Difficulty: Introductory to Intermediate | Prerequisites: Lewis structures, lone pairs, formal charge
Once you can draw Lewis structures, the next step is recognising that a single molecular formula can correspond to several different structures (isomers), and that a single molecule can sometimes be represented by more than one valid Lewis structure (resonance). Both concepts are tested heavily in organic chemistry and form the backbone of understanding stability, reactivity, and mechanism.
Structural (constitutional) isomers are different molecules with the same molecular formula but different connectivity. Resonance structures are different Lewis structures for the same molecule, where only electrons (not atoms) move. The real molecule is a blend of all contributors, weighted toward the major one.
Structural isomers (constitutional isomers)
Molecules that share the same molecular formula but differ in how the atoms are connected. They are genuinely different compounds with different physical and chemical properties.
Think of it as building different things from the same set of LEGO bricks.
Resonance structures (resonance contributors)
Two or more valid Lewis structures for the same molecule, differing only in the placement of electrons (not atoms). Connected by a double-headed arrow. No single structure fully describes the molecule; the real electron distribution is a weighted average (the resonance hybrid).
Think of it as different "snapshots" of the same molecule's electron arrangement, where reality is a blend of all of them.
Resonance hybrid
The actual electronic structure of a molecule that exhibits resonance. It is not any single contributor but rather the weighted average of all contributors.
Major resonance contributor
The resonance structure that contributes most to the hybrid. It is typically the one with more complete octets, fewer formal charges, and (when formal charges exist) negative charge on the more electronegative atom.
Minor resonance contributor
A valid resonance structure that contributes less to the hybrid, usually because it has more formal charges, incomplete octets, or charge placed on a less electronegative atom.
Curved arrow (electron-pushing arrow)
A notation showing the movement of an electron pair from one position to another. The arrow always starts from the electrons (a lone pair or a bond) and points to where those electrons move.
The worksheet covers three molecular formulas, each with multiple valid structures.
C3H6O (3 carbons, 6 hydrogens, 1 oxygen)
One isomer: propenol (allyl alcohol), CH2=CH-CH2-OH, with a C=C double bond and a hydroxyl group.
The Lewis structure shows a carbon-carbon double bond on one end and an oxygen with two lone pairs bonded to the terminal carbon and a hydrogen.
Other possible isomers for C3H6O include acetone (propan-2-one) and propanal, though the worksheet focuses on the allyl alcohol structure.
C6H15N (6 carbons, 15 hydrogens, 1 nitrogen)
Two isomers shown: one with a secondary amine (NH between two carbon groups, drawn as a branched structure), and one with a tertiary amine (three carbon chains attached to nitrogen, with lone pair on N).
The degree of unsaturation is zero, so there are no double bonds or rings.
C6H10O2 (6 carbons, 10 hydrogens, 2 oxygens)
One isomer contains a ring (cyclobutane ring) with an ester-like linkage.
Another isomer is drawn as a chain with a carbonyl (C=O) and an ether-type oxygen in the ring.
Degree of unsaturation = 2, which can be satisfied by two double bonds, one ring and one double bond, or two rings.
The worksheet covers three examples of resonance with major/minor contributor identification.
Example 4a: Dicarbonyl compound
Two Lewis structures are drawn, differing in the placement of the negative formal charge.
The major contributor places the negative charge where it is more stabilised (more electronegative atom or more substituted position).
Curved arrows show how a lone pair pushes into a pi bond, displacing electrons onto the adjacent atom.
Example 4b: Diazo compound (diazomethane-type)
The first structure shows a positive formal charge on nitrogen and a negative formal charge on the terminal nitrogen, connected by a double bond.
The major contributor is the one with the negative charge on the more electronegative or more terminal atom.
The alternative structure shows a triple bond between the two nitrogens with rearranged formal charges.
Example 4c: Amide resonance
The amine nitrogen's lone pair can delocalise into the adjacent C=O, forming a C=N bond and placing the negative charge on oxygen.
The major contributor is the one with a C=N bond and a negative formal charge on oxygen (the more electronegative atom).
This is the classic amide resonance that explains why amides are less basic than amines and why the C-N bond in amides has partial double-bond character.
The structure with complete octets on all atoms is preferred.
Fewer formal charges are better than more.
When formal charges are unavoidable, place negative charge on the more electronegative atom.
Charge separation (having both a positive and a negative formal charge) is less favourable than a neutral structure.
Degree of unsaturation (DoU) = (2C + 2 + N - H - X) / 2, where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Each degree corresponds to one double bond or one ring.
Formal charge = (valence electrons of free atom) - (lone pair electrons) - (1/2 bonding electrons)
Isomers matter in drug design: ibuprofen and its mirror image have vastly different biological effects, even though they share the same molecular formula. Resonance stabilisation explains why aspirin's carboxylate group readily donates a proton (it forms a resonance-stabilised conjugate base), and why amide bonds in proteins are rigid and planar, giving proteins their defined shapes.
Students often confuse structural isomers with resonance structures. In isomers, atoms are connected differently. In resonance, the atom connectivity is identical and only electron placement changes.
The resonance hybrid is not a rapid switching between structures. The molecule does not "flip" between contributors. It exists as a single, blended structure at all times.
A minor contributor is not "wrong" or invalid. It still contributes to the hybrid, just less than the major one.
Curved arrows show electron movement, not atom movement. Arrows always start from electrons (lone pair or bond), never from an atom.
Expect exam questions that give you a molecular formula and ask you to draw all possible structural isomers. You will also be asked to draw resonance structures, use curved arrows, and identify the major contributor with justification.
Amide resonance (Example 4c) appears repeatedly throughout the course, especially in amino acid and peptide chemistry.
True or False: Structural isomers have the same connectivity but different electron arrangements.
Fill in the blank: The degree of unsaturation for C6H10O2 is ___.
True or False: In the major resonance contributor, negative formal charge sits on the more electronegative atom.
Fill in the blank: Curved arrows always start from ___ and point to where those electrons ___.
True or False: A resonance hybrid is an average of all contributing structures.
Q: What is the degree of unsaturation for C3H6O? What structural features could account for it?
A: DoU = (2(3) + 2 - 6) / 2 = 1. This one degree can be either one double bond (C=C or C=O) or one ring.
Q: Draw two structural isomers of C3H6O and name them.
A: Propenol (allyl alcohol, CH2=CH-CH2OH) uses the degree of unsaturation as a C=C double bond. Acetone (CH3COCH3) uses it as a C=O double bond. Propanal (CH3CH2CHO) is another valid isomer.
Q: For an amide (R-CO-NH2), draw two resonance structures and identify the major contributor. Justify your answer.
A: Structure 1 has a C=O double bond with a lone pair on nitrogen. Structure 2 has a C=N double bond with a negative formal charge on oxygen. Structure 2 is the major contributor because oxygen is more electronegative than nitrogen and better stabilises the negative charge, and all atoms have complete octets.
Q: Why is the C-N bond in an amide shorter and more rigid than the C-N bond in an amine?
A: Resonance gives the amide C-N bond partial double-bond character. The lone pair on nitrogen delocalises into the carbonyl, making the bond stronger, shorter, and resistant to rotation.
Q: A student draws a resonance structure by moving a hydrogen atom from one carbon to another. Is this valid?
A: No. Resonance structures differ only in electron placement. Atoms cannot move between resonance contributors. Moving an atom creates a different isomer, not a resonance structure.
Structural isomers connect to nomenclature (naming organic compounds), which you will cover next. Resonance structures are essential for understanding acid-base chemistry (resonance stabilisation of conjugate bases), electrophilic aromatic substitution, and reaction mechanisms throughout the course. The degree of unsaturation formula comes back whenever you are given an unknown molecular formula on an exam or in spectroscopy problems.
Structural isomers, constitutional isomers, resonance structures, resonance contributors, resonance hybrid, major contributor, minor contributor, curved arrow, electron pushing, degree of unsaturation, index of hydrogen deficiency, formal charge, amide resonance, diazo compound, carbonyl, C=O, pi bond, lone pair delocalisation, CHM 25500, organic chemistry, Purdue