# How many diastereomers can be in any given molecule?

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Jan 11, 2017

The maximum number of diastereomers is ${2}^{n} - 2$.

You have probably learned that the maximum number of optical isomers is ${2}^{n}$, where $n$ is the number of chiral centres.

The number decreases if some of the optical isomers are meso compounds.

The number of diastereomers is less than ${2}^{n}$ because two of the isomers must be a pair of enantiomers.

However, every other optical isomer is a diastereomer of each enantiomer.

Thus, the maximum number of diastereomers is ${2}^{n} - 2$.

If $n = 1$,
${2}^{n} - 2 = {2}^{1} - 2 = \text{2 - 2} = 0$ (no diastereomers).

If $n = 2$,
${2}^{n} - 2 = {2}^{2} - 2 = \text{4 - 2} = 2$ (2 diastereomers).

If $n = 3$,
${2}^{n} - 2 = {2}^{3} - 2 = \text{8 - 2} = 6$ (6 diastereomers).

If $n = 4$,
${2}^{n} - 2 = {2}^{4} - 2 = \text{16 - 2} = 14$ (14 diastereomers).

For example, D-glucose has 4 chiral carbons, so there are 16 aldohexoses (8 D and 8 L).

L-Glucose is an enantiomer of D-glucose, and the other 14 aldohexoses are diastereomers of them.

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