= Solution
Use the proton labels in the original drawing below. Let $a$ denote the three methyl hydrogens, $b$ the crotonyl beta-CH, $c$ the crotonyl alpha-CH, $d$ the O-bound vinyl CH, and $e,f$ the terminal vinyl hydrogens trans and cis to $d$, respectively.
$$
\begin{array}{c|r|c|l}
\text{protons}&\delta/\mathrm{ppm}&\text{integral}&\text{multiplicity}\\\hline
a&1.9215&3&\text{doublet of doublets}\\
b&7.108&1&\text{doublet of quartets}\\
c&5.8765&1&\text{doublet of quartets}\\
d&7.343&1&\text{doublet of doublets}\\
e&4.906&1&\text{doublet of doublets}\\
f&4.576&1&\text{doublet of doublets}
\end{array}
$$
\b[The integrals sum to eight hydrogens.] Multiplet centres give the <chemical shifts>; a peak label at one edge of a multiplet is not its centre. The terminal vinyl hydrogens are chemically nonequivalent because one is cis and the other trans to the O-bound vinyl hydrogen. Each crotonyl CH couples to the other and to the methyl group; the resolved long-range interaction makes the methyl a doublet of doublets rather than a simple doublet.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-proton-couplings.png]
{title=Original structure showing all eight hydrogens, their environments and calculated <NMR coupling constants>}
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