复方止痛片的分离与鉴定.ppt

<p>单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,复方止痛片的分离与鉴定,答辩人：李璐,专业：,08,化学,指导教师：宋伟新,实验方法：,本实验主要采用了两种不同的方法对复方阿司匹林的三种活性成分进行分离与鉴定，即薄层层析法与紫外分光光度法。,薄层层析法主要讨论其不同活化条件下对分离效果的影响，并确定最佳反应条件。,紫外分光光度法作为对比实验，并比较其与薄层层析法的优缺点。,实验内容：,1,、薄层层析法：,a,、标准样品的制备（咖啡因为提取物、阿司匹林为合成物）,b,、层析板的制备（在不同活化条件下）,c,、点样,d,、展开（混合展开剂：苯,:,乙醚,:,冰醋酸,:,甲醇,=120:60:18:1,）,e,、鉴定并计算,Rf,值,（,Rf,=,斑点中心到原始点的距离,/,展开剂原始点中心到前沿的距离）,讨论：,本实验分别在干燥温度为,80,下活化,0.5,小时、,1,小时、,1.5,小时,;100,下活化,0.5,小时、,1,小时、,1.5,小时,;110,下活化,0.5,小时、,1,小时、,1.5,小时下进行讨论，得出最佳活化条件。,在干燥温度为,80,时，,对于复方阿司匹林的鉴定与分离效果不是很明显，改用红外灯干燥后，再活化,0.5,小时、,1,小时、,1.5,小时。平行进行三次。,复方阿司匹林,0.5h,相对误差,%,1h,相对误差,%,1.5h,相对误差,%,Rf1,0.41,0.28,0.39,0.22,0.38,0.19,0.45,0.41,0.39,0.22,0.37,0.16,0.42,0.31,0.41,0.28,0.38,0.19,Rf2,0.69,0.15,0.67,0.12,0.63,0.05,0.67,0.12,0.68,0.13,0.65,0.08,0.63,0.05,0.63,0.05,0.62,0.03,Rf3,0.82,0.05,0.79,0.01,0.74,0.05,0.83,0.06,0.74,0.05,0.73,0.06,0.79,0.01,0.73,0.06,0.71,0.09,由上表可得出，在,80,条件下，随着活化时间的增加，,Rf,值减小，相对误差都较大不可取。,干燥温度为,110,时，活化,0.5,小时、,1,小时、,1.5,小时。平行进行三次,复方阿司匹林,0.5h,相对误差,%,1h,相对误差,%,1.5h,相对误差,%,Rf1,0.39,0.22,0.38,0.16,0.33,0.03,0.37,0.16,0.33,0.03,0.31,0.03,0.34,0.22,0.35,0.09,0.30,0.06,Rf2,0.65,0.08,0.61,0.03,0.58,0.03,0.67,0.12,0.63,0.02,0.59,0.02,0.67,0.12,0.63,0.05,0.59,0.02,Rf3,0.80,0.03,0.78,0.00,0.69,0.12,0.81,0.04,0.77,0.01,0.73,0.06,0.75,0.04,0.74,0.05,0.70,0.10,由上表可得出，在,110,条件下，随着活化时间的增加，,Rf,值减小，在活化时间为,1,小时时，相对误差较小，为最佳干燥活化条件。,干燥温度为,150,时，活化,0.5,小时、,1,小时、,1.5,小时。平行进行三次,复方阿司匹林,0.5h,相对误差,%,1h,相对误差,%,1.5h,相对误差,%,Rf1,0.27,0.16,0.27,0.16,0.25,0.22,0.30,0.06,0.29,0.19,0.26,0.19,0.24,0.25,0.23,0.28,0.21,0.34,Rf2,0.57,0.05,0.58,0.03,0.57,0.05,0.60,0.00,0.59,0.02,0.58,0.03,0.58,0.03,0.57,0.05,0.55,0.08,Rf3,0.69,0.12,0.67,0.14,0.65,0.17,0.69,0.12,0.67,0.14,0.67,0.14,0.68,0.13,0.65,0.17,0.62,0.19,由上表可得出，在,150,条件下，随着活化时间的增加，,Rf,值减小，相对误差偏大，不可取,。,结果分析与讨论,本实验主要论证了薄层含水量与,Rf,值的关系。由实验数据可知，薄层含水量越多、,Rf,的值越大。同时与相同条件下的标样试剂比较，在,110,、活化,1,小时的条件下，复方阿司匹林中三种活性成分的平均相对误差分别为,0.06%,、,0.03%,、,0.02%,。,由此可以得出：用薄层层析法对复方阿司匹林进行坚定的最佳活化条件为干燥箱,110,、活化,1,小时。,2,、紫外分光光度法,a,、溶液的制备：,精密称取阿司匹林,30mg,非那西汀,20mg,，咖啡因,20mg,，分别置于,100ml,容量瓶中，加入,5ml,乙醇溶解，加蒸馏水定容，即得对照品溶液,另取磷酸二氢钾,6.8g,放入小烧杯中，加,0.1mol/LNaOH 290ml,使其溶解，转移至,100ml,容量瓶中，加蒸馏水定容，即得缓冲溶液,分别精确量取对照品溶液适量（以相当于复方阿斯匹林药片标量的,50%,、,80%,、,100%,、,120%,、,150%,称取）于,100ml,容量瓶中，加缓冲溶液定容。,b,、结果分析,取阿司匹林、非那西汀、咖啡因对照溶液和样品溶液分别在,200400nm,进行紫外扫描，得相应紫外吸收曲线。,确定在,200300nm,波长范围内的吸收光度值，并计算其相对偏差。,1.0,1.0,1.0,0.5,0.5,0.5,0.0,0.0,0.0,200 300 400 200 300 400 200 300 400,A B C,1.0,0.5,0.0,200 300 400,D,A.,阿司匹林,B.,非那西汀,C.,咖啡因,D.,复方阿司匹林,图,1,紫外吸收光谱,由,A,、,B,、,C,的紫外吸收光谱可知，阿司匹林、非那西汀、咖啡因的标准吸收光度分别为,0.62,、,0.83,、,0.68.,从合成样品中选取,5,组进行紫外分析，确定在,200300nm,波长范围内测定其吸收光度值。,序号,阿司匹林（,A,）,相对误差,%,非那西汀（,A,）,相对误差,%,咖啡因（,A,）,相对误差,%,1,0.58,0.06,0.78,0.06,0.63,0.07,2,0.55,0.11,0.84,0.12,0.62,0.09,3,0.56,0.09,0.79,0.05,0.61,0.11,4,0.55,0.11,0.76,0.08,0.59,0.13,5,0.65,0.05,0.86,0.04,0.73,0.07,平均相对误差,%,0.08,0.07,0.09,由对标准样品,A,、,B,、,C,得出的吸收光谱图可以看出，复方阿司匹林药片中三组分的吸收光度明显不同，其吸收光度值作为标准参考值。得出复方阿司匹林中各组分的平均相对偏差分别为,0.08%,、,0.07%,、,0,09%,。,结论,紫外分光光度法实验操作较为复杂,在本实验中，各组分的平均相对误差分别为,0.08%,、,0.07%,、,0,09%,。,薄层层析法设备简单，操作容易、快速，灵敏度较高，现象明显。在其最佳条件（,110,、,活化时间,1,小时,）下，各组分的平均相对误差分别为,0.06%,、,0.03%,、,0.02%,。,总体来比较，运用薄层层析法对复方阿司匹林进行分离与鉴定更为理想。,致谢,经过将近两个月的忙碌，本次毕业设计已经接近尾声，作为一个本科生的毕业设计，由于经验的匮乏，难免有许多考虑不周全的地方，如果没有导师的督促指导，想要完成这个设计是难以想象的。本课题在选题及研究过程中都得到了宋伟新宋老师的亲切关怀和悉心指导。宋老师学识渊博考虑，问题全面周到，帮助我解决了很多困难。同时还要感谢郭元平老师等对我论文的细心指导。他们细心指导我的学习与研究，在此，我要向诸位老师深深地鞠上一躬。,最后，向我所有的老师致谢，感谢他们对我的精心培养。,</p>