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HRID500392

反应详情

EQUATION

反应方程式

HRID 500392 的结构方程式

REACTANTS

反应物

2
结构：O=C(Cl)c1ccc(Cl)cc1
HCID8501

4-Chlorobenzoyl chloride

C7H4Cl2O

结构：CC(C)(C)OC(=O)NC1CCCCCC=CC2CC2(C(=O)NS(=O)(=O)C2CC2)NC(=O)C2CC(O)CN2C1=O
HCID68797769

未命名化合物

PRODUCTS

生成物

1
结构：CC(C)(C)OC(=O)NC1CCCCCC=CC2CC2(C(=O)NS(=O)(=O)C2CC2)NC(=O)C2CC(OC(=O)c3ccc(Cl)cc3)CN2C1=O
HCID72445364

未命名化合物

收率 55%

PROCEDURE

实验过程

Prepared by way of method I using 18-hydroxy-14-tert-butoxycarbonylamino-4-cyclopropylsulfonylaminocarbonyl-2,15-dioxo-3,16-diazatricyclo[14.3.0.04,6]-nonadec-7-ene (100 mg, 0.175 mmol) and 4-chlorobenzoyl chloride (45 μL, 0.35 mmol). The final trituration (diethyl ether/hexane) and filtration gave 68 mg (55%) of 18-(4-chlorobenzoyloxy)-14-tert-butoxycarbonylamino-4-cyclopropylsulfonylaminocarbonyl-2,15-dioxo-3,16-diazatricyclo-[14.3.0.04,6]-nonadec-7-ene as a white powder: 98.5% pure (HPLC), MS m/z 705 (M−H)−, HRMS: calcd. 707.2518 and found 707.2499 (M+H)+, mp 143-145° C.

WORKUP

后处理

  1. customPrepared by way of method I
  2. filtrationThe final trituration (diethyl ether/hexane) and filtration