The low-temperature 13C and 2H spectra and relaxation rates of methyl groups depend on the azimuth of B0 in the molecular frame.: Those of 1H do not.: Why?

It is shown that the answer to the question asked in the title is: Because the axial symmetry axes of the H-H dipolar coupling tensors in a-CH3 group are perpendicular to the (assumed) threefold axis of the group. By contrast, those of the C-13-H dipolar and H-2 quadrupolar coupling tensors are not. The use of "symmetry adapted" spin functions and of a symmetry adapted form of the (dipolar) coupling Hamiltonian greatly simplifies the analysis. (C) 2003 Elsevier Inc. All rights reserved.