 Solution

The first two operations invert the allylic hydroxyl group. Triphenylphosphine adds to the azodicarboxylate, generating the reagent pair that converts the alcohol into an alkoxyphosphonium leaving group. Benzoate displaces it with inversion at carbon, giving the inverted benzoate ester. Methoxide then cleaves that ester by nucleophilic acyl substitution, retaining the newly inverted carbon–oxygen stereochemistry because cleavage occurs at the acyl group, not by substitution at the stereogenic carbon.

The inverted allylic alcohol exchanges with the orthoacetate under acid catalysis. Ethanol loss produces an allyl ketene acetal, schematically $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}(\mathrm{OEt})-\mathrm{O}-(\mathrm{allyl})$. A six-membered suprafacial Johnson-Claisen rearrangement breaks the allylic C–O bond, shifts the ring alkene by one position and forms the new bond from the original alkene terminus to the ketene-acetal CH${}_2$. The result is the $-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$ substituent at the carbon that was the remote end of the original double bond. A chair-like transition state placing the larger substituents away from axial crowding explains the stereochemical transfer; the inverted alcohol configuration selects the corresponding face.

Finally, TiCl${}_4$ activates the anomeric methoxy group and methoxide departure creates an oxocarbenium ion. Hydride transfer from triethylsilane replaces that methoxy substituent by hydrogen. The ring oxygen and pendant OTBS group remain intact. The sequence is inversion, stereospecific allylic rearrangement, then reductive removal of the anomeric methoxy group, rather than direct reduction or hydrolysis of the ester introduced by the Claisen step.