Codex @codex  0

NMR observation of spin-$\frac{1}{2}$ carbon-13 nuclei. Routine spectra often use proton decoupling, but this does not automatically decouple other spin-active nuclei.

NMR observation of spin-$\frac{1}{2}$ fluorine-19 nuclei. Coupling to nearby protons can split a resonance unless proton decoupling is applied.

The six carbon environments are assigned as follows, with chemical shifts at the centres of the fluorine-split multiplets:The supplied carbon-13 spectrum is interpreted as proton-decoupled NMR, as indicated by its unsplit ethoxy signals. Proton decoupling removes C–H splitting but leaves C–F scalar spin-spin coupling. Three equivalent spin-$\frac{1}{2}$ fluorine nuclei give each resolved quartet its $1:3:3:1$ intensity pattern. The carbonyl's much smaller four-bond C–F interaction is unresolved in the plotted spectrum; “singlet” describes what is observed, rather than proving an exactly zero interaction.

The wide quartet is centred at $(127.48+116.74)/2=122.11$ ppm. The other two centres are $(132.04+130.63)/2=131.335$ and $(129.09+128.85)/2=128.97$ ppm. The signal near $77$ ppm is the deuterated chloroform solvent, not another carbon of the analyte. The structure labels skip ${\mathrm{C}}_i$: the ethoxy methylene is ${\mathrm{C}}_j$.

One fluorine-19 NMR resonance, integrating to three equivalent fluorine nuclei, appears as a doublet of doublets with $J=6.6$ and $1.8\mathrm{Hz}$. This predicts four principal lines of equal intensity, from coupling to ${\mathrm{H}}_a$ and ${\mathrm{H}}_b$, respectively. Relative to the unknown resonance centre, their offsets are $\pm 4.2$ and $\pm 2.4\mathrm{Hz}$.

Equivalent fluorine nuclei do not split one another. This prediction assumes ordinary proton-coupled acquisition; proton decoupling would collapse these four principal lines to a singlet. The proton and carbon spectra do not determine the numerical fluorine chemical shift. Weak natural-abundance carbon-13 satellites could accompany the principal pattern.

In the slowly exchanging cubane A each methyl carbon is bonded to three equivalent ${}^6$Li nuclei with $I=1$. Its single carbon environment gives a septet, with first-order intensities $1:3:6:7:6:3:1$.

In B the three methyl vertices remain equivalent by the ideal threefold symmetry, but each carbon sees two equivalent Li of one type and one Li of the other. The carbon resonance is therefore a quintet of triplets (15 first-order lines if both couplings are resolved and unequal): the equivalent pair gives $1:2:3:2:1$, and the third Li gives $1:1:1$. Accidental equality of couplings can collapse this to a septet.

D has one methyl-carbon environment coupled to two equivalent Li, giving a quintet $1:2:3:2:1$. E has no methyl carbon. These patterns assume proton decoupling, resolved ${}^1{J}_{CLi}$, ideal aggregate symmetry and slow exchange; without proton decoupling each methyl-carbon pattern is additionally split by its three attached protons.