| Abstract | The IR spectra of the O-H stretch for hydrogen bonds (HBs) arising from complex formation between the HB donor (HBD), 4-fluorophenol, and the HB acceptors, peroxides and ethers, frequently show asymmetry that appears to arise from two incompletely resolved bands from two different complexes, but the O-H HB bands with the HBD methanol are symmetric (M. Berthelot, F. Bessau, and C. Laurence. Eur. J. Org. Chem. 925 (1998)). The present studies show that this difference in O-H HB band shapes also is true for other phenols and alcohols. However with ethylene oxide, 4-fluorophenol gives an almost symmetric O-H HB band with a very broad maximum, while alcohols give symmetric O-H HB bands with well-defined maxima. It is shown by experiment that the unusual O-H HB band shapes for the phenols are not due to Fermi resonance and are unrelated to the enthalpies of HB complex formation. Theoretical exploration of the potential energy (PE) surfaces for complexes of 4-fluorophenol and methanol with tert-butyl methyl ether and ethylene oxide reveals that O-H HB band asymmetry or broadness cannot be ascribed to the presence of two different HB complexes. For this ether, the PE surfaces for rotation about the HB and for up-and-down motion of the HBD with respect to the COC plane of the ether are relatively symmetric for methanol, but are strongly asymmetric for 4-fluorophenol, hence the differences in the O-H HB band shapes. The PE surfaces for the epoxide are effectively symmetric, but the PE for rotation about the HB has a single broad minimum for methanol, whereas with 4-fluorophenol there are two minima owing to attractive interactions between the phenyl group and the CH₂ groups of the epoxide. The previously unknown βH₂ values for ethylene oxide and tetramethylethylene oxide are 0.36 and 0.58, respectively. |
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