How many diastereomers can be in any given molecule?

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Answer 1 · 2017-01-11T21:36:04

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 $n = 1$ ,
$2^n-2 = 2^1-2 = "2 - 2" =0$ (no diastereomers).

If $n = 2$ ,
$2^n-2 = 2^2-2 = "4 - 2" =2$ (2 diastereomers).

If $n = 3$ ,
$2^n-2 = 2^3-2 = "8 - 2" =6$ (6 diastereomers).

If $n = 4$ ,
$2^n-2 = 2^4-2 = "16 - 2" =14$ (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.