[Big-Daddy]

I've come to notice that in ¹H NMR spectra the ratio of the integrals under each peak represents the ratio of numbers of nuclei in each peak, but this does not occur in ¹³C NMR. For which spectra (i.e. of which isotopes) is it possible to find the relative numbers of nuclei in each peak by integration of their areas, and why is it not possible for spectra of the other isotopes?

I have heard that the peaks in ¹³C spectra are meaningless in terms of peak height and integral, except for the one little fact that carbon atoms with no hydrogen atoms directly attached show up as small peaks (i.e. small peak integral -> probably small peak height). Can you verify this and if it is true, explain it?

Then the same question applies to spin-spin coupling; why does it occur with ¹H-spectra (any H nuclei bonded to atoms, other than O and F, adjacent to the atom on which the H in question is held, cause coupling, i.e. create extra peaks, for the H nucleus in question and all other H nuclei in the same environment) but not with 13C spectra, and how can I tell for which isotope spectra it will be present to give me more information?

[Corribus]

I've come to notice that in ¹H NMR spectra the ratio of the integrals under each peak represents the ratio of numbers of nuclei in each peak, but this does not occur in ¹³C NMR. For which spectra (i.e. of which isotopes) is it possible to find the relative numbers of nuclei in each peak by integration of their areas, and why is it not possible for spectra of the other isotopes?

¹³C NMR isn't very quantitative because the ¹³C peaks are so affected by surrounding proton spins.  I believe there are ways to make ¹³C NMR spectra more quantitative by playing around with the pulse sequence (actually you can do a whole lot with ¹³C NMR this way) but that's beyond my knowledge level to describe in any detail.

Then the same question applies to spin-spin coupling; why does it occur with ¹H-spectra (any H nuclei bonded to atoms, other than O and F, adjacent to the atom on which the H in question is held, cause coupling, i.e. create extra peaks, for the H nucleus in question and all other H nuclei in the same environment) but not with 13C spectra, and how can I tell for which isotope spectra it will be present to give me more information?

It occurs with all nuclei. But: what is the natural abundance of ¹³C?

¹³C nuclei also spin-couple (very strongly, in fact), with proton nuclei.  You would normally think this would mean ¹³C NMR spectra would be an absolute mess, but there are ways to decouple ¹³C nuclei from ¹H nuclei again by modulating the radio frequency pulses.  And again beyond my knowledge level to explain in a whole lot of detail off the top of my head.

EDIT: I didn't just want to write something without checking it first, but I checked and proton spins are decoupled by continuously irradiating the sample with a low power proton absorption signal.  This scrambles the proton spins, which "washes out" the spin coupling.

[Big-Daddy]

¹³C NMR isn't very quantitative because the ¹³C peaks are so affected by surrounding proton spins.

Thanks. Which other nuclei are highly affected by surrounding spins? (Enough so that their integrals, generally, do not represent accurate ratios for the nuclei themselves?)

But: what is the natural abundance of ¹³C?

An apt point. Does this mean that, since ¹³C has low natural abundance, we cannot expect to get accurate values for the ratio of ¹³C nuclei in each environment anyway?

[Corribus]

Well, any nucleus which has a nonzero spin will couple to another nucleus with a nonzero spin, and that coupling will be observed if the two nuclei are close enough to each other.

Does this mean that, since 13C has low natural abundance, we cannot expect to get accurate values for the ratio of 13C nuclei in each environment anyway?

It's just a matter of probability.  For sake of argument let's say ¹³C has a natural abundance of 1%.  The probability that any carbon in a molecule is a ¹³C nuclei is the same (1/100).  For a large number of molecules, then, every carbon position in that molecule will have equal probability of having a spin active ¹³C signal.  That signal will be small because the abundance is low - which is why ¹³C NMR experiments take a long time - but you have equal probability of observing every carbon position (all other things being equal).  But to see splitting, you need two ¹³C nuclei in the same molecule, and indeed almost right next to each other.  The probability of this happening in a single molecule is several orders of magnitude lower, particularly two positions immediately adjacent to each other.  True, the law of averages means that for a large number of molecules there will be occasions where it happens, but the signal is going to be so much lower that you would need an incredibly sensitive instrument to see them.

[Big-Daddy]

Thanks for your help here.

[muse]

You can obtain quantitative 13C spectra, but you have to make sure your experiment is set up correctly. Check out this blog: http://u-of-o-nmr-facility.blogspot.no/2007/12/how-can-i-get-quantitative-13-c-nmr.html

To obtain highly accurate rations for you integration you also have to make sure you experiment setup for proton is correct, using long enough relaxation time, right number of scans to increase S/N, etc. The field of quantitative NMR (qNMR) is becoming widespread and can be used for purity checks.

Nuclei with spin 1/2 like 1H, 13C, 19F, 31P, and some more gives you your normal splitting pattern, eg neighbouring group CH gives doublet and CH2 gives triplet etc (n+1 multiplet). In some proton spectra you can see coupling between proton and carbon as small satelites on each side of your proton peak.
Coupling to nuclei with spin 1, as deuterium gives 1:1:1 multiplet, and this you see in the solvent peak of chloroform in your carbon spectra.
Especially 19F and 31P can give you a lot of coupling, as the have 100% abundance and spin 1/2, and they are fairly easy to get quantitative spectra.