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protonation of the alcohol converts its OH group into a water leaving group. Departure of water initially gives the secondary carbocation $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}^{+}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$. A 1,2-hydride shift from the neighboring methine group transfers its C–H bonding pair to the electron-deficient carbon:The resulting tertiary carbocation is the tert-amyl electrophile. An aromatic $\pi$ bond of the dimethoxybenzene attacks that carbon, producing a Wheland intermediate. Removal of the proton from the attacked ring carbon restores aromaticity and regenerates acid. Repeating this Friedel-Crafts alkylation at the symmetry-related second site gives the disubstituted product.

The hydride migration explains the new ethyl group and the absence of a benzylic methine group proton. Simply attaching the original alcohol skeleton at its OH-bearing carbon would not explain the twelve-proton NMR singlet.

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.

The new ring is formed by intramolecular Friedel-Crafts addition to the ketone. titanium tetrachloride coordinates the ketone oxygen of the cyclic beta-ketoester, increasing electrophilicity at that carbonyl carbon. The nearby electron-rich aryl ring attacks this activated ketone at the available position ortho to its ethyl tether.

The electron-flow sequence is: an aromatic $\pi$ bond forms the new aryl–carbonyl-carbon bond, the ketone $\pi$ bond moves onto oxygen, and loss of the aromatic proton restores aromaticity. The two-carbon tether and the shared quaternary carbon make the newly closed carbocycle six-membered. Hydrolytic work-up converts the titanium-bound alkoxide into the tertiary alcohol.

The lactone carbonyl survives. This is an addition to the ketone with retention of its oxygen as OH, rather than an acyl substitution that removes that oxygen. The pendant terminal alkene is not involved. These bond changes account for the fused aromatic ring system and the alcohol in C without needing to assign the new stereochemistry.

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.

For an amine-first pathway, use triethylamine as a nucleophile in a reversible conjugate addition. Its lone pair attacks C3 of the enone; electron shifts through C2 and the carbonyl give a zwitterionic enolate with a C3–$\mathrm{NE}{\mathrm{t}}_3^{+}$ bond.

The C2 nucleophilic site of that enolate attacks $\mathrm{B}{\mathrm{r}}_2$, installing C2–Br and releasing bromide. Removal of C2–H by base then reforms C2=C3 while the C3–N bond breaks back to neutral triethylamine. This addition–bromination–elimination sequence gives the same C2-brominated enone and regenerates the nucleophilic amine, with another base equivalent accepting the eliminated proton.

These are alternative formal sequences requested by the question, not a kinetic determination that one must dominate under unspecified conditions. In particular, the amine-first sequence does not require triethylamine to generate a high concentration of a vinylic carbanion by direct deprotonation.

Label the enamine carbon bonded to nitrogen C1 and its alkene partner C2; C6 is the other carbon next to C1. B is the C-alkylated iminium ion bromide. Its new substituent at C2 is $-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}(\mathrm{C}{\mathrm{O}}_2\mathrm{Et})=\mathrm{C}{\mathrm{H}}_2$, while C1 is double-bonded to the pyrrolidine nitrogen, which bears positive charge. bromide is the counterion, accounting for Br in the empirical formula without requiring a retained C–Br bond.

The C2 end of the enamine attacks the bromomethyl carbon by $S_{\mathrm{N}}2$ displacement. Simultaneously, donation of the nitrogen lone pair creates the C1=N bond. This is ordinary enamine C-alkylation, not N-alkylation.

proton transfer removes an alpha proton at C6, converting the iminium ion into the enamine regioisomer C1=C6. That enamine attacks the terminal $\mathrm{C}{\mathrm{H}}_2$ of the tethered acrylate in an intramolecular Michael addition. nitrogen donation restores the iminium ion, and protonation of the ester enolate completes the sequence.

The new bridge is $C2-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{C}{\mathrm{O}}_2\mathrm{Et})-\mathrm{C}{\mathrm{H}}_2-C6$. Together with the original ring and its C1 bridge this produces the bicyclo$[\mathrm{3.3.1}]$nonane iminium ion skeleton in C. It retains the pyrrolidine group and the ester; hydrolysis to a ketone is not part of the displayed reaction. The second panel of the original annulation figure shows the chemical atom connectivity and distinguishes ionic bromide in B from a covalent bromine substituent.

Both reactions are E2 eliminations requiring an axial chloride and an axial beta hydrogen in an antiperiplanar conformation relative to the C–Cl bond. The reacting C–H bond supplies the alkene $\pi$ bond as the C–Cl bond breaks; hydroxide accepts the proton.

For the trans-1,2 substrate, the low-energy chair has methyl and chlorine both equatorial. ring inversion is needed to place chlorine axial, but this also places methyl axial. The population of the reactive chair is consequently small, explaining the slow reaction. At the methyl-bearing neighboring carbon the axial position is occupied by methyl, not hydrogen. Only the other beta carbon has the needed anti axial H, so elimination gives 3-methylcyclohexene rather than the more substituted alkene.

For the cis substrate, a favorable chair has methyl equatorial and chlorine axial. It can react without paying the axial-methyl penalty, explaining the faster rate. Both adjacent carbons now supply an anti axial hydrogen. Removing the H on the methyl-bearing carbon gives 1-methylcyclohexene, the major product; removing the other gives 3-methylcyclohexene, the minor product. The former has the more substituted double bond.

Thus alkene stability selects between accessible pathways in the cis case, but it cannot create the missing antiperiplanar conformation H in the trans case.

sulfide first opens one epoxide by backside attack, giving a carbon-bound thiolate and an alkoxide. Using the same numbering, take this first attack at C2: the C2–S bond is down and C3–O remains up.

The attached sulfur still bears a nucleophilic lone pair and negative charge. It attacks C5 of the second epoxide intramolecularly, breaking C5–O and closing a bridge between C2 and C5. Work-up protonates the oxygen anions at C3 and C6. Thus B is 7-thiabicyclo$[\mathrm{2.2.1}]$heptane-2,5-diol, with both OH groups on the face opposite the sulfur bridge.

Both C–S bonds are formed by inversion at their attacked epoxide carbons. Because the original epoxides are cis, both new C–S bonds point to the same opposite face and one sulfur chemical atom can connect them. The 1,4 bridge gives two five-membered sulfur-containing circuits, unlike the more strained closure between nearby 1,3 positions.

The sulfur therefore replaces two separate external nucleophiles: two epoxides are opened but only one sulfur chemical atom is incorporated. Adding that sulfur and two work-up protons gives ${\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_2\mathrm{S}$, consistent with B.

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 alcohol oxygen attacks sulfur of methanesulfonyl chloride, chloride leaves, and triethylamine removes the O–H proton. This gives an O-mesylate with retention of the carbon configuration: the original carbon–oxygen bond has not been displaced.

An accessible chair places the mesylate axial. triethylamine then removes the anti axial hydrogen on the carbon between the ketone and mesylate. Its C–H bond forms the new C=C bond while the C–O bond expels mesylate. The result is the conjugated cyclohexenone.

The neighboring carbon on the acetonide side has its relevant axial position occupied by its oxygen substituent, rather than an anti axial hydrogen. The elimination toward the ketone is therefore favored both by the trans-diaxial geometry and by conjugation of the resulting alkene with C=O. The acetonide remains intact. No inversion of the original alcohol carbon is required during mesylation, and it becomes planar during elimination.

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.

The activated aromatic pi system polarizes bromine and attacks its electrophilic end while Br–Br cleaves. This makes a C–Br bond at C1 or C3 and gives a Wheland intermediate. Donation from phenolic oxygen stabilizes contributors with positive charge at its attached ring carbon. acetate or bromide then removes the proton from the attacked carbon; the C–H bond electrons restore the aromatic pi system. Repetition at the other activated ortho site gives the dibrominated product.

Each step replaces C–H by C–Br and restores aromaticity. No radical initiator or benzylic abstraction is needed, and no Lewis acid catalyst should be invented for this already activated phenolic substrate. The overall substitution balance is $\mathrm{Ar}{\mathrm{H}}_2+2B{\mathrm{r}}_2\to \mathrm{ArB}{\mathrm{r}}_2+2HBr$.

amide ion removes a methyl-side alpha proton of pinacolone. The C–H electrons form the enolate C=C bond as the carbonyl pi pair moves onto oxygen; ammonia is formed. The tert-butyl-side alpha carbon has no hydrogen, fixing the ordinary enolate regiochemistry.

The carbon end of D attacks the acyl chloride carbonyl carbon. Move the C=O pi pair onto oxygen, then collapse the tetrahedral intermediate with chloride departure. This nucleophilic acyl substitution gives $\mathrm{tCO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COt}$ and constructs the bond between the original enolate methylene and the incoming carbonyl. The observed product requires carbon acylation, rather than an O-acyl enol ester. A further equivalent of D removes the much more acidic central proton of the new 1,3-diketone, generating its extensively delocalized enolate and returning that second D equivalent to pinacolone. Aqueous acid workup protonates the product enolate; keto–enol equilibration gives the conjugated, internally stabilized by a hydrogen bond enol observed by NMR. The acyl chloride contributes its entire pivaloyl group; neither tert-butyl group is lost.

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 five-carbon unit is initially an open chain, not a completed carbocycle. Label it $P_1-P_2-P_3-P_4-P_5$, with phenyl-bearing $P_1$, carbonyl $P_3$, amide-substituted $P_4$, and the two ethylthio groups on $P_5$. Its double bonds are initially $P_1=P_2$ and $P_4=P_5$. Let $X$ be the nitro-bearing carbon of nitroethane; its methyl remains a substituent throughout.

DBU removes an alpha proton from nitroethane to give a nitronate. carbon attack at the Michael acceptor $P_1$ forms $X-P_1$, with electron redistribution through $P_2-P_3$ onto the carbonyl oxygen. proton transfer restores the ketone. Removing the remaining X–H proton gives another carbon-nucleophilic nitronate. It attacks the other conjugated end, $P_5$, intramolecularly, closing the six-membered circuit $P_1-P_2-P_3-P_4-P_5-X$. Addition to the ketene dithioacetal and ethylthiolate elimination restore the $P_4=P_5$ bond, leaving one SEt group at $P_5$.

base-assisted loss of the proton at $P_1$, with C–NO2 cleavage at X, creates $P_1=X$ and expels nitrite. Enolization at $P_2-P_3$ supplies the third aromatic double bond and the phenolic OH. proton transfers to the leaving anions complete the acid–base balance. nitroethane supplies the new ring carbon X and its methyl substituent; its nitro group is eliminated. Aromatization, rather than a separate oxidation of an intact six-membered starting ring, provides the final driving force. This is a proposed stepwise nitroalkane benzannulation; related DBU-promoted tandem Michael addition/cyclization/aromatization has direct primary precedent.^[ref]

Let $\mathrm{Ar}=4-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4$ and number the carbon substituent from oxygen: $\mathrm{ArO}-\mathrm{C}{\mathrm{H}}_2(1)-\mathrm{C}{\mathrm{H}}_2(2)-\mathrm{C}{\mathrm{H}}_2(3)-\mathrm{CN}$. sodium hydride removes a proton from C3, alpha to the nitrile, forming hydrogen and a nitrile-stabilized carbon anion. That carbon attacks the nitroaryl ipso carbon bearing O. The tether closes a five-membered cyclic Meisenheimer intermediate with sequence C3–C2–C1–O–aryl ipso carbon. The para nitro group stabilizes the negative charge generated by temporary loss of aromaticity.

Rearomatization breaks the aryl–O bond, releasing the tether oxygen as an alkoxide while retaining the new aryl–C3 bond. protonation gives $\mathrm{Ar}-\mathrm{CH}(\mathrm{CN})-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. This is a Truce–Smiles rearrangement: the aryl group migrates from oxygen to carbon. The nitrile remains attached to its original side-chain carbon; the original ether oxygen becomes the terminal alcohol oxygen. The substrate and product have identical atomic composition. Ordinary substitution at a saturated chain carbon would not produce this aryl–carbon bond or restore the terminal OH.

triethylamine removes an alpha proton from nitromethane to generate a nitronate population. Its carbon attacks the cyclic ketone carbonyl of E, forming a new C–CH2NO2 bond while the C=O pi pair moves onto oxygen. protonation by triethylammonium or another proton donor gives the tertiary alcohol F and regenerates base. This is a Henry reaction, also called a nitroaldol reaction: the former ketone carbon now bears both OH and CH2NO2.

The nitro group remains attached to its original nitromethane carbon; attack through oxygen would give a different connectivity. The methyl acetal and the original alcohol substituents need not be cleaved or displaced. This addition produces the isolated connectivity without requiring dehydration to a nitroalkene. The stated complex mixture is compatible with competing additions or stereoisomers, but the scheme supplies no basis for claiming F is the unique or kinetically dominant product. Only its formation is being explained.

The first base equivalent removes O–H; the second removes the alpha C–H. The first equivalent of lithium diisopropylamide converts the carboxylic acid to its lithium carboxylate. A further equivalent removes the remaining hydrogen on the carbon between the two allyl groups, giving a carboxylic acid dianion:The second negative charge has enolate-like delocalization into the carboxyl group; lithium coordination and aggregation are not represented by this simple charge-bookkeeping structure. The carbon nucleophile attacks methyl iodide with C–I bond cleavage, forming the new C–Me bond. acid workup reprotonates the carboxylate, giving $F=\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{Me})(\mathrm{L}{)}_2$. The two equivalents are therefore chemically necessary for the stated carbon alkylation rather than simply an arbitrary excess. A single equivalent is consumed predominantly by the much more acidic carboxylic proton. The product has formula ${\mathrm{C}}_9{\mathrm{H}}_{14}{\mathrm{O}}_2$ and still has two identical allyl groups, so this methylation does not create a stereocenter.

hydroxide attacks an ester carbonyl to form a tetrahedral intermediate. Collapse expels ethoxide, and acid–base transfer gives the carboxylate and ethanol. Repeating this nucleophilic acyl substitution at the other ester produces the malonate dianion; acid workup gives $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{L}{)}_2-\mathrm{C}{\mathrm{O}}_2\mathrm{H}$, with $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ as defined above.

On heating, one carboxylic acid group is lost through a six-membered cyclic proton-transfer/bond-cleavage arrangement. A carbonyl oxygen of the retained group accepts the proton from the departing carboxyl OH as its C–C bond breaks and carbon dioxide forms; electron redistribution gives the enol-like form of the retained acid. proton transfer/tautomerization then producesThe net decarboxylation is ${\mathrm{C}}_9{\mathrm{H}}_{12}{\mathrm{O}}_4\to {\mathrm{C}}_8{\mathrm{H}}_{12}{\mathrm{O}}_2+\mathrm{C}{\mathrm{O}}_2$. The central carbon gains a hydrogen, both allyl substituents survive, and only one carboxyl group is removed. It would be incorrect to draw hydrolysis directly giving the final acid without accounting for this carbon loss. The favourable cyclic pathway is characteristic of malonic acid decarboxylation.

The malonate carbon receives two allyl groups. With $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, attached through its first carbon, the product issodium hydride removes a methylene proton from diethyl malonate; molecular hydrogen formation accompanies production of an enolate stabilized by resonance in chemistry. Its carbon attacks the bromomethylene carbon of allyl bromide in an SN2 reaction, expelling bromide. The monoalkylated malonate still has one hydrogen between the carbonyls. Excess base removes that hydrogen, and a second equivalent of allyl bromide produces a second carbon–carbon bond. The final malonate carbon has no remaining hydrogen and cannot undergo a third analogous deprotonation there.

The two alkenes are preserved because the substitution occurs at the primary carbon bearing bromine. Counting the introduced groups gives ${\mathrm{C}}_{13}{\mathrm{H}}_{20}{\mathrm{O}}_4$ for the diallylated diethyl ester. Its two identical allyl groups also mean that the central carbon is not a stereocenter. This stage is the alkylation portion of the malonic ester synthesis.