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An enol has an OH group bonded to a carbon of a C=C bond. Protonation and pi-bond rearrangement can interconvert it with a carbonyl compound through keto-enol tautomerism.

Keto-enol tautomerism interconverts carbonyl and enol forms by proton transfer and pi-bond redistribution. Solvent-mediated relays avoid an unrealistic isolated intramolecular proton jump.

Use C1 for the carbonyl carbon, C2 for the adjacent carbon, and C3 for the beta carbon of the reactive enone fragment $\mathrm{R}-C1(=\mathrm{O})-C2H=C3H-{\mathrm{R}}^{\prime }$. water adds conjugately at C3, giving a beta-hydroxy ketone. The oxygen lone pair of neutral water attacks C3; move the C2=C3 pi pair toward C1–C2 and the carbonyl pi pair onto oxygen. This produces the charge-separated enolate–oxonium intermediateA second water molecule can relay a proton from the newly attached oxonium oxygen to the enolate system. Using the carbon-centered resonance structure, reform C1=O and protonate C2 while the donating O–H bond returns its pair to the added oxygen. Equivalently, oxygen protonation first gives an enol, followed by water-mediated keto-enol tautomerism. Either path yields a neutral beta-hydroxy ketone and regenerates the relay water. There is no need to introduce externally supplied H${}^{+}$ or OH${}^{-}$; solvent molecules perform the proton transfers. The uncatalyzed charge separation has a large barrier, so this mechanism describes a slow formal pathway, not a claim of rapid hydration in pure water.

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.

A consistent proposed sequence is desilylation, five-membered hemiketal formation, retro hetero-Diels–Alder opening, intramolecular aldol addition and hemiacetal closure. The essential carbon rearrangement is loss of the original ketone–side-chain bond and formation of a bond between its two neighbouring carbons; it is not a simple Michael addition.

For an explicit atom map, label the starting six-membered ring $A-B=C-D-E-O_r-A$, with the ketone at $D$, methyl substituents at $A$ and $B$, the hydroxy group at $A$, and the side chain $E-X-T-\mathrm{Et}$. Here $X$ is the methylene and $T$ is the carbon bearing the protected alcohol. Let $O_D$ be the original ketone oxygen and $O_T$ the protected alcohol oxygen. These labels identify atoms, not new substituents.

First, Fluoride removes the silyl protecting group from $O_T$. The resulting alcohol attacks the protonated ketone at $D$, forming the five-membered hemiketal $D-E-X-T-O_T-D$. The original oxygen $O_D$ is now a hydroxy group; the original pyran ring remains intact at this stage.

Next, The original pyran ring can undergo a retro hetero-Diels-Alder reaction. The three electron-pair movements are $B=C\to C=D$, $D-E\to E=O_r$ and $O_r-A\to A=B$: break $D-E$ and $O_r-A$, and form the three indicated double bonds. The result is the tethered aldehyde/dienol esterThe temporary $D-O_T$ bond keeps both reacting ends within the same molecule. At $D$, both oxygen substituents remain attached, and the new $D-O_T$ tether prevents separation into independent fragments.

Then The $C=D$ enol attacks the tethered aldehyde at $E$. Formation of $C-E$, restoration of $D=O_D$, and proton transfer give the six-membered lactone ring $C-D-O_T-T-X-E-C$, with $E-O_{r}H$. This is an intramolecular aldol reaction, and it supplies the crucial new carbon–carbon bond without losing any carbon atom.

Finally, Keto-enol tautomerism at $A=B$ gives the methyl ketone at $A$ and introduces the hydrogen at $B$. The $E-O_{r}H$ alcohol then attacks that ketone, closing the five-membered hemiacetal ring $A-B-C-E-O_r-A$. The resulting fused rings have exactly the product's connectivity: $D$ is the lactone carbonyl, $O_T$ is its ring oxygen, and $O_r$ is the five-membered-ring oxygen.

The stereocentre at $T$ survives throughout this sequence. The configuration initially at $E$ is erased at the aldehyde stage, and those at $A$, $B$, $C$ and $E$ arise during the later cyclization and proton-transfer steps. A folded intramolecular aldol reaction followed by reversible hemiacetal formation is consistent with the drawn cis ring junction, but connectivity and orbital bookkeeping alone do not prove that it is the unique selected diastereomer. In particular, a claim of simple retention at $E$ would contradict this atom map.

A related pyranone-to-fused-lactone rearrangement under aqueous hydrogen fluoride is experimentally reported; that account also explicitly qualifies the relative stereochemistry as assigned on mechanistic grounds. The sequence above is a proposed explanation of the examination structure, rather than a mechanism claimed to have been experimentally established in that paper.^[ref]