Difficulty: Intermediate | Prerequisites: General chemistry (Lewis structures, electronegativity, basic bonding).
Nomenclature, hybridisation, and stereochemistry form the language of organic chemistry. If you cannot name a molecule, assign its geometry, or tell two stereoisomers apart, every reaction chapter that follows will feel like reading a foreign language. This material sits right at the start of CHEM 2301, but it is tested on every exam, including the final. You should already be comfortable drawing Lewis structures and assigning formal charges from general chemistry.
IUPAC nomenclature gives every organic molecule a unique, systematic name built from its longest carbon chain, substituents, and stereodescriptors. Hybridisation (sp, sp², sp³) tells you the geometry around any atom. Stereoisomer relationships (enantiomers, diastereomers, identical, constitutional isomers) hinge on how two structures differ at their stereocentres, and getting these right is worth easy points on every exam.
IUPAC nomenclature
The standardised system for naming organic compounds maintained by the International Union of Pure and Applied Chemistry. Names are built from a parent chain, substituent prefixes, and locant numbers.
In simple terms, it is the official recipe for turning a structure into a name (and vice versa) so that every chemist worldwide means the same molecule.
Stereocentre (chiral centre)
An atom (usually carbon) bonded to four different groups, making the molecule non-superimposable on its mirror image.
Think of it as a carbon where swapping any two groups gives you a different molecule, the way your left and right hands are mirror images but cannot be overlaid.
R and S configuration (Cahn-Ingold-Prelog rules)
A labelling system that assigns absolute configuration at a stereocentre by ranking substituents by atomic number, then reading the 1-2-3 priority arc: clockwise = R, anticlockwise = S.
In simple terms, it is the naming convention that tells you whether the groups spiral right (R, from Latin rectus) or left (S, from Latin sinister) around a stereocentre.
E and Z configuration
Stereodescriptors for alkene geometry. Z (zusammen, together) means the two higher-priority groups sit on the same side of the double bond; E (entgegen, opposite) means they sit on opposite sides.
Think of it as the replacement for "cis" and "trans" that works even when the substituents are all different.
Enantiomers
Stereoisomers that are non-superimposable mirror images of each other. They have opposite configuration at every stereocentre.
In simple terms, a pair of molecules that are mirror images but cannot be stacked on top of one another, like left and right shoes.
Diastereomers
Stereoisomers that are not mirror images. They differ at some, but not all, stereocentres.
Think of them as stereoisomers that are "mismatched" rather than perfectly reflected.
Constitutional isomers (structural isomers)
Molecules with the same molecular formula but different atom-to-atom connectivity.
In simple terms, the atoms are the same, but they are plugged together in a different order.
Hybridisation
The mixing of atomic orbitals (one s with one, two, or three p orbitals) to form new hybrid orbitals that explain observed molecular geometries.
sp³: four hybrid orbitals, tetrahedral geometry (109.5°)
sp²: three hybrid orbitals, trigonal planar geometry (120°), one unhybridised p orbital
sp: two hybrid orbitals, linear geometry (180°), two unhybridised p orbitals
In simple terms, hybridisation tells you the shape around an atom. Count the number of "groups" (bonds + lone pairs) attached to it: four = sp³, three = sp², two = sp.
Degree of unsaturation (index of hydrogen deficiency)
A formula-based count of how many rings or pi bonds a molecule contains. For CₙHₘ (with adjustments for halogens, nitrogen, oxygen), the formula is (2n + 2 - m) / 2.
Think of it as a quick way to know how many double bonds or rings are hiding in a molecular formula before you even look at the structure.
Find the longest continuous carbon chain that includes the principal functional group. This chain gives the parent name (methane, ethane, propane, butane, pentane, hexane, etc.).
Number the chain so that the principal functional group gets the lowest possible locant. If there is a tie, give the lowest locant to the first point of difference among substituents.
Name and number each substituent. Substituents are listed in alphabetical order in the final name. Multiplying prefixes (di-, tri-, tetra-) do not count for alphabetical ordering, but "iso" and "cyclo" do when they are part of the substituent name itself.
Add stereodescriptors. Place (R), (S), (E), or (Z) in parentheses at the start of the name or immediately before the relevant locant.
Common parent names to know: "-ol" = alcohol, "-al" = aldehyde, "-one" = ketone, "-oic acid" = carboxylic acid, "-amine" / "-aniline" for amines, "-ene" = alkene, "-yne" = alkyne.
Work backwards from the name. Start with the parent chain length, place the functional group, add substituents at their numbered positions, then assign wedge/dash bonds for any (R)/(S) centres.
Example from the exam: (2S,3R)-2-bromo-3-hexanol
Parent chain: hexane (6 carbons)
Functional group: -ol on C3
Substituent: bromo on C2
Stereochemistry: S at C2, R at C3. Assign priorities at each centre, draw wedge/dash bonds so the 1-2-3 arc reads correctly with the lowest priority group going away from you.
p-chloroaniline: "para" means the chlorine is at the 1,4-positions on the benzene ring relative to the amino group (-NH₂). The name "aniline" means aminobenzene.
Count the steric number (number of atoms bonded + lone pairs) around the atom in question.
Steric number 4 = sp³. Four single bonds, or three bonds and one lone pair, etc. Tetrahedral or trigonal pyramidal geometry.
Steric number 3 = sp². Atom involved in one double bond (or has one lone pair participating in resonance). Trigonal planar.
Steric number 2 = sp. Atom involved in a triple bond or two double bonds. Linear.
Exam-tested tricky cases:
A carbon in a C=C double bond is sp².
The carbon of a carbocation is sp² (empty p orbital).
An oxygen with two bonds and two lone pairs is sp³ (e.g. in water or an alcohol), but an oxygen in a carbonyl (C=O) is sp².
A carbon bearing a lone pair and a negative charge attached to three groups total (as in a carbanion drawn with a lone pair, like :CH₂⁻ on a ring) is sp³ if it has four electron groups, or sp² if the lone pair sits in a p orbital for conjugation.
The indicated carbon in an allene-type system may be sp.
Aromatic ring carbons are sp².
Same molecular formula? If no, they are not isomers at all.
Same connectivity (atom-to-atom bonding)? If no, they are constitutional isomers.
Superimposable? If yes, they are identical (the same compound drawn differently).
Non-superimposable mirror images? If yes, they are enantiomers.
Non-superimposable and not mirror images? They are diastereomers.
Redraw both structures in the same orientation. Fischer projections, Newman projections, and wedge-dash structures can all represent the same molecule, so put them in a common format before comparing.
Assign R/S at every stereocentre in both molecules.
Compare the configurations:
All centres opposite = enantiomers
Some centres same, some opposite = diastereomers
All centres the same = identical (or you have drawn the same molecule twice)
For cyclopentane or cyclohexane rings with two substituents, draw the ring flat and note whether the substituents are on the same face (cis) or opposite faces (trans). Then assign R/S to determine the relationship.
Remember: in a Fischer projection, horizontal bonds come towards you (wedges) and vertical bonds go away (dashes). To compare two Fischer projections, either convert both to wedge-dash drawings, or use the rule that swapping any two groups on a Fischer projection inverts the configuration.
A molecule with stereocentres can still be achiral if it has an internal mirror plane. Such a compound is called meso. A meso compound and its "mirror image" are identical, not enantiomers. If one molecule of a pair is meso and the other has the same connectivity but different configuration, they are diastereomers.
Degree of unsaturation (DoU)
DoU = (2C + 2 + N - H - X) / 2
where C = carbons, H = hydrogens, N = nitrogens, X = halogens. Each ring or pi bond contributes one degree. Oxygen does not appear in the formula.
CIP priority rules (for R/S and E/Z)
Compare atoms directly attached to the stereocentre: higher atomic number = higher priority.
If tied, move outward to the next set of atoms and compare.
Double and triple bonds are "expanded": a C=O is treated as C bonded to (O, O) and O bonded to (C, C).
For R/S: orient the molecule so priority 4 points away, then read 1 → 2 → 3. Clockwise = R, anticlockwise = S.
For E/Z: compare the two substituents on each carbon of the double bond. Higher-priority groups on the same side = Z; opposite sides = E.
Stereochemistry is not just an academic exercise. The pharmaceutical industry depends on it: the (S)-enantiomer of ibuprofen is the active painkiller, while the (R)-enantiomer is largely inactive. Thalidomide is the most infamous example, where one enantiomer treated morning sickness and the other caused birth defects. IUPAC naming ensures that chemists, pharmacists, and regulatory bodies across the world refer to the same substance unambiguously.
"Cis/trans and E/Z are interchangeable." They are not. Cis/trans only works for disubstituted alkenes where the two groups on each carbon are the same. E/Z uses CIP priorities and works for all cases. On this exam, always use E/Z unless the question specifically asks for cis/trans.
"A molecule with stereocentres must be chiral." Not true. Meso compounds have stereocentres but are achiral because of an internal mirror plane.
"Rotating a Fischer projection proves two molecules are identical." Be careful. You may rotate a Fischer projection 180° in the plane and it stays the same, but rotating 90° inverts all configurations. Lifting it out of the plane and flipping it also changes configurations.
"If I see the same substituents, the molecules must be identical." Same substituents with different spatial arrangement can make enantiomers or diastereomers. Always assign R/S before deciding.
⚠️ Nomenclature questions appear on every CHEM 2301 exam. The final exam tested both directions: name → structure and structure → name, with stereochemistry required.
⚠️ Hybridisation questions ask you to identify sp, sp², or sp³ at a specific indicated atom. Lone pairs count as electron groups. Do not forget them.
⚠️ The isomers question (18 points on this practice final) is one of the highest-value sections. Assign R/S at every stereocentre before choosing your answer.
⚠️ Exam instructions state that unclear or illegible answers receive no credit. Draw wedge and dash bonds clearly.
True or false: A molecule with two stereocentres always has exactly four stereoisomers. (False, a meso compound reduces the count.)
Fill in the blank: An atom with three bonding groups and no lone pairs is ______ hybridised. (sp²)
True or false: (R)-2-bromobutane and (S)-2-bromobutane are diastereomers. (False, they are enantiomers.)
Fill in the blank: The IUPAC suffix for an alcohol is ______. (-ol)
True or false: In E/Z nomenclature, Z means the higher-priority groups are on the same side of the double bond. (True)
Q: Draw the structure of (2S,3R)-2-bromo-3-hexanol. Indicate stereochemistry.
A: Draw a six-carbon chain. Place -OH on C3, -Br on C2. At C2, assign groups by CIP priority (Br > chain > CH₃ > H) and arrange wedge/dash so the 1→2→3 arc reads anticlockwise (S). At C3, assign priorities (OH > chain > chain > H) and arrange for clockwise reading (R).
Q: What is the IUPAC name for a cyclohexane ring with a bromine substituent drawn in a specific position?
A: Identify the parent ring (cyclohexane), name the substituent (bromo), and assign the locant. If stereochemistry is shown, assign R or S. A typical answer might be (R)-1-bromocyclohexane or (S)-bromocyclohexane, depending on the drawn configuration.
Q: Determine the hybridisation of the oxygen atom in a carbonyl group (C=O).
A: The oxygen is bonded to one carbon via a double bond and has two lone pairs. Steric number = 3 (one sigma bond + two lone pairs). Hybridisation = sp².
Q: Two structures are drawn as Fischer projections. One shows (R,R) configuration, the other shows (R,S). What is their relationship?
A: They share the same connectivity but differ at one stereocentre while being the same at the other. They are diastereomers.
Q: A molecule has the formula C₆H₁₂. How many degrees of unsaturation does it have?
A: DoU = (2(6) + 2 - 12) / 2 = (14 - 12) / 2 = 1. One degree of unsaturation, meaning one ring or one double bond.
Stereochemistry feeds directly into reaction chemistry: SN2 reactions invert configuration at the stereocentre, while SN1 reactions produce racemic mixtures (a mix of both enantiomers). Understanding R/S assignments is essential for predicting product stereochemistry in addition reactions to alkenes (covered in the reactions study notes). Hybridisation determines bond angles and molecular shape, which in turn affects how a molecule interacts with reagents, the acidity of attached protons, and whether a system is aromatic.
IUPAC naming, organic nomenclature, systematic naming, R/S configuration, Cahn-Ingold-Prelog, CIP priority, E/Z alkene geometry, cis-trans isomerism, stereocentre, chiral centre, chirality, enantiomer, diastereomer, constitutional isomer, structural isomer, meso compound, Fischer projection, wedge-dash notation, hybridisation, sp3, sp2, sp, orbital hybridisation, steric number, degree of unsaturation, index of hydrogen deficiency, CHEM 2301, organic chemistry I, UMN, Salmon