高级检索

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

  • 摘要: 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 34 ,收率分别为44.1%和28.6%;路线2:20(S)-PPT经乙酰化、m-CPBA氧化和皂化反应,制备化合物 34 ,收率分别为16.4%和16.2%。化合物 34 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 34 ;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 34

     

    Abstract: 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.

     

/

返回文章
返回