Diisopropanolamine
133.192-Propanol, 1,1¢-iminobis-.
1,1¢-Iminodi-2-propanol
[110-97-4].» Diisopropanolamine is a mixture of isopropanolamines, consisting largely of diisopropanolamine. It contains not less than 98.0 percent and not more than 102.0 percent of isopropanolamines, calculated on the anhydrous basis as NH(C3H6OH)2.
Packaging and storage—
Preserve in tight, light-resistant containers. No specific storage conditions required.
Identification—
The IR absorption spectrum of a thin film exhibits regions of absorption between 2.8 µm and 4.0 µm, between 6.7 µm and 7.1 µm, and between 8.5 µm and 9.4 µm; and several characteristic peaks, the most pronounced being at about 7.3 µm, 7.5 µm, 8.3 µm, 9.6 µm, 10.4 µm, and 10.7 µm.
Water, Method I 
921
:
not more than 0.50%, a mixture of 5.0 mL of glacial acetic acid and 25 mL of methanol being used as the solvent.
921:
not more than 0.50%, a mixture of 5.0 mL of glacial acetic acid and 25 mL of methanol being used as the solvent.
Limit of triisopropanolamine—
Mixed indicator—
Dissolve about 150 mg of methyl orange and about 80 mg of xylene cyanole FF in water, and dilute with water to 100 mL.
Procedure—
Add 100 mL of methanol and 6 to 8 drops of Mixed indicator to a glass-stoppered, 500-mL conical flask, and neutralize with 0.1 N alcoholic sulfuric acid or 0.1 N alcoholic potassium hydroxide. The neutral solution is amber when viewed by transmitted light and is red-brown when viewed by reflected light. Add about 20 g of Diisopropanolamine, accurately weighed. Cautiously add 75 mL of acetic anhydride, and swirl to dissolve. Allow to stand at room temperature for 30 minutes. Cool to room temperature, if necessary. Titrate with 0.5 N alcoholic sulfuric acid VS. Perform a blank determination, and make any necessary correction. Each mL of 0.5 N alcoholic sulfuric acid is equivalent to 95.7 mg of triisopropanolamine: the limit is 1.0% by weight.
Assay—
Transfer about 2 g of Diisopropanolamine, accurately weighed, to a 250-mL conical flask, add 50 mL of water and bromocresol green TS, and titrate with 0.5 N hydrochloric acid VS. Perform a blank determination, and make any necessary correction (see Titrimetry 541). Each mL of 0.5 N hydrochloric acid is equivalent to 66.60 mg of isopropanolamines, expressed as NH(C3H6OH)2.
541). Each mL of 0.5 N hydrochloric acid is equivalent to 66.60 mg of isopropanolamines, expressed as NH(C3H6OH)2.
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 |
USP32–NF27 Page 1228
Pharmacopeial Forum: Volume No. 31(4) Page 1140