HRID1443212

反应详情

EQUATION

反应方程式

HRID 1443212 的结构方程式

REACTANTS

反应物

2

结构：BrB(Br)Br

Boron tribromide

BBr3

结构：COc1c(F)ccc2c1C(C)(C)CC(O)(C(F)(F)F)C2Nc1cccc2c(=O)n(C)ncc12

未命名化合物

PRODUCTS

生成物

1

结构：Cn1ncc2c(NC3c4ccc(F)c(O)c4C(C)(C)CC3(O)C(F)(F)F)cccc2c1=O

未命名化合物

PROCEDURE

实验过程

29 mg of the compound of Example 69 is reacted analogously to Example 1 with 0.13 ml of BBr3 (1N in dichloromethane) to form 18 mg of the title compound.