= Solution
Label the acetyl methyl $a$, the methylene $b$, and the terminal methyl $c$ in $\mathrm{CH_3^{a}C(=O)C(=O)CH_2^{b}CH_3^{c}}$. The assignments are
$$
\begin{array}{c|c|c|c}
\text{group}&\delta_H/\mathrm{ppm}&\text{integral}&\text{multiplicity}\\\hline
a&2.345&3&\text{singlet}\\
b&2.779&2&\text{quartet}\\
c&1.090&3&\text{triplet}
\end{array}
$$
The <NMR multiplet centre> of the <NMR quartet> is the mean of its outer lines, $(2.815+2.743)/2=2.779$ ppm. Its adjacent lines differ by 0.024 ppm; the <NMR triplet> has the same spacing. Therefore
$$
\boxed{{}^3J_{bc}=300(0.024)=7.2\,\mathrm{Hz}.}
$$
The two groups, through <vicinal coupling>, split one another into the expected $1:3:3:1$ <NMR quartet> and $1:2:1$ <NMR triplet>. The acetyl methyl has no resolved neighbouring <proton> across its adjacent <carbonyl> and appears as a <NMR singlet>. The integrals sum to eight <protons>.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1-diketone-protons.png]
{title=Original separate proton assignment for A with the measured methylene–methyl coupling}
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