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.

Let $\mathrm{Ar}$ denote the 2,5-dimethoxyphenyl group. A is the doubly C-alkylated beta-ketoesterThe central carbon has no hydrogen after the second step. The alkene, ester and tert-butyl ether remain intact.

carbonate reversibly removes a proton from the methylene group between the two carbonyl groups. The resulting resonance-stabilized enolate attacks the primary iodoethyl carbon of the aryl-containing reagent by an $S_{\mathrm{N}}2$ displacement, releasing iodide. The monoalkylated intermediate still has one proton at the same activated carbon. A second deprotonation followed by $S_{\mathrm{N}}2$ attack on the primary bromide introduces the but-3-en-1-yl group.

The beta-dicarbonyl position is much more acidic than the methylene group next to the ketone and ether at the other end, explaining the alkylation site. Counting the retained carbon skeleton gives ${\mathrm{C}}_{24}{\mathrm{H}}_{36}{\mathrm{O}}_6$, including the two carbonyl groups and one terminal alkene. The second panel of the original product drawing specifies this connectivity.

Number the cyclohexane consecutively, putting the top methylene group at C1 and the two original epoxides at C2/C3 and C5/C6. Both epoxide oxygens are up.

A is a diazidocyclohexanediol, with each epoxide replaced by one azide and one alcohol. A standard successive chair-favored opening gives N3 down at C2 and C5, OH up at C3 and C6. This is the 1,4-diazido/1,4-diol connectivity drawn in the bottom row of the original map.

For each opening, azide attacks from the face opposite oxygen, breaking one C–O bond with inversion at the attacked carbon. protonation of the two resulting alkoxides gives the two OH groups. After the first opening, the preferred conformation of the opened portion must be used for the second trans-diaxial opening; simply imposing one unchanged chair on both steps can select the wrong regioisomer.

The elemental change is addition of two ${\mathrm{N}}_3^{-}$ groups and two work-up protons to ${\mathrm{C}}_6{\mathrm{H}}_8{\mathrm{O}}_2$, giving ${\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{N}}_6{\mathrm{O}}_2$. The formula alone would not distinguish alternative positional isomers; the depicted assignment uses the ordinary successive-opening conformational model.

The first step is a vinylogous aldol condensation at C5 of the unsaturated lactam. Number nitrogen N1, the lactam carbonyl C2, and then the methoxy-substituted alkene C3=C4 and the methylene group C5.

hydroxide reversibly removes a C5 proton. The resulting vinylogous enolate is delocalized through C5–C4–C3–C2–O: it is not necessary to create an unstabilized isolated carbanion beside nitrogen. Its C5 nucleophilic contributor attacks acetone. protonation gives a C5-linked tertiary alcohol, $C5-\mathrm{C}(\mathrm{OH})(\mathrm{Me}{)}_2$. Further removal of the remaining C5 hydrogen and elimination through an aldol-dehydration sequence forms C5=C(Me)2, leaving the original methoxy enol ether in place.

In the second step, aqueous acid hydrolyses that enol ether to a ketone at C4. protonation at C3 gives an oxygen-stabilized cation at C4. water attacks C4, proton transfers form a hemiacetal-like intermediate, and loss of methanol followed by deprotonation produces C4=O and the C3 methylene group.

The lactam carbonyl and N–H group remain, as does the exocyclic isopropylidene group. This accounts for replacement of the C3=C4–OMe unit by C3H2–C4=O without inserting a new oxygen into the ring skeleton.

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.

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.

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.