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ethanol undergoes proton transfer with the methoxide produced. The oxygen lone pair of methoxide attacks the hydroxyl proton of ethanol, while the ethanol O–H bond pair returns to its oxygen:The alcohol acidities are similar, so this is an equilibrium, not an overwhelmingly one-way deprotonation determined solely by intrinsic acidity. The large ethanol reservoir and initially small methanol concentration nevertheless matter. This exchange means that lithium ethoxide need not be consumed stoichiometrically in the net transesterification. Neutral ethanol can also participate in proton shuttling; it is not acting as a reducing reagent or as a substrate for elimination under these conditions.

Aqueous base selectively converts the methyl ester to a carboxylate, and brief acidification gives the corresponding monoacid while retaining the tert-Butyl ester. At the methyl-ester end, a hydroxide lone pair attacks the acyl carbon and the carbonyl pi pair moves to oxygen. The tetrahedral intermediate collapses as the oxygen anion reforms C=O and the acyl C–OMe bond breaks to methoxide. proton transfer then gives the carboxylate and methanol. carboxylate formation makes the net saponification effectively irreversible under the basic conditions; protonation protonates that carboxylate.

The tert-Butyl ester reacts more slowly under these conditions because its bulky alkoxy group hinders acyl attack, and tert-butoxide is a stronger base and poorer leaving group than methoxide. The selectivity is kinetic, not an assertion that tert-butyl esters can never undergo base hydrolysis. The work-up should be sufficiently mild and brief to avoid the separate acid-labile pathway.

Anhydrous trifluoroacetic acid instead removes the tert-butyl group, leaving the methyl ester. protonation activates the ester; in a useful reactive protonation form the alkoxy oxygen bears the proton. The O–C(tert-butyl) bond then cleaves with its electron pair going to oxygen, producing the carboxylic acid and a tert-butyl-cation-like fragment. Loss of a methyl proton gives 2-methylpropene and regenerates the acid catalyst. Carbonyl-oxygen protonation and charge delocalization can precede cleavage; drawing alkoxy-oxygen protonation is a transparent way of tracking the bonds broken. The key distinction is alkyl–oxygen cleavage here versus acyl–oxygen cleavage in the base reaction. Methyl-cation formation is strongly unfavourable, and without water ordinary acid-catalysed methyl-ester hydrolysis is not available. No water is required to account for the tert-butyl deprotection or its alkene by-product.

Amine nitrogen binds the carbon end of CO${}_2$; Mg${}^{2+}$ binds through oxygen. The polarized C=O bonds make carbon electrophilic and oxygen electron rich. An amine nitrogen lone pair therefore donates toward carbon, potentially giving a carbamate-like adduct after proton transfer. The hard Lewis acid Mg${}^{2+}$ accepts an oxygen lone pair, either from coordinated CO${}_2$ or from an oxygen of a captured carbamate species. Simply reversing these assignments because nitrogen and oxygen both have lone pairs would put two electron-pair donors together without accounting for the electrophilic site.

The local binding pattern does not by itself determine the full proton-transfer state or whether cooperative chains form; those depend on the framework and amine arrangement. A documented diamine-appended framework shows cooperative carbon-dioxide insertion and ammonium carbamate chain formation, illustrating one possible implementation.^[ref]

To regenerate the material, lower the CO${}_2$ chemical potential using reduced pressure or a CO${}_2$-free purge, and, where binding is exothermic, heat the framework to favour desorption. The temperature must remain compatible with framework stability. This reverses adsorption or reversible chemisorption; it does not require permanently destroying the metal–organic framework.

The primary chloride undergoes an SN2 reaction; the triphenylmethyl chloride undergoes an SN1 reaction. The primary site is accessible to back-side attack, whereas three phenyl groups make back-side attack at the triphenylmethyl site extremely hindered and stabilize its carbocation through resonance.

For the SN2 reaction, a suitable transition state isThe entering C and leaving Cl lie on opposite sides of the reacting carbon, with partial C–C and C–Cl bonds. The total charge is $-1$. This transition state is not a discrete reaction intermediate.

For the SN1 reaction, the approximately planar $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ carbocation is a genuine reaction intermediate. Assuming its formation by C–Cl heterolysis is rate determining and trapping is rapid,Thus the limiting rate law is independent of ethanol concentration. The successive attack and proton transfer occur after the slow ionization; a cation-stabilizing environment and efficient trapping underpin this idealized SN1 description.

D is $(\mathrm{MeO}{)}_2\mathrm{CHC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Me}$, and E is $\mathrm{OHCC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$. D results from transesterification; E results from acid-catalysed acetal hydrolysis.

For D, methoxide attacks the ester carbonyl carbon and moves the carbonyl $\pi$ pair to oxygen. Collapse of the tetrahedral reaction intermediate reforms C=O and expels ethoxide. Proton transfer with the methanol solvent regenerates methoxide. Excess methanol favours replacement of the ethyl ester by a methyl ester; the dimethyl acetal survives these basic conditions.

For E, protonate one acetal oxygen, then break its C–O bond with the pair going to oxygen, releasing methanol and forming an oxocarbenium ion. Water attacks that carbon. Deprotonation gives a hemiacetal; protonation of its remaining methoxy oxygen permits loss of a second methanol. The hydroxyl oxygen's lone pair forms C=O during this loss, and deprotonation gives the aldehyde, regenerating the acid catalyst.

The spectroscopy is consistent with these products. D has one ester carbonyl near $1745\mathrm{c}{\mathrm{m}}^{-1}$ and six carbon-13 NMR environments: acetal CH, two distinct chain $\mathrm{C}{\mathrm{H}}_2$ sites, carbonyl, ester $\mathrm{OC}{\mathrm{H}}_3$, and the two equivalent acetal $\mathrm{OC}{\mathrm{H}}_3$ groups. Thus seven carbon atoms give six resonances. E has separate ester and aldehyde carbonyl bands, approximately $1745$ and $1730\mathrm{c}{\mathrm{m}}^{-1}$, and six distinct carbon atoms: two carbonyls, two chain methylenes and the two ethyl carbons.

Treat this as standalone organophosphorus chemistry. The phosphorus(III) lone pair attacks the terminal nitrogen of the organic azide. Cyclization of the initial phosphazide and loss of nitrogen give an iminophosphorane:The ortho amide provides a neighboring acidic NH proton and a carbonyl oxygen donor. A proton relay to the imino nitrogen and intramolecular O$\to$P attack generate a five-membered, O-bound pentacoordinate phosphorus intermediate. Water can then attack phosphorus, with proton transfer to nitrogen, followed by P–N cleavage and opening of the neighboring-group contact.

The overall hydrolysis products areThe carbonyl group assists the hydrolysis and is regenerated; the product is an aniline and the ortho-carboxamide phosphine oxide, rather than an amide-linked ligation product. The favorable five-membered intramolecular interaction and proton delivery help hydrolyze a normally rather persistent aryl iminophosphorane. This explanation concerns only the printed chemical transformation.

The first step is an endo-selective Diels-Alder reaction of the pyrone diene with the electron-poor acrylamide. Number the pyrone $O1-C2(=O)-C3(OH)=C4-C5=C6$, and the acrylamide carbons $C_{\alpha }$ and $C_{\beta }$, with $C_{\alpha }$ next to its carbonyl. The favored polar connection is C3–C${}_{\alpha }$ and C6–C${}_{\beta }$, leaving C4=C5 and the original C2–O1 lactone bridge intact.

Lithium allyloxide, generated from n-Butyllithium and allyl alcohol, attacks the bridging lactone carbonyl. Collapse of the tetrahedral intermediate cleaves its acyl–oxygen bond. proton transfer converts the former O1 bridge into an OH substituent at C6 and leaves the former pyrone carbonyl as an allyl ester attached to C3. The morpholine amide is retained.

Thus C is allyl 1,4-dihydroxy-2-(morpholine-4-carbonyl)cyclohex-5-ene-1-carboxylate, with the stereochemistry described below. In this product numbering C1 is the former pyrone C3, C2 is the acrylamide $C_{\alpha }$, C3 is $C_{\beta }$, and C4 is the former pyrone C6.

The bridge fixes the former C3 carboxyl group and C6 oxygen on the same face. An endo acrylamide places its amide substituent on that face too. Accordingly CO${}_2$allyl at C1, the C2 amide and the C4 OH are cis; the C1 OH is on the opposite face. The alternative face of attack gives the enantiomer, so achiral reagents give a racemate. The $1731\mathrm{c}{\mathrm{m}}^{-1}$ band is consistent with the newly formed ester, while the strong lower-frequency band is consistent with the retained, strongly polarized amide carbonyl. These bands support carbonyl-class retention; they do not independently prove the predicted regioisomer or stereochemistry.

The terminal alkene of D first undergoes Wacker oxidation, giving the side chain $\mathrm{Ar}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{COC}{\mathrm{H}}_3$. Alkene coordination to Pd(II), water attack, proton transfers and beta-hydride elimination give the ketone. Copper and oxygen regenerate the active palladium oxidation state.

Base removes a proton from the terminal methyl of that ketone. Its enolate attacks the neighboring aromatic aldehyde intramolecularly, closing a six-membered ring. Dehydration gives a conjugated cyclohexadienone-type intermediate, and keto-enol tautomerism furnishes aromatic naphthol E. The retained methoxy and MOM-protected oxygen groups follow the original aromatic-carbon positions.

F is the O-allyl ether of E. Carbonate forms the naphthoxide and allyl bromide undergoes an SN2 reaction at its terminal carbon. Heating F then causes a [3,3]-Claisen rearrangement: the O–allyl bond breaks, the terminal allyl carbon bonds to the adjacent naphthalene C1, and the phenolic carbon temporarily becomes a ketone. Rearomatization gives G, with a free phenolic OH and a C1 allyl group. The C1 position gives the favored alpha-substituted naphthol and avoids the alternative less favorable annulation pattern.

The final step is a dearomative Michael addition, not a Nef conversion of the nitro group. Hydrogen bonding from the thiourea activates nitroethylene. The nucleophilic C1 resonance form of G attacks the terminal carbon of $\mathrm{C}{\mathrm{H}}_2=\mathrm{CHN}{\mathrm{O}}_2$, and proton transfer converts the nitronate to a 2-nitroethyl substituent. The phenolic carbon becomes a ketone.

H is the corresponding 1-allyl-1-(2-nitroethyl)naphthalen-2-one, retaining the methoxy and MOM substituents of G. The second ring now has a C2=O group and a C3=C4 bond, while the other ring remains aromatic. The reported ${}^{13}$C signal near $202$ ppm is consistent with that newly formed ketone. Related dearomative additions establish this reaction class.^[ref]

C1 is now an all-carbon quaternary stereocenter bearing both allyl and 2-nitroethyl groups. Both the drawn thiourea and the starting materials are achiral, so H is racemic. A chiral catalyst would be needed to select one absolute configuration.

An aromatic–aliphatic polyamide and methanol are formed. Let $\mathrm{Ar}=p\text{-}{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{C}}_6{\mathrm{H}}_4\text{-}p$. The polymer repeat unit isEach amide linkage comes from nucleophilic acyl substitution: an amine attacks an ester carbonyl, the tetrahedral intermediate expels methoxide, and proton transfer produces methanol. There are approximately $2n$ molecules of methanol for $n$ complete repeat units in a long chain; the precise finite-chain count depends on its end groups. Removal of methanol drives this step-growth polymerization.

The carbon–phosphorus coupling in the ${}^{13}$C spectrum is consistent with deprotonation of the methyl group rather than loss of that group:Addition to the nitrile initially gives $\mathrm{P}{\mathrm{h}}_2\mathrm{P}(\mathrm{O})\mathrm{C}{\mathrm{H}}_2\mathrm{C}(\mathrm{tBu})=\mathrm{NLi}$. An intramolecular proton transfer then converts the imidoyl anion into a conjugated lithiated enamide:The phosphorus–carbon bond is retained. The rearrangement is a hydrogen migration, not a migration of the phosphoryl group to nitrogen. Addition of half an equivalent of TMEDA per Li gives the hemisolvated dimerThe tetrameric and dimeric nuclearities are established for this reaction system; the spectra in the question are consistent with them, but spectra alone would not uniquely prove every detail of the aggregates.^[ref]

Label the ynone carbons $\mathrm{Ar}(\mathrm{OH})-\mathrm{CO}-{\mathrm{C}}_{\alpha }\equiv {\mathrm{C}}_{\beta }-\mathrm{Ph}$. The N-oxide oxygen adds to the electrophilic $C_{\beta }$, giving an O-vinylated quinolinium species with an enolate-like nucleophilic $C_{\alpha }$.

The chain $C_{\alpha }-C_{\beta }-O-N-C{2}_{\mathrm{quinoline}}$ is now arranged for rearrangement: form a bond between $C_{\alpha }$ and quinoline C2 while cleaving N–O and forming the $C_{\beta }=O$ bond. This can be represented through a cyclic addition/rearrangement intermediate, followed by proton transfer and rearomatization. The resulting carbon skeleton is the heteroarylated 1,3-dicarbonyl systemFinally, the ortho phenolic oxygen attacks the distal carbonyl. Cyclodehydration forms the chromone ring and eliminates water. The product is 2-phenyl-3-(quinolin-2-yl)chromen-4-one. The original ynone carbonyl becomes the chromone 4-one, the phenyl-bearing alkyne carbon becomes C2, and the other alkyne carbon becomes C3 carrying the heteroaryl substituent.

This atom mapping explains both the new C–C bond and the loss of the N-oxide oxygen from the heteroarene; it is not a direct substitution on the phenyl ring.

Interconversion of constitutional forms by proton transfer accompanied by redistribution of pi bonds. Keto–enol interconversion is a familiar example.