2,5-Dichloronitrobenzene → 4-Chloro-1-fluoro-2-nitrobenzene

반응식

[Cl:1][C:2]1[CH:7]=[CH:6][C:5]([Cl:8])=[CH:4][C:3]=1[N+:9]([O-:11])=[O:10].[Cl-].C([N+](C)(C)C)CCCCCCCCCCCCCCCCC.[F-:35].[K+]>>[F:35][C:2]1[CH:7]=[CH:6][C:5]([Cl:8])=[CH:4][C:3]=1[N+:9]([O-:11])=[O:10].[Cl:1][C:2]1[CH:7]=[CH:6][C:5]([Cl:8])=[CH:4][C:3]=1[N+:9]([O-:11])=[O:10]

참가 물질

반응물 3

HCID 69772,5-Dichloronitrobenzene

HCID 8155Stearyltrimethylammonium chloride

HCID 522689Potassium Fluoride

생성물 2

HCID 69772,5-Dichloronitrobenzene수율 29.7%

HCID 676604-Chloro-1-fluoro-2-nitrobenzene수율 72.3%

실험 절차

2020 g (10.5 mol) of 2,5-dichloronitrobenzene, 70 g (0.2 mol) of octadecyltrimethylammonium chloride and 580 g (10.0 mol) of potassium fluoride were reacted analogously to Example 1. 936 g (72.3%, relative to 2,5-dichloronitrobenzene converted) of 2-fluoro-5-chloronitrobenzene and 600 g (29.7%, relative to 2,5-dichloronitrobenzene employed) of unreacted 2,5-dichloronitrobenzene were obtained.