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An ether with oxygen directly attached to an alkene carbon. Electrophilic activation and hydrolysis can convert an appropriate enol ether back to a carbonyl compound.

$\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, an ether formed by trapping a triphenylmethyl carbocation with ethanol, followed by deprotonation.

Intramolecular capture of an alkene-derived iodonium ion by an alcohol produces a cyclic ether and a carbon–iodine bond. Backside opening gives an anti relationship between the new C–O and C–I bonds.

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.

B is 1,3-dibromo-2-hydroxyxanthone: both bromines are on the hydroxy-substituted outer ring, at its two available positions ortho to OH. Define C2 as the OH-bearing carbon; C1 lies between it and the carbonyl-side ring junction, and C3 lies on its other side. These labels specify the substitution independently of the orientation of the printed drawing.

Phenolic lone pair donation strongly activates its ortho and para positions towards electrophilic substitution. The para position is a fused junction and has no replaceable hydrogen, leaving C1 and C3 as the two directly activated sites. C1 also benefits from the conjugative influence of the ring ether oxygen. Although other positions receive some activation from that oxygen, they lack the stronger local phenolic activation of the selected pair. Excess bromine can therefore replace both ortho hydrogens. Replacing two hydrogens in ${\mathrm{C}}_{13}{\mathrm{H}}_8{\mathrm{O}}_3$ gives ${\mathrm{C}}_{13}{\mathrm{H}}_6\mathrm{B}{\mathrm{r}}_2{\mathrm{O}}_3$, consistent with the stated composition. Addition across an outer-ring double bond would sacrifice aromaticity and would not account for the hydrogen loss.

For a monocyclic, approximately planar, continuously conjugated pi system, Hückel's rule predicts aromatic closed-shell filling for $4n+2$ pi electrons, with $n=0,1,2,\dots$. Every atom on the circuit must have an appropriately oriented orbital; counting double bonds without checking pi conjugation is insufficient.

The two outer benzene rings are clearly aromatic; the central pyrone ring has a weaker, polarized six-pi aromatic contribution. Each outer ring has the familiar six-electron circuit. For the central ring, a useful charge-separated contributor moves the exocyclic carbonyl pi pair onto oxygen and donates one endocyclic oxygen lone pair into the ring. Its four fused carbon p centres contribute four electrons and the ring-oxygen lone pair supplies two: a six-pi, pyrylium-like circuit, with positive charge on the ring oxygen and negative charge on the carbonyl oxygen. This does not make three independent benzene rings of equal aromatic stabilization; localized carbonyl bonding and the fused benzene circuits remain important contributors.

The three oxygens play different roles. The endocyclic ether oxygen donates a p-type lone pair to pi conjugation. The carbonyl oxygen accepts pi density in the charge-separated contributor. The phenolic oxygen donates into the outer ring as a substituent. Its in-plane lone pair is not an extra pair to add to every ring's electron count. In particular, neither the external C=O pair nor both lone pairs on a ring oxygen should be counted as independent extra ring electrons. Structural studies of xanthone support a largely planar conjugated framework, while the division into strongly aromatic outer rings and a more polarized central ring is an electronic-structure interpretation.^[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.

protonation of the secondary OH permits water departure. The neighbouring phenylthio sulfur can assist this departure by forming a bond to the adjacent alcohol carbon, generating a thiiranium ion spanning that carbon and the original sulfur-bearing cyclohexane carbon. Label the latter Q and the former L. The temporary Q–S–L bridge transfers the phenylthio attachment from Q to L when its Q–S bond subsequently breaks. sulfur attack opposite the departing OH accounts for the new C–S stereochemistry without requiring a freely rotating carbocation.

For the protected terminal alcohol, no free tethered OH is available to trap the bridge. Removal of a beta hydrogen from a cyclohexane carbon adjacent to Q forms the ring alkene while Q–S breaks. sulfur remains bonded to L, giving the allylic phenyl thioether. acid is regenerated by the deprotonation, so the catalyst need not be consumed.

For the free terminal alcohol, its oxygen attacks Q intramolecularly, opening Q–S instead. Count the new ether ring: O–CH2–CH(Me)–L–Q, a five-membered ring sharing Q with the cyclohexane. deprotonation gives the spirocyclic ether. This is capture of the same neighbouring-group-stabilized intermediate rather than migration of the methyl group. The pre-existing methyl-bearing carbon retains its configuration; its wedge/dash appearance changes when the product is redrawn in a different ring projection. The two substrates differ in whether a suitably placed oxygen nucleophile is available for intramolecular trapping.

base removes an alpha-amide proton from G, giving a carbon nucleophile. Its intramolecular attack on the nitroaryl ipso carbon, which currently bears the ether oxygen, forms a five-membered cyclic Meisenheimer complex. Count its atoms along the existing tether: alpha carbon, the original phenyl carbon bearing the side chain, its adjacent phenoxy carbon, ether oxygen, and attacked nitroaryl ipso carbon. The ortho nitro group again stabilizes the temporarily dearomatized anion.

Rearomatization now breaks the nitroaryl–O bond, leaving phenoxide on the original ring. proton transfer gives $(2-\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{CH}(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}$. This is a Truce-Smiles rearrangement: the nitroaryl group migrates from oxygen to carbon, while the original phenoxy oxygen stays on its original ring and becomes OH. No nitro-group migration or amide cleavage is needed. G and H have identical formulas; only the bond connections and a proton position change. carbonate need not produce a high equilibrium enolate concentration: a small population can be consumed by favourable intramolecular cyclization and rearomatization. The specified transformation alone does not establish its detailed proton-transfer sequence.

G is the diaryl ether $(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{O}-(2-\mathrm{C}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}{\mathrm{C}}_6{\mathrm{H}}_4)$. Here $N_{morph}$ denotes the nitrogen of the intact morpholine ring; the phenoxy oxygen and phenylacetamide chain are ortho on their original ring. potassium carbonate first generates phenoxide from F. Its oxygen attacks the fluorine-bearing carbon of 1-fluoro-2-nitrobenzene, forming a Meisenheimer complex. Delocalization of its negative charge into the ortho nitro group stabilizes this addition intermediate. Fluoride elimination restores aromaticity and completes nucleophilic aromatic substitution.

This first step forms an O–aryl bond, rather than immediately arylating the less acidic alpha carbon. The neutral product has formula ${\mathrm{C}}_{18}{\mathrm{H}}_{18}{\mathrm{N}}_2{\mathrm{O}}_5$, obtained from the two substrates by net loss of HF. The connectivity is shown beside B in the independently drawn diagram above. The amide-substrate O-to-C migration strategy has direct primary precedent.^[ref]

Write the reacting terminus as $R-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, where $R$ includes the pre-existing stereocentres and the intact $\mathrm{OC}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O})\mathrm{NM}{\mathrm{e}}_2$ substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as ${\mathrm{O}}^{-}-{\mathrm{O}}^{+}=\mathrm{O}$, the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments $R-\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ and formaldehyde.

Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes $R-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.

The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.

The initial nucleophile attacks silicon, not the nitrile or an alkene carbon. Each trimethylsilyl group has an electrophilic Si center. A methyl group from methyllithium attacks Si, passing through a pentacoordinate silicon arrangement; Si–O cleavage releases $\mathrm{SiM}{\mathrm{e}}_4$ and the lithium salt of the oxygen-containing fragment. Cleaving the silyl enol ether generates a lithium enolate; cleaving the aryl silyl ether generates a phenoxide. Neither operation needs a new deprotonation at a carbon stereocenter.Here $R-O^{-}$ represents either the phenoxide or the oxygen-centered enolate contributor. Excess reagent permits both cleavages. The pre-existing methyl/nitrile stereocenter retains its configuration: it has no hydrogen to remove, and neither bond attached to that center is broken.

After HMPA is added, each anionic oxygen attacks the carbonyl carbon of allyl chloroformate. The C=O pi electrons move to the carbonyl oxygen, giving a tetrahedral intermediate; collapse expels chloride. The enolate becomes an allyl enol carbonate, and the phenoxide becomes an aryl allyl carbonate, each containing $\mathrm{O}-\mathrm{C}(\mathrm{O})-\mathrm{OC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. oxygen acylation accounts for preservation of the enol double bond and for the two newly installed carbonate groups. carbon attack would give a different carbon skeleton. The two oxygens are trapped because they were selectively unmasked by desilylation, rather than because every ether oxygen in the molecule is equally reactive.

Lithium diisopropylamide removes an alpha proton from the ethyl side of the ketone. The C–H electrons form the C=C part of a lithium enolate, while the C=O pi electrons move to oxygen. The phenyl side has no competing alpha C–H site. Diisopropylamine is the conjugate-acid byproduct.The carbon end of this enolate attacks the primary carbon bearing iodine in $\mathrm{IC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OMe}$. The C–I bond breaks to iodide in an SN2 reaction, and carbonyl reformation givesThus a three-carbon methoxypropyl chain is attached at the ketone alpha carbon. The remote methyl ether survives. The hindered amide base is chosen to generate the enolate rather than to add to the carbonyl. No absolute stereochemical preference is specified for the new stereocenter under these achiral conditions.

An enolate is an ambident nucleophile: the negative charge is concentrated toward oxygen, while its occupied conjugated orbital also has substantial reactive amplitude at the alpha carbon. An alkyl iodide presents an accessible, polarizable carbon electrophile. carbon attack through an SN2 reaction is generally kinetically favorable, while the more strongly charged oxygen is often coordinated to the counterion or solvated. The carbonyl reforms as the new C–C bond is made.carbon alkylation restores the carbonyl and makes the new C–C bond. oxygen alkylation instead gives an enol ether. The two outcomes are not resonance contributors of one product, and their ratio is not determined just by where the largest formal negative charge is drawn. electrophile, counterion, solvent and aggregation can change selectivity; hard silicon electrophiles and suitable acylating reagents often favor oxygen trapping. Product stability alone would not explain an irreversible kinetic choice.

Oxazolination occurs para to oxygen on the ring that does not bear bromine. The product is 2-[4-(4-bromophenoxy)phenyl]-2-Oxazoline, shown in the final panel above. The brominated ring retains its para bromine; the other ring receives the new oxazolinyl group at its open para position.

The diaryl ether oxygen activates both rings toward electrophilic aromatic substitution. On the brominated ring, its least crowded para position is blocked, and bromine also decreases ring reactivity inductively. The unbrominated ring has an accessible para position, favored over its ortho sites by steric effects. This is substitution of an aromatic hydrogen, not displacement of the aryl bromide.

The formula provides a useful consistency check: ${\mathrm{C}}_{12}{\mathrm{H}}_9\mathrm{BrO}$ loses one aromatic H and acquires ${\mathrm{C}}_3{\mathrm{H}}_4\mathrm{NO}$, giving ${\mathrm{C}}_{15}{\mathrm{H}}_{12}\mathrm{BrN}{\mathrm{O}}_2$. It cannot by itself distinguish positional isomers; the directing and steric arguments select the connectivity.

The alkene undergoes concerted oxygen transfer from the terminal peroxide oxygen of mCPBA. In the cyclic electron-flow description, the alkene pi bond begins one C–O bond, the peroxide oxygen supplies the second, O–O breaks, and proton transfer/carboxyl rearrangement gives m-chlorobenzoic acid. Both C–O bonds are installed from the same alkene face, so the alkene's relative substituent geometry is retained in the epoxide. There is no freely rotating carbocation intermediate.

The neighboring free OH hydrogen-bonds to the approaching peracid and favors a cyclic, organized transition-state geometry. oxygen is delivered from the OH-directed face, giving the indicated syn relationship between epoxide oxygen and the allylic hydroxyl. In the supplied projection, the epoxide C–O bonds and the original C–OH bond are on the rear/hashed face. Steric interactions still influence the preferred hydrogen-bonded conformer, but “attack from the least hindered face” alone misses the positive directing role of OH. The distant benzyl ether is not the directing free hydroxyl.

The classical experimental comparison of allylic alcohols and protected derivatives established this directing principle; it supports the mechanistic interpretation without itself measuring the selectivity of this acyclic substrate.^[ref]

Apply the Felkin-Anh model to the stereocentre alpha to the ketone. Its large group is the chain continuing through the phenyl-bearing carbon, its medium group is methyl, and its small group is hydrogen. In the preferred transition-state arrangement, put the large group approximately perpendicular to C=O and approach from the opposite face, passing the small hydrogen along a Bürgi-Dunitz trajectory. The remote ether cannot form the usual small alpha-alkoxy chelate, and the coordinating solvent further supports using an open-addition prediction.^[ref]

The predicted new stereocentre is $S$. In the question's zigzag projection, this corresponds to the new OH bond directed behind the page and the newly introduced methyl toward the reader, with the original stereocentres unchanged. The new-centre priority order is OH $>$ phenyl $>$ the substituted alpha-carbon chain $>$ methyl. This is a model-based major-product prediction, rather than a quantitative diastereomeric ratio.

A is dimethyl ether, $\mathrm{C}{\mathrm{H}}_3\mathrm{OC}{\mathrm{H}}_3$, with NaCl as the corresponding salt byproduct. Backside methoxide attack on methyl chloride gives an SN2 substitution:The rate is bimolecular because both reactants participate in the elementary bond-forming/bond-breaking event.

Build the phenethyl grignard reagent, use one portion to make a phenethyl-substituted ester, and add two further equivalents to that ester. Write $R=\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2$; the target is ${\mathrm{R}}_3\mathrm{COH}$, so all three carbon substituents must be phenethyl, not benzyl.

hydroboration followed by oxidation of styrene gives $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$: use $\mathrm{B}{\mathrm{H}}_3$ in a suitable ether, followed by alkaline hydrogen peroxide. The Anti-Markovnikov addition regiochemistry is essential; acid-catalysed Markovnikov addition of water would give the wrong alcohol. Convert the primary alcohol to $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$ with phosphorus tribromide, then form $\mathrm{RMgBr}$ with magnesium in dry ether. The organomagnesium reagent must be prepared and used without a protic solvent.

carboxylate one portion with carbon dioxide and acidify to obtain $\mathrm{RC}{\mathrm{O}}_2\mathrm{H}$, then esterify with ethanol under acid catalysis to give $\mathrm{RC}{\mathrm{O}}_2\mathrm{Et}$. Treat the isolated ester with at least two equivalents of another portion of $\mathrm{RMgBr}$ and then perform an aqueous acidic work-up:The first addition to the ester gives a ketone after ethoxide elimination; the second addition gives the tertiary alkoxide. Work-up supplies its OH proton. The extra central carbon comes from carbon dioxide, and each original styrene contributes one intact eight-carbon phenethyl unit. The resulting carbon count is $1+3\times 8=25$, matching the target skeleton.

The products are hexanenitrile, $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}\equiv \mathrm{N}$, and ethyl triphenylmethyl ether, $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The nitrile carbon comes from cyanide, so replacing chloride in the five-carbon chain gives a six-carbon product.

For the primary chloride, the carbon-end lone pair of $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted nucleophilic substitution step forms the C–C bond and releases $\mathrm{C}{\mathrm{l}}^{-}$. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$. An ethanol oxygen lone pair then attacks the carbocation, giving $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-{\mathrm{O}}^{+}(\mathrm{H})\mathrm{Et}$. Finally pyridine accepts that O–H proton, with the O–H bond pair returning to oxygen, giving the neutral ether and pyridinium chloride.