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2-Aminothiazole → 3-(2-Methyl-allyloxy)-4-nitrotoluene

HRID 131949
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HRID 131949 반응식
[CH2:1]([OH:5])[C:2](=[CH2:4])[CH3:3].F[C:7]1[CH:8]=[C:9]([CH3:16])[CH:10]=[CH:11][C:12]=1[N+:13]([O-:15])=[O:14].[CH3:17][C:18]1[CH:24]=[CH:23][C:21]([NH2:22])=[C:20]([O:25][CH2:26][C:27]([CH3:29])=[CH2:28])[CH:19]=1.[NH2:30][C:31]1[S:32][CH:33]=[CH:34][N:35]=1>>[CH3:4][C:2](=[CH2:3])[CH2:1][O:5][C:7]1[CH:8]=[C:9]([CH3:16])[CH:10]=[CH:11][C:12]=1[N+:13]([O-:15])=[O:14].[CH3:17][C:18]1[CH:24]=[CH:23][C:21]([NH:22][C:1]([NH:30][C:31]2[S:32][CH:33]=[CH:34][N:35]=2)=[O:5])=[C:20]([O:25][CH2:26][C:27]([CH3:29])=[CH2:28])[CH:19]=1
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참가 물질

6

반응물 4

HCID 21552-AminothiazoleHCID 10557Methallyl alcoholHCID 67966Benzene, 2-fluoro-4-methyl-1-nitro-HCID 450912454-Methyl-2-[(2-methylprop-2-en-1-yl)oxy]aniline

생성물 2

HCID 597330181-[4-Methyl-2-(2-methyl-allyloxy)-phenyl]-3-thiazol-2-yl-urea수율 70%HCID 60357844
3-(2-Methyl-allyloxy)-4-nitrotoluene
수율 75%

실험 절차

3-(2-Methyl-allyloxy)-4-nitrotoluene (0.78 g, 75%) was prepared from methallyl alcohol and 3-fluoro-4-nitrotoluene (0.77 g, 5.0 mmol) following the general procedure G. This was reduced to 4-methyl-2-(2-methyl-allyfoxy)aniline (0.47 g, 70%) following general procedure B. 1-[4-Methyl-2-(2-methyl-allyloxy)-phenyl]-3-thiazol-2-yl-urea (210 mg, 70%) was prepared from 4-methyl-2-(2-methyl-allyloxy)aniline (179 mg, 1.0 mmol) and 2-aminothiazole (100 mg, 1.0 mmol) following the general procedure D.

출처 및 근거

ORD 기록
ord-25995062b3c74c8cb170a8f08ba7950e
출처 데이터셋
uspto-grants-2008_06
DOI
10.6084/m9.figshare.5104873.v1
특허 번호
US07384967B2

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