HRID531517

反应详情

EQUATION

反应方程式

REACTANTS

反应物

HCID10340

Carbonic acid sodium salt (1:2)

CNa2O3

结构：Cc1nn(Cc2cc(F)cc(F)c2)c(C)c1B1OC(C)(C)C(C)(C)O1

HCID90031220

未命名化合物

2 次

结构：COc1ccc(-c2cc3c(I)cn(C(=O)OC(C)(C)C)c3nc2C2CC2)cc1N(C(=O)OC(C)(C)C)S(C)(=O)=O

HCID131735664

未命名化合物

2 次

PRODUCTS

生成物

结构：COc1ccc(-c2cc3c(-c4c(C)nn(Cc5cc(F)cc(F)c5)c4C)cn(C(=O)OC(C)(C)C)c3nc2C2CC2)cc1N(C(=O)OC(C)(C)C)S(C)(=O)=O

HCID131735663

未命名化合物

收率 42.9%

AUXILIARIES

试剂、催化剂与溶剂

结构：Cl[Pd](Cl)([P](c1ccccc1)(c1ccccc1)c1ccccc1)[P](c1ccccc1)(c1ccccc1)c1ccccc1

催化剂HCID178418476

未命名化合物

溶剂HCID18180031

未命名化合物

PROCEDURE

实验过程

Using similar reaction conditions as described in step-ii of example-1, tert-butyl 5-(3-(N-(tert-butoxycarbonyl)methylsulfonamido)-4-methoxyphenyl)-6-cyclopropyl-3-iodo-1H-pyrrolo[2,3-b]pyridine-1-carboxylate (intermediate 33) (150 mg, 0.21 mmol) was coupled with 1-(3,5-difluorobenzyl)-3,5-dimethyl-4-(4,4,5,5-tetramethyl-1,3,2-dioxaborolan-2-yl)-1H-pyrazole (intermediate 24) (91.69 mg, 0.24 mmol) in sodium carbonate (66.77 mg, 0.483 mmol), Pd(PPh3)2Cl2 (7.36 mg, 0.01 mmol), 1,2-dimethoxyethane/water (10/2 ml) to give 70 mg (41% yield) of titled compound. MS: m/z=778.0 (M+1).

WORKUP

后处理

customreaction conditions

来源与证据

来源类型: Open Reaction Database
ORD 记录: ord-72246be209ef406cb53ebda5f9fe19eb
来源数据集: uspto-grants-2016_05
DOI: 10.6084/m9.figshare.5104873.v1
专利: US09353107B2
来源说明: uspto-grants-2016_05

反应 SMILES

[C:1]([O:5][C:6]([N:8]([C:13]1[CH:14]=[C:15]([C:21]2[CH:22]=[C:23]3[C:32](I)=[CH:31][N:30]([C:34]([O:36][C:37]([CH3:40])([CH3:39])[CH3:38])=[O:35])[C:24]3=[N:25][C:26]=2[CH:27]2[CH2:29][CH2:28]2)[CH:16]=[CH:17][C:18]=1[O:19][CH3:20])[S:9]([CH3:12])(=[O:11])=[O:10])=[O:7])([CH3:4])([CH3:3])[CH3:2].[F:41][C:42]1[CH:43]=[C:44]([CH:62]=[C:63]([F:65])[CH:64]=1)[CH2:45][N:46]1[C:50]([CH3:51])=[C:49](B2OC(C)(C)C(C)(C)O2)[C:48]([CH3:61])=[N:47]1.C(=O)([O-])[O-].[Na+].[Na+]>Cl[Pd](Cl)([P](C1C=CC=CC=1)(C1C=CC=CC=1)C1C=CC=CC=1)[P](C1C=CC=CC=1)(C1C=CC=CC=1)C1C=CC=CC=1.COCCOC.O>[C:1]([O:5][C:6]([N:8]([C:13]1[CH:14]=[C:15]([C:21]2[CH:22]=[C:23]3[C:32]([C:49]4[C:48]([CH3:61])=[N:47][N:46]([CH2:45][C:44]5[CH:62]=[C:63]([F:65])[CH:64]=[C:42]([F:41])[CH:43]=5)[C:50]=4[CH3:51])=[CH:31][N:30]([C:34]([O:36][C:37]([CH3:40])([CH3:39])[CH3:38])=[O:35])[C:24]3=[N:25][C:26]=2[CH:27]2[CH2:29][CH2:28]2)[CH:16]=[CH:17][C:18]=1[O:19][CH3:20])[S:9]([CH3:12])(=[O:11])=[O:10])=[O:7])([CH3:4])([CH3:3])[CH3:2]