二氧化硅负载硅钨钼酸催化合成乳酸正戊酯  

Catalytic Synthesis of n-Amyl Lactate with H_4SiW_6Mo_6O_(40)/SiO_2

作　　者：钟立群[1] 余金霞[1] 喻莉[1] 杨水金[1] 

机构地区：[1]湖北师范学院化学与环境工程学院,湖北黄石435002

出　　处：《化工中间体》2012年第7期50-52,共3页

摘　　要：以二氧化硅负载硅钨钼酸H4SiW6Mo6O40/SiO2为催化剂,乳酸和正戊醇为原料合成乳酸正戊酯。探讨H4SiW6Mo6O40/SiO2对酯化反应的催化活性,较系统地研究了酸醇物质的量比、催化剂用量、带水剂用量、反应时间等因素对产物收率的影响。实验表明:H4SiW6Mo6O40/SiO2是合成乳酸正戊酯的良好催化剂,在固定乳酸的物质量为0.1mol,n(乳酸):n(正戊醇)=1:2.5,催化剂用量为0.6g,带水剂环己烷的用量为5mL,反应时间40min的最佳条件下,乳酸正戊酯的收率可达75.2%。n-Amyl lactate was synthesized from lactic acid and n-amyl alcohol in the presence of H4SiW6Mo6O40/SiO2 as catalyst.The factors influencing this reaction were discussed through orthogonal experiments.The experimental results show that the optimal condition is:molar ratio of lactic acid to n-amyl alcohol is1:2.5,catalyst dosage is 0.6g(lactic acid 0.1 mol),volume of cyclohexane as the water-carrying agent is 5 mL and reaction time is 40 min.The yield of n-amyl lactate can reach 75.2%.

关 键 词：乳酸正戊酯 H4SiW6Mo6O40/SiO2 催化 

分 类 号：TQ2[化学工程—有机化工]