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I am told that all anomers are diastereomers. I think this is BS. If there is a carbohydrate with only one chiral center, then changing the configuration of the anomeric carbon gives us enantiomers.

Or is it impossible to have carbohydrates with only one chiral center?

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  • $\begingroup$ It could be possible to have a pair of enantiomeric anomers if you started from an achiral sugar. But this is not possible. Another way would be to extend the definition of anomers. Then, using jerepierre's theoretical example, DL-glyceraldehyde would give a mixture of 4 stereoisomers including 2 pairs of enantiomeric "anomers" $\endgroup$ – K_P Oct 22 '14 at 23:27
  • $\begingroup$ Sugars are by definition chiral then @K_P? $\endgroup$ – Dissenter Oct 23 '14 at 1:18
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    $\begingroup$ I don't think the definition of sugars prohibits chirality. Obviously aldoses can not be achiral because of their inherent asymmetry. But after your comment I thought of one possible achiral sugar: a symmetric ketopentose. This has 2 stereocentres but is a meso compound and thus achiral. IF it exists and IF it can form anomers then it should give enantiomeric anomers. $\endgroup$ – K_P Oct 23 '14 at 7:39
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The only saccharides I know of that have only a single chiral centre are the trioses L- and D-glyceraldehyde, but, at least according to the Gold Book definition, anomers are specifically diastereomers of cyclic forms of sugars. Trioses are too short for a cyclic form to normally arise and therefore can't anomerize.

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Adding a bit to Michael's answer. Although it would be highly strained, let's look at what happens if glyceraldehyde cyclizes. Since glyceraldehyde has one stereocenter, the anomeric position becomes a second, giving diastereomers.

enter image description here

So in order for the anomers to be enantiomers, the compound actually must have zero stereocenters in the open form. If you consider glycoaldehyde a carbohydrate and you consider the three-membered ring cyclic hemiacetal a reasonable structure, then you have found a set of anomers that are enantiomers.

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  • $\begingroup$ Ah, you're quite right. I hadn't considered that. Since the modern biochemical definition of saccharide seems to preclude anything shorter than trioses (and those would be even harder to cyclize), I guess we're okay. $\endgroup$ – Michael DM Dryden Oct 22 '14 at 20:48

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