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HRID1462424

反应详情

EQUATION

反应方程式

HRID 1462424 的结构方程式

REACTANTS

反应物

3
结构：CCOC(=O)CBr
HCID7748

Ethyl bromoacetate

C4H7BrO2

结构：[H-].[Na+]
HCID24758

Sodium hydride

HNa

结构：O=C(CNC(=O)c1ccc(Cl)cc1)c1ccsc1
HCID13105428

未命名化合物

PRODUCTS

生成物

1
结构：CCOC(=O)CC(NC(=O)c1ccc(Cl)cc1)C(=O)c1ccsc1
HCID13105432

未命名化合物

收率 71.9%

PROCEDURE

实验过程

15.0 g of N-(4-chlorobenzoyl)-N-[(3-thienylcarbonyl)methyl]amine, 3.1 g of 50% sodium hydride and 10 g of ethyl bromoacetate are treated in the same manner as described in Preparation 1-(4). 14.1 g of ethyl 3-(4-chlorobenzoylamino)-3-(3-thienylcarbonyl)propionate are thereby obtained. Yield: 71.4%

WORKUP

后处理

  1. customas described in Preparation 1-(4)