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 13C-H dipolar and 2H 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.
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