A 1H NMR titration study on the binding constants for D- and l-tryptophan inclusion complexes with 6-O-α-D-glucosyl-β-cyclodextrin. Formation of 1:1 and 2:1 (host:guest) complexes

•1:1 and 2:1 (host:guest) complexes of D-, l-tryptophan with G1-β-CD are formed.•1H NMR titration method revealed the two step inclusion process.•The 1st inclusion is on indole-ring, and the 2nd is on the β-carbon of tryptophan.•The 2nd inclusion shows stronger chiral recognition for the enantiomers than the 1st.

A 1H NMR titration study revealed that 6-O-α-D-glucosyl-β-cyclodextrin (G1-β-CD) forms 1:1 and 2:1 (host:guest) inclusion complexes with D- and l-tryptophan in alkaline D2O solutions (pD 11.0). The binding constants (K1K1’s) for the 1:1 complexes of D-isomer at 298 K (59 mol−1 dm3) were virtually equal to that of L-isomer (54 mol−1 dm3). On the other hand, the K2 values for 2:1 complexes of D-isomer (42 mol−1 dm3) were larger than that of L-counterpart (12 mol−1 dm3). These facts suggest that the first CD molecule includes the indole ring moiety of tryptophan, followed by inclusion with the second CD molecule in the vicinity of chiral center, α-carbon of the guest, to result in the difference in K2K2’s for two enantiomers. Two-dimensional NMR measurement (Rotating-frame nuclear Overhauser Effect SpectroscopY, ROESY) supported this interpretation.

Graphical abstractTwo-step inclusion between 6-O-α-D-glucosyl-β-cyclodextrin and D-, l-tryptophan is found. The second host includes chiral-center of the guests (asterisk, *).Figure optionsDownload full-size imageDownload as PowerPoint slide