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Diastereomers and resolution of racemic mixture

Short answer

Diastereomers are stereoisomers that are not mirror images, while enantiomers are mirror-image stereoisomers. Hence all diastereomers are stereoisomers, but enantiomers are not diastereomers. To resolve a racemic acid, react it with an optically pure base to give diastereomeric salts; separate these by fractional crystallization, then treat each salt with HCl to recover the pure enantiomers.

All the diastereomers are stereoisomers but all the stereoisomers are not diastereomers. Explain. Describe the chemical method for resolution of racemic mixture.

Diastereomers and stereoisomers

Stereoisomers are isomers with the same constitution but a different arrangement of atoms in space. They are of two types: enantiomers (mirror images) and diastereomers (not mirror images). So every diastereomer is a stereoisomer, but not every stereoisomer is a diastereomer: a pair of enantiomers are stereoisomers of each other but not diastereomers. Example: for (2R,3R)-tartaric acid, the (2S,3S) form is its enantiomer and the meso form is its diastereomer. Both are stereoisomers of it, but only the meso form is a diastereomer.

Chemical resolution of a racemic mixture

Resolution is the separation of a racemic mixture into its pure enantiomers. Enantiomers have identical physical properties, so they are first converted into diastereomers.

Chemical (diastereomeric salt) method for a racemic acid (±)-A:

  1. React it with an optically pure base such as (−)-brucine, strychnine or (R)-1-phenylethylamine: (+)-A·(−)-B and (−)-A·(−)-B are formed. These two salts are diastereomers.
  2. They have different solubilities, so separate them by fractional crystallisation.
  3. Treat each separated salt with a strong acid (HCl) to liberate the pure (+)-A and (−)-A, and recover the base.
(±)-acid + (−)-base --> [(+)acid·(−)base] + [(−)acid·(−)base]
                          fractional crystallisation
                       --HCl--> (+)-acid        --HCl--> (−)-acid

A racemic base is resolved in the same way using an optically active acid such as (+)-tartaric acid.

✓ Answer checked by Claude Opus 5.5 (high effort)

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