= Solution
\b[The large vinyl couplings are 17.1 Hz trans and 10.5 Hz cis; the O-methylene-to-vinyl coupling is 6.0 Hz.] At this field, convert every separation by $J/\mathrm{Hz}=300\,\Delta\delta/\mathrm{ppm}$. Comparing the centres of corresponding terminal-proton submultiplets gives
$$
\begin{aligned}
{}^3J_{cd}&=300(5.350-5.293)=\boxed{17.1\,\mathrm{Hz}},\\
{}^3J_{ce}&=300(5.260-5.225)=\boxed{10.5\,\mathrm{Hz}},\\
{}^3J_{bc}&=300(4.585-4.565)=\boxed{6.0\,\mathrm{Hz}}.
\end{aligned}
$$
The repeated 0.005-ppm fine spacing from <allylic scalar coupling> and <geminal coupling> gives the magnitudes
$$
\boxed{|{}^2J_{de}|\simeq|{}^4J_{bd}|\simeq|{}^4J_{be}|\simeq1.5\,\mathrm{Hz}.}
$$
For example, splitting by a geminal <proton> and two equivalent long-range-coupled methylene <protons>, all with the same magnitude, produces overlapping lines with intensities $1:3:3:1$. The 6-Hz splitting of $b$ creates two such small <NMR triplets> centred at 4.580 and 4.560 ppm. An independent check comes from the internal-vinyl multiplet: a centre at 5.919 ppm with splittings 17.1, 10.5 and 6.0 Hz, the last to two equivalent <protons>, reproduces all twelve labelled line positions. The spectrum determines these coupling magnitudes, not their algebraic signs.
Back to article page
