Determination of the sign of a one-bond 15N–15N spin–spin coupling constant for a trans-diazene by selective 15N decoupling experiments in 13C nuclear magnetic resonance spectroscopy

Abstract

The sign of the one-bond ¹⁵N–¹⁵N spin–spin coupling constant, ¹J(N–N), of a trans-diazene (4-acetyl-aminoazobenzene) has been determined to test the theoretical prediction that ¹J(N–N) would be positive in sign when nitrogens have s-hybridized lone pairs and take a trans conformation. The sign was first determined relative to those of ¹³C–¹⁵N spin-spin coupling constants, ⁿJ(C–N), by selective ¹⁵N decoupling experiments and then finally referenced to that of ¹J(C–H) in the benzene ring, which undoubtedly bears a positive sign, by selective ¹⁵N and ¹H decoupling experiments. It was found that the sign of ¹J(N–N) of the trans-diazene is negative in contrast to the theoretical prediction. This discrepancy is attributed to the large negative contribution from the orbital–dipole term which was not correctly evaluated in previous calculations for the ¹J(N–N) coupling of the NN bond.