Decoquinate
;)
3-Quinolinecarboxylic acid,6-(decyloxy)-7-ethoxy-4-hydroxy-,ethyl ester.
Ethyl 6-(decyloxy)-7-ethoxy-4-hydroxy-3-quinolinecarboxylate [18507-89-6].
»Decoquinate contains not less than 99.0percent and not more than 101.0percent of C24H35NO5,calculated on the dried basis.
Packaging and storage—
Preserve in tight containers.
Labeling—
Label it to indicate that it is for veterinary use only.
Identification—
A:
Infrared Absorption á197Kñ.
B:
Transfer about 40mg of it,accurately weighed,to a 100-mLvolumetric flask,add 10mLof hot chloroform,swirl to dissolve,and while still warm add about 60mLof dehydrated alcohol.Allow to cool,dilute with dehydrated alcohol to volume,and mix.Promptly transfer 10.0mLof this solution to a second 100-mLvolumetric flask,dilute with dehydrated alcohol to volume,and mix.Transfer 10.0mLof this solution to a third 100-mLvolumetric flask,add 10mLof 0.1Nhydrochloric acid,dilute with dehydrated alcohol to volume,and mix:the UVabsorption spectrum of this solution exhibits maxima and minima at the same wavelengths as that of a similar solution of USP Decoquinate RS,concurrently measured,and the respective absorptivities,calculated on the dried basis,at the wavelength of maximum absorption at about 265nm do not differ by more than 2.5%.
Loss on drying á731ñ—
Dry it at 105
to constant weight:it loses not more than 0.5%of its weight.
to constant weight:it loses not more than 0.5%of its weight.
Residue on ignition á281ñ:
not more than 0.1%.
Ordinary impurities á466ñ—
Test solution:
chloroform,prepared with the aid of heat.
Standard solution:
chloroform,using dilutions of the Test solution.
Eluant:
a mixture of chloroform,dehydrated alcohol,and anhydrous formic acid (85:10:5).
Visualization:
1.
Tolerances:
no impurity exceeds 1%,and the total does not exceed 2%.
Assay—
Dissolve about 1000mg of Decoquinate,accurately weighed,in 100mLof glacial acetic acid,with the aid of gentle heat.Allow to cool,add 1drop of crystal violet TS,and titrate with 0.1Nperchloric acid VSto a green endpoint.Perform a blank determination,and make any necessary correction.Each mLof 0.1Nperchloric acid is equivalent to 41.76mg of C24H35NO5.
Auxiliary Information—
Staff Liaison:Ian DeVeau,Ph.D.,Senior Scientist
Expert Committee:(VET)Veterinary Drugs
USP28–NF23Page 574
Phone Number:1-301-816-8178