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Write $t=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3{)}_3$. D is the sodium enolate $\mathrm{t}-\mathrm{C}({\mathrm{O}}^{-})=\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{a}}^{+}$. Its two resonances in the alkene-proton region belong to the two hydrogens of the terminal methylene group. One is cis to oxygen and the other trans to oxygen, so they are not related by a symmetry operation even though both belong to the same carbon. Replacing one or the other gives different alkene geometries, so they are diastereotopic. Their weak or unresolved geminal coupling is consistent with the reported singlet descriptions. Without an additional spatial assignment, it is not justified to decide uniquely which numerical shift belongs to which of those two orientations. The nine-proton singlet is the tert-butyl group. The formula ${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{ONa}$ follows from removing one alpha hydrogen from pinacolone.

E is the enol-dominated 1,3-diketone $\mathrm{tCO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COt}$, observed as $\mathrm{tC}(\mathrm{OH})=\mathrm{CH}-\mathrm{COt}$. The very downfield broad signal belongs to the enol OH in a six-membered intramolecular O–H···O chelate. The one-proton signal in the vinyl region is its central enol CH; a predominantly diketo structure would instead require a two-proton central CH2 signal and no enol OH. The eighteen-proton singlet accounts for both tert-butyl groups. Rapid exchange between the two degenerate enol orientations can average their different local environments; accidental unresolved equality cannot be excluded by the supplied spectrum alone. Thus one should not infer a permanently symmetric single enol Lewis structure from that singlet. The total integration is $1+1+18=20$, agreeing with ${\mathrm{C}}_{11}{\mathrm{H}}_{20}{\mathrm{O}}_2$.

The bridged, sterically hindered biphenyl framework is twisted and does not rapidly interconvert the two faces of each methylene group on the NMR timescale. Label one hydrogen on each CH2 $H_A$ and the other $H_B$. Replacing $H_A$ versus $H_B$ by a test substituent gives diastereomeric environments, so these hydrogens are diastereotopic and need not have equal chemical shifts even in an achiral solvent. Molecular symmetry relates the two $H_A$ sites to each other and the two $H_B$ sites to each other.

Each methylene hydrogen undergoes geminal coupling to its partner, giving the two doublets, each accounting for two hydrogens overall. Their separation is $0.21$ ppm; at operating frequency ${\nu }_0$ in MHz it corresponds to $0.21{\nu }_0$ Hz. This separation is not the geminal coupling constant, which must be measured from the line spacing within a doublet. If the shift separation is not large compared with that coupling, an AB pattern with second-order distortion is expected rather than ideal first-order doublets. A racemic pair of twisted conformations still has these signals: enantiomers have the same spectrum in an achiral medium, while the two hydrogen environments within each molecule remain different.

P is H2, the acetal methine between two oxygens; its only resolved coupling is to the attached methyl, giving the quartet. S is that C2 methyl, whose doublet has the same 6.9 Hz coupling. T is tert-butyl, integrating to nine equivalent hydrogens with no adjacent carbon-bound methine proton across its quaternary attachment.

U is equatorial H5, coupled to the four hydrogens at C4 and C6. The two equivalent equatorial methylene hydrogens are Q, each a doublet of doublets with a 12.3 Hz geminal coupling to its axial partner and a 1.3 Hz coupling to H5. The two equivalent axial methylene hydrogens are R, each with the same 12.3 Hz geminal coupling and a 3.7 Hz vicinal coupling to H5. Equivalence between C4 and C6 makes two pairs rather than four separately shifted methylene hydrogens.Coupling U to two equivalent R hydrogens and two equivalent Q hydrogens produces its triplet of triplets. The different small vicinal couplings are consistent with oblique H–C–C–H torsions for equatorial H5; the 12.3 Hz value is geminal and must not be mistaken for a large vicinal trans-diaxial coupling.

A: ethoxy CH${}_2$ ↔ CH${}_3$, $J\simeq 7\mathrm{Hz}$; vinyl CH ↔ each alkene methyl, small ${}^{4}J\simeq 1$–$1.5\mathrm{Hz}$.

C: ethoxy CH${}_2$ ↔ CH${}_3$, about $7\mathrm{Hz}$; the two OCH${}_2$ proton types have a geminal coupling of about $9$–$10\mathrm{Hz}$; acetal CH ↔ C≡CH has ${}^{4}J\simeq 2\mathrm{Hz}$.

B: internal vinyl CH ↔ trans terminal H, about $17\mathrm{Hz}$; ↔ cis terminal H, about $10\mathrm{Hz}$; terminal CH${}_2$ geminal coupling about $1$–$2\mathrm{Hz}$. Vinyl CH ↔ OCH${}_2$ is about $6\mathrm{Hz}$. Both vicinal links along the butanoate alkyl chain are about $7$–$8\mathrm{Hz}$.

Use $J=300\Delta \delta$ when a separation is measured in ppm. Thus $0.057\mathrm{ppm}$ corresponds to about $17\mathrm{Hz}$, whereas $0.023\mathrm{ppm}$ corresponds to about $7\mathrm{Hz}$. These are approximate readings, not a fitted spin simulation.

The small four-bond splittings illustrate allylic scalar coupling and acetal-alkyne long-range coupling.

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.

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.

The large vinyl couplings are 17.1 Hz trans and 10.5 Hz cis; the O-methylene-to-vinyl coupling is 6.0 Hz. At this field, convert every separation by $J/\mathrm{Hz}=300\Delta \delta /\mathrm{ppm}$. Comparing the centres of corresponding terminal-proton submultiplets givesThe repeated 0.005-ppm fine spacing from allylic scalar coupling and geminal coupling gives the magnitudesFor example, splitting by a geminal proton and two equivalent long-range-coupled methylene protons, all with the same magnitude, produces overlapping lines with intensities $1:3:3:1$. The 6-Hz splitting of $b$ creates two such small NMR triplets centred at 4.580 and 4.560 ppm. An independent check comes from the internal-vinyl multiplet: a centre at 5.919 ppm with splittings 17.1, 10.5 and 6.0 Hz, the last to two equivalent protons, reproduces all twelve labelled line positions. The spectrum determines these coupling magnitudes, not their algebraic signs.

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.