1 Kan et smog-kammer bruges til at vise at en isbjørn indeholder 10 ng/g PFOA? Ole John Nielsen Copenhagen Center for Atmospheric Research (CCAR) Department of Chemistry University of Copenhagen

2 Smog kammer kvartetten Mads Andersen, Mike Hurley, Timothy J. Wallington, Ole John Nielsen

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4 Detektion af perfluorerede organiske syrer Lange perfluorerede syrer (8-20 kulstof) Whitehorse, YT (mink) Kuujjuarapik, QC (isbjørn, fisk) Arviat, NU (polar ræv) Holman, NWT (sæler) Prince Leopold Isl., NU (fugle) Grise Fjord, NU (sæler)

5 C 10 F 21 C(O)OH

6 Perfluorerede forbindelser i isbjørne Box plots show the outliers, 10th, 25th, median, 75 th and 90 th centile. n.d. <0.5 Vi er % PFOS ! 3M voluntarily PFOS phased-out completely in 2002

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8 Hvad er problemerne ? Det miljømæssige problem: Øgende koncentrationer af ikke-nedbrydelige, bio- akumulerbare, skadelige forbindelser. Det videnskabelige problem: Hvorfor finder vi disse forbindelser i fjerntliggende egne, når der ikke er kilder ? Hypotese: Der er andre fluorerede forbindelser som omdannes i atmosfæren til PFOA og kan nå til f.eks. Arktis

9 Check hypotese Kan vi finde et udgangsstof ? Lever stoffet længe nok til at kunne komme til Arktis ? Omdannes stoffet til perfluorerede syrer ? Kan en eventuel mekanisme forklare de koncentrationer man finder i Arktis ?

10 C x F 2x+1 CH 2 CH 2 OH (x = even) 4:2 FTOH = C 4 F 9 CH 2 CH 2 OH 6:2 FTOH = C 6 F 13 CH 2 CH 2 OH 8:2 FTOH = C 8 F 17 CH 2 CH 2 OH 10:2 FTOH = C 10 F 21 CH 2 CH 2 OH -Eller længere – kunne de blive til: -C 7 F 15 COOH FTOH = FluoroTelomer alkohol 12x10 6 kg/yr (2005) globalt Bruges i mange produkter mod pletter

11 4. Experimental apparatus and setup Lever det længe nok til at kunne komme til Arktis ? Hvor hurtigt reagerer FTOH i atmosfæren ?

12 Lever det længe nok til at kunne komme til Arktis ? Hvor hurtigt reagerer FTOH i atmosfæren ? Levetid ~ 20 dage

13 5m/s vind og 20 dages levetid, FTOH kan transporteres 8500 km 20 dage… Lang tid nok ? København til Detroit = 6500 km

14 Check hypotese Kan vi finde et udgangsstof ? Lever det længe nok til at kunne komme til Arktis ? Omdannes det til perfluorerede syrer ? Sker det i mængder der kan forklare hvad man finder i Arktis ?

15 Omdannes det til perfluorerede syrer ? Lidt kompliceret

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17 Check hypotese Kan vi finde et udgangsstof ? Lever det længe nok til at kunne komme til Arktis ? Omdannes det til perfluorerede syrer ? Sker det i mængder der kan forklare hvad man finder i Arktis ?

18 3D Global Atmosfærekemi og transport model (ACTM) Gasfase kemi af alle vigtige troposfæriske forbindelser – mere end 90 stoffer og 300 kemiske reaktioner. + alle kinetiske + mekanistiske data Antagelse: 1000 tons/år 8:2 FTOH emission (flux), fordelt globalt som propan PFOA 50 m højde (enhed = molekyler cm -3 ) Jan July Koncentrationen af PFOA fra FTOH afhænger af tid og sted Integration over o N  0.4 tons/år for PFOA deposition til Arktis. Samme størrelsesorden som PCBer PFCAs detekteres i isbjørne i samme koncentrationer som PCBer (  ng/g)

19 Konklusion 1.Data viser at atmosfærisk nedbrydning af FTOHs bidrager signifikant til deposition af perfluorerede syrer i Arktis. 2.Disse resultater er blevet publiseret i ES&T i Ja - et smog-kammer bruges til at vise at en isbjørn kan indeholde 10 ng/g PFOA

20 Ekstra slides

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23 IR spectra obtained before (A) and after (B) 55 s of irradiation of a mixture of 18.8 mTorr C 2 F 5 C(O)H, 218 mTorr Cl 2 and 2.8 Torr H 2 in 700 Torr of air. The consumption of C 2 F 5 C(O)H was 63%.

24 PFCAs are products of C x F 2x+1 C(O)O 2 + HO 2 reaction Offers reasonable explanation of observed PFCA formation in 4:2 FTOH expts.

25 Branching ratios in reactions of RC(O)O 2 with HO 2 radicals under ambient conditions ( Torr, 296  2K). RC(O)O 2 Products Reference RC(O)OOH+O 2 RC(O)OH+O 3 RC(O)O+O 2 +OH CH 3 C(O)O    0.16[21] CF 3 C(O)O    0.05This work C 2 F 5 C(O)O 2 <   0.04[27] C 3 F 7 C(O)O 2 <  0.02  0.90This work C 4 F 9 C(O)O 2 <  0.02  0.90This work

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28 Three necessary conditions: (1) Do FTOHs survive atmospheric transport? YES (2) Do FTOHs degrade to give PFCAs? YES (3) Magnitude of PFCA formation must be significant Does atmospheric oxidation of FTOHs contribute significantly to PFCA burden in remote locations?

29 FTOH flux into Northern Hemisphere = t yr -1 Assume molar PFCA yield from FTOH of 1-10% Hence, PFCA flux = t yr -1 Assume even spatial distribution Hence, PFCA flux to Arctic = t yr -1 Persistent organochlorine pesticides arctic loading =1.8 t yr -1 Organochlorine pesticides detectable in polar bears at a similar concentration to PFCAs (  ng/g) Order of magnitude calculations suggest atmospheric oxidation of FTOHs is plausible explanation of PFCAs in remote areas.

30 Three necessary conditions: (1) Do FTOHs survive atmospheric transport? YES (2) Do FTOHs degrade to give PFCAs? YES (3) Magnitude of PFCA formation must be significant Looks plausible … more work … Does atmospheric oxidation of FTOHs contribute significantly to PFCA burden in remote locations?

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32 Concentration of PFOA (in molecule cm-3) at 50 m. altitude in the University of Michigan 3D model (IMPACT) for January and July. The color scale extends from (A) 0 to 1.2x10 3 and (B) 0 to 3x10 3 molecule cm -3. UIUC 2D model

33 Conclusions 1.The available evidence suggests, that the atmospheric oxidation of FTOHs makes a significant contribution to the PFCA burden in remote locations. 2.These results have been written up and a paper submitted to Science February This is just the tip of the ice berg 4.The automobile industry uses large quantities of fluoropolymers but little, if any, FTOHs. Vehicles do not appear to be a source of PFCAs

34 The Atmospheric Science Group Ch F Dk Fin Dk Rus D Est US

35 PolyfluoroAlcohols are highly volatile!!! HC data from Daubert & Danner; FTOH data from Lei et al, submitted J Chem Eng Data and Stock et al, ES&T in press. 8:2 FTOH = 212 Pa

36 FTOH based coatings heavily used in consumer products; *TRP Presentation to USEPA OPPT. Nov 25, 2002 US Public Docket AR x10 6 kg/yr 40% in North America 80% are in polymers*

37 8:2 FTOH = C 8 F 17 CH 2 CH 2 OH PFNA = C 8 F 17 C(O)OH PFOA = C 7 F 15 C(O)OH Three necessary conditions: (1)FTOH survive atmospheric transport (2)FTOH degrade to give PFCAs (3)Magnitude of PFCA formation must be significant Use a FTIR Smog chamber Research Question: Does atmospheric oxidation of FTOHs contribute significantly to PFCA burden in remote locations?

38 FTIR SMOG CHAMBER o140 L Pyrex chamber oX/Cl 2 /N 2 /O 2 /black-lamps oX/CH 3 ONO/NO/air/black-lamps 296 K, 700 Torr

39 Three necessary conditions: (1)Do FTOHs survive atmospheric transport? Measurement of k(OH+FTOH) (2) Do FTOHs degrade to give PFCAs? (3) Magnitude of PFCA formation must be significant? Does atmospheric oxidation of FTOHs contribute significantly to PFCA burden in remote locations?

40 UV irradiation of FTOH/reference/CH 3 ONO/NO/air mixtures FTOH = 4:2 FTOH, 6:2 FTOH, or 8:2 FTOH reference = C 2 H 2 or C 2 H 4 CH 3 ONO  CH 3 O + NO CH 3 O + O 2  HCHO + HO 2 HO 2 + NO  OH + NO 2 OH + FTOH  products(1) OH + reference  products(2)

41 OH + FTOH  products(1) OH + reference  products(2) Integration gives: FTOH and reference have equal exposure to OH radicals, hence:

42 Loss of FTOH (squares = 4:2; circles = 6:2; triangles = 8:2) versus C 2 H 2 and C 2 H 4 on exposure to OH radicals in 700 Torr of air diluent at 296 K. No discernable difference in reactivity of OH radicals towards 4:2, 6:2, and 8:2 FTOH

43 OH + C n F 2n+1 CH 2 CH 2 OH → products (10) OH + C 2 H 2 → products (11) OH + C 2 H 4 → products (12) Linear fits give k 10 /k 11 = 1.18±0.15 and k 10 /k 12 = 0.131± Using k 11 = 8.5 x and k 12 = 8.66 x gives k 10 = (1.00±0.13) x and (1.13±0.16) x cm 3 molecule -1 s -1. Final value, k 10 = (1.07±0.22) x cm 3 molecule -1 s -1.

44 Assuming: atmospheric lifetime* for CH 3 CCl 3 = 5.7 years k(CH 3 CCl 3 + OH) = 1.0 x cm 3 molecule -1 s -1 then atmospheric lifetime* of F(CF 2 ) n CH 2 CH 2 OH  (1.0x )/(1.1x ) x 5.7 x 365  20 days. * with respect to reaction with OH radicals FTOH Lifetime Estimate

45 Other loss mechanisms? Photolysis – should be negligible Rainout – estimated to be negligible Dry deposition – lifetime estimated to be 8 years Homogeneous reactions other than with OH - unlikely Atmospheric lifetime determined by reaction with OH and is approximately 20 days.

46 Ramifications of Lifetime (1)Estimate flux of t yr -1 necessary to sustain observed atmospheric concentration. (2)FTOH have negligible GWP (3)Spatial distribution will be inhomogeneous (4)FTOH will be transported to remote locations. Global average wind speed = 5 m s -1, 20 days = 8500 km.

47 Three necessary conditions: (1) Do FTOHs survive atmospheric transport? YES (2) Do FTOHs degrade to give PFCAs? (3) Magnitude of PFCA formation must be significant Does atmospheric oxidation of FTOHs contribute significantly to PFCA burden in remote locations?

48 FTIR study of 4:2 FTOH oxidation CF 3 (CF 2 ) 3 CH 2 CHO is the major primary product from Cl atom and OH radical initiated oxidation of 4:2 FTOH

49 FTOH Oxidation mechanism C n F 2n+1 CH 2 CH 2 OH + OH  C n F 2n+1 CH 2 C()HOH + H 2 O C n F 2n+1 CH 2 C()HOH + O 2  C n F 2n+1 CH 2 CHO + HO 2

50 C n F 2n+1 CH 2 CHO is reactive … Gives secondary products …

51 Secondary products: CF 3 (CF 2 ) 3 CHO, CF 3 (CF 2 ) 3 CH 2 COOH, CF 3 (CF 2 ) 3 C(O)OOH

52 FTOH Oxidation mechanism C n F 2n+1 CH 2 CH 2 OH + OH  C n F 2n+1 CH 2 C()HOH + H 2 O C n F 2n+1 CH 2 C()HOH + O 2  C n F 2n+1 CH 2 CHO + HO 2 C n F 2n+1 CH 2 CHO + OH + O 2  C n F 2n+1 CH 2 C(O)OO + H 2 O C n F 2n+1 CH 2 C(O)OO + NO  C n F 2n+1 CH 2 C(O)O + NO 2 C n F 2n+1 CH 2 C(O)O  C n F 2n+ 1 CH 2 + CO 2 C n F 2n+ 1 CH 2 + O 2  C n F 2n+ 1 CH 2 O 2 C n F 2n+ 1 CH 2 O 2 + NO  C n F 2n+ 1 CH 2 O + NO 2 C n F 2n+ 1 CH 2 O + O 2  C n F 2n+ 1 CHO + HO 2

53 Secondary products: C 4 F 9 CHO, C 4 F 9 CH 2 COOH C 4 F 9 C(O)OOH Secondary products are reactive …

54 Tertiary products include: COF 2, CF 3 OH C 4 F 9 COOH Conclusion of FTIR experiments: simulated atmospheric oxidation of 4:2 FTOH (in absence of NO x ) gives a small (few %) yield of C 4 F 9 COOH

55 FTIR data shows that in gas phase: in absence of NO x 4:2 FTOH   C 4 F 9 CHO  C 4 F 9 COOH Likely explanation, presence of HO 2 radicals in absence of NO x Well established that CH 3 C(O)O 2 + HO 2 gives acetic acid and peracetic acid,, presumably C x F 2x+1 C(O)O 2 + HO 2 reaction gives C x F 2x+1 COOH and C x F 2x+1 COOOH. Product study of C x F 2x+1 C(O)O 2 + HO 2 (x=1-4) to test this idea. in presence of NO x 4:2 FTOH   C 4 F 9 CHO  C 4 F 9 COOH

56 C n F 2n+1 C(O)O 2 and HO 2 radicals generated by UV irradiation of C n F 2n+1 CHO/H 2 /Cl 2 mixtures in Torr of air at 296±2 K: Cl 2 + h  2Cl Cl + C n F 2n+1 CHO  C n F 2n+1 CO + HCl C n F 2n+1 CO + O 2 + M  C n F 2n+1 C(O)O 2 + M Cl + H 2  H + HCl H + O 2 + M  HO 2 + M C n F 2n+1 C(O)O 2 + HO 2  products C n F 2n+1 C(O)O 2 + C n F 2n+2 C(O)O 2  products As [H 2 ] o /C n F 2n+1 CHO] o , products/products , Method

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58 Taves, D. R., Nature 211, 192 (1966) Taves, D. R., Nature 215, 1380 (1967) Taves, D. R., Nature 217, 1050 (1968)

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