6:2 Fluorotelomer alcohol biotransformation in an aerobic river sediment system

Abstract

The 6:2 FTOH [F(CF2)6CH2CH2OH] is a major raw material being used to replace 8:2 FTOH [F(CF2)8CH2CH2OH] to make FTOH-based products for industrial and consumer applications. A novel aerobic sediment experimental system containing 20g wet sediment and 30mL aqueous solution was developed to study 6:2 FTOH biotransformation in river sediment. 6:2 FTOH was dosed into the sediment to follow its biotransformation and to analyze transformation products over 100d. The primary 6:2 FTOH biotransformation in the aerobic sediment system was rapid (T1/2<2d). 5:3 acid [F(CF2)5CH2CH2COOH] was observed as the predominant polyfluorinated acid on day 100 (22.4mol%), higher than the sum of perfluoropentanoic acid (10.4mol%), perfluorohexanoic acid (8.4mol%), and perfluorobutanoic acid (1.5mol%). Perfluoroheptanoic acid was not observed during 6:2 FTOH biotransformation. The 5:3 acid can be further degraded to 4:3 acid [F(CF2)4CH2CH2COOH, 2.7mol%]. This suggests that microbes in the river sediment selectively degraded 6:2 FTOH more toward 5:3 and 4:3 acids compared with soil. Most of the observed 5:3 acid formed bound residues with sediment organic components and can only be quantitatively recovered by post-treatment with NaOH and ENVI-Carb™ carbon. The 6:2 FTCA [F(CF2)6CH2COOH], 6:2 FTUCA [F(CF2)5CF=CHCOOH], 5:2 ketone [F(CF2)5C(O)CH3], and 5:2 sFTOH [F(CF2)5CH(OH)CH3] were major transient intermediates during 6:2 FTOH biotransformation in the sediment system. These results suggest that if 6:2 FTOH or 6:2 FTOH-based materials were released to the river or marine sediment, poly- and per-fluorinated carboxylates could be produced.