[1-(5-氟戊基)-1H-吲哚-3-基](2-碘苯基)甲酮的合成

合成了具有较强的医学活性的[1-(5-氟戊基)-1H-吲哚-3-基](2-碘苯基)甲酮(AM-694)。首先以吲哚为起始原料,苯磺酰氯保护吲哚的1-位,AlCl3催化在吲哚3-位傅克酰基化的进行,避免了文献中通过格氏镁试剂(RMgX)对1-位的保护对无水、无氧的高要求,再将1-位保护脱掉,然后在1-位进行N-烷基化、酯基水解、甲磺酯化,最后用四正丁基氟化铵氟化。通过液相色谱(HPLC)进行含量分析,且利用核磁(NMR)和质谱(MS)进行了表征,反应总收率高达41.8%。与以往路线相比,文中方法步骤简单,成本较低,收率很高。

Synthesis of ( 1- (5-fluoropentyl) -1H-indol-3-yl ) ( 2-iodophenyl )methanone

The compound(1-(5-fluoropentyl)-1H-indol-3-yl)(2-iodophenyl) methanone(AM-694) had strong biological activity.It had been synthesized from indole.In order to make sure the Friedel Crafts acylation catalysted by AlCl3 happened in 3-position of the indole,the benzenesulfonyl chloride was used to protect 1-position of the indole.In this way,it could avoid the high anhydrous and anaerobic requirements when protected by RMgX reagent.Then,further deprotection,N-alkylation,ester hydrolysis,in 1-position of the indole methylsulfonyl esterification were carried out.Finally,compound AM-694 was obtained through tetrabutylammonium fluoride.The content was analyzed by high performance liquid chromatography(HPLC) characterized by nuclear magnetic(NMR) and Mass Spectrometry(MS).The total yield reached 41.8%.Compared with the previous route,the method had advantages of simple,low cost and high yield.