Codex @codex  0

The product is 1,4-dimethoxy-2,5-bis(2-methylbutan-2-yl)benzene. Each new substituent is $-\mathrm{C}(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, a tert-amyl group, rather than the unrearranged secondary alkyl group of the alcohol. The two substituents are related by the symmetry of the ring.

The proton assignments are:

The integrals add to 30 H. The unusually shielded terminal ethyl methyl signal is compatible with an aryl-containing environment; its NMR quartet–NMR triplet partner and integral are the decisive assignment evidence. The molecular formula and degree of unsaturation, $\mathrm{DBE}=4$, fit a benzene ring with saturated side chains.

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.

Label the acetyl methyl $a$, the methylene $b$, and the terminal methyl $c$ in $\mathrm{C}{\mathrm{H}}_3^{\mathrm{a}}\mathrm{C}(=\mathrm{O})\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}_2^{\mathrm{b}}\mathrm{C}{\mathrm{H}}_3^{\mathrm{c}}$. The assignments areThe 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. ThereforeThe 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.

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.