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Stereoisomers that are not mirror images are diastereomers: pairs differing at some but not all centres, and cis/trans alkenes; they differ in physical properties and can be separated, and a symmetrical molecule's meso form lowers the stereoisomer count.
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You will classify pairs of configurations as identical, enantiomers or diastereomers, count stereoisomers when a meso form exists, decide whether stereoisomers can be separated, and place cis/trans alkenes.
You know enantiomers are mirror images with identical ordinary properties. With more than one stereocentre, some pairs of stereoisomers are not mirror images at all.
Diastereomers are stereoisomers that are not enantiomers. A meso compound has stereocentres but an internal mirror plane, so it is achiral and is its own mirror image.
2,3-dichloropentane has stereocentres at C2 and C3, so four combinations: (2R,3R), (2S,3S), (2R,3S), (2S,3R).
Compare any two centre by centre:
Diastereomers are not mirror images, so their shapes genuinely differ, and so do their melting points, boiling points and solubilities. That means they can be separated by ordinary distillation or crystallization, which enantiomers cannot.
Another way: steps
To classify a pair of stereoisomers:
When a molecule's two halves are identical, as in 2,3-dibromobutane, the (R,S) form has a mirror plane through its middle. It is its own mirror image — (R,S) and (S,R) are one compound — and it is achiral despite having two stereocentres. That is a meso compound, and it takes the count down from four to three: a pair of enantiomers plus the meso form. Tartaric acid, from grapes, is the classic case.
Diastereomers need not have stereocentres. (E)- and (Z)-but-2-ene are stereoisomers that are not mirror images, so they are diastereomers, with different boiling points.
Any two stereoisomers are enantiomers. Only mirror-image pairs are.
Diastereomers have identical properties. They differ, and can be separated.
Two stereocentres always give four stereoisomers. A meso form lowers the count.
A molecule with stereocentres must be chiral. A meso compound is achiral.
C2: R against S — differs. C3: S against S — the same.
Centre by centre.
One of two centres differs.
Some, not all.
So they are diastereomers, with different boiling points.
Separable by distillation.
How many centres differ?
Both.
What is the relationship?
Enantiomers: the chiral pair, alongside the meso form.
2,3-dichloropentane has stereocentres at C2 and C3. How are the (2S,3S) and (2R,3S) forms related?
Match each pair to its relationship.
| enantiomers | diastereomers | structural isomers | |
|---|---|---|---|
| (R)- and (S)-butan-2-ol | |||
| (2R,3R)- and (2R,3S)-2,3-dichloropentane | |||
| (E)- and (Z)-but-2-ene | |||
| butan-1-ol and butan-2-ol |
2,3-dibromobutane, CH3CHBrCHBrCH3, has two stereocentres, and its two halves are identical. How many stereoisomers does it actually have?
Answer:
A reaction gives a mixture of the (2R,3R) and (2R,3S) forms of a compound. Can they be separated by ordinary distillation or crystallization?
(Z)-but-2-ene boils at 4 °C and (E)-but-2-ene at 1 °C. Neither has a stereocentre. What is their relationship?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For 2,3-dichloropentane, give the relationship of each pair.
| relationship | |
|---|---|
| (2R,3R) and (2S,3S) | |
| (2R,3R) and (2S,3R) | |
| (2S,3R) and (2R,3S) | |
| (2R,3S) and (2R,3R) |
You can tell diastereomers from enantiomers. Tell someone why 2,3-dibromobutane has only three stereoisomers. Next: how a reaction's mechanism decides the stereochemistry of its product.
8. Your turn: (2R,3R)- and (2S,3S)-2,3-dibromobutane, step 3