Sodium Hydroxide → 4-Chloro-6-(trifluoromethoxy)quinoline
Reaction equation
[F:1][C:2]([F:16])([F:15])[O:3][C:4]1[CH:5]=[C:6]2[C:11](=[CH:12][CH:13]=1)[NH:10][CH:9]=[CH:8][C:7]2=O.[OH-].[Na+].O=P(Cl)(Cl)[Cl:21]>>[Cl:21][C:7]1[C:6]2[C:11](=[CH:12][CH:13]=[C:4]([O:3][C:2]([F:16])([F:15])[F:1])[CH:5]=2)[N:10]=[CH:9][CH:8]=1Participants
4
Conditions
- Temperature
- 130 °C
Procedure
In a 25 mL round bottom flask was dissolved 6-(trifluoromethoxy)quinolin-4(1H)-one (0.50 g, 2.18 mmol) in POCl3 (3.05 mL). A reflux condenser was attached and the reaction mixture was stirred at 130° C. for 3 hours. The reaction turned dark brown and became homogeneous. The reaction was cooled to RT, poured slowly into ice bath while stirring. A thick gum resulted, which was basified with 6N NaOH to give a dark brown solution. The solution was extracted into ethyl acetate, washed 1× water, 1×NaCl, dried with Na2SO4, filtered through flitted funnel, concentrated down. The crude material was purified using an Isco silica gel chromatography system eluting with a solvent gradient of 0-100% CH2Cl2:MeOH(90:10)/CH2Cl2. The product fractions were concentrated down to yield 4-chloro-6-(trifluoromethoxy)quinoline as an orange oil. MS [M+H]=248.0; Calc'd 247.6 for C10H5ClF3NO.