碘帕醇的合成工艺研究

Studies on the synthesis of lopamidol

以L-乳酸为原料,经两步反应,以55%的收率得到S-2-乙酰氧基丙酰氯。以5-氨基-2,4,6-三碘-1,3-苯二甲酸为原料,使用二氯亚砜转化得到5-氨基-2,4,6-三碘-1,3-苯二甲酰氯,再与S-2-乙酰氧基丙酰氯反应得到5-[(2S)-(2-乙酰氧基)丙酰胺基]-2,4,6-三碘-1,3-苯二甲酰氯,接着与2-氨基-1,3-丙二醇反应得到N,N′-双[2-羟基-(1-羟甲基)乙基]-5-[(2S)(2-乙酰氧基)丙酰胺基]-2,4,6-三碘-1,3-苯二甲酰胺,在碱性条件下脱乙酰基,得到非离子型X射线造影剂碘帕醇,这四步反应收率为50%。

S-2-Acetoxypropionyl chloride was synthesized from L-lactic acid in two steps with a yield of 55%. 5-Amino-2,4,6-triiodo-1,3-phthalic acid was converted to the acid chloride and was then coupled with S-2-acetoxypropionyl chloride to give 5-[（2S）-2-acetoxypropionylamino]-2,4,6-triiodo-1,3-phthaloylchloride.Further reaction with 2-amino-1,3-propanediol afforded N,N＇-bis[2-hydroxy-（1-hydroxymethyl）ethyl]-5-[（2S）-（2-acetoxypropionamido）] -2,4,6-triiodo-1,3-phthalamide.Hydrolysis to remove the acetyl group yielded Iopamidol with an overall yield of 50%for four steps.