Difficulty: Intermediate | Prerequisites: IUPAC nomenclature, Newman projections, chair conformations
Stereochemistry is about the three-dimensional arrangement of atoms in a molecule and how that arrangement affects the molecule's properties. This set of notes covers how to identify chiral centres, assign R/S configurations using Cahn–Ingold–Prelog rules, distinguish enantiomers from diastereomers from meso compounds, predict optical activity, and calculate degree of unsaturation.
Chiral / chirality
A molecule is chiral if it cannot be superimposed on its mirror image. The most common source of chirality in organic chemistry is a carbon with four different substituents (a stereogenic centre).
Think of it as left-hand vs right-hand: same parts, different spatial arrangement, and you cannot make one fit on top of the other.
Stereogenic centre (stereocentre, chiral centre)
A carbon atom bonded to four different groups. Each stereogenic centre can have two possible configurations (R or S).
R and S configuration (Cahn–Ingold–Prelog rules)
A labelling system for the absolute configuration at a stereogenic centre. Assign priorities 1–4 to the four substituents by atomic number (higher atomic number = higher priority). Orient the molecule so the lowest-priority group (4) points away from you. If the 1→2→3 sequence runs clockwise, the centre is R (Latin: rectus). If anticlockwise, it is S (Latin: sinister).
Enantiomers
A pair of stereoisomers that are non-superimposable mirror images of each other. They have identical physical properties (melting point, boiling point, solubility) except for the direction they rotate plane-polarised light.
In simple terms, enantiomers are the left-hand and right-hand versions of the same molecule.
Diastereomers
Stereoisomers that are not mirror images of each other. They arise when a molecule has two or more stereogenic centres and some, but not all, of the configurations are inverted. Diastereomers have different physical properties.
Meso compound
A molecule that contains stereogenic centres but is achiral overall because it has an internal plane of symmetry. The two "halves" of the molecule are mirror images, cancelling out the optical activity.
Think of it as a molecule that looks like it should be chiral but actually is not, because its own symmetry cancels the chirality.
Optical activity
The ability of a chiral substance to rotate the plane of plane-polarised light. A single enantiomer is optically active. A racemic mixture (equal amounts of both enantiomers) shows zero net rotation.
Racemic mixture (racemate)
A 50:50 mixture of two enantiomers. The equal and opposite rotations cancel, giving no observed optical activity.
Specific rotation ([α])
The measured rotation of plane-polarised light by a chiral compound under standard conditions. It is an intrinsic property of the compound: positive (+) means dextrorotatory, negative (−) means levorotatory.
Degree of unsaturation (index of hydrogen deficiency, DoU)
A formula that tells you how many rings and/or pi bonds a molecular formula implies. For a molecule CₙHₘNₓOᵧXᵤ (X = halogen):
DoU = (2n + 2 − m + x − u) / 2
Each ring or pi bond accounts for one degree of unsaturation.
In simple terms, it is a quick count of how far the molecule departs from being a fully saturated, open-chain compound.
A molecule is chiral if it has at least one stereogenic centre and lacks an internal plane of symmetry.
To check chirality quickly: look at each sp³ carbon and ask whether all four groups are different. If yes, that carbon is a stereogenic centre.
A molecule can have stereogenic centres yet still be achiral if it is a meso compound (internal mirror plane cancels the chirality).
Exam example (Q6): you may be shown several structures and asked "which is chiral?" Check each carbon systematically. Watch for molecules that look asymmetric but have a hidden symmetry plane.
Step 1: Identify the four groups attached to the stereogenic centre.
Step 2: Rank them by Cahn–Ingold–Prelog priority. Compare atoms at the first point of difference. Higher atomic number wins. If tied, move outward to the next atom.
Step 3: Orient the molecule so the lowest-priority group (group 4) points away from you (behind the page).
Step 4: Trace the path 1→2→3. Clockwise = R. Anticlockwise = S.
If group 4 is pointing towards you in the given drawing, you can still assign R/S and then reverse the result (what looks clockwise is actually S, and vice versa).
Enantiomers: every stereogenic centre is inverted between the two molecules. They are mirror images. Equal and opposite optical rotation.
Diastereomers: some, but not all, stereogenic centres are inverted. Not mirror images. Different physical properties.
Meso compounds: contain stereogenic centres but have an internal plane of symmetry that makes the overall molecule achiral. A meso compound has zero optical rotation even though it contains stereocentres.
Exam example (Q8): four structures of a diol are shown. Identify the meso form. Look for the structure where you can draw a mirror plane through the middle, with the stereogenic centres on either side having opposite configurations.
A pure enantiomer rotates plane-polarised light by a characteristic angle (its specific rotation).
An equal mixture of two enantiomers (racemic mixture) shows zero rotation, because the rotations cancel.
Two structures that do not rotate light when mixed in equal amounts are enantiomers of each other.
Exam example (Q9): "which two structures do not rotate polarised light when in equal amounts?" Those two must be an enantiomeric pair.
Formula: DoU = (2C + 2 + N − H − X) / 2, where C = carbons, H = hydrogens, N = nitrogens, X = halogens. Oxygen and sulfur do not appear in the formula (they do not change hydrogen count).
Each degree corresponds to one ring or one pi bond. A benzene ring accounts for four degrees of unsaturation (three C=C pi bonds + one ring).
Exam example (Q7): a complex drug structure is shown. Count carbons, hydrogens, nitrogens, oxygens and halogens from the molecular formula, then apply the formula. Alternatively, count rings and pi bonds directly from the structure.
When asked "what is the relationship between these two structures," work through a decision tree:
Same molecular formula? If no → different compounds.
Same connectivity? If no → constitutional isomers.
Mirror images? If yes → enantiomers (or same compound, if superimposable). If no → diastereomers.
Superimposable? If yes → identical (same molecule).
Exam example (Q11): two structures with OH and Br substituents. Check whether inverting all stereocentres in one gives the other (enantiomers) or only some (diastereomers).
Students often assume that any molecule with stereogenic centres must be chiral. Meso compounds are the key exception: they contain stereocentres but are achiral due to an internal mirror plane.
R/S configuration has nothing to do with the direction of optical rotation. R does not mean (+) and S does not mean (−). The relationship between configuration and rotation can only be determined experimentally.
Students confuse enantiomers with diastereomers. The test is straightforward: if every stereocentre is inverted, the pair are enantiomers. If only some are inverted, they are diastereomers.
Degree of unsaturation: students sometimes include oxygen in the DoU formula. Oxygen (and sulfur) do not change the hydrogen count and are left out of the calculation entirely.
⚠️ Identifying stereogenic centres in complex drug molecules is a reliable exam question (Problem II on this exam). Practice spotting them in large, intimidating structures by checking each sp³ carbon systematically.
⚠️ R/S assignment will appear. Know the CIP priority rules cold, and be comfortable with the "group 4 pointing towards you" shortcut (assign and flip).
⚠️ Meso identification is a favourite exam trap. If you see a molecule with two stereogenic centres that is symmetric in the middle, check for meso.
⚠️ The "what is the relationship" question type appears repeatedly. Practise the decision tree: same formula → same connectivity → mirror image → superimposable.
True or False: A meso compound has no stereogenic centres. (False. It has stereogenic centres, but an internal mirror plane makes it achiral overall.)
Fill in the blank: If the 1→2→3 priority sequence runs anticlockwise (with group 4 pointing away), the configuration is ___. (S)
True or False: Enantiomers have different melting points. (False. Enantiomers have identical physical properties except the direction of optical rotation.)
Fill in the blank: The degree of unsaturation for benzene (C₆H₆) is ___. (4)
True or False: Diastereomers are mirror images of each other. (False. Diastereomers are stereoisomers that are not mirror images.)
Q: Four structures (I–IV) are shown, each with different spatial arrangements of F, CH₃, and H around one or more sp³ carbons. Which is chiral?
A: Check each structure for a carbon bonded to four different groups (a stereogenic centre) and for any internal mirror planes. A structure with a stereogenic centre and no mirror plane is chiral. A structure that is symmetric despite apparent stereocentres (e.g. two identical halves) is meso and therefore achiral.
Q: A complex drug structure is shown. You are asked to mark all stereogenic centres. How do you avoid losing points?
A: Go carbon by carbon through the structure. At each sp³ carbon (no double bonds), list the four groups. If all four are different, mark it. Be careful with ring junctions: the two "paths" around the ring count as different groups if they differ in any way. An incorrect mark cancels a correct one, so do not guess.
Q: Four diol structures (I–IV) are shown. Which is the meso form?
A: Look for the structure with two stereocentres whose configurations are opposite (one R, one S) and an internal mirror plane through the middle of the molecule. That structure is meso.
Q: Two structures with OH and Br are shown. What is their relationship?
A: Check connectivity first (same atoms, same bonds → stereoisomers, not constitutional isomers). Then compare each stereocentre. If all are inverted → enantiomers. If some but not all → diastereomers. If identical at every centre → same molecule.
Q: What is the degree of unsaturation in a drug with molecular formula C₂₃H₂₅NO₅?
A: DoU = (2(23) + 2 + 1 − 25) / 2 = (46 + 2 + 1 − 25) / 2 = 24 / 2 = 12. Oxygen does not appear in the formula. Each ring and each pi bond counts as one.
Stereochemistry feeds directly into substitution and elimination reactions. SN2 reactions invert the configuration at the stereogenic centre (Walden inversion), while SN1 reactions produce racemic mixtures because the carbocation intermediate is planar. Knowing R/S lets you predict and verify the stereochemical outcome of every substitution reaction you will meet.
Meso compounds and optical activity also connect to spectroscopy: if your product is optically inactive, you need to consider whether you made a racemic mixture or a meso compound, and these have different NMR signatures.
Stereochemistry, chirality, chiral centre, stereogenic centre, stereocentre, asymmetric carbon, R configuration, S configuration, Cahn–Ingold–Prelog, CIP rules, CIP priority, enantiomer, diastereomer, meso compound, achiral, optical activity, optically active, plane-polarised light, specific rotation, racemic mixture, racemate, racemisation, degree of unsaturation, index of hydrogen deficiency, DoU, IHD, unsaturation number, constitutional isomers, structural isomers, stereoisomers, mirror image, superimposable, non-superimposable, CHEM 225, organic chemistry II, University of Michigan