摘要
以邻苯二胺和丙二酸为主要原料,3 mol·L-1盐酸水溶液为反应介质,合成了双(1H-苯并咪唑-2-基)甲烷。采用元素分析、FT-IR、UV-Vis和1H NMR对产物进行了结构表征和确认,并对产物的溶液荧光性能和固体荧光性能进行了测试研究。结果表明,产物在DMF、DMSO和Et OH三种溶剂中均具有315 nm和360 nm附近较强的紫外吸收峰,具有良好的抗日光UVA和UVB紫外线性能。溶液荧光测试表明,产物具有λmax,em=476~483 nm的蓝绿色荧光发射,且溶液荧光具有AIE效应。固体荧光测试发现产物的最大激发波长λmax,ex=252 nm,具有λmax,em=537 nm的黄绿色荧光发射。
The bis(1H-benzimidazol-2-yl) methane was synthesized in 3 mol·L-1HCl solution by using o-phenylenediamine and malonic acid as main raw materials. The target compound was characterized by means of elemental analysis,FT-IR,UV-vis and1 H NMR spectra. UV-Vis,solution fluorescence and solid fluorescence spectroscopy were used to study its photophysics properties. The results show that the compound has two UV absorption peaks at 315 nm and 360 nm in DMSO,DMF and Et OH(1×10-5mol·L-1),and possesses good performance on absorbing sunlight ultraviolet(UVA and UVB). The blue-green fluorescence emission(λmax,em=476-483nm) can be obtained under the excitation wavelength 320 nm in the above three solvents. Besides,a significant AIE effect of the titled compound is found through solution fluorescence tests with different VolDMSO∶VolDCM. Solid fluorescence indicates that the product has strong yellow-green fluorescence emission(λmax,em=537 nm) with the excitation wavelength at 252 nm.
引文
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