Packaging and storage
Preserve in well-closed containers.
Specific rotation 781S:
80 mg per mL, in 6 N hydrochloric acid.
pH 791 :
between 5.5 and 7.0, in a solution (1 in 20).
A 0.73-g portion shows no more chloride than corresponds to 0.50 mL of 0.020 N hydrochloric acid (0.05%).
A 0.33-g portion shows no more sulfate than corresponds to 0.10 mL of 0.020 N sulfuric acid (0.03%).
0.25-mm layer of chromatographic silica gel mixture.
Dissolve an accurately weighed quantity of Valine in 2 N hydrochloric acid to obtain a solution having a concentration of 10 mg per mL. Apply 5 µL.
Dissolve an accurately weighed quantity of USP L-Valine RS in 0.1 N hydrochloric acid to obtain a solution having a known concentration of about 0.05 mg per mL. Apply 5 µL. [NOTEThis solution has a concentration equivalent to about 0.5% of that of the Test solution. ]
System suitability solution
Prepare a solution in 0.1 N hydrochloric acid containing 0.4 mg each of USP L-Valine RS and USP L-Phenylalanine RS per mL. Apply 5 µL.
Dissolve 0.2 g of ninhydrin in 100 mL of a mixture of butyl alcohol and 2 N acetic acid (95:5).
Developing solvent system
Prepare a mixture of butyl alcohol, glacial acetic acid, and water (60:20:20).
Proceed as directed for Thin-Layer Chromatography
under Chromatography 621
. After air-drying the plate, spray with Spray reagent
, and heat between 100
for about 15 minutes. Examine the plate under white light. The chromatogram obtained from the System suitability solution
exhibits two clearly separated spots. Any secondary spot in the chromatogram obtained from the Test solution
is not larger or more intense than the principal spot in the chromatogram obtained from the Standard solution:
not more than 0.5% of any individual impurity is found; and not more than 2.0% of total impurities is found.
Organic volatile impurities, Method I 467 :
meets the requirements.
Transfer about 110 mg of Valine, accurately weighed, to a 125-mL flask, dissolve in a mixture of 3 mL of formic acid and 50 mL of glacial acetic acid, and titrate with 0.1 N perchloric acid VS, determining the endpoint potentiometrically. Perform a blank determination, and make any necessary correction. Each mL of 0.1 N perchloric acid is equivalent to 11.72 mg of C5H11NO2.