Source: Lecture notes / Organic Chemistry (The Ohio State University)
Tags: resonance problem solving, drawing resonance structures, resonance patterns, lone pair to pi bond, pi bond to lone pair, resonance tips, hybrid structures, organic chemistry
Difficulty: Intermediate Prerequisites: Resonance Fundamentals study notes (Part 1), Lewis structures, formal charge, curved arrow notation.
Knowing the theory of resonance is one thing; drawing correct resonance structures under exam pressure is another. This set of notes focuses on the practical side: the two core electron-movement patterns, a step-by-step method for drawing structures, and worked examples showing how to identify the major resonance contributor (MRC) and sketch the hybrid. If you have read Part 1 (Resonance Fundamentals), you have the conceptual foundation. This is where you learn to apply it.
There are two main patterns in resonance: lone pairs becoming pi bonds, and pi bonds becoming lone pairs (or new pi bonds). To draw resonance structures reliably, always start where the electrons are, push arrows in a consistent direction, check your formal charges, and verify your Lewis structures at the end. Most exam mistakes come from moving sigma bonds, losing track of arrow count, or skipping the formal charge check.
Lone pair to pi bond pattern
A resonance pattern where a lone pair on one atom donates into an adjacent bond position, forming a new pi bond while the existing pi bond shifts or a charge changes. Think of it as the lone pair "joining" the pi system.
Pi bond to lone pair pattern
A resonance pattern where a pi bond breaks, and its electrons become a lone pair on one of the bonded atoms. The reverse of the first pattern. Together these two patterns account for nearly every resonance arrow you will draw.
Hybrid structure
The averaged representation of all resonance contributors, showing partial bonds (dotted lines) and partial charges (δ+, δ−) rather than full bonds and full formal charges. This is what the molecule actually looks like.
Equal MRCs
Resonance structures that are identical in energy because they are mirror images of each other or have the same pattern of charges and bonding. When contributors are equal, both contribute equally to the hybrid.
Allylic system
A three-atom arrangement where a p orbital (bearing a lone pair, radical, or charge) is adjacent to a pi bond. This is the minimal unit for resonance. In simple terms, "allylic" means "next door to a double bond."
Nearly every resonance structure you draw will use one of two electron movements. Learning to recognise them quickly is the single most useful exam skill for this topic.
Pattern 1: Lone pairs become pi bonds
An atom with a lone pair donates two electrons into the adjacent bond, forming a new pi bond. This typically places a positive formal charge on the donating atom (it gave up electrons) and removes a positive charge or adds a negative charge on the receiving end.
Typical scenario: a lone pair on O, N, or a carbanion sits next to a C that is either part of a pi bond or bears a positive charge. The lone pair flows in, creating a new pi bond.
Pattern 2: Pi bonds become lone pairs (or shift to new pi bonds)
A pi bond breaks and its electrons either become a lone pair on one atom or shift over to form a new pi bond on the adjacent atoms. This is how electron density "walks" along a conjugated chain.
Typical scenario: in a conjugated system (alternating single and double bonds), the pi electrons shift one position over, converting one double bond into a single bond and one single bond into a double bond. At the end of the chain, the electrons may become a lone pair.
These two patterns often occur together in the same structure. For example, in an amide (like formamide), the nitrogen lone pair donates into the C–N bond (Pattern 1), while the C=O pi bond breaks and the electrons become a lone pair on oxygen (Pattern 2).
These tips come directly from the source material and represent the method that avoids the most common errors.
1. Identify the issue first.
Before you start pushing arrows, ask: why am I drawing resonance here? Is there a lone pair next to a pi bond? A positive charge adjacent to electron density? A polarised pi bond? Knowing the trigger tells you where to start.
2. Start where the electrons are.
Curved arrows always begin at a source of electron density: a lone pair or a pi bond. They do not start from a positive charge or from empty space. The arrow tail sits on the electrons; the arrow head points to where those electrons are going.
3. Arrows flow in the same direction.
All curved arrows in a single resonance step should push electrons in one consistent direction (electron-rich to electron-poor, or along the conjugated chain). Arrows pointing in opposite directions within the same step usually signal an error.
4. Same number of arrows each time.
If you used two curved arrows to get from structure A to structure B, you should use two curved arrows to get from B to C (if the system allows it). Consistency in arrow count is a good self-check.
5. Always move one atom or one bond away.
Electrons never skip atoms or teleport across a molecule. Each arrow must land on an atom or bond that is directly adjacent to the source. If you find yourself drawing an arrow across two bonds, something has gone wrong.
6. Check formal charges and Lewis structures.
After drawing the new structure, count the electrons on every atom that changed. Verify formal charges. Verify that no second-row atom exceeds an octet. This is the step students skip under time pressure, and it is the step that catches the most errors.
7. Remember the point of what you are doing.
Resonance structures explain electron distribution and reactivity. If your new structure does not help you understand where electron density sits, step back and reconsider. Are you drawing resonance, or have you accidentally drawn a reaction?
The following examples walk through the reasoning you would use on an exam. Each uses the three MRC criteria from Part 1 (octets, electronegativity of charged atoms, charge separation).
Example 1: Nitrile-containing system (two equivalent structures)
Two resonance structures of a nitrile-adjacent anion may be equal MRCs if the charge sits on atoms of equal electronegativity and both have full octets. In that case, both contribute equally to the hybrid.
For equal MRCs, the hybrid shows the charge and bond order evenly split between the two positions. Dotted lines replace the bonds that differ between the two forms, and each atom that shares the charge gets a δ− label.
Example 2: Amide nitrogen donating into a carbonyl
In an amide, nitrogen's lone pair can donate into the C–N bond while the C=O pi electrons shift onto oxygen. This gives two resonance structures:
Structure A: C=O double bond, C–N single bond, lone pair on N.
Structure B: C–O single bond (with negative charge on O), C=N double bond (with positive charge on N).
MRC determination: Structure A is the MRC. It has no formal charges (least charge separation wins) and all atoms still have octets. Structure B has charge separation, which costs energy.
The hybrid shows partial double-bond character in both C–O and C–N bonds, a partial negative charge on O, and a partial positive charge on N. This explains why amide C–N bonds are shorter and more rigid than typical C–N single bonds, and why amide nitrogen is a poor base compared with amine nitrogen.
Example 3: Carbocation adjacent to a lone pair
A carbocation (positive charge on carbon) next to an atom bearing a lone pair (e.g. nitrogen with NH₂ or oxygen with a lone pair) is stabilised by resonance. The lone pair donates to form a pi bond with the carbocation.
Structure A: positive charge on C, lone pair on the adjacent heteroatom.
Structure B: pi bond between C and the heteroatom, positive charge shifts to the heteroatom.
MRC determination: if the heteroatom achieves an octet by donating the lone pair (and the carbon also gets an octet), the form with the most complete octets is the MRC. Typically the form where every atom has an octet is the MRC, even if that means placing a positive charge on nitrogen or oxygen.
The hybrid shows partial positive charge on both carbon and the heteroatom, with a partial pi bond between them.
Example 4: Conjugated diene with a charged end
In a longer conjugated system (four or more atoms in a row of alternating single and double bonds), pi electrons can shift all the way along the chain. Each intermediate form is a valid resonance structure.
When the two end structures are mirror images (e.g. in a symmetric pentadienyl cation), they are equal MRCs. The hybrid distributes the charge symmetrically.
When they are not symmetric, rank them using the usual three criteria. The MRC with the best octet count and least charge separation dominates.
Example 5: Electronegativity-based MRC (e.g. enolate)
In an enolate ion, two resonance structures place the negative charge on different atoms:
One form has the charge on carbon (carbanion).
The other has the charge on oxygen (alkoxide).
Both have complete octets. The form with negative charge on oxygen is the MRC because oxygen is more electronegative than carbon and stabilises negative charge better. The hybrid has more electron density on oxygen than on carbon, but the carbon still has significant negative character, which is why enolates react at carbon in many reactions.
"I should draw as many resonance structures as possible." Draw only valid structures. An invalid resonance form (one that breaks an octet for a second-row element without necessity, or moves a sigma bond) does not count and will lose marks.
"Curved arrows start from the positive charge." They do not. Arrows always start from electrons (a lone pair or a pi bond) and point toward where those electrons are going. The positive charge is the destination, not the origin.
"If the atom has a lone pair and is next to a pi bond, resonance always happens." Check that the lone pair is in a p orbital aligned with the pi system. An sp3 lone pair perpendicular to the pi system, or an sp2 lone pair in the plane of the ring (as in pyridine), does not participate.
"All resonance structures contribute equally." Only structures that are equivalent by symmetry contribute equally. Non-equivalent structures contribute unequally, with the MRC contributing the most.
⚠️ "Draw all resonance structures and identify the MRC" is one of the most common exam question types for this unit. Practise it repeatedly with different functional groups.
⚠️ You will be asked to draw the resonance hybrid. Remember to use dotted lines for delocalised bonds and δ+/δ− for partial charges.
⚠️ Arrow direction is marked. Arrows that start from a positive charge, skip atoms, or point in inconsistent directions will cost marks even if the final structure is correct.
⚠️ Resonance problem solving feeds directly into later topics: it determines the regiochemistry of electrophilic aromatic substitution, the acidity of carboxylic acids versus alcohols, and the stability of intermediates in nearly every mechanism.
1. True or false: curved arrows in resonance always start from an electron source (lone pair or pi bond).
True.
2. Fill in the blank: the two core resonance patterns are (1) lone pairs become ______ and (2) pi bonds become ______ or shift to new pi bonds.
(1) pi bonds, (2) lone pairs.
3. True or false: in an amide, the resonance structure with no formal charges is the major resonance contributor.
True. The form with no charge separation (C=O intact, lone pair on N) is the MRC.
4. Fill in the blank: when two resonance structures are mirror images and identical in energy, they are called ______ MRCs.
Equal.
5. True or false: if you need three curved arrows to go from structure A to structure B, you should expect to need three arrows for each subsequent resonance step in that system.
True, as a general self-check. The number of arrows should stay consistent within the same system.
Q: You are given a molecule with a lone pair on oxygen directly adjacent to a C=C double bond. Describe the resonance pattern you would use and what the second resonance structure looks like.
A: This is Pattern 1 (lone pair becomes a pi bond). The oxygen lone pair donates into the O–C bond, forming a new pi bond (O=C). Simultaneously, the C=C pi bond breaks and its electrons become a lone pair or shift further down the chain. In the new structure, oxygen bears a positive formal charge (it donated electrons) and the terminal carbon bears a negative charge or gains a lone pair.
Q: In a set of three resonance structures, structure A has all octets and no formal charges, structure B has all octets but charge separation, and structure C has an incomplete octet on carbon. Rank them from greatest to least contribution to the hybrid.
A: A > B > C. Structure A dominates (no charges, full octets). Structure B contributes less (charge separation costs energy). Structure C contributes least (incomplete octet is the worst penalty).
Q: What is the most common mistake students make when drawing curved arrows for resonance?
A: Starting the arrow from a positive charge or an atom without electrons, rather than from a lone pair or pi bond. The arrow must originate where the electrons are.
Q: Why is the C–N bond in an amide shorter and more rigid than a typical C–N single bond?
A: Resonance gives the C–N bond partial double-bond character. The nitrogen lone pair partially delocalises into the C–N bond (forming a partial pi bond), which shortens the bond and restricts rotation around it.
Q: An enolate has two resonance structures. One places the negative charge on carbon, the other on oxygen. Which atom does the enolate react at, and why might this seem contradictory given the MRC?
A: The MRC places the charge on oxygen (more electronegative), but enolates frequently react at carbon. This is because the carbon site, while a minor resonance contributor, is a softer nucleophile and reacts preferentially with many electrophiles (especially in C–C bond-forming reactions). The MRC tells you about electron density distribution, not necessarily about which atom is the reactive site in every scenario.
The problem-solving patterns here apply directly to electrophilic aromatic substitution (drawing resonance structures of the arenium ion intermediate to predict ortho/meta/para selectivity), to enolate chemistry (understanding which atom reacts), and to radical chemistry (where single-headed arrows and radical resonance stabilisation follow the same logic). If you are comfortable with the two core patterns and the MRC criteria, you have the toolkit for every resonance-based exam question in the rest of the course.
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