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Abstract

The focus of this study is to understand the origin of the chiral recognition for a host–guest system containing complexes with different stoichiometries. Each enantiomer of 2-naphthyl-1-ethanol forms two different 1:1 complexes with β-cyclodextrin, leading to the formation of three different 2:2 complexes. One of these 2:2 complexes leads to excimer emission of the guest. Fluorescence studies were employed to determine the binding isotherms for the 1:1 and 2:2 complexes. No chiral discrimination was directly observed for the formation of the 1:1 complexes, while higher equilibrium constants (29% from binding isotherms and 40% from kinetic studies) were observed for the formation of the 2:2 complexes with (R)-2-naphthyl-1-ethanol when compared to the formation of the 2:2 complexes formed from (S)-2-naphthyl-1-ethanol. The relaxation kinetics was studied using stopped-flow experiments. The formation of the 2:2 complexes was followed by detecting the excimer emission from one of the 2:2 complexes. The relaxation kinetics was faster for (S)-2-naphthyl-1-ethanol, where a higher dissociation rate constant, by 47%, was observed, suggesting that the chiral discrimination occurs because the interaction between two cyclodextrins is more favorable for the complexes containing (R)-2-naphthyl-1-ethanol when compared to (S)-2-naphthyl-1-ethanol. The same overall equilibrium constants were observed for the 1:1 complexes with both enantiomers showing that at a given cyclodextrin concentration the sum of the two types of 1:1 complexes is the same for both enantiomers. However, analysis of the binding isotherms indicates that the ratio between the two different 1:1 complexes for each enantiomer was different for (R)- and (S)-2-naphthyl-1-ethanol.