【作者】 张皋； 王民昌； 齐珠钗； 刘红妮；

【Author】 Zhang Gao~1 Wang Minehang~1 Qi Zhuchai~1 Liu Hongni 1 Xi’an Modern Chemistry Research Institute,China, Xi’an,710065

【机构】 西安近代化学研究所；

【摘要】 <正>兰索拉唑化学名称为2-[[[3-甲基-4-(2,2,2-三氟乙氧基)-2-吡啶基]甲基]亚磺酰基]-1H- 苯并咪唑。它是胃粘膜壁细胞 H~+/K~+-ATP 酶质子泵的抑制剂,被成功的应用于治疗胃溃疡、十二指肠溃疡、反流性食道炎。有关其 NMR 研究的文章见文献,室温下,氢谱与碳谱中均更多还原

【Abstract】 Lansoprazole is the generic name of 2-[3-methyl-4-（2,2,2-trifluoroethoxy）-2-pyridyl methyl]sulfinyl-1H-benzimidazole,it is a proton pump inhibitor that reduces gastric aced secretion and has successfully been used to heal and relieve symptoms of gastric or duodenal ulcers and gastro-esophagal reflux.The ¹H NMR of Lansoprazole and its assignments can be seen in reference At room temperature,there are several broad low lines in spectrums shown in Fig.1,there are six broad low lines（δ_H7.33 and δ_H7.70）while two in Fig.2,Why is it?The authors suggest Lansoprazole exists as two tautomerides shown in Fig.1 and transforms into each other by rearrangement of carbon-nitrogen double bond in DMCO at 25℃.When double bond（C₈=N₇）at B ring is located at crs position of thionyl（S=O）.Lansoprazole exists as configuration I.On the other hand configuration Ⅱ provides transformation of double bonds（C₈=N₉）and thionyl（S=O）.The signals of H-2 and H-5 in I form are combined into broad low lines at δ_H7.70 approximately （expressed by∽δ_H7.70 in the following）.So do the signals of H-2’ and H-5’ in Ⅱ form.The two forms changing into each other instantly give an average δ_H value.Thus,four signals of H-2（H-5’） and H-5（H-2’）overlap at δ_H7.70 approximately that observed as broad low lines.For the same explain,the signals of H-3（H-4’）and H-4（H-3’）are observed as broad low lines at δ_H7.33.In I form C₂ resonates at∽δ113.14 and the C₅ at∽δ120.94.The ¹³C NMR signal of C_5’ overlap with C₂ at∽δ113.14,C_2’ and C₅ at∽δ120.94.Both of the signals are broad low lines.So it’s very obvious that C₃ and C_4’ at∽δ123.59,Cyand C₄ at∽δ124.77 also are broad low lines.Of the quaternary carbon signals assignment of the ones at δ135.72 and δ145.20 can be attributed to C1 and C6 because of its broad and weak intensity due to the same reason which above-mentioned and the sharp single peak at δ155.60 to C8 because of equivalence of C8 in two tautomerides.The tautomerism between Ⅰ and Ⅱ form has no effect on the carbon chain from 10^#,so in the ¹H NMR and ¹³C NMR spectrum of 10^#∽20^# no broad low lines are observed any more.更多还原