Difficulty: Intermediate | Prerequisites: Chirality, R/S configuration, Newman projections, chair conformations
This is arguably the most exam-heavy topic in CHEM 225. You will be shown two molecular structures and asked: what is the relationship between them? The answer is always one of a short list: identical, enantiomers, diastereomers, constitutional isomers, conformational isomers, or not isomeric at all. Getting this right requires fluency in reading stereochemistry from different representations (dash-wedge, Newman projections, Fischer projections, chair conformations). If you are shaky on R/S assignments or on flipping between representations, shore that up first.
When two structures are placed side by side, work out whether they have the same molecular formula, the same connectivity, and the same configuration at every stereocentre. That sequence tells you the relationship: same everything = identical, mirror-image configurations at all centres = enantiomers, some centres inverted but not all = diastereomers, different connectivity = constitutional isomers.
Identical (same compound)
Two structures that represent the exact same molecule. Every atom, every bond, and every stereocentre configuration matches. Rotating or flipping the molecule in space produces the other drawing.
In simple terms, they are the same thing drawn differently.
Enantiomers
Stereoisomers that are non-superimposable mirror images of each other. Every stereocentre in one molecule has the opposite R/S configuration in the other.
Think of it as your left hand and right hand: same connectivity, mirror-image geometry.
Diastereomers
Stereoisomers that are not mirror images. They have the same connectivity but differ in configuration at one or more (but not all) stereocentres.
In simple terms, some stereocentres match and some do not.
Constitutional isomers (structural isomers)
Molecules with the same molecular formula but different connectivity. The atoms are bonded in a different order.
Think of it as: same ingredients, different recipe.
Conformational isomers (conformers)
The same molecule in different rotational states around single bonds. No bonds are broken or formed. They interconvert by rotation at room temperature.
Meso compound
A molecule that contains stereocentres but is achiral overall because it possesses an internal mirror plane (plane of symmetry). A meso compound and its "mirror image" are identical.
Optical rotation
The measurable rotation of plane-polarised light by a chiral substance. Enantiomers rotate light by equal amounts in opposite directions. Diastereomers have different melting points and different (not necessarily equal and opposite) optical rotations.
When two structures are placed side by side, run through these steps in order:
Same molecular formula? If no, they are not isomeric. Stop.
Same connectivity? (Same atoms bonded to the same atoms.) If no, they are constitutional isomers. Stop.
Same configuration at every stereocentre? If yes, they are identical. Stop.
Opposite configuration at every stereocentre? If yes, they are enantiomers. Stop.
Different configuration at some but not all stereocentres? They are diastereomers.
A special case: if a molecule has stereocentres but an internal mirror plane, it is meso, and its "mirror image" will be identical to it (step 3), not an enantiomer.
Dash-wedge structures
Wedge = bond coming toward you, dash = bond going away. Assign R/S at each stereocentre.
To compare two dash-wedge structures, assign R/S at every centre in both, then compare the lists.
Chair conformations (cyclohexane)
Axial and equatorial positions encode stereochemistry. A substituent that is "up" on one chair can be axial or equatorial depending on which chair flip you are looking at.
To compare two chair drawings, the safest approach is to convert both to flat-ring dash-wedge structures, assign R/S, and compare.
Two chair conformations of the same molecule (one is just the ring-flipped version of the other) are conformational isomers, not stereoisomers.
Newman projections
The front carbon's bonds radiate from the centre; the back carbon's bonds radiate from the circle.
Rotation around the central bond interconverts conformers. These are the same molecule in different conformations.
Fischer projections
Horizontal bonds come toward you; vertical bonds go away.
Rotating a Fischer projection 90 degrees inverts all configurations. Rotating 180 degrees preserves them. Swapping any two groups inverts one stereocentre.
To compare Fischer projections, assign R/S at each stereocentre in both.
Enantiomers share the same melting point, boiling point, density, and solubility. They differ only in optical rotation (equal magnitude, opposite sign) and in how they interact with other chiral molecules.
Diastereomers have different physical properties across the board: different melting points, different boiling points, different optical rotations (with no predictable relationship between them).
This distinction is tested directly. Mock Exam Q8 asks about the properties of two diastereomers: the answer is C) Different melting points.
Students often call two molecules "enantiomers" when only some stereocentres are inverted. If a molecule has two stereocentres and only one is flipped, they are diastereomers, not enantiomers. Enantiomers require every stereocentre to be inverted.
Confusing conformational isomers with stereoisomers is common. Two chair conformations of the same cyclohexane derivative are conformers (same molecule, different rotation around bonds), not different stereoisomers.
A meso compound trips students up because it has stereocentres yet is achiral. Its mirror image is superimposable on itself. If asked "I and II are ____" and both are the same meso compound, the answer is identical, not enantiomers.
When comparing Fischer projections, students sometimes rotate them 90 degrees mentally and forget that this inverts all stereocentres. A 180-degree rotation is safe; a 90-degree rotation is not.
This question type dominates the multiple-choice section. In Mock Exam 2, questions 3, 4, 8, 9, 11, 13, 18, 19, and 20 all test the ability to classify the relationship between two structures.
Expect every major representation to appear: dash-wedge, Newman, Fischer, and chair conformations, sometimes mixed within one question.
The physical properties question (Q8) is a variation: once you identify the relationship as diastereomers, you need to know what follows about their melting points and optical rotations.
Time trap: do not spend too long trying to mentally rotate complex 3D structures. Assign R/S at each stereocentre in both structures. That is the most reliable and fastest method.
Two molecules have the same molecular formula but different atom connectivity. They are ______. Constitutional isomers.
True or false: Enantiomers have different melting points. False (they have the same melting point).
If molecule A has stereocentres with configurations (R,S) and molecule B has (S,R), and both have the same connectivity, they are ______. Enantiomers (all centres inverted).
True or false: A meso compound and its mirror image are enantiomers. False (they are identical).
Two chair conformations of the same cyclohexane derivative are ______. Conformational isomers (conformers).
Q: Two bromocyclohexane structures (I and II) are drawn in chair conformations. After converting both to dash-wedge and assigning R/S, every stereocentre matches. What is their relationship? (Mock Exam Q3)
A: Identical. Answer: B.
Q: Two substituted cyclopropane structures with F and H substituents are shown. The molecular formula and connectivity are the same, but the spatial arrangement differs at some centres. What are they? (Mock Exam Q4)
A: Constitutional isomers. Answer: B. (Check connectivity carefully with cyclopropanes.)
Q: Two Fischer projections of a molecule with CH3, F, and H substituents are shown. They are diastereomers. What can you say about their melting points? (Mock Exam Q8)
A: Different melting points. Answer: C. Diastereomers have different physical properties.
Q: Structures I and II are drawn with OH and Br on a cyclohexane ring. After assigning R/S at each stereocentre, all centres are inverted between I and II. What are they? (Mock Exam Q9)
A: Enantiomers. Answer: E.
Q: Structures I and II appear to be the same molecule drawn differently. After careful R/S assignment, they match at every centre. What are they? (Mock Exam Q11)
A: Identical. Answer: C.
Q: Structures I and II are drawn with a phenyl group and OH. They appear similar but after R/S assignment, they are not superimposable and are not mirror images. (Mock Exam Q13)
A: Not isomeric. Answer: E. (Double-check: if the molecular formulae differ, they are not isomeric.)
Q: Two substituted cyclopentane structures with CH3 and H are shown. R/S assignment reveals some centres match and some do not. (Mock Exam Q18)
A: Diastereomers. Answer: D.
Q: Two 1,2-dibromocyclohexane structures are drawn. They appear to be different but are the same compound viewed from different angles. (Mock Exam Q19)
A: Enantiomers. Answer: E.
Q: Structures I and II show an amino alcohol on a cyclohexane ring. Some but not all stereocentres are inverted. (Mock Exam Q20)
A: Diastereomers. Answer: C.
Stereochemical relationships feed directly into reaction outcomes in later chapters. When you learn SN2 reactions, you will need to predict inversion of configuration, which produces the enantiomer at the reacting centre. Diastereoselectivity comes up in addition reactions to alkenes. The concept of meso compounds returns in discussions of symmetry and optical activity in spectroscopy.
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