HBr Addition to Alkenes and its “Regiochemistry” Chemistry 125: Lecture 40 January 14, 2011 Reactivity-Selectivity “Principle” HBr Addition to Alkenes and its “Regiochemistry” Rates of Chain Reactions This For copyright notice see final page of this file

Reactivity/Selectivity “Principle” 13.6 More Selective k2/k1 ~ 4 k2/k1 ~ 35 11.1 G‡  0.8 G‡  4.5 Less Reactive Less Selective 2 -2.1 -4.6 DGCl• 7 13.6 11.1 DGBr• -2.1 1° abstraction 2° abstraction More Reactive -4.6

Reactivity/Selectivity “Principle” 13.6 More Selective How can transition states be more different in energy than products are? k2/k1 ~ 35 11.1 G‡  4.5 G  2.5! Br H •R Br• H R Relative to H-CH3 •CH3 H•••CH3 H CH3CCH3 CH3CCH3 H 7 13.6 11.1 DGBr• ? R+ character can be helpful near the transition state. Br H• R+ Plus a 3rd resonance structure 1° 2° It is irrelevant in RH and R• As we shall see next week, substituting CH3 for H stabilizes cations a LOT.

(We’ll see dramatic examples when we discuss pericyclic reactions.) Sometimes factors involved in stabilizing Transition States can be different from those involved in stabilizing either starting materials or products. In such cases we can’t easily rationalize activation energy changes on the basis of exothermicities. http://commons.wikimedia.org/wiki/Image:Gefahrenzeichen_10.svg (We’ll see dramatic examples when we discuss pericyclic reactions.)

Two Problems for Wednesday (work in groups if you wish) 1. If monochlorination of propane gives n-propyl chloride (43%) and isopropyl chloride (57%), and monochlorination of 2-methylpropane gives t-butyl chloride (36%) and 1-chloro-2-methylpropane (64%), predict the products from monochlorination of 2-methyl-butane. HINT : Consult your textbooks on free-radical halogenation to be sure you are taking the statistics of H numbers into account properly. 2. Solve the puzzle on Slide 18.

Radical-Chain H-X Addition to Alkenes C=C cyclic machinery • C-C X • X C-C X H H-X

Radical-Chain H-X Addition to Alkenes 10 -10 -20 -30 -40 -50  Only HBr works fast enough in both steps. But only HBr works. Why?   146 X-C-C• (+ X-H) 83 83 99 X• + C=C (+ X-H) X-C-C-H + X• 83 99 116 Average Bond Energies F Cl Br I 146 135 281 298 = 17

“Regiospecificity” in Addition H2C CH3 H H-X C CH2 CH3 H X C CH2 CH3 H X for X = any halogen occasionally anti-Markovnikov but only with X = Br Markovnikov “Orientation” (1870) Understood in terms of initial addition of H+ (1930s) (ionic mechanism to be discussed in three weeks) Traced to peroxide catalysis (1933) “Initiator” of radical chain

“Regiospecificity” in Addition H2C CH3 H R-O-O-R 2 R-O• R-O-H •Br H-Br secondary radical “more stable” C •CH2 CH3 H Br C• CH2 CH3 H Br For discussion of radical chain addition see texts (e.g. JF pp. 481-490) Traced to peroxide catalysis (1933) “Initiator” of radical chain

Catalytic Cycle Rate Law R-H X-H k1 [RH] [X•] cyclic machinery • X • R k2 [X2] [R•] R-X X-X k1 [RH] [X•] = k2 [X2] [R•] “this is a real democracy, catalysis” (K. B. Sharpless, 12/3/08)

But the two fluxes (cars/min) must be equal at steady state! “If you get that [barrier] down, the rate goes way up. And if you get them all the same height, you're really rolling.” “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.” (Sharpless, 12/3/08) But the two fluxes (cars/min) must be equal at steady state! E-ZPass Low rate constant High rate constant Cash Low concentration High concentration k [ ] = k [ ]

Catalytic Cycle Rate Law 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. R-H X-H small k1 [RH] [X•] ~1 Rate  [RH]? cyclic machinery ( Rate  k1 ) X R • • ~0 Rate  [X2]? ( Rate insensitive to k2 ) Rate  [R• + X•]1 k2 [X2] [R•] R-X X-X Suppose X• is dominant Double [X2] Double [RH] large [R•] [X•] k2 [X2] k1 [RH] = [X•]  [R•]  2 98 1 99 4 96 k1 [RH] [X•] = k2 [X2] [R•] k1 [RH] [X•] grows = 1.96 fold. 2 • 96 1 • 98 k2 [X2] [R•] grows ~ fold; not at all ! 2 • 1 1 • 2 Assuming [R•] + [X•] = Const

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

Both Radical-Molecule “Propagation” Reactions of the Chain “Machine” must be Fast, or X• and R• will find partners. “Termination”

Kinetic Order in Initiator 1/2 Rate  [RO-OR] ? ki Initiation : RO-OR 2 RO• Rate of forming radicals  [RO-OR] kt Termination : 2 R’• R’-R’ Rate of destroying radicals  [R’•]2 at Steady State : [R’•]2  [RO-OR] 1/2

An Ionic Effect in Free-Radical Substitution Termination by H• transfer (i-Pr)2NH i-Pr -N -C(CH3)2 • H H R-H (i-Pr)2N-H slow =C(CH3)2 ~100 kcal/mole ~92 kcal/mole (i-Pr)2N add H • R 2 Cl-Cl 58 Br-Br 46 R-Br 73 H-Cl 103 H-Br 87.5 Charge keeps dominant radicals apart; inhibits termination. ~46 kcal/mole dominant ~85 kcal/mole (i-Pr)2N-Cl R-Cl

Chlorination Selectivity 20 40 60 80 100 Chlorination Selectivity Isomer Percent 50% H2SO4 • ROH half protonated 60% H2SO4 • 70% H2SO4 • ROH fully protonated H + HO CH2 CH3 ? i-Pr HN +• i-Pr HN +• N.C. Deno, et al. (1971)

30% H2SO4 gave a completely different product: CH2 CH3 C O 92% yield! which probably was formed from aldehyde CH2 CH3 C O H mostly formed by HOMO/LUMO (non-radical) “oxidation” HO CH2 CH3 ? Puzzle: Propose mechanisms to form aldehyde involving both substitution (radical or HOMO/LUMO) and elimination (reaction with base) i-Pr N + H Cl CH3 HN +• N.C. Deno, et al. (1971)

End of Lecture 40 Jan. 14, 2011 Copyright © J. M. McBride 2011. 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). 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