Levorphanol Tartrate
(lee vor' fa nol tar' trate).
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C17H23NO·C4H6O6·2H2O 443.49

Morphinan-3-ol, 17-methyl-, [R-(R*,R*)]-2,3-dihydroxybutanedioate (1:1) (salt), dihydrate.
17-Methylmorphinan-3-ol, tartrate (1:1) (salt) dihydrate [5985-38-6].

Anhydrous 407.47 [125-72-4].
» Levorphanol Tartrate contains not less than 99.0 percent and not more than 101.0 percent of C17H23NO·C4H6O6, calculated on the anhydrous basis.
Packaging and storage— Preserve in well-closed containers. Store at 25, excursions permitted between 15 and 30.
USP Reference standards 11
USP Levorphanol Tartrate RS Click to View Structure
A: Infrared Absorption 197K—Obtain the test specimen as follows. Dissolve 50 mg in 25 mL of water in a 125-mL separator. Add 2 mL of 6 N ammonium hydroxide, extract with 25 mL of chloroform, and filter the chloroform extract through a layer of 4 g of granular anhydrous sodium sulfate supported on glass wool into a 125-mL conical flask. Evaporate the chloroform extract on a steam bath with the aid of a stream of nitrogen to dryness. Dissolve the residue in 1 mL of acetone, and evaporate to dryness. Dry in vacuum at 90 for 1 hour. Proceed as directed with the dried levorphanol so obtained and a similar preparation of USP Levorphanol Tartrate RS.
B: Ultraviolet Absorption 197U
Solution: 130 µg per mL.
Medium: 0.1 N hydrochloric acid.
Absorptivities at 279 nm, calculated on the anhydrous basis, do not differ by more than 3.0%.
Specific rotation 781S: between 14.7 and 16.3.
Test solution: 30 mg per mL, in water. Heat on a water bath or sonicate to dissolve 750 mg in 20 mL of water in a 25-mL volumetric flask, dilute with water to volume, and mix.
Water, Method I 921: between 7.0% and 9.0%.
Residue on ignition 281: not more than 0.1%.
Ordinary impurities 466
Test solution: water.
Standard solution: water.
Eluant: a mixture of hexanes, dehydrated alcohol, and ammonium hydroxide (80:25:1).
Visualization: 17; then view immediately under short-wavelength UV light.
Assay— Dissolve about 900 mg of sample, accurately weighed, in 85 mL of glacial acetic acid, warming slightly if necessary. Titrate with 0.1 N perchloric acid VS and determine the endpoint potentiometrically. Perform a blank determination and make any nececssary corrections. Each mL of 0.1 N perchloric acid consumed by the sample is equivalent to 40.75 mg of C17H23NO·C4H6O6.
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Monograph Clydewyn M. Anthony, Ph.D.
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(SM22010) Monographs - Small Molecules 2
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USP35–NF30 Page 3675
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