HRID561635

反应详情

EQUATION

反应方程式

HRID 561635 的结构方程式

PROCEDURE

实验过程

The title compound was prepared starting from 462 mg (2.61 mmol) of the compound from Example 8A and 600 mg (2.61 mmol) of the compound from Example 139A in analogy to the synthesis of the compound from Example 184. 273 mg (27% of theory) of the target compound were obtained.

WORKUP

后处理

  1. custom273 mg (27% of theory) of the target compound were obtained