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

The 6:2 FTOH [F(CF(2))(6)CH(2)CH(2)OH] is a major raw material being used to replace 8:2 FTOH [F(CF(2))(8)CH(2)CH(2)OH] to make FTOH-based products for industrial and consumer applications. A novel aerobic sediment experimental system containing 20 g wet sediment and 30 mL 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 100 d. The primary 6:2 FTOH biotransformation in the aerobic sediment system was rapid (T(1/2)<2d). 5:3 acid [F(CF(2))(5)CH(2)CH(2)COOH] was observed as the predominant polyfluorinated acid on day 100 (22.4 mol%), higher than the sum of perfluoropentanoic acid (10.4 mol%), perfluorohexanoic acid (8.4 mol%), and perfluorobutanoic acid (1.5 mol%). Perfluoroheptanoic acid was not observed during 6:2 FTOH biotransformation. The 5:3 acid can be further degraded to 4:3 acid [F(CF(2))(4)CH(2)CH(2)COOH, 2.7 mol%]. 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(CF(2))(6)CH(2)COOH], 6:2 FTUCA [F(CF(2))(5)CF=CHCOOH], 5:2 ketone [F(CF(2))(5)C(O)CH(3)], and 5:2 sFTOH [F(CF(2))(5)CH(OH)CH(3)] 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.