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At $300\mathrm{MHz}$, $J/\mathrm{Hz}=300\Delta \delta /\mathrm{ppm}$. The six resolved scalar spin-spin couplings areFor example, the two quartets of $c$ are $\{5.842,5.848,5.854,5.860\}$ and $\{5.893,5.899,5.905,5.911\}$ ppm. Their origins differ by $0.051$ ppm, while the spacing within either quartet is $0.006$ ppm. This separates the trans alkene interaction from the smaller allylic methyl interaction.

The four $d$ lines independently give $(7.377-7.330)300=14.1$ and $(7.377-7.356)300=6.3\mathrm{Hz}$. The large interaction identifies $e$ as trans to $d$; $f$ has the smaller cis interaction. The double-headed arrows in the structural map identify the actual interacting nuclei. Coupling to $a$ means the same coupling to each of its three equivalent hydrogens.

Use the NMR coupling constant conversion $J/\mathrm{Hz}=(\Delta \delta /\mathrm{ppm})({\nu }_0/\mathrm{MHz})$. At $300\mathrm{MHz}$, separating the overlapping quartets givesThe ${\mathrm{H}}_b$ doublet separation independently gives $(6.513-6.461)300=15.6\mathrm{Hz}$. For ${\mathrm{H}}_a$, the two quartets are $\{6.728,6.750,6.772,6.794\}$ and $\{6.780,6.802,6.824,6.846\}$ ppm; simply taking adjacent lines in their combined ordering would mix different scalar spin-spin couplings.

The methylene quartet gives an average adjacent spacing $(4.317-4.244)/3=0.02433$ ppm, or $7.3\mathrm{Hz}$. Thus its printed rounded positions agree with the methyl-derived value to about $0.1\mathrm{Hz}$; quote the ethyl NMR coupling constant as approximately $7.2$–$7.3\mathrm{Hz}$. The large vinylic value supports the trans arrangement. The arrows below join the interacting nuclei; each arrow to ${\mathrm{F}}_3$ represents the same interaction with each equivalent fluorine nucleus.

The assignments are ${\mathrm{H}}_a$: $6.787$ ppm, $1$H, doublet of quartets; ${\mathrm{H}}_b$: $6.496$ ppm, $1$H, doublet of quartets; ${\mathrm{H}}_c$: $4.281$ ppm, $2$H, quartet; ${\mathrm{H}}_d$: $1.331$ ppm, $3$H, triplet. The chemical shift of a multiplet is its centre, rather than its most downfield line.

Each vinylic proton has scalar spin-spin coupling to the other vinylic proton and to three equivalent spin-$\frac{1}{2}$ fluorine nuclei. The first interaction produces a doublet, and the second a $1:3:3:1$ quartet: eight lines overall, partly interleaved for ${\mathrm{H}}_a$. The two chemical shifts differ because the vinylic sites have different substituents.

The ethoxy $\mathrm{OC}{\mathrm{H}}_2$ protons are deshielded by oxygen and split into a $1:3:3:1$ quartet by the three methyl protons. The methyl protons are split into a $1:2:1$ triplet by the two equivalent methylene protons. The relative NMR signal integrals are $1:1:2:3$.

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 ${\mathrm{C}}_j$.

Irradiation that removes resolved scalar spin-spin coupling to protons from another nucleus’s spectrum. Other couplings, such as carbon–fluorine coupling in a proton-decoupled carbon-13 NMR spectrum, can remain.