Water Purification Using Choline-Amino Acid Ionic Liquids: Removal of Amoxicillin

使用胆碱-氨基酸离子液体净化水:去除阿莫西林

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作者:Pedro Velho, Catarina Lopes, Eugénia A Macedo

Abstract

Antibiotics are the main active pharmaceutical ingredients (APIs) for the treatment and prevention (prophylaxis) of bacterial infections, for which they are essential for health preservation. However, depending on the target bacterial strain, an efficient treatment may imply weeks of continuous intake of antibiotics, whose unmetabolized fraction ends up in the wastewater system by human and animal excreta. The presence of these chemical compounds in wastewater is known to damage aquatic ecosystems and cause antibiotic resistance of pathogenic agents, which threatens the future application of these medicines. Aqueous two-phase systems (ATPSs), an emergent extraction technology for biomolecules such as proteins and vitamins, could provide more eco-friendly and cost-effective extractive alternatives given their nontoxicity and low energetic requirements. Moreover, choline-amino acid ([Ch][AA]) ionic liquids (also known as CAAILs or ChAAILs) are considered one of the greenest classes of ionic liquids due to their favorable biocompatibility, biodegradability, and ease of chemical synthesis. In this work, partition studies of amoxicillin were performed in three ATPSs containing dipotassium hydrogen phosphate (K2HPO4) and the CAAILs (cholinium l-alaninate, [Ch][Ala]; cholinium glycinate, [Ch][Gly]; and cholinium serinate, [Ch][Ser]) at 298.15 K and 0.1 MPa. To better characterize the extract and reduce errors in quantification, the effect of pH on the intensity and stability of the UV-vis spectra of amoxicillin was studied prior to the partition studies, and computational chemistry was used to validate the molecular structure of the synthesized ionic liquids. During experimental determinations, it was observed that the extraction of amoxicillin was favored by less polar ionic liquids, achieving maximum partition coefficients (K) and extraction efficiencies (E) of K = (16 ± 6)·101 and E / % = 97 ± 2, respectively, for {[Ch][Gly] (1) + K2HPO4 (2) + Water (3)} in the longest tie-line.

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