1. Chemistry 125: Lecture 41 January 20, 2010 Rates of Chain Reactions and Electronegativity This For copyright notice see final page of this file

2. “Regiospecificity” in Addition CH2CH2C CH 3 H H-X CCH 2 CH 3 H X H CCH 2 CH 3 H H X for X = any halogen Markovnikov “Orientation” (1870) occasionally anti-Markovnikov but only with X = Br Traced to peroxide catalysis (1933) “Initiator” of radical chain Understood in terms of initial addition of H + (1930s) (ionic mechanism to be discussed later)

3. “Regiospecificity” in Addition CH2CH2C CH 3 H H-Br CCH 2 CH 3 H Br CCH 2 CH 3 H Br Traced to peroxide catalysis (1933) “Initiator” of radical chain R-O-O-R 2 R-O R-O-H Br secondary radical “more stable” [for discussion of radical chain addition see text pp. 481-490]

4. R Catalytic Cycle Rate Law X-HR-H X-X R-X X cyclic machinery k 1 [RH] [X] k 2 [X 2 ] [R] “this is a real democracy, catalysis” (K. B. Sharpless, 12/3/08) k 1 [RH] [X] = k 2 [X 2 ] [R]

5. E-ZPass Cash Low rate constant High rate constant High concentration Low concentration “If there's a slow step, there's 99.9% of the titaniums that you need, stuck before this one mountain, that goes way up like Mount Everest.” “If you get that [barrier] down, the rate goes way up. And if you get them all the same height, you're really rolling.” But the two fluxes (cars/min) must be equal at steady state! k [ ] = k [ ] (Sharpless, 12/3/08)

6. k 1 [RH] [X] grows = 1.96 fold. 2 96 1 98 Catalytic Cycle Rate Law X-HR-H X-X R-X X R cyclic machinery k 1 [RH] [X] k 2 [X 2 ] [R] k 1 [RH] [X] = k 2 [X 2 ] [R] [R] [X]k 2 [X 2 ] k 1 [RH] = 2 98 Rate  [RH] ? 4 96 1 99 Rate  [X 2 ] ? Rate  [R + X] 1 ~1 ~0 Suppose X is dominant Double [RH] Double [X 2 ] [X]  [R]  small large ( Rate  k 1 ) ( Rate insensitive to k 2 ) Assuming [R] + [X] = Const Summary: When one step in a cycle is much slower than the others, the rate of cycling is pretty insensitive to the rate constant and concentration of incoming reagent for the fast step(s), because concentration of the minor form of cycling reagent adjusts to compensate. Fractional changes in concen- tration of the dominant radical are much more modest. k 2 [X 2 ] [R] grows ~ fold; not at all ! 2 1 1 2

7. and Termination of Radical Chains Typically involves breaking a weak bond with light, heat, or e - ~30 kcal/mole Cl-Cl h 2 Cl O-O  2 O HOOH e - HO - OH Initiation

8. Both Radical-Molecule “Propagation” Reactions of the Chain “Machine” must be Fast, or X and R will find partners. “Termination” (END OF THE MACHINE)

9. Kinetic Order in Initiator Rate  [RO-OR] ? 1/2 at Steady State : [R’] 2  [RO-OR] Initiation : RO-OR 2 RO Termination : 2 R’ R’-R’ Rate of forming radicals  [RO-OR] Rate of destroying radicals  [R’] 2 1/2 kiki ktkt

10. dominant R-H H H H An Ionic Effect in Free-Radical Substitution R R-Cl (i-Pr) 2 N-H (i-Pr) 2 N-Cl (i-Pr) 2 N 2 (i-Pr) 2 NH i-Pr -N add H Termination by H transfer ~46 kcal/mole =C(CH 3 ) 2 ~92 kcal/mole Charge keeps dominant radicals apart; inhibits termination. ~85 kcal/mole ~100 kcal/mole -C(CH 3 ) 2 H slow

11. i-Pr HN + HO CH 2 CH 3 CH 2 70% H 2 SO 4 60% H 2 SO 4 50% H 2 SO 4 20 40 60 80 100 0 Chlorination Selectivity Isomer Percent N.C. Deno, et al. (1971) i-Pr HN + H + ROH half protonated ROH fully protonated ?

12. HO CH 2 CH 3 CH 2 30% H 2 SO 4 gave a completely different product: N.C. Deno, et al. (1971) CH 3 HN + RO CH 2 CH 3 C CH 2 O 92% yield! CH 2 CH 3 C CH 2 O H which probably was formed from aldehyde mostly formed by HOMO/LUMO (non-radical) “oxidation” i-Pr N + H Cl Puzzle: Propose mechanisms to form aldehyde involving both substitution ( radical or HOMO/LUMO) and elimination (reaction with base) (Hint: a little different from Wiki 227) ?

13. 3. Consider the chlorination reaction : i-Pr 2 NCl + RH  i-Pr 2 NH + RCl and these approximate bond dissociation energies (kcal/mole): N-Cl (46), R-Cl (85), N-H (92), R-H (100). G.(3 min) Why should the BDEs of N-Cl and R-Cl be so different, when those for N-H and R-H are so similar? From 2009 Exam

14. Could e-Pair Repulsion Explain BDEs? 1.79 Å 1.73 Å 1.40 Å 1.53 Å H 3 C CH 3 HO OHH 2 N Cl H 3 C Cl 84 BDE 90 kcal/mole 51 46 Total e-Density Contour (a 0 -3 ) 0.002 0.01 0.05 0.25 Drawing proton(s) away from nucleus removes OMO-OMO e-density from overlap region. isoelectronic

15. from Wikipedia Lone pair repulsion seems a plausible explanation for weakening O-O vs. C-C or N-Cl vs. C-Cl. But might electronegativity help explain stronger C-Cl than N-Cl ? C + Cl -

16. Which Bond is Stronger N-Cl or C-Cl? Cl N Electron Energy separate C Compared to What? N-Cl stronger if forming Ions (N + Cl - ) C-Cl stronger if forming Radicals (Cl C) together mismatch aids Heterolysis mismatch hinders Homolysis BDE

17. “Electronegativity” and Bond Strength First use in English (O.E.D.) 1837 J. D. Dana Syst. Mineral. 82 When chemistry has so far advanced, that the relative electro-negativity, (if I may so call it,) or electro-positivity, of the several elements, is fully known,..we shall probably be able to construct a natural arrangement of minerals on chemical principles. J. D. Dana 1813-1895 Silliman’s son-in-law Dana House 1849

18. “Electronegativity” and Bond Strength

19. H-X “normal” (average of H-H and X-X) H + X - 1932 Pauling was pushing resonance. Why not use  to measure resonance stabilization? in Pauling’s theory (electron volt = 23.06 kcal/mole)  Actually F-F is 38 kcal/mole, when Pauling thought 63.6. Observed or A:B = (A:A * B:B) 1/2

20. “Normal”     BDE (obs) (units of electron volts) 1932 Pauling was pushing resonance. ~ additive nothing special O-OO-F F-F actually 1.65

21. Relative to H & F 0.48 Relative to O Relative to C Relative to H 1.00 1.48 0.58 1932 Pauling was pushing resonance. Is it surprising that bond strength should correlate with Pauling electronegativity differences? No, his  P  scale was defined by differences in bond strength.

22. from Wikipedia Mulliken Electronegativity (1934) average of Atomic Ionization Potential and Electron Affinity Pauling Electronegativity We expect energy-mismatch to strengthen bonds, so crude correlation of  P with IP and EA is hardly surprising.

23. End of Lecture 41 Jan. 20, 2010 Copyright © J. M. McBride 2010. Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0