从L-苹果酸出发,经过与苄胺缩合反应得到(S)-1-苄基-3-羟基吡咯烷-2,5-二酮,再通过硼氢化钠-碘体系还原制得重要的医药中间体(S)-1-苄基-3-羟基吡咯烷。采用红外光谱法对原料、中间体及产物进行了测试并对硼氢化钠-碘体系还原酰亚胺的机理进行了研究。通过比较原料、中间体及产物相应特征吸收峰的消失或是出现,可知所得的产物为(S)-1-苄基-3-羟基吡咯烷;还原机理研究表明,硼氢化钠在碘催化下生成的硼烷与酰亚胺中的羰基形成四元环状过渡态,同时另外一分子硼烷与酰亚胺中的氮原子形成N-BH3复合物,然后羰基还原完全,得到(S)-1-苄基-3-羟基吡咯烷-硼烷复合物,最后在甲醇作用下脱去硼烷得到最终产物。
The title compound was synthesized by reducing(S)-1-benzyl-3-hydroxypyrrolidine-2,5-dione with sodium borohydride-iodin.The raw material,intermediate,and the end product were characterized by IR spectra and the mechanism of reduction of imide by sodium borohydride-iodin was also studied by IR spectra.According to the IR spectra,it was concluded that the product was(S)-1-bezyl-3-hydroxypyrrolidine.The mechanism of the reduction is that borane,formed in situ by reacting sodium borohydride with iodine,partly was conjugated to carbonyl to form four-member-ring intermediate and partly conjugated to nitrogen.(S)-1-benzyl-3-hydroxypyrrolidine/BH3 complex was gained when the reduction finished.The title compound was obtained by removing borane from the complex in methanol.