HRID1055727

反应详情

EQUATION

反应方程式

HRID 1055727 的结构方程式

PROCEDURE

实验过程

50 mg (0.17 mmol) of the compound of Example 6A and 52.9 mg (0.17 mmol) of the compound of Example 19A were reacted and worked up analogously to the procedure of Example 19. This gave 49 mg (49% of theory) of the title compound.