HRID1055718

反应详情

EQUATION

反应方程式

HRID 1055718 的结构方程式

PROCEDURE

实验过程

30 mg (0.10 mmol) of the compound of Example 6A and 29 mg (0.10 mmol) of the compound of Example 25A were reacted and worked up analogously to the procedure of Example 16. This gave 32 mg (55% of theory) of the title compound.