Difficulty: Intermediate | Prerequisites: General chemistry (atomic orbitals, Lewis structures, acid-base basics)
Nomenclature is the grammar of organic chemistry: if you cannot name a molecule or draw one from its name, every later topic stalls. Structure, bonding and reactivity then explain why molecules behave as they do, covering acidity, aromaticity, conformational analysis, hybridisation and the nucleophile/electrophile logic that drives substitution reactions. Master these two sections first and the mechanism and synthesis material downstream becomes far more manageable.
IUPAC nomenclature
The systematic method for naming organic compounds endorsed by the International Union of Pure and Applied Chemistry. You find the longest continuous carbon chain, number it so substituents get the lowest possible locants, then list substituents alphabetically before the parent name.
In simple terms, this is the universal "address system" for molecules so that any chemist, anywhere, draws the same structure from the same name.
Substituent
An atom or group of atoms that replaces a hydrogen on the parent chain (e.g. methyl, ethyl, bromo, chloro).
Think of it as a side branch hanging off the main highway of carbons.
Stereodescriptor (R/S)
Labels assigned to chiral centres using the Cahn-Ingold-Prelog priority rules. R (rectus) means the priority sequence 1 to 2 to 3 runs clockwise; S (sinister) means anticlockwise, when the lowest-priority group points away from you.
In simple terms, R and S are like "left-handed" and "right-handed" labels for a carbon with four different groups.
E/Z isomerism
A naming system for alkene stereochemistry. E (entgegen, "opposite") means the two highest-priority groups sit on opposite sides of the double bond; Z (zusammen, "together") means they sit on the same side.
Think of Z as "zee zame zide."
Ortho, meta, para substitution
Positional labels for disubstituted benzene rings. Ortho (o) means the two groups are on adjacent carbons (1,2). Meta (m) means they are separated by one carbon (1,3). Para (p) means they sit directly across the ring from one another (1,4).
In simple terms, ortho is next door, meta skips one, para is straight across.
Identify the parent chain from the root name (heptane = 7 carbons, pentene = 5 carbons with a double bond).
Add substituents at the numbered positions. Dimethyl means two methyl groups; specify each locant (e.g. 2,5-dimethylheptane).
If a stereodescriptor is present, set the geometry:
(S) or (R) at a chiral centre requires you to arrange four different groups around that carbon using CIP priorities, then orient the lowest-priority group away from you and check the 1-2-3 direction.
(E) or (Z) at a double bond requires you to assign CIP priorities to the two groups on each carbon of the double bond, then check whether the higher-priority groups are on the same side (Z) or opposite sides (E).
Find the longest continuous chain that includes the principal functional group.
Number the chain so the principal group gets the lowest locant. If there is a tie, substituents break it.
Name each substituent with its locant. List them alphabetically (bromo before chloro before methyl). Prefixes like di- and tri- do not count for alphabetical ordering.
Example from the exam: a structure with a bromine on carbon 1 and a methyl on carbon 2 of propane is 1-bromo-2-methylpropane (not 2-bromo-1-methylpropane, because the parent chain is propane numbered to give the bromo group the lower locant when the methyl is a substituent).
Ortho (1,2-disubstituted): the two groups are on neighbouring ring carbons.
Meta (1,3-disubstituted): one ring carbon separates them.
Para (1,4-disubstituted): they sit directly opposite on the ring.
On the exam, you may be shown several disubstituted benzene rings and asked to identify the ortho isomer. Count the ring positions between the two substituents.
Acidity (Bronsted-Lowry)
A measure of how readily a compound donates a proton (H+). In organic chemistry, the more stable the resulting conjugate base, the more acidic the proton.
In simple terms, the easier it is for a hydrogen to leave, the more acidic it is.
Aromaticity
A cyclic, planar, fully conjugated system with (4n + 2) pi electrons, where n is a non-negative integer (Huckel's rule). Aromatic compounds are unusually stable compared to similar non-aromatic structures.
Think of it as a molecular VIP pass: if a ring is flat, fully conjugated and has the right electron count (2, 6, 10, 14...), it gets special stability.
Newman projection
A way of viewing a molecule along a specific C-C bond axis. The front carbon is drawn as a dot, the back carbon as a circle, and substituents radiate from each.
In simple terms, you are looking straight down the barrel of a carbon-carbon bond.
Hybridisation (sp, sp2, sp3)
The mixing of atomic orbitals to form new hybrid orbitals. sp3 gives tetrahedral geometry (four groups, 109.5 degrees), sp2 gives trigonal planar geometry (three groups, 120 degrees) and sp gives linear geometry (two groups, 180 degrees).
Think of it as: count the sigma bonds plus lone pairs on an atom. Four = sp3, three = sp2, two = sp.
Nucleophile
An electron-rich species that donates an electron pair to an electron-poor centre. Stronger nucleophiles react faster in SN2 reactions.
In simple terms, a nucleophile is the "electron donor" that attacks a positive or partially positive site.
Electrophile
An electron-poor species that accepts an electron pair. In SN2 reactions, the electrophile is the substrate bearing the leaving group.
Think of it as the molecule with the "Help Wanted" sign for electrons.
Leaving group
The atom or group that departs with the bonding electrons during a substitution or elimination reaction. Better leaving groups are weaker bases (e.g. Br minus is a better leaving group than OH minus).
In simple terms, the leaving group is whatever gets kicked out when the nucleophile arrives.
Meso compound
A molecule with chiral centres whose mirror image is superimposable on itself due to an internal plane of symmetry. A meso compound is achiral despite having stereocentres.
Think of it as a molecule that "cancels out" its own chirality.
The most acidic proton on a molecule is the one whose removal produces the most stable conjugate base.
Factors that stabilise conjugate bases (and therefore increase acidity):
Electronegativity of the atom bearing the charge (O-H more acidic than N-H more acidic than C-H, across a row).
Resonance delocalisation of the negative charge.
Inductive effects from nearby electronegative atoms.
Size and polarisability of the atom (going down a group: H-I more acidic than H-F).
Exam example: given Ha (on an sp carbon adjacent to a double bond), Hb (on an sp3 carbon) and Hc (on another sp3 carbon), the answer was Ha, because an sp-hybridised carbon is more electronegative than sp3 and can better stabilise the conjugate base.
Requirements: the ring must be cyclic, planar, fully conjugated (every atom in the ring has a p orbital) and have (4n + 2) pi electrons.
Common aromatic species: benzene (6 pi electrons, n = 1), cyclopentadienyl anion (6 pi electrons), cycloheptatrienyl cation (tropylium, 6 pi electrons).
Antiaromatic species have 4n pi electrons in a cyclic, planar, conjugated system and are destabilised. Cyclobutadiene (4 pi electrons) is the textbook example.
Non-aromatic species fail one or more criteria (not fully conjugated, not planar, not cyclic).
Watch for charged species. Lone pairs that sit in a p orbital and join the conjugated system count toward the pi electron total. A lone pair in an sp2 hybrid orbital perpendicular to the ring does not.
Most stable conformation: large groups are anti (180 degrees apart) in a staggered arrangement.
Least stable conformation: large groups are eclipsed, ideally synperiplanar (0 degrees).
For a phenyl-substituted butane (as on the exam), the most stable Newman projection has the two largest groups (e.g. phenyl and methyl) anti and staggered. The least stable has those groups eclipsed.
To assess chirality from a Newman projection, identify the four groups on each stereocentre and assign R or S. A compound with two stereocentres of opposite configuration (R,S) and an internal mirror plane is meso.
Assign hybridisation by counting sigma bonds plus lone pairs (the steric number) on the atom of interest.
sp3: 4 electron domains, tetrahedral, 109.5 degrees.
sp2: 3 electron domains, trigonal planar, 120 degrees.
sp: 2 electron domains, linear, 180 degrees.
Atoms in aromatic rings and attached to double bonds are typically sp2.
Exam example (iBET762): atoms within the aromatic/conjugated system (C1, C2, C3, N1, O1) were all sp2 with trigonal planar geometry, reflecting their participation in a planar, delocalised pi system.
Higher bond order means shorter bond length: triple bonds are shorter than double bonds are shorter than single bonds.
Within the same bond order, bonds to smaller atoms are shorter.
Resonance can give intermediate bond orders. A bond that is "one and a half" (as in benzene) is shorter than a pure single bond but longer than a pure double bond.
A stronger nucleophile attacks faster in SN2.
In polar aprotic solvents, nucleophilicity follows basicity across a row.
Sulphur nucleophiles (e.g. CH3S minus) are generally stronger than their oxygen counterparts (e.g. CH3O minus) in protic solvents because sulphur is larger, more polarisable and less tightly solvated.
Charged species are stronger nucleophiles than their neutral conjugate acids (CH3O minus stronger than CH3OH).
Steric bulk on the nucleophile reduces SN2 rate (tert-butoxide is a poor SN2 nucleophile despite being a strong base).
Exam answer: CH3S minus (option d) was the strongest nucleophile for SN2.
SN2 requires backside attack, so steric hindrance at the electrophilic carbon is critical.
Methyl > primary > secondary substrates for SN2. Tertiary substrates do not undergo SN2.
The leaving group matters: I minus is better than Br minus is better than Cl minus (larger, weaker base, more polarisable). OH is a poor leaving group because hydroxide is a strong base.
Exam answer: CH3Cl (option a, a primary alkyl chloride with minimal steric hindrance) was the most reactive electrophile for SN2.
Students often confuse E/Z with cis/trans. E/Z uses CIP priority rules and works for all alkenes; cis/trans only works cleanly when both carbons of the double bond each carry one hydrogen. Use E/Z as your default.
Students often think that the longest chain in a molecule must run in a straight line across the page. It does not. The longest continuous chain can bend, and you must trace every possible path to find the true parent chain.
Students often assume that a charged species is always aromatic or always antiaromatic. Charge alone does not determine aromaticity. You still need to check all four Huckel criteria (cyclic, planar, fully conjugated, 4n + 2 pi electrons).
Students often forget that nucleophilicity and basicity are not the same thing. A bulky strong base (e.g. tert-butoxide) is a terrible SN2 nucleophile. Nucleophilicity depends on how well the species can reach the electrophilic carbon, not just on how readily it donates electrons to a proton.
True or false: (E)-3-chloropent-2-ene has the chlorine and the longest carbon chain on the same side of the double bond.
Answer: False. E means the highest-priority groups are on opposite sides.
Fill in the blank: A molecule with two stereocentres and an internal plane of symmetry is called a ____ compound.
Answer: meso
True or false: The tropylium cation (cycloheptatrienyl cation) is aromatic.
Answer: True. It is cyclic, planar, fully conjugated and has 6 pi electrons (4n + 2, n = 1).
Fill in the blank: In an SN2 reaction, the rate increases when steric hindrance at the electrophilic carbon ____.
Answer: decreases
True or false: CH3O minus is a stronger nucleophile than CH3S minus in a protic solvent.
Answer: False. Sulphur is larger and more polarisable, making CH3S minus the stronger nucleophile in protic solvents.
Q: Draw the structure of (S)-2,5-dimethylheptane. What is the parent chain length, and which carbon is the stereocentre?
A: The parent chain is heptane (7 carbons). Methyl groups sit on C2 and C5. C2 is the stereocentre (it bears four different groups: H, CH3, the longer chain toward C3-C7, and the terminal CH3 at C1). The (S) descriptor sets the configuration at that centre.
Q: You are given a disubstituted benzene with Br and Cl. How do you determine whether it is ortho, meta or para?
A: Count the ring carbons between the two substituents along the shorter path. One carbon apart (adjacent) = ortho. Two carbons apart (one empty position between them) = meta. Three carbons apart (directly across) = para.
Q: Rank the following in order of increasing acidity: a C-H on cyclohexane, an O-H on ethanol, an O-H on acetic acid.
A: Cyclohexane C-H (least acidic) < ethanol O-H < acetic acid O-H (most acidic). The carboxylate conjugate base of acetic acid is stabilised by resonance across two equivalent oxygens, making it far more acidic than ethanol.
Q: Why is CH3S minus a better nucleophile than CH3O minus in an SN2 reaction in a protic solvent?
A: Sulphur is larger and more polarisable than oxygen, so it is less tightly solvated by protic solvent molecules and its electron density is more accessible for backside attack on the electrophilic carbon.
Q: A molecule has the formula C4H10 with two chiral centres and an internal mirror plane. Is it chiral or meso? What is its specific rotation?
A: It is meso. Its specific rotation is zero because the two chiral centres cancel each other's optical activity.
Nomenclature was worth 12 points (8% of the exam). Structure/bonding/reactivity was worth 28 points (nearly 19%). Together these two sections account for over a quarter of the total marks.
Expect to draw a structure from an IUPAC name and to name a structure from a drawing. Both directions are tested.
Aromaticity identification (marking aromatic species from a set) appears regularly.
Newman projection questions tend to ask for both the most stable and least stable conformations, plus a chirality or meso determination.
Hybridisation and geometry are often tested through a complex molecule (like iBET762) where you assign sp, sp2 or sp3 and the corresponding geometry to several labelled atoms.
Nomenclature feeds directly into every synthesis and mechanism problem: if you misread a name, you draw the wrong starting material and the entire answer chain collapses. Acidity concepts return in elimination reactions (E1cb, E2 base strength) and in enolate chemistry. Aromaticity is central to electrophilic aromatic substitution, which often appears in multistep synthesis. Nucleophilicity and leaving-group ability govern whether a reaction proceeds by SN1, SN2, E1 or E2, which is the backbone of the mechanisms section.
IUPAC naming, organic nomenclature, CIP priority rules, Cahn-Ingold-Prelog, R and S configuration, E and Z alkenes, cis trans isomers, ortho meta para, substituent naming, Newman projection, conformational analysis, staggered eclipsed anti gauche, hybridisation, sp sp2 sp3, molecular geometry, bond angle, Huckel rule, 4n+2, aromatic antiaromatic nonaromatic, nucleophile electrophile, SN2 reaction, leaving group ability, meso compound, chiral centre, stereocentre, conjugate base stability, CHEM 2301, organic chemistry I, Pomerantz, University of Minnesota