= 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>, \b[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 $\mathrm{CH_2=C(OEt)-O-(allyl)}$. 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 $\mathrm{-CH_2CO_2Et}$ 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. \b[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.
Back to article page
