HRID1765392

反应详情

EQUATION

反应方程式

REACTANTS

反应物

HCID10796

Cesium carbonate

CCs2O3

结构：CC1(C)c2cccc(P(c3ccccc3)c3ccccc3)c2Oc2c(P(c3ccccc3)c3ccccc3)cccc21

HCID636044

1,1'-(9,9-Dimethyl-9H-xanthene-4,5-diyl)bis(1,1-diphenylphosphine)

C39H32OP2

HCID737200

1-tert-Butyl-3-methyl-1H-pyrazol-5-amine

C8H15N3

结构：CC(C)(C)c1ccc(-n2nc(Br)c3cc(Br)ccc3c2=O)cc1

HCID21106542

未命名化合物

PRODUCTS

生成物

结构：Cc1cc(Nc2nn(-c3ccc(C(C)(C)C)cc3)c(=O)c3ccc(Br)cc23)n(C(C)(C)C)n1

HCID21106578

未命名化合物

收率 194.9%

AUXILIARIES

试剂、催化剂与溶剂

结构：O=C(/C=C/c1ccccc1)/C=C/c1ccccc1.O=C(/C=C/c1ccccc1)/C=C/c1ccccc1.O=C(/C=C/c1ccccc1)/C=C/c1ccccc1.[Pd].[Pd]

催化剂HCID9811564

Bis(dibenzylideneacetone)palladium

C51H42O3Pd2

CONDITIONS

反应条件

温度: 100 °C

PROCEDURE

实验过程

11 mg 4,6-Dibromo-2-(4-tert-butyl-phenyl)-2H-phthalazin-1-one (preparation see ZA-1) were reacted with 39.0 mg 1-tert.-butyl-3-methyl-1H-pyrazol-5-ylamine, 132 mg cesium carbonate, 7.0 mg tris-(dibenzylideneacetone)-dipalladium and 8.8 mg 4,5-bis(diphenylphosphino)-9,9-dimethylxanthene were stirred under argon at 100° C. for 20 hrs until HPLC indicated complete conversion. The solvent was removed under vacuum, the residue was taken up in dichloromethane and washed dilute HCl. Removal of the dichloromethane and chromatography on silica in first heptane and then heptan/ethyl acetate 8:2 yielded 25 mg of the title product.

WORKUP

后处理

customThe solvent was removed under vacuum
washwashed dilute HCl
customRemoval of the dichloromethane and chromatography on silica in first heptane

来源与证据

来源类型: Open Reaction Database
ORD 记录: ord-8f2513b0cdab450e928774948b5a21ac
来源数据集: uspto-grants-2007_06
DOI: 10.6084/m9.figshare.5104873.v1
专利: US07226923B2
来源说明: uspto-grants-2007_06

反应 SMILES

Br[C:2]1[C:11]2[C:6](=[CH:7][CH:8]=[C:9]([Br:12])[CH:10]=2)[C:5](=[O:13])[N:4]([C:14]2[CH:19]=[CH:18][C:17]([C:20]([CH3:23])([CH3:22])[CH3:21])=[CH:16][CH:15]=2)[N:3]=1.[C:24]([N:28]1[C:32]([NH2:33])=[CH:31][C:30]([CH3:34])=[N:29]1)([CH3:27])([CH3:26])[CH3:25].C(=O)([O-])[O-].[Cs+].[Cs+].C1(P(C2C=CC=CC=2)C2C3OC4C(=CC=CC=4P(C4C=CC=CC=4)C4C=CC=CC=4)C(C)(C)C=3C=CC=2)C=CC=CC=1>C1C=CC(/C=C/C(/C=C/C2C=CC=CC=2)=O)=CC=1.C1C=CC(/C=C/C(/C=C/C2C=CC=CC=2)=O)=CC=1.C1C=CC(/C=C/C(/C=C/C2C=CC=CC=2)=O)=CC=1.[Pd].[Pd]>[Br:12][C:9]1[CH:10]=[C:11]2[C:6](=[CH:7][CH:8]=1)[C:5](=[O:13])[N:4]([C:14]1[CH:19]=[CH:18][C:17]([C:20]([CH3:22])([CH3:23])[CH3:21])=[CH:16][CH:15]=1)[N:3]=[C:2]2[NH:33][C:32]1[N:28]([C:24]([CH3:27])([CH3:26])[CH3:25])[N:29]=[C:30]([CH3:34])[CH:31]=1