How does the presence of various NMR active nuclei in a compound affect coupling and signal multiplicity?

Question

I am basically wondering how the presence of various NMR active nuclei present in a organic molecules affect the shift values and coupling states normally seen in a molecule lacking these nuclei.

Can NMR active metallic moieties in organometallic compounds such as magnesium 25 in a Grignard reagents couple with the spin of protons in H-NMR or carbon 13 atoms in C-NMR? Do the peaks show up in the same range and how are the shift values affected?

In a structure containing many active NMR nuclei (such as a structure containing appropriate isotopes of chlorine, phosphorus, and nitrogen) is special decoupling software or supplemental instrumentation needed to interpret these spectra or are there other techniques to readily discern structure?

Answer 1

I'd like to expand on the other answer a bit (at least on the question of coupling). An NMR active nuclei X can couple with (and detectably split peaks of) another NMR active nuclei Y if X is both not too rare (as discussed in the other answer) AND does not have an extremely short relaxation time.

I'll focus on the common elements that an organic chemist might work with: $\ce{H}$, $\ce{C}$, $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$, $\ce{P}$, and toss in $\ce{Mg}$ (as mentioned in the question), $\ce{Li}$ and $\ce{Si}$.

"The good"

NMR active isotopes with high natural abundance and not extremely short relaxation times: in our list $\ce{^1H}$, $\ce{^19F}$, $\ce{^31P}$, and, marginally, $\ce{^29Si}$ and $\ce{^6Li}$. These all are decent NMR nuclei and abundant, so they definitely split peaks of nearby NMR active nuclei.

• $\ce{^1H}$: 99.98% natural abundance, spin 1/2 (so 0 quadrupolar moment), T1 around 10 seconds for the standard.

• $\ce{^19F}$: 100% natural abundance, spin 1/2, sharp peaks (thus
  relaxation times not too short).

• $\ce{^29Si}$: A bit marginal. Around 5% natural abundance (so the 'side
  peaks' caused by splitting come from only 5% of the molecules and
  thus will be small, called satellites), spin 1/2.

• $\ce{^6Li}$: Around 8% natural abundance, spin 1, but is an honorary spin 1/2 as it has a very small quadrupolar moment. Relaxation times are long.

These nuclei will split nearby NMR active nuclei (not many bonds away). This is either a bug or a feature. The splitting pattern can tell you a lot about the molecule along with other advantages you can get from internuclear coupling (I have a colleague who uses this all the time on fluorocarbons). If you don't want the effect, though, you will have to use a decoupling pulse on these nuclei.

For instance, in the text "Organic Structures from 2D NMR Spectra" (Field, Li, Magill) the examples include many uncoupled experiments (Note: Y{X} means Y is observed while uncoupling X): $\ce{^13C}\{\ce{^1H}\}$ (often), $\ce{^19F}\{\ce{^1H}\}$, $\ce{^13C}\{\ce{^1H, ^19F}\}$, $\ce{^1H}\{\ce{^31P}\}$, $\ce{^31P}\{\ce{^1H}\}$.

"The bad"

There are nuclei that, if enriched, would split nearby peaks (having not-too-short relaxation times), but have a low natural abundance. These can be analyzed quite well using NMR, and may have peaks split by other active nuclei, but won't usually cause visible splitting in other nuclei. Out of our list: $\ce{^13C}$, $\ce{^15N}$.

• $\ce{^13C}$: Only 1% natural abundance, spin 1/2. While it is often examined either alone or in conjunction with other active nuclei, it won't cause anything but the tiniest satellites in other nuclei.
• $\ce{^15N}$: Even worse. Spin 1/2 but only 0.4% natural abundance.

"The ugly"

These might or might not have high natural abundance, but they have short relaxation times. When relaxation times are very short, the nuclei is, in effect, self-decoupling. Nuclei with very short relaxation times have, though the Heisenberg Uncertainty Principal, broad lines and thus often are pretty hard to work with in any NMR context. They certainly don't show visible splitting of the signal of nearby nuclei. From our list, we have: $\ce{^14N}$, $\ce{^33S}$, $\ce{^35Cl}$, $\ce{^37Cl}$, $\ce{^79Br}$, $\ce{^81Br}$, $\ce{^127I}$, $\ce{^17O}$ and $\ce{^25Mg}$. At least you don't have to use decoupling pulses on these nuclei!

• $\ce{^14N}$: Too bad. Even though it's 99.6% abundant (the rest is $\ce{^15N}$), it has spin 1 and a reasonably large quadrupolar moment, thus a short relaxation time (often around 20 ms). It might broaden the peaks of nearby nuclei, but won't show coupling in any but the most symmetric environments.
• $\ce{^33S}$: Ouch. 0.8% natural abundance, spin 3/2, high quadrupolar moment, only NMR active sulfur nuclei.
• $\ce{^35Cl}$, $\ce{^37Cl}$: Between these, 100% abundant. Both are spin 3/2 with VERY large quadrupolar moments and short relaxation times.
• $\ce{^79Br}$, $\ce{^81Br}$: Same as the sad fate of chlorine.
• $\ce{^25Mg}$: Only active NMR nuclei of Mg, 6% abundance, spin 5/2, VERY large quadrupolar moment.
• $\ce{^17O}$: The only NMR active oxygen nuclei. Very rare (0.04% natural abundance), spin 5/2 with a fairly large quadrupolar moment.
• $\ce{^127I}$: 100% natural abundance, but VERY large quadrupolar moment. Maybe also prone to CSA relaxation (I'm not sure).

COMMENTS on short relaxation in nuclei

Nuclei with very short relaxation times often get these via two relaxation processes. One is quadrupolar relaxation (only for spin > 1/2) and chemical shift anisotropy (CSA) relaxation.

For high spin (>1/2) nuclei, relaxation can occur because charge on the nuclei is not symmetric (thus a quadrupolar moment) so that energy can be exchanged with an environment that is not highly symmetric. The effectiveness of this type of relaxation rises as the square of the quadrupolar moments times the square of the applied magnetic field. Spin 1/2 nuclei have quadrpolar moment zero. Quadrupolar moments range all the way from values such as 0.0028 ($\ce{^2H}$), -0.0008 ($\ce{^6Li}$) to values such as 0.31 ($\ce{^79Br}$).

CSA: Also caused by tumbling in an asymmetric environment. Larger for nuclei with large shift ranges, proportional to gyromagnetic ratio squared times magnetic field strength squared. Not really predominant in protons or any carbon but Quaternary carbon.

These relaxation mechanisms don't work if the environment of the nuclei is very (especially tetrahedral or octahedral) symmetric, in which case relaxation times can be long.

The perverse thing about these relaxation mechanisms is that they get worse (more decoupling) with higher field strength. Usually anyone working with NMR dreams night and day of getting a higher field spectrometer. However, if you are interested in a nuclei prone to CSA relaxation (say, $\ce{^199Hg}$ or $\ce{^195Pt}$, which are spin 1/2 and not prone to quadrupolar relaxation), and you want to see splitting of peaks (coupling) in nearby nuclei, you might have to move to a LOWER field NMR.

References

A good site with lots of information on NMR active nuclei is the NMR periodic table.

Answer 2

Indeed those NMR active nuclei will effect the coupling of intended signal, which is generally depend on the natural abundance of "foreign" NMR active nuclei. For example, $\ce{^{13}C}$-nuclei would not affect $\ce{^1H}$-NMR signals because of the low natural abundance of $\ce{^{13}C}$ ($\mathrm {\approx 1.12\%}$) compared to that of $\ce{^1H}$. Also keep in mind that the average width of a $\ce{^1H}$-NMR (if not working on specific compounds with higher range) is $\pu{10 ppm}$. You may able to see $\ce{^{13}C}$ satellite peaks only when you are working on concentrated samples. However, when you are taking $\ce{^{13}C}$-NMR, the weight of the effect is reversed and you'd see $\ce{^{13}C-^1H}$ couplings on $\ce{^{13}C}$-NMR spectrum, if it was taken without applying decoupling techniques. The modern NMR techniques can resolve this problem for you, and even worked it for your advantage. See following three $\ce{^{13}C}$-NMR spectra of ethylbenzene (last spectrum shows only aliphatic region). First spectrum shows only singlet because it was subjected to regular decoupling. The last one shows gated-decoupled spetrum so that it coupled only with protons, which are directly attached to the relevant carbon. If you know the $n+1$ rule, you'd know carbon resonance at $\pu{15.5 ppm}$ is the methyl carbon of ethyl group and the resonance at $\pu{28.8 ppm}$ represents the methylene carbon of the same.