Chromatographic purity
Dissolve a quantity of Norfloxacin in a mixture of methanol and methylene chloride (1:1) to obtain a test solution containing 8.0 mg per mL. Dissolve 4.0 mg of
USP Norfloxacin RS in 1 mL of glacial acetic acid, add 4 mL of methanol, and mix. To 1 mL of this Standard stock solution add 9 mL of the mixture of methanol and methylene chloride (1:1) to obtain
Comparison solution A. Dilute a portion of this solution with an equal volume of the mixture of methanol and methylene chloride (1:1) to obtain
Comparison solution B. Separately apply 5 µL of the test solution, 1, 1.5, and 2 µL of
Comparison solution A, and 5 µL of
Comparison solution B to a suitable high-performance thin-layer chromatographic plate (see
Chromatography 621) coated with a 0.25-mm layer of silica gel mixture, previously washed with methanol and air-dried. The spots of
Comparison solutions A and
B are equivalent to 0.2, 0.3, 0.4, and 0.5% of impurities, respectively. Place the plate in a paper-lined chromatographic chamber previously equilibrated with a solvent system consisting of a mixture of chloroform, methanol, toluene, diethylamine, and water (40:40:20:14:8). Seal the chamber and allow the chromatogram to develop until the solvent front has moved about nine-tenths of the length of the plate. Remove the plate from the chamber, mark the solvent front, allow the solvent to evaporate, and examine the plate under both short- and long-wavelength UV light. Compare the intensities of any secondary spots observed in the chromatogram of the test solution with those of the principal spots in the chromatograms of
Comparison solutions A and
B: the sum of the intensities of secondary spots obtained from the test solution corresponds to not more than 0.5% of impurities.
Assay
Dissolve about 460 mg of Norfloxacin, accurately weighed, in 100 mL of glacial acetic acid. Titrate potentiometrically with 0.1 N perchloric acid VS using a suitable anhydrous electrode system (see
Titrimetry 541).
[NOTERemove any aqueous solution in the electrode(s), render anhydrous, and fill with 0.1 N lithium perchlorate in acetic anhydride.
] Perform a blank determination, and make any necessary correction. Each mL of 0.1 N perchloric acid is equivalent to 31.93 mg of C
16H
18FN
3O
3.