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Write the proton-labelled structure as $\mathrm{C}{\mathrm{H}}_3^{\mathrm{a}}\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}^{\mathrm{b}}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3^{\mathrm{c}}$. The assignments areThe NMR quartet is centred at $(4.371+4.302)/2=4.3365$ ppm; each adjacent line differs by $0.023$ ppm. The NMR doublet has the same spacing, $1.618-1.595=0.023$ ppm. At 300 MHz,Three equivalent methyl protons split $b$ into a $1:3:3:1$ NMR quartet; the single methine proton splits $c$ into a $1:1$ NMR doublet. No resolved long-range scalar coupling splits the acetyl methyl, giving its NMR singlet. The integrals total seven protons.

Use $a$ for the acetyl methyl, $b$ for the O-bound methylene group pair, $c$ for the internal vinyl proton, and $d,e$ for the terminal vinyl protons respectively trans and cis to $c$. The assignments areThe two terminal protons are diastereotopic: one is cis and the other trans to $c$, so they have different shifts and different vicinal couplings. Each terminal proton also couples through geminal coupling to the other and through long-range scalar coupling to the two $b$ protons. When those small couplings coincide, an underlying doublet of doublets of NMR triplets collapses to the observed NMR doublet of apparent NMR quartets. The O-bound protons are enantiotopic and have chemical equivalence in NMR in this achiral environment; they are split by $c$ and by the two terminal protons. The integrated terminal envelope contains two protons, not two equivalent protons at one shift.