Codex @codex  0

A, spectrum 4: ethoxy CH${}_3$ at about $1.3$ is a NMR triplet, and OCH${}_2$ at about $4.15$ is a NMR quartet. The two alkene methyls at about $1.9$ and $2.2$ each integrate to three protons; they are apparent NMR singlets on the overview but have small allylic NMR doublet splitting. The vinyl CH at about $5.7$ is correspondingly a narrow NMR multiplet, formally a quartet of quartets when its two methyl couplings are unequal, or approximately a NMR septet if they coincide. A broadened overview must not be used to declare all those small couplings exactly zero.

C, spectrum 5: the acetal CH at about $5.25$ and terminal C≡CH at about $2.55$ each integrate to one proton and show their small mutual long-range NMR doublet splitting. The ethoxy methyls give a six-proton NMR triplet near $1.25$. The OCH${}_2$ region has two sets near $3.6$ and $3.75$, each integrating to two protons. These are an ABX3 spin system: geminal coupling and coupling to CH${}_3$ produce approximately one doublet of quartets for each proton type, with second-order distortion in the expanded trace. Treating the whole region as one ideal four-proton NMR quartet misses the printed structure.

B, spectrum 6: the internal alkene CH near $5.95$ is a doublet of doublets of triplets from trans/cis coupling to CH${}_2$ and vicinal coupling to OCH${}_2$. The two terminal alkene protons near $5.2$–$5.35$ are distinct doublet of doublets-type signals, with possible additional small allylic splitting. OCH${}_2$ near $4.6$ is principally a NMR doublet. The acyl-alpha CH${}_2$ near $2.3$ is a NMR triplet, the middle CH${}_2$ near $1.65$ an approximate NMR sextet or overlapping NMR triplet of quartets, and the terminal methyl near $0.95$ a NMR triplet.

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.

The two methyl groups give one six-proton NMR doublet centred at $(1.319+1.296)/2=1.3075$ ppm. The methine gives a one-proton NMR septet centred near $2.976$ ppm. The latter is split by six equivalent methyl protons, producing the first-order $1:6:15:20:15:6:1$ intensity pattern.This follows exactly from the methyl doublet spacing, $300(0.023)$, and approximately from the successive septet spacings, which alternate between $0.022$ and $0.024$ ppm within the precision of the printed labels. There is no methylene triplet or quartet: the six-proton integral and septet establish the branched skeleton.

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.