 Solution

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]