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

HRID 2276245Notes

Reaction equation

HRID 2276245 reaction equation
[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].NC1C=CC=CC=1.[CH3:24][C:25]1[CH:31]=[CH:30][C:28]([NH2:29])=[C:27]([O:32][CH2:33][C:34]([CH3:36])=[CH2:35])[CH:26]=1.[NH2:37][C:38]1[S:39][CH:40]=[CH:41][N:42]=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].[CH3:24][C:25]1[CH:31]=[CH:30][C:28]([NH:29][C:1]([NH:37][C:38]2[S:39][CH:40]=[CH:41][N:42]=2)=[O:5])=[C:27]([O:32][CH2:33][C:34]([CH3:36])=[CH2:35])[CH:26]=1

Participants

7

Reactant 5

HCID 21552-AminothiazoleHCID 6115AnilineHCID 10557Methallyl alcoholHCID 67966Benzene, 2-fluoro-4-methyl-1-nitro-HCID 450912454-Methyl-2-[(2-methylprop-2-en-1-yl)oxy]aniline

Product 2

HCID 59733018HCID 59733018, compoundYield 70%HCID 603578443-(2-Methyl-allyloxy)-4-nitrotolueneYield 75%

Procedure

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-allyloxy)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.

Sources & evidence

ORD record
ord-09351cba112541aea856cc373f053718
Source dataset
uspto-grants-2015_09
Patent number
USRE045670E1

Reaction notes

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