HRID85197

反应详情

EQUATION

反应方程式

HRID 85197 的结构方程式

PROCEDURE

实验过程

Analogously to Example 4A, 1.50 g (4.13 mmol) of the compound from Example 3A are reacted with 704 mg (4.54 mmol) of 2-fluoro-5-nitrotoluene. This gives 570 mg (28% of theory) of the title compound.