–O⁻. Rotating the double bond among the three oxygens gives three equivalent structures. In the hybrid, every C–O bond is the same and each oxygen carries −2/3.

Question 8For the cation CH₃–O–CH₂⁺, which contributor is more important?

Show solution

Arrow: a lone pair on O forms a C=O π bond. In CH₃–O⁺=CH₂, oxygen has 3 bonds and 1 lone pair (+1), and every atom has an octet. Because the octet rule outranks electronegativity, this contributor is major. It's why oxygen next to a carbocation stabilizes it so strongly.

Answer key

  1. C, acetone and its enol: an atom moved, so they're tautomers.
  2. D, +1: 6 − 2 − 3 = +1.
  3. B, CH₂=C(CH₃)–O⁻: negative charge on the more electronegative atom.
  4. A: amide resonance gives C–N partial double-bond character.
  5. B: the π bond moves toward the cationic carbon.
  6. B, CH₃–N(=O)=O: nitrogen would have ten electrons.
  7. C, 3: the double bond can sit on any of the three oxygens.
  8. A, CH₃–O⁺=CH₂: full octets come first.

Frequently asked questions

What is the difference between resonance structures and isomers?

Resonance structures have the same atoms in the same places and differ only in where electrons sit. They describe one real molecule. Isomers, including tautomers like keto and enol forms, have atoms in different places and are different compounds.

Is resonance on the MCAT?

Yes. Resonance, formal charge, and delocalization appear in the AAMC MCAT content outline under bonding and organic chemistry. They show up mostly as tools for explaining acidity, stability, and reactivity, such as why carboxylic acids are acidic or why amides are unreactive.

Does a molecule switch back and forth between resonance structures?

No. The molecule is always the hybrid, a single structure with electrons spread over several atoms. The individual contributors are just drawings, because one Lewis structure can't show delocalized electrons.

Why can't I break a single bond in a resonance structure?

Breaking a sigma bond would separate atoms or change which atoms are connected, and that makes a different species. Resonance only moves pi electrons and lone pairs, which are spread over p orbitals lined up next to each other.

Sources

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