The hydrogens adjacent to the chirality center are not chemically equivalent. A simple test is to replace them, for example with fluorine. The result would be an RR, RS, and SS, SR mixture.
In this case the chemical shifts are sufficiently different that you can see each hydrogen clearly. That may not be the case. See the 2-chloro butane shown here: http://faculty.tlu.edu/nmr/fac.html.
Thanks for the reply. The link was helpful. I also found
this page to be helpful for explaining diastereotopic hydrogens in NMR, too.
The spectrum is run in TFA, so the NH2 splitting shows a time averaging effect resulting in the NH2 peak being broadened.
Are you sure this is from the TFA? Shouldn't the peak be further downfield?
An article states it's usually >10 ppm.For the splitting, deuterium does not split and should simplify a spectrum. Because the CH2-hydrogens are different, they can, and in this case do have different J-values.
Why doesn't coupling with the the CH
2-hydrogens split the CH peak?
orgopete, I am not sure what you mean about deuterium not splitting. It is occasionally difficult to observe splitting between deuterium and protium because the coupling constants are small, but in some cases it is quite apparent.
I also thought Deuterated samples elicit J
D-H coupling. I think maybe the coupling is too small to resolved here.
One source says J
D-H coupling is about 1/6th the magnitude of J
H-H coupling, which on the scale of this spectrum would probably mean it isn't resolved , given how closely spaced the J
H-H splitting appears in the other peaks. That also might explain the broad singlet for CH. At least, that's the best I can reason. What do you think?