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Ephedrine Hydrochloride
C10H15NO·HCl 201.69

Benzenemethanol, -[1-(methylamino)ethyl]-, hydrochloride, [R-(R*,S*)]-.
()-Ephedrine hydrochloride [50-98-6].
» Ephedrine Hydrochloride contains not less than 98.0 percent and not more than 100.5 percent of C10H15NO·HCl, calculated on the dried basis.
Packaging and storage— Preserve in well-closed, light-resistant containers.
Identification—
A: Dissolve 100 mg in 5 mL of water, add 1 mL of potassium carbonate solution (1 in 5), and extract with 2 mL of chloroform: the IR absorption spectrum of the chloroform extract so obtained exhibits maxima only at the same wavelengths as that of a similar preparation of USP Ephedrine Sulfate RS.
B: A solution of it responds to the tests for Chloride 191.
Melting range, Class I 741: between 217 and 220.
Specific rotation 781S: between 33.0 and 35.5.
Test solution: 50 mg per mL, in water.
Acidity or alkalinity— Dissolve 1.0 g in 20 mL of water, and add 1 drop of methyl red TS. If the solution is yellow, it is changed to red by not more than 0.10 mL of 0.020 N sulfuric acid. If the solution is pink, it is changed to yellow by not more than 0.20 mL of 0.020 N sodium hydroxide.
Loss on drying 731 Dry it at 105 for 3 hours: it loses not more than 0.5% of its weight.
Residue on ignition 281: not more than 0.1%.
Sulfate— Dissolve 50 mg in 40 mL of water, and add 1 mL of 3 N hydrochloric acid and 1 mL of barium chloride TS: no turbidity develops within 10 minutes.
Ordinary impurities 466
Test solution: alcohol.
Standard solution: alcohol.
Eluant: a mixture of isopropyl alcohol, ammonium hydroxide, and chloroform (80:15:5).
Visualization: 1, followed by 4.
Organic volatile impurities, Method I 467: meets the requirements.
Residual solvents 467: meets the requirements.
(Official January 1, 2007)
Assay— Dissolve about 500 mg of Ephedrine Hydrochloride, accurately weighed, in 25 mL of glacial acetic acid. Add 10 mL of mercuric acetate TS and 2 drops of crystal violet TS, and titrate with 0.1 N perchloric acid VS to an emerald-green endpoint. Perform a blank determination, and make any necessary correction. Each mL of 0.1 N perchloric acid is equivalent to 20.17 mg of C10H15NO·HCl.
Auxiliary Information— Staff Liaison : Daniel K. Bempong, Ph.D., Scientist
Expert Committee : (MDPS05) Monograph Development-Pulmonary and Steroids
USP29–NF24 Page 803
Pharmacopeial Forum : Volume No. 30(3) Page 839
Phone Number : 1-301-816-8143