Chiral recognition by cyclic oligosaccharides. Enantioselective complexation of binaphthyl derivatives with cyclodextrins
Chiral recognition of binaphthyl derivatives, such as 1,1′-bi-2-naphthol (1), 1,1′-binaphthyl-2,2′-diyl hydrogen phosphate (2), and 2,2′-dihydroxy-1,1′-binaphthyl-3,3′-dicarboxylic acid (3), by cyclodextrins (CDxs) has been studied. The S enantiomers of 1 and 2 are bound to heptakis(2,3,6-tri-O-meth...
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Veröffentlicht in: | Supramolecular chemistry 1993-07, Vol.2 (2-3), p.137-143 |
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Hauptverfasser: | , , , , , , |
Format: | Artikel |
Sprache: | eng |
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Zusammenfassung: | Chiral recognition of binaphthyl derivatives, such as 1,1′-bi-2-naphthol (1), 1,1′-binaphthyl-2,2′-diyl hydrogen phosphate (2), and 2,2′-dihydroxy-1,1′-binaphthyl-3,3′-dicarboxylic acid (3), by cyclodextrins (CDxs) has been studied. The S enantiomers of 1 and 2 are bound to heptakis(2,3,6-tri-O-methyl)-β-CDx (TMe-β-CDx) as well as β-CDx more strongly than the R enantiomers. The molecular mechanics and molecular dynamics calculations for the 1:1 complex of 1 and β-CDx suggest that more effective van der Waals contacts and intermolecular hydrogen bonding stabilize the complex of S-1 compared with that of R-1. Meanwhile the R enantiomer of 3 is the preferable guest for β- and TMe-β-CDxs. Circular dichroism spectroscopy suggests that the complex of S-3 is more unstable than that of R-3 because the dihedral angle of the naphthalene planes of S-3 needs to be reduced for forming the inclusion complex. The enantiomers of the guest binaphthyls are completely separated by means of capillary zone electrophoresis (CZE) when TMe-β-CDx is used as a separating agent. The results of the CZE correspond well with those of the binding constants of the inclusion complexes. |
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ISSN: | 1061-0278 1029-0478 |
DOI: | 10.1080/10610279308038308 |