Ocotillol及其差向异构体的合成与形成机制

Synthesis and formation mechanism of ocotillol and its epimer

以20(S)-原人参三醇[20(S)-PPT]为原料,采用两条路线合成ocotillol(3)及其差向异构体(4),并探讨其形成机制。路线1:以m-CPBA氧化20(S)-PPT,制备化合物3和4,收率分别为44.1%和28.6%;路线2:20(S)-PPT经乙酰化、mCPBA氧化和皂化反应,制备化合物3和4,收率分别为16.4%和16.2%。化合物3和4的形成机制推断如下:1)20(S)-PPT的分子内氢键导致C24(25)双键两侧的化学环境不同,两侧被氧化的概率不等,进而得到不等量的化合物3和4;2)20(S)-PPT被乙酰化后,不能形成分子内氢键,C24(25)双键平面两侧的化学环境近乎相同,经m-CPBA氧化反应、分子内SN2反应和皂化反应,最终得到几乎等物质量的化合物3和4。

Ocotillol（3）and its epimer（4）have been synthesized from 20（S）-protopanaxatriol ［20（S）-PPT］ via two routes, and their formation mechanism has been speculated. Route 1： Compounds 3 and 4 were obtained from 20（S）-PPT by oxidation with m-CPBA at the yield of 44. 1% and 28. 6%, respectively. Route 2： Compounds 3 and 4 were prepared from 20（S）-PPT by acetylation, oxidation and saponification at the yield of 16. 4% and 16. 2%, respectively. The formation mechanism of compounds 3 and 4 is speculated as below： 1）The chemical environments of both sides of C24（25）double bond in 20（S）-PPT are different due to the existence of intramolecular hydrogen bond, which led to the different oxidation ratio of the two sides, and the different yields of compounds 3 and 4. 2）There is not intramolecular hydrogen bond in acetylated 20（S）-PPT, and the chemical environments of both sides of C24（25）double bond are similar, which resulted in almost equal yields of compounds 3 and 4 synthesized through oxidation with m-CPBA, intramolecular SN2 and saponification.