HRID85197
反应详情
EQUATION
反应方程式
REACTANTS
反应物
PRODUCTS
生成物
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.