HRID1055715

反应详情

EQUATION

反应方程式

HRID 1055715 的结构方程式

PROCEDURE

实验过程

50 mg (0.17 mmol) of the compound of Example 6A and 79.6 mg (0.17 mmol) of the compound of Example 17A were reacted and worked up analogously to the procedure of Example 16. This gave 82 mg (77% of theory) of the title compound.