2-Aminothiazole → 1-(4-Methoxyphenoxy)-2-nitrobenzene

HRID 559835Notes

Reaction equation

HRID 559835 reaction equation
[CH3:1][O:2][C:3]1[CH:8]=[CH:7][C:6]([OH:9])=[CH:5][CH:4]=1.F[C:11]1[CH:16]=[CH:15][CH:14]=[CH:13][C:12]=1[N+:17]([O-:19])=[O:18].[CH3:20][O:21][C:22]1[CH:35]=[CH:34][C:25]([O:26][C:27]2[CH:33]=[CH:32][CH:31]=[CH:30][C:28]=2[NH2:29])=[CH:24][CH:23]=1.[NH2:36][C:37]1[S:38][CH:39]=[CH:40][N:41]=1>>[CH3:1][O:2][C:3]1[CH:8]=[CH:7][C:6]([O:9][C:11]2[CH:16]=[CH:15][CH:14]=[CH:13][C:12]=2[N+:17]([O-:19])=[O:18])=[CH:5][CH:4]=1.[CH3:20][O:21][C:22]1[CH:35]=[CH:34][C:25]([O:26][C:27]2[CH:33]=[CH:32][CH:31]=[CH:30][C:28]=2[NH:29][C:6]([NH:36][C:37]2[S:38][CH:39]=[CH:40][N:41]=2)=[O:9])=[CH:24][CH:23]=1

Participants

6

Reactant 4

HCID 21552-AminothiazoleHCID 90154-MethoxyphenolHCID 738951-Fluoro-2-nitrobenzeneHCID 3612862-(4-methoxyphenoxy)aniline2×

Product 2

HCID 6681501-(4-Methoxyphenoxy)-2-nitrobenzeneYield 80%HCID 59733128HCID 59733128, compoundYield 75%

Procedure

2-(4-Methoxyphenoxy)-1-nitrobenzene (0.98 g, 80%) was prepared from 4-methoxyphenol (0.62 g, 5.0 mmol) and 1-fluoro-2-nitrobenzene (0.71 g, 5.0 mmol) following the general procedure A. This was reduced to 2-(4-methoxyphenoxy)aniline (0.32 g, 60%, 2.5 mmol scale) following the general procedure B. N-[2-(4-Methoxyphenoxy)phenyl)-N′-(thiazol-2-yl)urea (256 mg, 75%) was prepared from 2-(4-methoxyphenoxy)aniline (215 mg, 1.0 mmol) and 2-aminothiazole (100 mg, 1.0 mmol) following the general procedure D.

Sources & evidence

ORD record
ord-c06a9194021d46bb84fc3dc1a732ab25
Source dataset
uspto-grants-2011_11
Patent number
US08063081B2

Reaction notes

Loading…