HRID1055734

反应详情

EQUATION

反应方程式

HRID 1055734 的结构方程式

PROCEDURE

实验过程

35 mg (0.12 mmol) of the compound of Example 6A and 47 mg (0.12 mmol) of the compound of Example 21A were reacted and worked up analogously to the procedure of Example 16. This gave 25 mg (95% pure, 38% of theory) of the title compound.