HRID2105027

反应详情

EQUATION

反应方程式

REACTANTS

反应物

结构：CC(C)(C)OC(=O)N(C(=O)OC(C)(C)C)c1nccc2cc(N)ccc12

HCID58717298

未命名化合物

PRODUCTS

生成物

结构：CCOc1cc(CC)cc(C(Nc2ccc3c(N(C(=O)OC(C)(C)C)C(=O)OC(C)(C)C)nccc3c2)c2nc(-c3ccccc3Br)cn2C(c2ccccc2)(c2ccccc2)c2ccccc2)c1F

HCID86607943

1,3-Bis(1,1-dimethylethyl) 2-[6-[[[4-(2-bromophenyl)-1-(triphenylmethyl)-1H-imidazol-2-yl](3-ethoxy-5-ethyl-2-fluorophenyl)methyl]amino]-1-isoquinolinyl]imidodicarbonate

C58H57BrFN5O5

PROCEDURE

实验过程

According to the procedure for the preparation of Intermediate 206.4, Intermediate 217.3 (424 mg, 0.641 mmol) was converted to the chloro intermediate, then displaced with di-tert-butyl (6-aminoisoquinolin-1-yl)imidodicarbonate to afford after flash chromatography (0 to 60% EtOAc/hexanes gradient) 615 mg of intermediate 217.4 as a pale yellow solid.

来源与证据

来源类型: Open Reaction Database
ORD 记录: ord-aab145f98ea04cbaa9918bb09ab3bd66
来源数据集: uspto-grants-2009_08
DOI: 10.6084/m9.figshare.5104873.v1
专利: US07576098B2
来源说明: uspto-grants-2009_08

反应 SMILES

BrC1N=C(C([NH:38][C:39]2[CH:40]=[C:41]3[C:46](=[CH:47][CH:48]=2)[C:45]([N:49]([C:57]([O:59][C:60]([CH3:63])([CH3:62])[CH3:61])=[O:58])[C:50]([O:52][C:53]([CH3:56])([CH3:55])[CH3:54])=[O:51])=[N:44][CH:43]=[CH:42]3)C2C=C(CC)C=C(OC)C=2F)N(C(C2C=CC=CC=2)(C2C=CC=CC=2)C2C=CC=CC=2)C=1.[Br:64][C:65]1[CH:70]=[CH:69][CH:68]=[CH:67][C:66]=1[C:71]1[N:72]=[C:73]([CH:95]([C:97]2[CH:102]=[C:101]([CH2:103][CH3:104])[CH:100]=[C:99]([O:105][CH2:106][CH3:107])[C:98]=2[F:108])O)[N:74]([C:76]([C:89]2[CH:94]=[CH:93][CH:92]=[CH:91][CH:90]=2)([C:83]2[CH:88]=[CH:87][CH:86]=[CH:85][CH:84]=2)[C:77]2[CH:82]=[CH:81][CH:80]=[CH:79][CH:78]=2)[CH:75]=1.BrC1N=C(C(Cl)C2C=C(CC)C=C(OC)C=2F)N(C(C2C=CC=CC=2)(C2C=CC=CC=2)C2C=CC=CC=2)C=1.NC1C=C2C(=CC=1)C(N(C(OC(C)(C)C)=O)C(OC(C)(C)C)=O)=NC=C2>>[Br:64][C:65]1[CH:70]=[CH:69][CH:68]=[CH:67][C:66]=1[C:71]1[N:72]=[C:73]([CH:95]([NH:38][C:39]2[CH:40]=[C:41]3[C:46](=[CH:47][CH:48]=2)[C:45]([N:49]([C:50]([O:52][C:53]([CH3:56])([CH3:55])[CH3:54])=[O:51])[C:57]([O:59][C:60]([CH3:61])([CH3:62])[CH3:63])=[O:58])=[N:44][CH:43]=[CH:42]3)[C:97]2[CH:102]=[C:101]([CH2:103][CH3:104])[CH:100]=[C:99]([O:105][CH2:106][CH3:107])[C:98]=2[F:108])[N:74]([C:76]([C:83]2[CH:88]=[CH:87][CH:86]=[CH:85][CH:84]=2)([C:77]2[CH:82]=[CH:81][CH:80]=[CH:79][CH:78]=2)[C:89]2[CH:90]=[CH:91][CH:92]=[CH:93][CH:94]=2)[CH:75]=1