Source: PSO Worksheet 1, CHM 25500, Purdue University
Difficulty: Intermediate | Prerequisites: Lewis structures, lone pairs, formal charge (see Part 1 study notes).
This material picks up where Lewis structures leave off. Resonance explains why some molecules cannot be described by a single Lewis structure, and why electron density is sometimes spread across several atoms rather than locked in one place. Functional groups are the vocabulary of organic chemistry: recognising them on sight lets you predict reactivity without memorising every reaction. Hybridisation and bond angles connect the flat drawings you have been making to the three-dimensional shapes molecules actually adopt.
Resonance structures are alternative Lewis structures for the same molecule, connected by curved-arrow (electron-pushing) notation. Functional groups are specific arrangements of atoms (like –OH, C=O, –NH₂) that determine a molecule's chemical behaviour. Hybridisation (sp, sp², sp³) determines the geometry and bond angles around each atom.
Resonance structures (contributing structures, canonical forms)
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 it as: the molecule does not flip between these structures. It is all of them at once, blended together.
Resonance hybrid
The true electronic structure of the molecule, which is a blend of all resonance contributors. No single Lewis structure captures it perfectly.
Curved-arrow notation (arrow pushing / electron pushing)
Curved arrows show the movement of electron pairs from one position to another within a structure. The tail starts where the electrons are; the head points to where they go. A full curved arrow moves two electrons. A fishhook arrow (half-headed) moves one electron.
Major contributor
The resonance structure that contributes most to the hybrid. It is usually the one with the most atoms carrying full octets, the fewest formal charges, and negative charges on the most electronegative atoms.
Functional group
A specific grouping of atoms within a molecule that has its own characteristic chemical behaviour. Recognising functional groups lets you predict what reactions a molecule can undergo.
In simple terms, functional groups are the "moving parts" of the molecule.
Hydroxyl group (–OH)
An oxygen bonded to a hydrogen and to the rest of the molecule. Found in alcohols and carboxylic acids.
Carbonyl group (C=O)
A carbon double-bonded to an oxygen. The parent group in aldehydes, ketones, carboxylic acids, esters, and amides.
Amide (–C(=O)–NH–)
A carbonyl bonded to a nitrogen. Found in proteins and many drugs, including clindamycin.
Thioether (R–S–R)
A sulphur atom bonded to two carbon atoms. The sulphur analogue of an ether.
Hybridisation (sp³, sp², sp)
The mixing of atomic orbitals to form new hybrid orbitals that determine molecular geometry. sp³ gives a tetrahedral shape (109.5°), sp² gives trigonal planar (120°), and sp gives linear (180°).
In simple terms, hybridisation tells you the shape around an atom based on how many things (bonds + lone pairs) are attached to it.
Bond angle
The angle formed between two bonds emanating from the same atom. Determined by the hybridisation of the central atom.
Resonance structures differ only in electron placement. The atoms stay in exactly the same positions. If you have to move an atom, it is not a resonance structure.
Use curved arrows to show how you get from one contributor to the next. The arrow tail starts at a lone pair or a bond (where the electrons are now) and the arrow head lands on a bond or an atom (where the electrons are going).
Rules for valid resonance structures:
Do not exceed the octet for second-row elements (C, N, O, F). These atoms cannot have more than eight electrons.
Every structure must have the same net charge and the same number of total electrons.
Atoms do not move, only electrons.
Evaluating which contributor is the major one:
The structure with more complete octets is more important.
The structure with fewer formal charges is more important.
If formal charges are unavoidable, negative charges on more electronegative atoms make a better contributor.
Charge separation (having both a + and a – in the same structure) is less stable than no charge separation.
Worksheet examples (Q4):
Ester (methyl ester with two oxygens): the major contributor has the C=O double bond intact, with full octets on both oxygens and no formal charges. The minor contributor places a lone pair from the single-bonded oxygen into the C–O bond, creating a C=O on that side and breaking the original C=O to a single bond. This minor contributor has charge separation (+1 on one O, –1 on the other).
Epoxide with a positive oxygen: resonance possible by pushing a lone pair from the positively charged oxygen into an adjacent bond. Evaluate which structure leaves the positive charge on the least electronegative atom.
Conjugated system with nitrogen: the nitrogen lone pair can delocalise into the pi system. The major contributor keeps a full octet on nitrogen.
Vinyl amine with a positive charge: the nitrogen lone pair delocalises into the double bond system.
Phenyl anion (benzene ring with a carbanion): the lone pair on carbon delocalises into the ring.
Cyclohexadiene: if there is no heteroatom with a lone pair adjacent to the pi system, and no charge, resonance may not be possible. Write N/A if you cannot draw a valid second structure without violating the rules.
The worksheet asks you to circle and label functional groups in clindamycin. This is a pattern-recognition exercise. The functional groups present in clindamycin include:
Hydroxyl groups (–OH): clindamycin has multiple hydroxyl groups on the sugar portion. Oxygen bonded to hydrogen and to a carbon.
Amide (–C(=O)–NH–): a carbonyl directly bonded to a nitrogen. This is the linkage connecting the sugar to the rest of the molecule.
Thioether (C–S–C): a sulphur atom bonded to two carbons. Found in the ring system.
Chloroalkane (C–Cl): a carbon bonded to a chlorine atom.
Tertiary amine (–NR₂): a nitrogen bonded to carbon substituents (the pyrrolidine ring nitrogen in clindamycin carries alkyl groups).
Ether-like oxygen (C–O–C): in the sugar ring, an oxygen sits between two carbons.
When identifying functional groups, work systematically: scan every heteroatom (anything that is not C or H), note what it is bonded to, and match it to the standard list. Do not skip atoms in the middle of complex structures.
Count the number of electron groups (bonds + lone pairs) around an atom to determine its hybridisation. A double or triple bond counts as one electron group.
4 electron groups = sp³ = tetrahedral geometry = 109.5° bond angles.
3 electron groups = sp² = trigonal planar geometry = 120° bond angles.
2 electron groups = sp = linear geometry = 180° bond angles.
Applying this to the worksheet molecule (Q6):
The molecule has an alkyne, a conjugated diene, and a carbonyl. Working through the labelled bond angles:
Angle a: at a carbon in the triple bond (alkyne). sp hybridised. Bond angle = 180°.
Angle b: at a carbon in the double bond region (alkene). sp² hybridised. Bond angle = 120°.
Angle c: at the carbonyl carbon (C=O). sp² hybridised. Bond angle = 120°.
Angle d: at a tetrahedral carbon (four single bonds). sp³ hybridised. Bond angle = 109.5°.
Identifying hybridised carbons in the structure:
Circle all sp² carbons: these are every carbon in a double bond (alkene carbons, the carbonyl carbon).
Box all sp carbons: these are the carbons in the triple bond (alkyne carbons).
All other carbons are sp³ (not circled or boxed).
Hybridisation quick reference:
sp³: 4 electron groups, tetrahedral, 109.5°. Example: CH₄, any saturated carbon.
sp²: 3 electron groups, trigonal planar, 120°. Example: C=C (alkene), C=O (carbonyl).
sp: 2 electron groups, linear, 180°. Example: C≡C (alkyne), C≡N (nitrile).
Rules for identifying the major resonance contributor:
All atoms have complete octets (most important).
Fewest formal charges.
Negative formal charges on most electronegative atoms.
Least charge separation.
Resonance is why the peptide bond in proteins is planar: the nitrogen's lone pair delocalises into the adjacent C=O, giving the C–N bond partial double-bond character. This rigidity is fundamental to protein structure.
Functional group recognition is the basis of medicinal chemistry. Clindamycin's amide bond, for example, is the pharmacophore linkage. Changing functional groups on a drug scaffold is how chemists tune its potency, solubility, and metabolism.
Hybridisation determines molecular shape, which determines how a drug fits into its target protein. The 120° angles around a carbonyl carbon vs. the 109.5° around a saturated carbon make a real difference to binding.
Students often think resonance structures are different molecules that interconvert rapidly. They are not. The molecule does not switch between them. It exists as a single hybrid that is a blend of all contributors simultaneously.
Students sometimes move atoms when drawing resonance structures. Only electrons move. If you have to break and reform sigma bonds to different atoms, you are drawing a different molecule, not a resonance structure.
Students frequently forget that a double or triple bond counts as only one electron group for hybridisation. A carbon with one double bond and two single bonds has three electron groups (sp²), not four.
Students sometimes label an amide as an amine. An amide has a carbonyl (C=O) bonded to the nitrogen. An amine is just nitrogen bonded to carbons and hydrogens with no adjacent carbonyl. The distinction matters because their chemistry is very different.
⚠️ Resonance structure questions appear on virtually every organic chemistry exam. You will be asked to draw all reasonable contributors, use arrow pushing, and identify the major contributor with a justification.
⚠️ Functional group identification is tested both directly ("circle and label") and indirectly (predicting reactivity based on which groups are present).
⚠️ Hybridisation and bond angle questions are quick marks if you know the three patterns. Expect to identify sp, sp², and sp³ carbons in a complex molecule and assign the correct angle.
⚠️ Arrow pushing is not just a drawing exercise. It is the language used to explain every reaction mechanism for the rest of the course. Get comfortable with it now.
True or False: Resonance structures can differ in the positions of atoms. (False, only electron positions change)
Fill in the blank: An sp² hybridised carbon has ____ electron groups and bond angles of approximately ____. (3, 120°)
True or False: A carbon involved in a triple bond is sp hybridised. (True)
Fill in the blank: The major resonance contributor is the one with the most complete ____ and the fewest ____. (octets, formal charges)
True or False: An amide and an amine are the same functional group. (False, an amide has a carbonyl next to the nitrogen)
Q: An ester has two resonance contributors. In one, the C=O double bond is intact and the single-bonded oxygen has two lone pairs. In the other, a lone pair from the single-bonded oxygen forms a new pi bond to carbon, breaking the original C=O to a single bond. Which is the major contributor and why?
A: The structure with the C=O double bond intact is the major contributor. It has no formal charges and all atoms have full octets. The other contributor has charge separation (+1 on one oxygen, –1 on the other), which is less stable.
Q: What hybridisation does the carbon in a C≡N (nitrile) group have, and what is the bond angle?
A: sp hybridised, 180° (linear). The carbon has two electron groups: one triple bond to nitrogen and one single bond to the adjacent carbon.
Q: Name four functional groups present in clindamycin.
A: Hydroxyl (–OH), amide (–C(=O)–NH–), thioether (C–S–C), and chloroalkane (C–Cl). Also acceptable: tertiary amine, cyclic ether.
Q: A molecule has the structure CH₃–C≡C–CH=CH–C(=O)–CH₃. Identify the hybridisation of each carbon.
A: The two carbons in the C≡C are sp. The two carbons in the C=C are sp². The carbonyl carbon (C=O) is sp². The two CH₃ carbons and any saturated carbons are sp³.
Q: Why can you not draw a resonance structure for cyclohexane?
A: Cyclohexane has no pi bonds, no lone pairs adjacent to pi systems, and no charges. There is no electron pair that can be delocalised to create a valid second structure. The answer is N/A.
Resonance is the gateway to understanding reaction mechanisms. Every time you draw curved arrows in an acid-base reaction, nucleophilic substitution, or elimination, you are using the same electron-pushing logic practised here.
Functional group recognition becomes critical in spectroscopy (IR, NMR, mass spec), where you identify groups from absorption patterns. It also underpins retrosynthetic analysis later in the course.
Hybridisation links to molecular orbital theory and to the concept of sigma vs. pi bonds, which are essential for understanding addition reactions of alkenes and alkynes.
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