Question

Organic Chemistry. Draw mechanisms for the following.

Scheme III. Preparation of the Lactone Portion QAc OAc OH OH T56% Meon g.h Meo OAc OTr OTr Отr 29 21 22 OH OH MeO O Meo MeO O

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Answer #1

Mechanism:

In starting material three ester group are present. In the presence of NaOMe base ester hydrolysis takes place which will give three alcohol substituted product.

In the second step, Oxymercuration reaction takes place. In this reaction three-membered ring-opened by nucleophilic methanol solvent. Then sodium borohydride act as the hydride donor which is used to substitute the mercury by hydrogen.

In the third step, the SN1 reaction takes place. The tri-phenyl chloride creates more stable carbocation. The formed carbocation attacked by specifically primary alcohol, (other two alcoholic groups are secondary)  

In the fourth step, tosylation happen (Tosylates are a good leaving group) followed by epoxidation takes place which is an SN2 reaction so, backside attack takes place.

The formed epoxide ring-opened by LiAlH4(hydride donor). Then Birch reduction takes place on an aromatic ring.

PhOCSCl has more electrophilic carbon. So nucleophilic oxygen attack that carbon and give the substituted product.

alkyl-substituted Sn easily generates radical in the presence of Light. Then radial substitution reaction takes place.

Finally, silly production takes place on oxygen by using BuSiMe2Cl in presence of base-imidazole. BuSiMe2 indicated as R in the final product.   

Si one ea Ac OH 4) Na Ome b) Hglom) Meow Ester hydrolysis Meow Come Starting i sme material SN2 | c) Nad OH OH SNI W OW all N:0 H y si Bume, Si Bume Busine, u ме от Men ] Medis Final product

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