Asparagine
150.13 l-Asparagine
l-
-Aminosuccinamic acid, monohydrate
[5794-13-8].Anhydrous 132.12
[70-47-3].» Asparagine is anhydrous, or contains one molecule of water of hydration. It contains not less than 98.0 percent and not more than 101.5 percent of C4H8N2O3, as l-asparagine, calculated on the dried basis.
Packaging and storage—
Preserve in well-closed, light-resistant containers. Store at room temperature.
Labeling—
Label it to indicate whether it is anhydrous or the monohydrate.
Identification,
Infrared Absorption 
197K
. [note—Use USP Asparagine Anhydrous RS and USP Asparagine Monohydrate RS for the evaluation of the anhydrous and monohydrate forms of Asparagine, respectively.]
197K. [note—Use USP Asparagine Anhydrous RS and USP Asparagine Monohydrate RS for the evaluation of the anhydrous and monohydrate forms of Asparagine, respectively.]
Specific rotation 781S:
between +33.0
and +36.5, measured at 20.
781S:
between +33.0 and +36.5, measured at 20.
Test solution:
10 mg per mL in 6 N hydrochloric acid.
Microbial enumeration tests 61 and Tests for specified microorganisms 62—
The total aerobic microbial count does not exceed 1000 cfu per g, and the total combined molds and yeasts count does not exceed 100 cfu per g.
61 and Tests for specified microorganisms 62—
The total aerobic microbial count does not exceed 1000 cfu per g, and the total combined molds and yeasts count does not exceed 100 cfu per g.
Loss on drying 731—
Dry it at 130 for 3 hours: the anhydrous form loses not more than 1.0% of its weight, and the monohydrate loses between 11.5% and 12.5% of its weight.
731—
Dry it at 130 for 3 hours: the anhydrous form loses not more than 1.0% of its weight, and the monohydrate loses between 11.5% and 12.5% of its weight.
Residue on ignition 281:
not more than 0.1%, determined on 1.0 g.
281:
not more than 0.1%, determined on 1.0 g.
Lead 251—
Prepare a Test Preparation using a 1-g portion of Asparagine, and use 5 mL of Diluted Standard Lead Solution (5 µg of Pb) for the test: the limit is 5 µg per g.
251—
Prepare a Test Preparation using a 1-g portion of Asparagine, and use 5 mL of Diluted Standard Lead Solution (5 µg of Pb) for the test: the limit is 5 µg per g.
Chromatographic purity—
Adsorbent:
0.25-mm layer of chromatographic silica gel mixture.
Test solution:
10 mg per mL.
Standard solution—
Prepare a solution of USP Asparagine Anhydrous RS or USP Asparagine Monohydrate RS in water having a known concentration of about 0.05 mg per mL. [note—Use USP Asparagine Anhydrous RS and USP Asparagine Monohydrate RS for the evaluation of the anhydrous and monohydrate forms of Asparagine, respectively.]
Application volume:
5 µL.
Developing solvent system:
a mixture of butyl alcohol, water, and glacial acetic acid (3:1:1).
Spray reagent—
Dissolve 0.2 g of ninhydrin in 100 mL of a mixture of butyl alcohol and glacial acetic acid (95:5).
Procedure—
Proceed as directed for Thin-Layer Chromatography under Chromatography 621, and then dry the plate at 80 for 30 minutes. Spray the plate with the Spray reagent, heat at 80 for 10 minutes, and examine under white light: no secondary spot in the chromatogram obtained from the Test solution is larger or more intense than the principal spot in the chromatogram obtained from the Standard solution (0.5%); and not more than 1.0% of total impurities is found.
621, and then dry the plate at 80 for 30 minutes. Spray the plate with the Spray reagent, heat at 80 for 10 minutes, and examine under white light: no secondary spot in the chromatogram obtained from the Test solution is larger or more intense than the principal spot in the chromatogram obtained from the Standard solution (0.5%); and not more than 1.0% of total impurities is found.
Assay—
Dissolve about 130 mg of Asparagine, accurately weighed, in 3 mL of formic acid and 50 mL of glacial acetic acid, and titrate with 0.1 N perchloric acid VS (see Titrimetry 541), determining the endpoint potentiometrically. Perform a blank determination, and make any necessary corrections. Each mL of 0.1 N perchloric acid is equivalent to 13.21 mg of C4H8N2O3, calculated on the dried basis.
541), determining the endpoint potentiometrically. Perform a blank determination, and make any necessary corrections. Each mL of 0.1 N perchloric acid is equivalent to 13.21 mg of C4H8N2O3, calculated on the dried basis.
Auxiliary Information—
Please check for your question in the FAQs before contacting USP.
| Topic/Question | Contact | Expert Committee |
| Monograph | Robert H. Lafaver, B.A.
Scientist 1-301-816-8335 |
(EM105) Excipient Monographs 1 |
| Reference Standards | Lili Wang, Technical Services Scientist 1-301-816-8129 RSTech@usp.org |
|
| 61 |
Radhakrishna S Tirumalai, Ph.D.
Senior Scientist 1-301-816-8339 |
(MSA05) Microbiology and Sterility Assurance |
| 62 |
Radhakrishna S Tirumalai, Ph.D.
Senior Scientist 1-301-816-8339 |
(MSA05) Microbiology and Sterility Assurance |
USP32–NF27 Page 1168
Pharmacopeial Forum: Volume No. 31(1) Page 87