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.
