as organic matter e d l v s o d i s h h i g a - u l - PowerPoint PPT Presentation

of reactivity and e u r u c t t r S as organic matter e d l v s o d i s h h i g a - u l t r determined by electrospray resolution y t r m e r o c t p e mass s n i o z a t o n i i Kim n h w a u n g S r c h e h a t G. i c k a t r Adviser: P University, State Ohio The Chemistry, o f Department 43210 , OH b u s l u m C o

w r v i e O v e d organic l v e i n e d i s s o h e r i v e r of t Understanding i t t l e is t however l r t a n is i m p o D O M ) matter ( molecular l e v e l . s material in a b o u t t h i known a w b o n f l o major source of c a r is a D O M e r i n e R i v i n the t role p o r t a n a n i m oceans and plays to t h e carbon. of global cycle

r b o n cycle in a g l o b a l c a A riddle • 0.3 Gt of carbon is large enough to sustain the turnover of the entire ocean • However, a little resemblance was found between two carbon pools. • Low amount of lignin phenol (terrestrial bio-marker) exists in oceanic DOC pool. isotope signatures are • 6 13 C different. Adapted from Houghton et. al. Climate change 1994 (1995), Cambridge Press, New York (units: Gt) * Meyers-Schuttle & Hedges, Nature (1986), 321, 61-63

of DOM from I s o l a t i o n samples natural river w a t e r

How to isolate the DOM? N Final elution i n c l u d e s a l t d i a l n y s i s e c e s s a r y X A D extraction — Established method L a r g e volume o f chemical F r a c t i o n a t i o n L a r g e amount o f s a m p l e Ultrafiltration possible Requires instrument Takes a l o n g time Not-selective Freezedrying Concentrate salts too

H o w t o isolate DOM from river water? Disk type C 18 solid p h a s e extraction Benefits 1. F a s t e r e x t r a c t i o n 2. S i m p l e experiment setup 3. easily handle s m a l l amount o f sample Questions to be answered 1 . Retention rate? 2. D o e s c o l l e c t e d sample reflect DOM in r i v e r ? L o u c h o u a r n , P.; Opsahl, S.; Benner, R.;(2000), Anal. C h e m . 72, 2780-2787

Retention rate? Retention rate of 70 % 0.8 0.4 eluent 0- 250 400 300 350 450 500 Wavelength (nm)

Does final eluent reflect the DOM in river? H2O \J DOM acquired by freeze drying CH 3 OH DOM acquired by SPE I Ipp1n

E S I - M S data on D O M (quadrupole time of flight mass spectrometry) M S spectrum from control ( u l t r a - h i g h purity w a t e r p r o c e s s e d with same p r o c e d u r e ) 200 0 aa 800 1 0 l2 1 4 1600 2000 - - 912 313 1 M S s p e c t r u m from D O M i s o l a t e d b y C 18 SDE 0 -r.--- . n h 1 1 r I l - s . I . , J l T . n - I o ia 300 M / z 2 0 0 0

S p e c t D r O u m M of f r o Rio m TempisquitO Costa at R i c a q37 050 l0 218 344) Sm(SG 23 0 0 ) , 011(37323) 4 3 0 1 . 1 8 6 149051 224 347.199375 .A03.233 ..417.251 2 8 9 . 1 9 7 433257 I IIi iI 287.191k N 219.163 I N 1 0 _ 2000

_ _ High resolution mass spectrum Resolution “ 80,000 dê âk&M 384 388 392 396

_ p m 2D N M R ( T O C S Y ) d a t a o n D O M Sample by freezedrying Sample by 18 C SDE 2.%?4 3 6 5 3 2 1 (1) a r o m a t i c s ; (2) s u g a r s ; aliphatic units bridging lignin aromatics; amino acids couplings); (3) (a-13 methylene units adjacent to ethers, e s t e r s , and hydroxyls in aliphatic chains; amino acids couplings); (4) (a-3-’y methylene in aliphatic chains; and methyl units in amino acids and aliphatic chains.

Conclusion • 1 . D O M c a n b e e x t r a c t e d w i t h C 18 s o l i d p h a s e d i s k e x t r a c t i o i n n a s h o r t e r t i m e w i t h s i m p l e e q u i p m e n t . • 2. Extracted D O M w a s successfully analyzed by high resolution ESI-MS resulting high resolution spectra. Sunghwan Kim a! submitted et. to O r g a n i c G e o c h e m i s t r y

R e s o l v i n g p o w e r is v e r y critical 100 1 — Q - T O F analysis of D O M Reso’ving power — 1 0 , 0 0 0 I agi 7 T FT-ICR analysis of D O M Resolving power —80,000 r I, 390.3 390.5 390.7 390.9 3 9 1 . 1 391.3 391.5 391.7

U l t r a - h i g h resolution spectrum of DOM 14 Resolving power 200,000 8 17 P r o p o s e d m o l e c u l a r T h e o r e t i c a l D i f f e r e n c e from P e a k # 10 O b s e r v e d v a l u e s f o r m u l a v a l u e s T h e o r e t i c a l v a l u e ( p p m ) 6 91 I 10 10 C 25 H 0 I 469.02018 4 6 9 . 0 2 0 1 2 - 0 . 1 4 1 5 2 1 2 C 0 H 1 4 2 2 4 6 9 . 0 4 1 1 8 4 6 9 . 0 4 1 2 5 0.1 1 3 23 18 7 3 9 0 H 26 C 14 4 6 9 . 0 5 6 4 6 469.05651 16 0.1 UJIiAiL 1 2A j r 1 4 23 C 18 H 11 0 4 6 L J 9 . 0 7 7 6 3 4 6 9 . 0 7 7 6 4 0 A S 5 C 0 18 27 H - 4 6 9 . 0 9 2 8 8 4 6 9 . 0 9 2 8 9 0 469 074 469 115 469 156 6 C 24 H 22 0 10 4 6 9 . 1 1 4 0 1 4 6 9 . 1 1 4 0 2 0 7 7 22 0 H 28 C 4 6 9 . 1 2 9 3 4 6 9 . 1 2 9 2 8 0 8 26 C 25 H 0 9 4 6 9 . 1 5 0 4 2 4 6 9 . 1 5 0 4 1 0 9 26 O 6 H 29 C 4 6 9 . 1 6 5 7 6 4 6 9 . 1 6 5 6 6 - 0 . 2 1 0 C 22 H 30 11 4 6 9 . 1 7 1 5 1 4 6 9 . 1 7 1 5 4 0 . 1 11 C 26 H 30 0 8 4 6 9 . 1 8 6 8 1 4 6 9 . 1 8 6 7 9 0 12 H 5 0 30 30 C 4 6 9 . 2 0 2 0 1 4 6 9 . 2 0 2 0 5 0 . 1 13 C 23 H 34 0 10 4 6 9 . 2 0 7 8 9 4 6 9 . 2 0 7 9 2 0 . 1 III 1 4 C 27 H 34 0 7 4 6 9 . 2 2 3 1 6 4 6 9 . 2 2 3 1 8 0 34 0 4 H 31 C 15 4 6 9 . 2 3 8 3 8 4 6 9 . 2 3 8 4 3 0 . 1 16 9 0 H 38 24 C 4 6 9 . 2 4 4 2 3 4 6 9 . 2 4 4 3 1 0 . 2 1 7 C 28 H 38 0 6 4 6 9 . 2 5 9 4 9 4 6 9 . 2 5 9 5 6 0 . 2 18 42 0 5 H 29 C 469.29584 4 6 9 . 2 9 5 9 5 0 . 2 m l z

9.4 T FT-ICR at NHMFL

Trees or forest? 14 8 17 6 2 12J 469.033 469.074 469.115 469.156 469.197 469.238 469.279 mlz II,’, I.. Ii