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HRID 461462Erstellt: 25. Oktober 2026

Sodium hydride → 1-[2,4-Dichloro-beta-(4-thiazolylmethoxy) phenethyl]imidazole hydrochloride

https://huagongshe.com/reaction/461462

Reaktionsgleichung

Beteiligte Stoffe

RolleNameHCIDEinwaage
EduktSodium hydrideHCID 24758—
Edukt(+-)-alpha-(2,4-Dichlorophenyl)-1H-imidazole-1-ethanolHCID 32238—
Edukt4-(Chloromethyl)-1,3-thiazoleHCID 2763290—
LösungsmittelWaterHCID 962—
LösungsmittelTetrahydrofuranHCID 8028—
Produkt1-[2,4-Dichloro-beta-(4-thiazolylmethoxy) phenethyl]imidazole hydrochlorideHCID 8673528192,6%

Reaktionsbedingungen

Temperatur70 °C

Reaktionsschritte

A solution of 1-(2,4-dichloro-phenyl)-2-(1-imidazolyl) ethanol (2.4 g, 9.4 mmole) in dry tetrahydrofuran (20 ml) was added to sodium hydride (0.62 g, as an 80% dispersion in oil, 0.02 mole) and warmed at 70° C for 90 minutes. The solution was cooled to 0° C and 4-chloromethylthiazole (1.5 g, 11.2 mmole), dissolved in a little dry tetrahydrofuran, was added. The mixture was stirred at 0° C for 1 hour and at room temperature for 6 hours. Further 4-chloromethylthiazole (0.25 g) was added and the stirring continued for 36 hours. The reaction mixture was then diluted with water (70 ml) and extracted several times with ether. The ether extracts were combined, dried over MgSO4 and evaporated. The oily residue was taken up in dry ether (50 ml) and treated with a solution of HCl in ether. The solid precipitate was collected and recrystallized from a mixture of methanol and di-isopropyl ether to give 1-[2,4-dichloro-β-(4-thiazolylmethoxy) phenethyl]imidazole hydrochloride (1.7 g, 46%) m.p. 183°-184°. (Found: C, 46.3; H, 3.7; N, 10.7. C15H13N3Cl2OS.HCl requires C, 46.1; H, 3.6; N, 10.7%).

Herkunft & Nachweise

DOI10.6084/m9.figshare.5104873.v1
PatentnummerUS04062966
Quelldatensatzuspto-grants-1977_12
ORD IDord-def45323cb3d433197a0a26f4a50ada9
Quellennotizuspto-grants-1977_12