13C - 14N J coupling

It has always puzzled me how students so willingly accept the fact that ¹³C is J coupled to a spin I = 1 nucleus like ²H (eg. the multiplets for the ¹³C spectra of deuterated solvents) however they do not question the fact that ¹³C almost never shows a J coupling to ¹⁴N (another spin I = 1 nucleus). The reason that ¹³C exhibits J coupling to ²H is that the relaxation among the three energy levels of ²H is rather slow and each ¹³C "sees" the ²H in each of its three Zeeman states, splitting the ¹³C resonance into 3 lines of equal intensity. The efficiency of the relaxation among the energy levels of quadrupolar nuclei (among other things) depends on the magnitude of the quadrupolar coupling constant - the larger the quadrupolar coupling constant, the faster the relaxation. Unlike ²H, which has a very small quadrupolar coupling constant, ¹⁴N (and most other quadrupolar nuclei) has a substantial quadrupolar coupling constant and therefore the relaxation among its energy levels is very fast. The ¹³C therefore "sees" the ¹⁴N in a single "average" state and as a result is a singlet. There are however compounds where the ¹⁴N is in a very symmetric environment. The high symmetry make the quadrupolar coupling constant much smaller and the relaxation therefore much slower. In these cases one can observe the ¹³C - ¹⁴N J coupling. Depending on the relaxation of the ¹⁴N, the ¹³C lines can vary from being a sharp 1:1:1 triplet, to a broad unresolved triplet, to a very sharp singlet. Below is an example of a case where the ¹⁴N is in a very symmetric environment and the ¹³C - ¹⁴N J couplings are resolved.