IUPAC Nomenclature, Structure and Bonding, CHEM 210 – Study Notes
Difficulty: Foundational | Prerequisites: General chemistry (atomic orbitals, Lewis structures, electronegativity)
Before you can follow any reaction in organic chemistry, you need two things: the ability to name a molecule on sight and a mental picture of how its atoms are bonded. This set of notes covers IUPAC nomenclature (the universal naming system), hybridisation and molecular geometry, resonance, and aromaticity. Everything here feeds directly into reaction mechanisms later in the course, because whether a molecule reacts, and how, depends on the shape and electron distribution you learn to read here. If you skipped general chemistry's coverage of orbitals and Lewis structures, review those first.
IUPAC names follow a strict algorithm: find the longest chain, number it, name substituents alphabetically, and assign stereochemistry where needed. Hybridisation tells you shape (sp3 = tetrahedral, sp2 = trigonal planar, sp = linear). Resonance structures are bookkeeping for delocalised electrons, and aromaticity is a special stability that kicks in when a cyclic, planar, fully conjugated ring has 4n+2 pi electrons.
IUPAC nomenclature
The International Union of Pure and Applied Chemistry naming system. A set of rules that gives every organic molecule exactly one unambiguous name. Think of it as the molecule's legal name, the one everyone agrees on.
Parent chain
The longest continuous carbon chain in a molecule, which determines the root name (meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec-). In simple terms, this is the backbone you build the name around.
Locant
A number that indicates the position of a substituent or functional group along the parent chain. You number the chain to give the lowest possible set of locants.
E/Z (entgegen/zusammen) isomers
A naming system for alkene stereochemistry based on Cahn-Ingold-Prelog priority rules. E means higher-priority groups are on opposite sides of the double bond; Z means they are on the same side. Think of Z as "zee same side."
Hybridisation
The mixing of atomic orbitals (s and p) into new hybrid orbitals that explain observed molecular geometry. sp3 gives four equivalent orbitals pointing to the corners of a tetrahedron; sp2 gives three in a plane with one unhybridised p orbital; sp gives two along a line with two unhybridised p orbitals.
Resonance structures
Alternative Lewis structures for the same molecule that differ only in electron placement (not atom positions). The real molecule is a weighted average (hybrid) of all valid resonance structures. Think of them as snapshots that, blended together, show where the electrons really spend their time.
Resonance hybrid
The actual electronic structure of a molecule, which is a blend of all contributing resonance structures. Bond lengths and charges in the hybrid are intermediate between those shown in any single contributor.
Aromaticity
A special thermodynamic stability found in cyclic, planar, fully conjugated systems with 4n+2 pi electrons (Huckel's rule). Aromatic compounds are far more stable than you would predict from their structure alone. In simple terms, the electrons are delocalised around the entire ring, creating an unusually low-energy arrangement.
Antiaromatic
A cyclic, planar, fully conjugated system with 4n pi electrons (4, 8, 12...). These are less stable than the non-conjugated version would be. The molecule is actively destabilised by its conjugation.
Cahn-Ingold-Prelog (CIP) priority rules
A system for ranking substituents by atomic number (higher atomic number = higher priority). Used to assign E/Z on double bonds and R/S at stereocentres. When two atoms are the same, you move outward to the next set of atoms until you find a difference.
Find the longest continuous carbon chain. This determines the parent name: meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C), hex- (6C), hept- (7C), oct- (8C), non- (9C), dec- (10C).
Number the chain from the end nearest the first branch point, giving the lowest set of locants.
Name substituents alphabetically, each with its locant. Ignore multiplying prefixes (di-, tri-, tetra-) when alphabetising. Example: ethyl comes before methyl, even if there are two methyl groups (dimethyl).
Halide substituents: fluoro-, chloro-, bromo-, iodo-.
Cycloalkanes: prefix "cyclo-" to the parent name. Number substituents to give the lowest set of locants.
Replace "-ane" with "-ene" (double bond) or "-yne" (triple bond).
Number from the end nearest the pi bond.
Assign E/Z stereochemistry when the double bond has substituents on both carbons. E = higher-priority groups on opposite sides (entgegen). Z = higher-priority groups on the same side (zusammen).
Priority assigned by CIP rules: higher atomic number wins; if tied, move outward to the next atom.
Carboxylic acids > anhydrides > esters > acid halides > amides > aldehydes > ketones > alcohols > thiols > amines > alkenes > alkynes > alkanes.
The highest-priority group gets the suffix. All others become prefixes (oxo-, hydroxy-, amino-, etc.).
Alcohols: suffix "-ol"; number so that OH gets the lowest locant.
Ethers: named as alkoxy substituents on the parent chain (methoxy-, ethoxy-) or by common names (diethyl ether).
Amines: suffix "-amine"; N-substituents carry the N- prefix (e.g. N-methylpropylamine).
Aldehydes: suffix "-al"; C1 is always the carbonyl carbon.
Ketones: suffix "-one"; lowest locant to C=O.
Carboxylic acids: suffix "-oic acid".
Esters: named as alkyl alkanoate (e.g. ethyl ethanoate = ethyl acetate).
Amides: suffix "-amide".
Acid chlorides: suffix "-oyl chloride".
Anhydrides: name both acid components + "anhydride".
Bicyclo[x.y.z]alkane: the numbers in brackets list bridge sizes from largest to smallest. Numbering starts at one bridgehead and traverses the longest bridge first.
Count the number of electron domains (bonds + lone pairs) around an atom to determine its hybridisation.
4 domains = sp3. Tetrahedral geometry, bond angles ~109.5 degrees. Example: methane (CH4), any saturated carbon.
3 domains = sp2. Trigonal planar geometry, bond angles ~120 degrees. Example: carbons in a C=C double bond, carbonyl carbons.
2 domains = sp. Linear geometry, bond angles 180 degrees. Example: carbons in a C triple bond, CO2.
Each unhybridised p orbital either participates in pi bonding or holds a lone pair available for resonance.
Lone pairs compress bond angles slightly. VSEPR repulsion strength: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. This is why water's H-O-H angle is ~104.5 degrees rather than the tetrahedral 109.5 degrees.
Hybridisation determines the geometry of drug molecules, which in turn determines whether they fit into an enzyme's active site. Medicinal chemists use hybridisation to predict how a new compound will interact with its biological target.
Only electrons move. Atoms never change position between resonance structures.
Curved arrows start at a source of electrons (a lone pair or a pi bond) and point to where those electrons go.
Every valid resonance structure must obey valence rules: H max 2 electrons, C/N/O max 8 electrons.
Structures with more bonds and less charge separation are more stable and contribute more to the hybrid (major contributors).
Negative charge on a more electronegative atom = more stable contributor.
Never break a sigma bond in resonance. Only pi bonds and lone pairs move.
The real molecule is a hybrid of all resonance structures. Actual bond lengths and energies are intermediate between what any single structure shows.
A lone pair adjacent to a pi bond (allylic or vinylic position).
A carbocation adjacent to a pi bond (allylic carbocation).
An anion adjacent to a C=O group (enolate).
Full delocalisation around an aromatic ring.
A lone pair on an atom directly bonded to C=O. This is amide resonance: the C-N bond gains partial double-bond character, which is why amide bonds are planar and relatively rigid.
Amide resonance is the reason peptide bonds in proteins are planar. The partial double-bond character restricts rotation and gives proteins their characteristic backbone geometry.
A compound is aromatic if and only if it meets all four:
Cyclic.
Planar (all atoms in the ring lie in one plane).
Fully conjugated (every atom in the ring has a p orbital).
Contains 4n+2 pi electrons, where n = 0, 1, 2, 3... This gives allowed counts of 2, 6, 10, 14 pi electrons.
If all four criteria are met but the system has 4n pi electrons (4, 8, 12...), the compound is antiaromatic and is destabilised relative to a non-conjugated reference.
If any criterion is not met, the compound is nonaromatic, meaning ordinary conjugation rules apply with no special aromatic stability or instability.
Benzene (6 pi electrons, n=1). The textbook example.
Pyridine (6 pi electrons). The nitrogen lone pair is NOT in the pi system; it sits in an sp2 orbital in the plane of the ring and is available as a base.
Pyrrole (6 pi electrons). The nitrogen lone pair IS in the pi system, which is why pyrrole is a very weak base: donating that lone pair would break aromaticity.
Furan (6 pi electrons). One oxygen lone pair is in the pi system.
Cyclopentadienyl anion (6 pi electrons). Aromatic.
Cycloheptatrienyl cation / tropylium (6 pi electrons). Aromatic.
Cyclobutadiene (4 pi electrons). Antiaromatic. Extremely unstable.
Aromatic rings are everywhere in pharmaceuticals. Roughly two-thirds of all approved drugs contain at least one aromatic ring, because the flat, rigid geometry is ideal for binding to protein targets.
Students often think resonance structures represent different molecules that flicker back and forth. They do not. The molecule exists as one fixed hybrid; the individual structures are just our imperfect way of drawing delocalised electrons.
Students often confuse E/Z with cis/trans. E/Z is the more general system based on CIP priority, and it always works. Cis/trans only applies when the groups on either side of the double bond are identical or very simple. Use E/Z by default.
Students sometimes assume that any ring with alternating double bonds is aromatic. It is not. The ring must also be planar and have the correct electron count (4n+2). Cyclooctatetraene has alternating double bonds but is tub-shaped and nonaromatic.
Students often count all lone pairs on a heteroatom when determining aromaticity. Only one lone pair per atom can be in the pi system. In pyridine, the nitrogen lone pair is in the plane of the ring (not in the pi system), while in pyrrole it is in a p orbital (in the pi system).
⚠️ Naming questions are among the easiest marks on an exam. The algorithm is mechanical: follow the steps and you will get the right answer.
⚠️ Hybridisation questions often appear as "what is the hybridisation of atom X in this molecule?" Count electron domains, not bonds. A double bond = one domain.
⚠️ Resonance is tested by asking you to draw all contributing structures or identify the major contributor. Remember: more bonds, less charge separation = major.
⚠️ Aromaticity is frequently tested by giving you a molecule and asking whether it is aromatic, antiaromatic, or nonaromatic. Apply all four criteria systematically.
⚠️ Pyrrole vs. pyridine is a classic exam question. Know which nitrogen lone pair is in the pi system and which is not, and why that affects basicity.
True or false: A double bond counts as two electron domains for hybridisation purposes. (False. A double bond is one domain.)
Fill in the blank: A molecule is aromatic if it is cyclic, planar, fully conjugated, and has ______ pi electrons. (4n+2)
True or false: In resonance, atoms can move between structures. (False. Only electrons move.)
Fill in the blank: The IUPAC suffix for a ketone is ______. (-one)
True or false: Cyclobutadiene is aromatic because it is cyclic, planar, and fully conjugated. (False. It has 4 pi electrons, which is 4n, making it antiaromatic.)
Q: Name the following compound: a six-carbon chain with a double bond at C-2 and a methyl group at C-4. The higher-priority groups on the double bond are on the same side.
A: (Z)-4-methylhex-2-ene.
Q: What is the hybridisation of the nitrogen atom in pyrrole, and why is pyrrole a poor base?
A: The nitrogen is sp2 hybridised. Its lone pair occupies a p orbital and is part of the aromatic pi system (contributing to the 6-pi-electron count). Donating that lone pair to a proton would destroy aromaticity, so pyrrole is a very weak base.
Q: Draw the major resonance contributor for the acetate anion (CH3COO-). Explain your choice.
A: Both resonance structures (negative charge on either oxygen) are equivalent and contribute equally. The resonance hybrid has each C-O bond as 1.5 bonds, with the negative charge shared equally between both oxygens.
Q: A compound is cyclic, planar, fully conjugated, and has 8 pi electrons. Is it aromatic, antiaromatic, or nonaromatic?
A: Antiaromatic. It meets all four criteria for aromaticity except the electron count: 8 = 4n (n=2), which corresponds to antiaromaticity rather than aromaticity (4n+2).
Q: Place these functional groups in order from highest to lowest IUPAC naming priority: ketone, carboxylic acid, amine, alcohol.
A: Carboxylic acid > ketone > alcohol > amine.
This connects to stereoisomerism because E/Z naming at double bonds uses the same CIP priority rules you will use for R/S assignments at stereocentres. Hybridisation and resonance feed directly into understanding reaction mechanisms: knowing where electron density sits tells you where nucleophiles and electrophiles will attack. Aromaticity is the foundation for the entire unit on electrophilic aromatic substitution.
IUPAC naming, systematic nomenclature, organic nomenclature, parent chain, longest chain, substituent naming, functional group priority, E/Z isomers, cis/trans, Cahn-Ingold-Prelog, CIP rules, sp3, sp2, sp, hybridisation, hybridization, bond angles, VSEPR, electron domains, resonance structures, resonance hybrid, curved arrows, electron delocalisation, major contributor, aromaticity, Huckel's rule, 4n+2, antiaromatic, nonaromatic, benzene, pyrrole, pyridine, furan, tropylium, cyclopentadienyl anion