Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
In the lecture our professor told us that nuclei with $S>1/2$ have a quadrupole and are therefore detected with NQR instead of NMR. The majority of all elements have nuclei with $S>1/2$.
I wondered why he used $S$ (electron spin) instead of $I$ (nuclear spin) in the slides?
The nomenclature you mean has nothing to do with the electron spin S, it is something completely different. In heteronuclear experiments with two spins it is common to call them $I$ and $S$, with $I$ being the more sensitive nucleus. They're still both spin 1/2 nuclei.
$I$ is one spin, and $S$ is a different spin in this nomenclature, with a scalar coupling between them.
This nomenclature is usually introduced with the INEPT transfer and experiments like HSQC or HMQC.
Mad Scientist has provided a nice answer to your question about S and I. I'd like to comment on another aspect of your post.
our professor told us that nuclei with S>1/2 have a quadrupole and are therefore detected with NQR instead of NMR
"Instead of nmr" might be too strong. Most nuclei with a nuclear spin >1/2 can also be observed by nmr. Some of these nuclei may have a low relatively sensitivity to the nmr experiment, or due to nuclear quadrupole relaxation might produce signals where information has been lost, in which case NQR might be preferred. While there are many cases where the NQR experiment would be more informative than the NMR experiment, there are also many cases where NMR would work just fine.
An interesting example related to proton-nmr compares the following 3 compounds, pentadeuterioacetone, N-methylaniline and chloroform.
Deuterium, nitrogen and chlorine are quadrupolar nuclei. Deuterium and nitrogen-14 (the major isotope of nitrogen) both have a nuclear spin of 1, while chlorine-35 and -37 (the 2 predominant isotopes of chlorine) have a nuclear spin of 3/2. If one examines the proton-nmr of pentadeuterioacetone, one can observe the coupling between the deuterium and hydrogen nuclei. By the same token, if one examined the deuterium-nmr of this compound, one would observe the same H-D coupling. If one examines the proton-nmr of N-methylaniline, the coupling between nitrogen and the various protons is washed out, broadened signals where the coupling information has been lost are observed. In the case of chloroform, the proton-nmr shows only a sharp singlet.
Why the change in coupling to a proton as we change the coupled nucleus from D to N to Cl? The situation is very similar to physical exchange phenomena. If a proton is attached to a nucleus, then coupling can be observed between that proton and adjacent protons. If the proton is exchanging with the environment at a rate similar to the timescale of the nmr experiment, then the coupling will broaden. If the proton is exchanging very rapidly compared to the timescale of the nmr experiment, then the proton is effectively decoupled from the system and a sharp singlet will be observed.
In the case of the 3 compounds discussed above, instead of physical exchange, the proton is being relaxed (decoupled) at varying rates by the attached quadrupolar nuclei. The rate of quadrupolar relaxation caused by a coupled deuterium atom is slow on the nmr timescale, so all coupling is preserved and observed. The quadrupolar relaxation rate of nitrogen-14 is comparable to the timescale of the nmr experiment and so the coupling is beginning to wash out. The chlorine nuclei undergo rapid quadrupolar relaxation on the nmr timescale and are effectively decoupled from other nuclei.
Thanks to Mad Scientist's answer, I understand the topic better.
An electron can has a spin of $s=1/2$ (lower case s). The total spin quantum number $S=1/2$ (upper case S) is the sum of the single spins.
Although protons are nuclei, they are elementary particles (which have usually a spin of $s=1/2$). Protons have only a spin $s=1/2$ and therefor $S=1/2.$
In contrast to the mentioned two cases of elementary particles examples an atom (or ion with at least one electron; a proton is also an ion) have a spin quantum number $s$ and a angular momentum quantum number $l.$ So the two quantum numbers sum up to the nuclear spin which is usually determined by $I$ as the sum over all $s$ plus the sum over all $l.$ Very often we find as well instead of $I$ the letter $J = L + S$ in spectroscopy.
I also agree with long's comment
I don't think this is why the lecturer is using S instead of I. I think it is being used simply as an abbreviation for the term spin. Although I is the correct symbol for spin, a lot of NQR papers tend to refer to Spin-1 (or Spin-3/2 etc) nuclei. I think it is simply an extension of that terminology.
and I think that the lecturer wrote $S$ as an abbreviation of spin without distinguishing $I$ und $S.$
