摘要
研究了微波辐射下用二氧化硅负载磷钨酸作催化剂直接催化合成苯乙酸β-苯乙酯的反应。考察了影响酯化率的因素,确定了该反应的优化条件:当苯乙酸用量为0.05 mol时,β-苯乙醇与苯乙酸摩尔比为1.6:1,磷钨酸负载量为24.8%的催化剂用量1.2 g,带水剂环己烷用量6 mL,微波辐射功率500 W,反应时间8 min。在此条件下,酯化率可达91.7%。用红外光谱对产物结构进行了表征。
β-Phenylethyl phenylacetate was as catalyst under microwave irradiation. The synthesized using silicon dioxide supported phosphotungstic acid factors affecting the yield of esterfication were investigated. The optimum conditions were as follows： when the dosage of phenylacetic acid was 0.05 mol, the molar ratio of β-phenethyl alcohol to phenylacetic acid was 1.6 ： 1, amount of catalyst supported phosphotungstic acid 24.8% was 1.2 g, microwave power was 500 W, and microwave irradiation time was 8 min. The yield of esterification reached to 91.7% under the optimum condition. The product was characterized bv IR.
出处
《精细石油化工进展》
CAS
2009年第8期27-30,共4页
Advances in Fine Petrochemicals
基金
湖南省自然科学基金资助项目(07JJ3017)
湖南省教育厅优秀青年科研基金项目(06B089)
关键词
苯乙酸β-苯乙酯
二氧化硅负载磷钨酸
微波辐射
催化酯化
β-phenylethyl phenylaeetate, silicon dioxide supported phosphotungstic acid, microwave irradiation, catalytic esterification
