Sodium hydride → 2-Chloro-4-(2,6-dimethylphenylthio)pyridine

반응식

[CH3:1][C:2]1[CH:7]=[CH:6][CH:5]=[C:4]([CH3:8])[C:3]=1[SH:9].[H-].[Na+].[Cl:12][C:13]1[CH:18]=[C:17]([N+]([O-])=O)[CH:16]=[CH:15][N:14]=1>>[Cl:12][C:13]1[CH:18]=[C:17]([S:9][C:3]2[C:4]([CH3:8])=[CH:5][CH:6]=[CH:7][C:2]=2[CH3:1])[CH:16]=[CH:15][N:14]=1

참가 물질

반응물 3

HCID 24758Sodium hydride

HCID 610452,6-Dimethylthiophenol

HCID 7351522-Chloro-4-nitropyridine

생성물 1

HCID 866347342-Chloro-4-(2,6-dimethylphenylthio)pyridine수율 86%

실험 절차

Using the method of Example 3, Step A, 2,6-dimethylbenzenethiol (0.872 g, 6.31 mmol), 60% sodium hydride in mineral oil (252 mg, 6.31 mmol), and 2-chloro-4-nitropyridine (1.00 g, 6.31 mmol) were reacted to provide 2-chloro-4-(2,6-dimethylphenylthio)pyridine (1.35 g, 86% yield) as a white solid. 1H NMR (CDCl3) δ 8.07 (d, 1H), 7.22-7.34 (m, 3H), 6.75 (s, 1H), 6.71 (d, 1H), 2.40 (s, 6H).