Chemistry 125 Lecture 40 January 14 2011 ReactivitySelectivity

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” k 2/k 1 ~ 4 G‡ 0. 8 More Selective k 2/k 1 ~ 35 G‡ 4. 5 13. 6 11. 1 Less Reactive Less Selective -2. 1 More Reactive -4. 6 2 7 -2. 1 1° abstraction 13. 6 -4. 6 2° abstraction 11. 1 GCl • GBr •

Reactivity/Selectivity “Principle” How can transition states be more different in energy than products are? Br • Br Plus a 3 rd resonance structure Br H R H • R More Selective k 2/k 1 ~ 35 G‡ 4. 5 Relative to H-CH 3 H CH 3 CCH 3 13. 6 11. 1 G 2. 5! H • • • CH 3 H CH 3 CCH 3 • CH 3 CCH 3 7 13. 6 + H • R It is irrelevant in RH 2°and R • 11. 1 As we shall see next week, substituting GBr • CH for H stabilizes cations a LOT. 3 R+ character can be helpful near 1° the transition state. ?

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. (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 -methylbutane. 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 X • XH C-C cyclic machinery X • C-C H-X

Radical-Chain H-X Addition to Alkenes 10 0 Only HBr works fast enough in both steps. -10 -20 -30 -40 -50 But only HBr works. Why? 83 146 X • + C=C (+ X-H) X-C-C • (+ X-H) 83 146 99 135 116 281 298 = 17 83 99 X-C-C-H + X • Average Bond Energies F Cl Br I

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

“Regiospecificity” in Addition CH 3 R-O-O-R H 2 C 2 R-O-H R-O • • Br H-Br C H CH 3 • CH 2 C Br H CH 3 Br CH 2 C • H secondary radical “more stable” For discussion of radical chain Traced to peroxide catalysis (1933) addition see texts (e. g. JF pp. 481 -490) “Initiator” of radical chain

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

E-ZPass “If you get that [barrier] down, “If there's a slow step, there's 99. 9% of the rate goes way up. And if you the titaniums that you need, stuck before get them all the same height, this one mountain, that goes way up like you're really rolling. ” (Sharpless, 12/3/08) Mount Everest. ” But the two fluxes (cars/min) must be equal at steady state! Low rate constant Cash te a r h ig H t n w a t o s L n n o o c i t tra n e c con High concentration k [ ] = k[ ]

Catalytic Cycle Rate Law Summary: R-H small X-H 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 concentration of the dominant radical are much more modest. X • k 1 [RH] [X • ] ~1 Rate [RH]? cyclic machinery R • Rate [R • + X • ]1 X-X large k 1 [RH] [X • ] = k 2 [X 2] [R • ] [X • ] [R • ] Rate [X 2]? ~0 ( Rate insensitive to k 2 ) k 2 [X 2] [R • ] R-X ( Rate k 1 ) = k 2 [X 2] k 1 [RH] 2 • 196 fold; not at all ! k 2 k 1[X[RH] ] [R • ] grows ~ [X • ] grows 2 1 • 298 = 1. 96 fold. Suppose Double X • is [X 2] [RH] dominant [X • ] 98 [R • ] 2 99 1 96 4 Assuming [R • ] + [X • ] = Const

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

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

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

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

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

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

End of Lecture 40 Jan. 14, 2011 Copyright © J. M. Mc. Bride 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-Non. Commercial-Share. Alike 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. Mc. Bride, Chem 125. License: Creative Commons BY-NC-SA 3. 0