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.