Glutamine
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C5H10N2O3 146.14

L-Glutamine.
2-Aminoglutaramic acid [56-85-9].
»Glutamine contains not less than 98.5percent and not more than 101.5percent of C5H10N2O3,calculated on the dried basis.
Packaging and storage— Preserve in well-closed containers,and store at controlled room temperature.
Identification,Infrared Absorption á197Kñ.
Specific rotation á781Sñ: between +6.3and +7.3,determined at 20.
Test solution— In a suitable flask,accurately prepare a solution of about 40mg per mLin water at about 40.Cool and dilute with water to volume before use.
Loss on drying á731ñ Dry it at 105for 3hours:it loses not more than 0.3%of its weight.
Residue on ignition á281ñ: not more than 0.3%.
Chloride á221ñ A0.7-g portion shows no more chloride than corresponds to 0.50mLof 0.020Nhydrochloric acid:not more than 0.05%is found.
Sulfate á221ñ A0.8-g portion shows no more sulfate than corresponds to 0.25mLof 0.020Nsulfuric acid:not more than 0.03%is found.
Iron á241ñ: 0.003%.
Chromatographic purity—
Adsorbent: 0.25-mm layer of chromatographic silica gel mixture.
Test solution: 10mg per mL,in water.
Standard solution— Prepare a solution of USP Glutamine RSin water having a known concentration of about 0.05mg per mL.
Application volume: 5µL.
Developing solvent system: a mixture of butyl alcohol,water,and glacial acetic acid (3:1:1).
Spray reagent— Dissolve 0.2g of ninhydrin in 100mLof a mixture of butyl alcohol and 2Nacetic acid (95:5).
Procedure— Proceed as directed for System Suitabilityunder Chromatography á621ñ,then dry the plate at 80for 30minutes.Spray the plate with the Spray reagent,heat at 80for 10minutes,and examine under white light:no secondary spot in the chromatogram obtained from the Test solutionis larger or more intense than the principal spot in the chromatogram obtained from the Standard solution(0.5%).
Organic volatile impurities,Method Iá467ñ: meets the requirements.
Assay— Transfer about 150mg of Glutamine,accurately weighed,to a 125-mLflask,dissolve in 3mLof formic acid and 50mLof glacial acetic acid,and titrate with 0.1Nperchloric acid VS,determining the endpoint potentiometrically.Perform a blank determination,and make any necessary correction (see Titrimetry á541ñ).Each mLof 0.1Nperchloric acid is equivalent to 14.615mg of C5H10N2O3.
Auxiliary Information— Staff Liaison:Lawrence Evans,III,Ph.D.,Scientist
Expert Committee:(DSN)Dietary Supplements:Non-Botanicals
USP28–NF23Page 909
Phone Number:1-301-816-8389