1. Motivation for using Radiation Chemistry to Study Free Radicals

2. In photoinduced electron transfer, oxidation and reduction occur together In photoinduced electron transfer, both radical cations and radical anions are produced. If there are overlapping absorptions, clean kinetics can be difficult. D + A + h  D + + A 

3. Radiation Chemistry can selectively oxidize or reduce Although both oxidizing and reducing radicals are produced in solvents by ionizing radiation, one or the other can usually be selectively scavenged. Scavenger (N 2 O) converts reducing species (e aq  ) into an oxidizing one (O  ). e aq  + N 2 O  N 2 + O 

4. Problem with concentration gradients with photo-induced transients For quantitative work, free-radical concentrations are needed, but reliable extinction coefficients are not easy to obtain from photochemical methods. Examples were all the disadvantages we saw in measuring triplet-triplet extinction coefficients. In addition, Beer’s law is an ever-present problem producing non-constant concentrations of optically-generated transients.

5. Ionizing radiation generates uniform concentrations of transients Ionizing radiation generates a constant concentration of transients if the energies are sufficient to penetrate the sample. For certain solvents, in particular water, radiation chemists have carefully measured yields of primary radicals coming from the solvent.

6. Radiolytic yields of radicals are well- characterized Radiation chemists have also characterized scavenging efficiencies for secondary radical formation. The Schuler formula can be used to determine the yield of radicals in water Radiation Chemical Yields G is number of free radicals formed from 100 eV of energy absorbed

7. Schuler formula The Schuler formula gives the yield of secondary radicals formed from the reaction of hydroxyl radicals and the scavenger S at concentration [S]  OH + S  secondary radical k S

8. Selectivity in multi-component systems Photolytic methods have the advantage of selective excitation of species as long as there is an appropriate chromophore Radiation chemists have a similar, but not as selective technique. They can use redox potentials.

9. Radiation Chemistry provides selectivity through redox potentials An example is to use Br 2  to selectively oxidize tryptophan and tyrosine residues in proteins. Tryptophan and tyrosine are easily oxidized amino acid residues. However, Br 2  cannot oxidize most amino acids, whereas  OH can.

10. Example Free Radicals in Biology

11. Radiolysis of Liquid Water Thanks to Professor Chantel Houée-Lévin

12. Global scheme of the first stages H2OH2O H 2 O* H 2 O + + e — H 3 O + + OH H 2 O H + OH O + H 2 e — ps e — aq 0 10 —14 10 —12 Radiation

13. Physical stage The first act in the radiolysis of water: Ionization: H 2 O  H 2 O + + e — having a excess kinetic energy Excitation : H 2 O  H 2 O *. This process is minor w/ ionization These two entities H 2 O + and H 2 O * are very unstable and disappear in ~10 —14 s : H 2 O + + H 2 O  OH + H 3 O + (acid-base reaction) H 2 O * de-excites or cleaves: H 2 O *  OH + H is the most probable cleavage, but there are other possible cleavage paths (O + H 2 ) e — slows and then thermalizes with the solvent in ~2 stages.

14. Physical-chemical stage During the first 10 -12 s, the molecules do not have time to move significantly. The environment therefore consists principally of heterogeneous regions (spurs) notably at the end of the ionization track of the secondary electrons. Thes spurs contain the surviving species e aq -, OH, H 3 O + and H.

15. Physical-chemical stage from S.M. Pimblott’s NDRL Web Site End of track of a secondary electron

16. Yields in this stage 1 ps after the primary radiolytic act, the quantification of the destruction of water is, per 100 eV absorbed: 5.7 H 2 O = 4.78 e — aq + 4.78 H 3 O + + 0.62 H + 0.15 H 2 + 5.7 OH These data are the yields in molecules/100 eV: 1 mol J —1 = 1.036 10 —7 molecules/100 eV Problem: origin of H 2 at this stage.

17. Recombination reactions Recombination reactions are favored by the large local concentrations. They are in competition with diffusion (producing a homogenation of the media)

18. There is formation of molecular species: H 2 et H 2 O 2.

19. Chemical stage The homogenation of the media is complete after about 10 ns. At this time, the media has become a homogeneous solution of free radicals and molecular products: H 2 O ~  H 3 O +, OH, e — aq, H, H 2 O 2, OH -, H 2 At this stage, the yields are (  from 60 Co, accelerated electrons) : g(OH) = G OH = 0.29 µmol J -1 g(H) = G H = 0.057 µmol J -1 g(e - aq ) = G e = 0.28 µmol J -1 g(H 2 ) = G H2 = 0.046 µmol J -1 g(H 2 O 2 ) = G H2O2 = 0.072 µmol J -1

20. Chemical balance in water radiolysis Destruction of water: G(-H 2 O) = 2 g(H 2 ) + g(H) + g(e — aq ) = 2 g (H 2 O 2 ) + g(OH) 1. Conservation of O : G(-H 2 O) = 2 g (H 2 O 2 ) + g(OH) 2. Conservation of H : G(-H 2 O) = g (H 2 ) + 1/2 g(H) + 1/2 g (e — aq ) + g (H 2 O 2 ) + 1/2 g(OH)  2 g (H 2 O 2 ) + g(OH) = g (H 2 ) + 1/2 g(H) + 1/2 g (e — aq )

21. Effect of temperature on the radiolysis of water A priori, there should not be temperature effects at the level of the primary acts of radiolysis Physical-chemical stage (ps):  The temperature has an effect on the rate constants and on the diffusion constants: Arrhenius behavior  The activation energies are low (10-20 kJ mol -1 ) In total, the yields are little modified (0 — 200°C)

22. Effect of Linear Energy Transfer (LET) The distribution of energy deposits vary: the probably of short tracks increases with LET depending on the placement of the clusters of radicals (spurs). The recombinations in the heterogenous zones are much more probable as the LET is elevated because the concentration of free radicals is very high. As a consequence, the yields of radicals decrease as the yield of molecular produces increase with LET

23. Effect of Linear Energy Transfer (LET) from J.A. LaVerne’s NDRL Web Site

24. Yields Yields (  mol J -1 ) in water at pH 7 LET (keV  m -1 ) e — aq OHHH2H2 H2O2H2O2 HO 2 0.20.260.270.0550.0450.0680 610.07 2 0.0910.0420.0961.000.005

25. Formation of oxygen Irradiation with high LET radiation produces the appearance of superoxide ions O 2 — and/or the hydroperoxyl radical HO 2 In the absence of solutes, these free radicals undergo dismutation: HO 2 + O 2 - + H +  O 2 + H 2 O 2 Hence the formation of oxygen.

26. Effect of pH Experimentally: the yields are constant between pH 3 and 10. Nevertheless, there are acid-base equilibria : But the equilibria are not always established

27. Reactions of e — aq In acidic media: these reactions are in competition  e — aq + H 3 O +  H + H 2 O2.3 x 10 10 M -1 s -1 (1)  e — aq + H  H 2 + OH — 2.5 x 10 10 M -1 s -1 (2)  e - aq + OH  OH — 3 x 10 10 M -1 s -1 etc … In summary, g(e — aq )  0 and g(H)  for pH < 3: and under pH ~2, g(e — aq ) = 0. The more, reaction (1) occurs in the heterogeneous regions : g(H) 

28. Basic media Competitions for e — aq :  H + OH —  e - aq + H 2 O 2.3 × 10 7 M -1 s -1  H + H  H 2 etc… Conclusion : It remains the same in basic media. OH gives birth to its basic form O -.

29. Yields Radiolytic yields (µmol J -1 ) for water irradiated with  rays or with electrons having energies between 1-20 MeV.

30. Conclusion The values of the yields result:  From the competition between diffusion- recombination  From the distribution of heterogeneous zones (LET). low LET (X, , accelerated electrons): [free radicals] > [molecular products] high LET: the opposite. At high LET: novel species (HO 2 ) appear.

31. Instrumentation and Example Linear Accelerator (Linac) at Notre Dame –8 MeV pulsed electrons –Down to 2 ns pulse widths Example of Lycosyme using Br 2  to selectively produce Trp  and TyrO  and to watch the repair of Trp  by TyrOH residues

32. Linac

33. Linac with Detection System

34. Kinetic Traces on Scope

35. Linac’s Computer Screen

36. Pulse radiolysis electron accelerator electron pulse computer time light absorption Reaction kinetics pulse monochromator light detector light source lenses digital oscilloscope shutter quartz cell

37. Yield of Radicals as a Function of Scavenger Concentration The Schuler formula gives the yield of secondary radicals formed from the reaction of hydroxyl radicals and the scavenger S at concentration [S]  OH + S  secondary radical k S

38. Creation of oxidizing radicals N 2 O + e aq   N 2 + O  O  + H 2 O   OH + OH   OH + Br   OH  + Br  at [Br  ] = 100 mM, bromide will scavenge most of the hydroxyl radicals [lysozyme] = 70  M and cannot compete with Br  for hydroxyl radicals Br  + Br  Br 2   

39. Redox Potential of Br 2  is only sufficient to oxidize TrpH and TyrOH An example is to use Br 2  to selectively oxidize tryptophan and tyrosine residues in proteins. Tryptophan and tyrosine are easily oxidized amino acid residues. However, Br 2  cannot oxidize most amino acids but  OH can.

40. Reaction Scheme Br 2  + TrpH(lysozyme)  Trp  (lysozyme) + 2Br  Trp  (lysozyme) + TyrOH(lysozyme)  TrpH(lysozyme) + TyrO  (lysozyme) max = 360 nm max = 510 nm max = 406 nm

41. Repair of Tryptophan Radical