摘要
以硬脂酸与乙醇胺为原料 ,采用二步法合成硬脂酸单乙醇酰胺的反应优化条件是 :硬脂酸与乙醇胺的总的物质的量比为 1∶1 1 ;第一步反应中 ,硬脂酸与乙醇胺的物质的量比控制在 1∶0 7,温度为 1 70℃ ,反应 4h ;第二步反应中 ,加入剩余的乙醇胺 ,温度为 1 0 0℃ ,催化剂KOH用量为 0 0 76mol mol硬脂酸 ,反应 4h。硬脂酸单乙醇酰胺的产率可达到 86 9% ,终产物的熔点为 88～ 90℃。
Stearic acid monoethanolamide was prepared by two step reaction of stearic acid with ethanolamine in such a proportion that the total mole ratio of stearic acid to ethanolamine was 1.0∶11.In the first step, mole ratio of stearic acid to ethanolamine was controlled at 1.0∶07. The time of 4 h was essential to this reaction at 170 ℃. In the second step,the process was carried out in the presence of catalyst(KOH)in an amount of 0076 mol/mol stearic acid and the remaining amount of ethanolamine was added and the reaction proceeded at 100 ℃ for 4 h. The yield of stearic acid monoethanolamide could reach 869%. The melting point of the final products was 88-92 ℃.
出处
《精细化工》
EI
CAS
CSCD
北大核心
2000年第4期191-193,210,共4页
Fine Chemicals
