Difficulty: Intermediate | Prerequisites: Lewis structures, formal charge, lone pairs, electronegativity
Resonance and curved arrows are the core language of organic reaction mechanisms. Resonance explains why electrons in a molecule are not always confined to one bond or one atom; they can be spread (delocalised) across several atoms, which affects stability, acidity, bond lengths, and reactivity. Curved arrows show where electrons move during a chemical reaction, step by step. Together, these two tools let you predict products, explain selectivity, and understand why reactions happen the way they do. Every subsequent chapter in the course uses them.
Resonance structures are different Lewis structures for the same molecule that differ only in electron placement (not atom positions). The major contributor is the one with the most bonds, fewest formal charges, and negative charges on electronegative atoms. Curved arrows always flow from an electron source (lone pair or bond) toward an electron sink (an atom or bond that can accept them). In Bronsted acid-base reactions, the base's lone pair attacks the acidic proton, and the bonding electrons of the H-X bond move onto X.
Resonance structures (resonance contributors)
Two or more valid Lewis structures for the same molecule that differ only in the placement of electrons (not atoms). The real molecule is a weighted average (hybrid) of all contributors.
Think of them as different "snapshots" of where the electrons could be. The true molecule is a blend of all of them.
Major resonance contributor
The resonance structure that contributes most to the actual electronic character of the molecule. It is the most stable contributor, determined by rules of thumb (see Core Content below).
Resonance hybrid
The real molecule, which is a weighted average of all resonance structures. It is not any single contributor; it is the blend.
Curved arrow (electron-pushing arrow)
A notation showing the movement of a pair of electrons from a source (lone pair or bond) to a sink (atom or bond). The tail starts at the electrons; the head points to where they go.
In simple terms, the arrow is a pair of electrons moving from where they are to where they end up.
Bronsted acid
A proton (H+) donor.
Bronsted base
A proton (H+) acceptor. The base uses a lone pair to form a new bond with the proton.
Nucleophile
An electron-rich species that donates a pair of electrons to form a new bond. In Bronsted acid-base chemistry, the base acts as the nucleophile attacking the proton.
Electrophile
An electron-poor species that accepts a pair of electrons. The acidic proton in an acid-base reaction is the electrophile.
Leaving group
The species that departs with a pair of electrons when a bond breaks. In the reaction of an alcohol with HCl, water acts as the leaving group after protonation.
Only electrons move. Atoms stay in the same positions. If you have to move an atom, you are drawing a constitutional isomer, not a resonance structure.
Curved arrows show where the electrons go: tail at the source (lone pair or pi bond), head at the destination.
Every resonance structure must be a valid Lewis structure (correct number of bonds for each atom, no more than 8 electrons on second-row elements).
Common electron movements: lone pair into an adjacent bond (creating a double bond), pi bond electrons moving to an adjacent atom (becoming a lone pair).
Rank contributors using these criteria (in order of importance):
Most bonds wins. A structure with more covalent bonds is more stable than one with fewer.
Fewest formal charges wins. A structure with no formal charges is better than one with separated charges.
Negative charges on electronegative atoms. If formal charges are unavoidable, the structure where the negative charge sits on the most electronegative atom is preferred.
All atoms have octets. A contributor where every atom (especially C, N, O) has a full octet beats one with an incomplete octet.
The exam showed a nitrosoamine with three resonance contributors:
Contributor 1 (given): N=O with a formal positive charge on N and formal negative on O, plus a lone pair on the second N.
Contributor 2: The lone pair on the second nitrogen donates into the N-N bond, forming a double bond there. The N=O double bond breaks to a single bond, pushing electrons onto oxygen.
Contributor 3: Further delocalisation creates an N=N double bond with a positive charge on the second nitrogen and a negative charge on oxygen.
The major contributor is the one with the most bonds and the negative charge on the most electronegative atom (oxygen). From the exam, the circled structure was the one with the fewest formal charges or the one where oxygen carries the negative charge and nitrogen the positive.
Atoms involved in resonance must have p orbitals available to overlap. This means they are sp2 or sp hybridized.
In the nitrosoamine example, both nitrogen atoms are sp2 hybridized. The lone pair on the second nitrogen sits in a p orbital that overlaps with the pi system.
Arrows always flow from electron-rich to electron-poor: from a lone pair or a bond (the source) toward an atom that can accept electrons (the sink).
A full curved arrow represents the movement of two electrons (a pair).
A half-headed arrow (fishhook) represents one electron. For this exam, you only need full arrows.
Every arrow must start from electrons (a lone pair or a bond) and end at an atom or between two atoms (forming a new bond).
The exam showed a three-step reaction of benzyl alcohol with HCl to form benzyl chloride:
Step 1: Protonation (Bronsted acid-base)
The oxygen lone pair of the alcohol attacks the H of HCl (curved arrow from O lone pair to H).
Simultaneously, the H-Cl bond breaks, and the electrons go to Cl (curved arrow from the H-Cl bond to Cl).
Products: protonated alcohol (oxonium ion, with +1 charge on O) and Cl- (with -1 charge).
Step 2: Loss of leaving group
Water departs as a leaving group. The electrons in the C-O bond move onto the oxygen (curved arrow from C-O bond to O).
This generates a carbocation (C with +1 charge) and water.
Step 3: Nucleophilic attack
Cl- (the nucleophile) donates a lone pair to the carbocation (curved arrow from Cl lone pair to C+).
This forms the new C-Cl bond, giving the final product: benzyl chloride.
Always draw arrows starting from the electron source. An arrow from Cl to C means Cl's electrons are forming the new bond.
An arrow pointing in the wrong direction (from the electron-poor atom to the electron-rich atom) is a common error and earns zero credit.
Show formal charges on every intermediate. If you protonate an oxygen, show the +1 charge. If Cl leaves as Cl-, show the -1.
Each step should balance: the total charge on the left equals the total charge on the right.
Curved arrows are used to explain every organic reaction mechanism you will encounter. This includes substitution (SN1, SN2), elimination (E1, E2), addition reactions, and more complex multi-step syntheses. Learning to push arrows correctly now saves enormous time later.
Students often draw curved arrows pointing from the electrophile to the nucleophile (the wrong direction). Arrows always go from the electron source to the electron sink. If the arrow starts at the proton, it is backwards.
Confusing resonance structures with equilibrium. Resonance structures are not different molecules in equilibrium. They are different descriptions of one molecule. The double-headed resonance arrow (↔) is not the same as equilibrium arrows (⇌).
Forgetting to include formal charges on resonance structures. If you move electrons but do not update the formal charges, the structure is incomplete and typically marked wrong.
Assuming the structure with the most separated charges is the major contributor. The opposite is usually true: fewer formal charges means more stable, and therefore a bigger contributor.
⚠️ Resonance questions are worth 17 points on this exam (the second-highest section). You must be able to draw all valid contributors and identify the major one.
⚠️ Curved arrow direction is strictly graded. An arrow pointing the wrong way earns -1 per arrow, and on a multi-step mechanism that adds up fast.
⚠️ The hybridization of atoms in resonance systems is a common follow-up question. If an atom participates in resonance (donates or accepts pi electrons), it is sp2.
⚠️ Multi-step mechanisms (like the three-step alcohol-to-chloride reaction) require you to show every intermediate with correct formal charges.
True or False: Resonance structures differ in the positions of atoms.
False. Only electrons move between resonance structures; atoms stay put.
Fill in the blank: The major resonance contributor is the one with the most ______ and the fewest ______.
Bonds; formal charges.
True or False: A curved arrow can start at a hydrogen atom with no lone pair.
False. Arrows must start from electrons (a lone pair or a bond), not from an atom with no available electrons.
Fill in the blank: In a Bronsted acid-base reaction, the curved arrow for proton transfer starts at the ______ of the base.
Lone pair.
True or False: Both nitrogen atoms in the nitrosoamine are sp2 hybridized.
True. Both participate in the pi system and have three electron groups.
Q: A nitrosoamine has the structure O=N-N(CH3)2 with a negative formal charge on O and a positive on the first N. Draw the two other resonance contributors.
A: Contributor 2: Move the lone pair from the second N into the N-N bond (making it a double bond) while breaking the N=O double bond to a single bond (electrons go to O). This gives N-N=double bond, O with two lone pairs and -1 charge, second N with +1 charge. Contributor 3: Further push electrons so that the second N donates fully, giving a structure with a C-N single bond, N=N double bond, and negative charge on O.
Q: Which resonance contributor of the nitrosoamine is the major contributor, and why?
A: The structure with the most covalent bonds, fewest formal charges, and negative charge on the most electronegative atom (oxygen) is the major contributor. This is typically the original structure with the N=O double bond, or the one that maximises bonds while keeping the negative charge on O.
Q: For the reaction of benzyl alcohol with HCl, draw the curved arrows for the first step (protonation).
A: Arrow 1: from the oxygen lone pair of the alcohol to the H of HCl (forming O-H bond). Arrow 2: from the H-Cl bond to the Cl (breaking the bond, Cl takes both electrons). Products: protonated alcohol (O has +1 charge, three bonds) and Cl-.
Q: What is the hybridization of each nitrogen atom in the nitrosoamine?
A: Both N1 and N2 are sp2. N1 has a double bond to O, a single bond to N2, and a lone pair (3 electron groups). N2 has a bond to N1, two bonds to CH3 groups, and a lone pair in a p orbital participating in resonance (3 electron groups in the plane).
Q: In the mechanism of benzyl alcohol reacting with HCl, why does water leave in step 2 rather than chloride?
A: After protonation, the oxygen carries a positive charge and the C-O bond is weakened. Water (H2O) is a good leaving group because it is a stable, neutral molecule. Chloride left in step 1 but returns in step 3 as the nucleophile.
Resonance is the bridge between Lewis structures and reaction mechanisms. The stability of resonance-stabilised intermediates (carbocations, carbanions, radicals) determines the outcome of nearly every reaction you will study. Curved arrow mechanisms extend from Bronsted acid-base reactions here to SN1, SN2, E1, E2, and addition reactions in the coming chapters.
The acidity trends covered in the Structure and Bonding notes rely heavily on resonance stabilisation of the conjugate base, so these two topics reinforce each other.
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