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Let $a$ be the internal vinylic proton, $b$ the terminal proton trans to $a$, $c$ the terminal proton cis to $a$, $d$ the oxygen-bound methylene and $e$ the chlorine-bound methylene. The assignments areFor the $a$ doublet of doublets, pairing corresponding lines gives $J_{ab}\simeq 300[(6.516-6.468)+(6.492-6.445)]/2=14.25$ Hz and $J_{ac}\simeq 300[(6.516-6.492)+(6.468-6.445)]/2=7.05$ Hz. The larger value identifies trans vinylic coupling; the smaller identifies cis vinylic coupling. The $b$ resonance confirms the large splitting, approximately $14.25$ Hz, and has a small geminal coupling to $c$.

The small spacings are not perfectly reciprocal if treated as exact First-order NMR multiplets: the $b$ lines imply $|J_{bc}|\simeq 2.55$ Hz, whereas the $c$ lines imply $|J_{bc}|\simeq 2.10$ Hz and $J_{ac}\simeq 6.60$ Hz. Thus report $J_{ab}\simeq 14.3$ Hz, $J_{ac}\simeq 7$ Hz and $|J_{bc}|\simeq 2$–$3$ Hz, rather than asserting incompatible exact constants. Printed precision and departures from a simple first-order pattern limit this line-picking estimate; a full spin-system fit would be needed for refined values. Splittings alone do not establish the signs of these couplings.

The two methylene groups split one another into approximate NMR triplets, with $J_{de}\simeq 300(3.727-3.708)=5.7$ Hz; the oxygen-bound triplet gives approximately the same value from its rounded line labels. The three vinylic protons plus two methylenes give the required seven-proton integral. The terminal vinylic protons are diastereotopic, because replacing one or the other creates different alkene geometries.