The Time Dependence of NMR Spectroscopy

We expect the signal for the methyl protons adjacent to the diastereotopic hydrogens of 2-bromobutane to be a doublet of doublets as a result of splitting by the nonequivalent diastereotopic hydrogens. The signal, however, is a triplet. Use a splitting diagram to explain why it is a triplet rather than a doublet of doublets. 

We have seen that the three methyl hydrogens of ethyl bromide produce just one signal in the NMR spectrum because they are chemically equivalent due to rotation about the C—C bond. At any given instant, however, the three hydrogens can be in quite different environments. For example, one can be anti to the bromine, one can be gauche to the bromine, and one can be eclipsed with the bromine. 

The average time to take an NMR spectrum is about one second. Any process (in this case rotation about a C—C bond) that happens faster than once every second will cause the signals to average. Therefore, we see only one signal in the H NMR spectrum for the three methyl hydrogens of ethyl bromide. The signal represents an average of their environments. 

A series of H NMR spectra of oyolohexane-d-n obtained as the temperature is lowered from -49 °C to-89 °C. 

---