[Li+].[CH2:2]1[C:5](=[O:6])[N:4]2[C@@H:7]([C:13]([O-:15])=[O:14])/[C:8](/[O:12][C@H:3]12)=[CH:9]/[CH2:10][OH:11].C(=O)([O-])[O-].[K+:20].[K+].C1C(=O)N2[C@@H](C(O)=O)/C(/O[C@H]12)=C/CO.C(=O)([O-])[O-].[Li+].[Li+]>O.CC(C)=O>[CH2:2]1[C:5](=[O:6])[N:4]2[C@@H:7]([C:13]([O-:15])=[O:14])/[C:8](/[O:12][C@H:3]12)=[CH:9]/[CH2:10][OH:11].[K+:20]
Lithium clavulanate (10.3 g; 0.05 mole) was dissolved in water (90 ml) and cooled in an ice bath. Acetone (60 ml) was then added. A solution of potassium carbonate (3.5 g; 0.025 mole in 10 water) was added slowly to the clavulanate solution. The first 3 drops caused the pH to reach 10.3. A few milligrams of lithium carbonate solid were added to seed out the lithium carbonate formed during the reaction. At this point the pH started to decrease. The remaining potassium carbonate solution was then added at a rate such that the pH remained about 10. Halfway through this addition a small volume of acetone (30 ml) was added. After a further 1/4 hour the mixture was made up slowly to 650 ml) with acetone. The lithium carbonate was then filtered off. (Weight obtained 1.95 g.) The filtrate was made up to 1 liter with acetone and filtered through celite (3 g) thereby removing most of the yellow colour which had formed. The resulting liquor was slowly added to 30 volumes acetone (3 liters) over 3/4 hour, left 1 hour and then filtered. The resulting solid washed with acetone and dried under vacuum at room temperature to give potassium clavulanate (very slight yellow tint) (8.5 g). (This material assayed as 81.5% pure expressed as pure free acid and contained 0.22% lithium.)
불러오는 중…
