“Symmetrical” and asymmetrical (NH ⋯ N)+ hydrogen bonds. Infrared investigations

Abstract

(N₍₁₎H ⋯ N₍₂₎)⁺CIO^–₄ complexes of 5- and 6-membered aromatic rings and of bicyclic non-planar N-base molecules in acetonitrile solutions are investigated by i.r. spectroscopy. Studied are “symmetrical” complexes (N₍₁₎=N₍₂₎) and asymmetrical complexes (N₍₁₎≠N₍₂₎). The stability of “symmetrical” complexes increases with increasing pK_a. In the region 2700-1900 cm^–1, two broad bands are found with aromatic compounds, but only one with bicyclic ones. It is shown that Fermi resonance and not proton tunnelling is responsible for the band splitting. In this resonance effect the NH⁺ stretching vibration in the hydrogen bonds and a combination vibration, in which the NH⁺ bending vibration and a ring vibration take part, are involved. In addition, a continuous absorption is observed which begins at these bands and extends toward smaller wave numbers. This continuum is also caused by (NH⋯N)⁺ bonds and demonstrates that these bonds are easily polarisable, i.e., they have a double minimum energy surface when isolated from their environment. The fact that a continuum as well as the band pair are found in the spectra of solutions with (NH⋯N)⁺ hydrogen bonds is explained.

If one partner is an asymmetrical complex is non-planar and the proton is present on this base molecule, only one band is observed in the region 2800–2000 cm^–1 instead of the band pair which is observed with complexes formed by two planar N-base molecules. The band splitting with asymmetrical complexes may also be explained by Fermi resonance. The dependence of the relative intensities on the ΔpK_a may also be understood by this assumption. When the ΔpK_a between the two N-base molecules becomes smaller, i.e., the degree of asymmetry of the hydrogen bonds decreases, a continuous absorption occurs, demonstrating that the asymmetrical (NH⋯N)⁺ hydrogen bonds may become easily polarisable, too.