Just for clarity, I am curious about the small sattelite peaks, on either side of the main CDCl3 peak in proton NMR of CDCl3... Not in C13 NMR. I understand the peak splitting.
I would guess that it is for the same reason that all other peaks proton signals have satellite peaks. About 1% of the carbon in the sample is 13C, so that would split the singlet into a doublet since there is normally no 13C decoupling for proton spectra.
Oh yes.. true.. 13C has a natural abundance just over 1% and the major isotope (12C) is not NMR active so very little of the proton signal is coupled. The coupled signal appears as small satellite signals either side of the main uncoupled signal
you can verify the 13C satellites by their distance from one another. It should be ca. 209 Hz. This is the coupling constant 1JCH. The centre of this doublet should be at a slightly lower frequency than the CHCl3 signal (isotope effect).
In Fourier transform NMR there are no spinning side bands so it is easier to identify the 13C satellites. You should also note that the 1JC-H coupling constant [209 Hz] is related to the 1JC-D coupling {32 Hz] seen in the 13C spectrum of CDCl3 by a factor of 6.51 [the ratio of the gyromagnetic constants of H vs D.]. In addition the !J C-H coupling is related to the hybridization of C in CHCl3. The index of p = 500/[1JC-H] -1 which works out to 500/209 - 1 =1.4. The hybridization of C in CHCl3 is thus sp^1.4, not sp^3 as stated in many General Chemistry textbooks.
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Ludger Ernst
Technische Universität Braunschweig
As to Donald Clark‘s answer: it is not true that there are no spinning side bands in FT NMR. It is only true if you do not spin the sample tube during the acquisition of the FID.
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