Acyl-palladium-alkoxide

The last possibility for ester formation (20, Figure 12.15) comprises the reductive elimination of esters from acyl-alkoxy-palladium complexes 17, formed by deprotonation of the alcohol adducts 16. Clearly, it requires cis coordination of the alkoxide and acyl fragment. Since monodentates have a preference for ester formation, it was thought that this mechanism was very unlikely. 

Scheme 4.4 details a short synthesis by Trost of an intermediate employed in the Gerlach synthesis of 9. ° Protected aldol 19 was reacted with a-lithio ethyl vinyl ether and the resulting alkoxide intermediate acylated to provide the protected diol 20 in 75% yield. Chain extension by way of a palladium-mediated allylic displacement with isopropyl phenylsulfonylacetate followed by ester exchange afforded a 62% yield of 21. Hydrolysis of the enol ether and 3-elimination of the sulfonyl group produced 22 in 66% yield. Ketalization of 22 then gave 16, which had been previously converted to the natural product in 50% yield. 1 ... 

There are two possibilities for formation of the products from the acylpalladium species. One is the direct attack of the nucleophile on the acyl ligand and the other involves coordination of the NuH to the palladium center. The palladium-bound NuH ligand, such as alcohol, is deprotonated by a base to give an acylpalladium alkoxide, which releases the ester as the reductive elimination product. Recent studies on model complexes provided some evidence supporting a route via reductive elimination [59]. 

The most difficult step in the catalytic cycle is the first an oxidative insertion of the palladium catalyst a into the C—O bond of the carboxylic ester 3c, leading to the acyl complex b. This was unprecedented even for activated esters, and is increasingly endothermic with increasing basicity of the alkoxide leaving group. In analogy to the de Vries process, an... 

---