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The isomeric SMILES for D-glucose is C([C@@H]1[C@H]([C@@H]([C@H](C(O1)O)O)O)O)O.

The Haworth projection looks like this:

Haworth project of D-Glucose

I can parse the SMILES and draw the diagram, but the question is how to translate the CW/ACW of isomeric SMILES into up/down in the Haworth projection.


In the example given, the SMILES @@/@/@@/@ matches up/down/up/down, reading anti-clockwise around the ring, but I am far from confident that will always be the case.

[The SMILES string does not treat the last carbon as a stereo centre.]

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  • $\begingroup$ Isn't this question basically the same as "How to draw a Haworth projection"? Because I believe the clockwise/anticlockwise notation uses the same convention as the Fischer projection $\endgroup$
    – Szgoger
    Aug 3, 2022 at 14:00
  • $\begingroup$ @Szgoger: Perhaps it is, but I haven't been able to confirm this is so. See edit. $\endgroup$
    – david.pfx
    Aug 4, 2022 at 13:45

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Wikipedia link states:

Configuration at tetrahedral carbon is specified by @ or @@. Consider the four bonds in the order in which they appear, left to right, in the SMILES form. Looking toward the central carbon from the perspective of the first bond, the other three are either clockwise or counter-clockwise. These cases are indicated with @@ and @, respectively (because the @ symbol itself is a counter-clockwise spiral).

So I suppose your question is equivalent to asking if a SMILES string always maps to a Haworth projection in such a way that switching between @ and @@ strictly corresponds to alternating between opposite 'sides' of the pseudo-3D representation, and vice versa.

I am not sure, either.
What I can show you however (hence my answer) is that if you use python rdkit.Chem to translate your SMILES to a molecule, you get this:

m = Chem.MolFromSmiles('C([C@@H]1[C@H]([C@@H]([C@H](C(O1)O)O)O)O)O')

enter image description here

But if you then translate m back to SMILES (which makes it 'canonical' by default):

s = Chem.MolToSmiles(m)

'OC[C@H]1OC(O)[C@H](O)[C@@H](O)[C@@H]1O'

Now there are two consecutive @@ notations, although no consecutive carbons in the ring have OH substituents on the same side of the 'plane' (the hemiacetal OH in your projection should be flat, not down, as its configuration is indeed not specified in the SMILES).

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  • $\begingroup$ Thank you for your answer. My sense is that the SMILES rule based on the order in which elements are added does not reliably translate to any particular geometric configuration. I think your example tends to confirm that, but doesn't really solve my problem. $\endgroup$
    – david.pfx
    Aug 26, 2022 at 12:47

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