Chemistry 125: Lecture 40 January 15, 2010 Predicting Rate Constants, and Reactivity - Selectivity Relation. Rates of Chain Reactions. This For copyright notice see final page of this file

R-X BDE : Alkyl Variation in Detail X R Me-R Et-R i Pr-R t Bu-R H-R If this trend is due to radical stabilization by substitution, other X-R bond strengths should show the same trend. BDE relative to CH 3 X (kcal/mole) t Bu-R Molecular Mechanics Strain Energies in Starting Material

t-Bu van der Waals Energy26.9kcal/mole5.2 “Idealized” Bond Lengths and Angles “Relaxed” Structure Crunch! “steric hindrance”

van der Waals Energy drop by 16.8 to 5.2 kcal/mole

comes at the expense of bond stretching and bending. van der Waals Energy drop by 16.8 to 5.2 kcal/mole

1.52Å  1.57Å 109.5° 112.3°  comes at the expense of bond stretching and bending. (which total 4.8 kcal/mole) van der Waals Energy drop: 16.8 to 5.2 kcal/mole Resultant Total Strain (incl. 2.2 torsion) 12.2 kcal/mole

R-X BDE : Alkyl Variation in Detail X R For X = alkyl almost all of the Me to t-Bu change is due to strain energy in the starting material. Me-R Et-R i Pr-R t Bu-R H-R If this trend is due to radical stabilization by substitution, other X-R bond strengths should show the same trend. BDE relative to CH 3 X (kcal/mole) t Bu-R Me-R molecular mechanics strain energies kcal/mole diff. in initial Strain  whole 8.9 kcal/mole diff. in BDE Ditto 1.5  2.6 But not for H-R

R-X BDE : Corrected for R-X Strain X R Me-R Et-R i Pr-R t Bu-R BDE corr relative to CH 3 X (kcal/mole) Alternative to hypothesis of radical stabilization by substitution Intrinsic C-C bond strength (corrected for strain) is practically insensitive to substitution.

R-X BDE : Corrected for R-X Strain X R Me-R Et-R i Pr-R t Bu-R H-R Alternative to hypothesis of radical stabilization by substitution But C-H bonds are weakened by alkylation of the carbon. Intrinsic C-C bond strength (corrected for strain) is practically insensitive to substitution. BDE corr relative to CH 3 X (kcal/mole)

R-X BDE : Corrected for R-X Strain X R Intrinsic C-C bond strength (corrected for strain) is practically insensitive to substitution. Me-R Et-R i Pr-R t Bu-R H-R But C-H bonds are weakened by alkylation of the carbon. Alternative to hypothesis of radical stabilization by substitution While C-Cl and C-Br are strengthened by alkylation of the carbon. Cl-R Br-R I -R BDE corr relative to CH 3 X (kcal/mole) No one I know of understands this, but the textbooks are clearly wrong.

Can we use energies of stable structures that we “understand” to infer the energies of other structures ( e.g. transition states ), so as to predict reactivity? How can we predict activation energy? Might exothermic reactions be faster than endothermic ones?

How can we predict activation energy? This is no easy task a priori, especially when interaction with solvent is important. But often one can say something sensible about relative values of E a (or  G ‡ ). Compared to What?

The Hammond Postulate (1955) George S. Hammond ( ) “If two states, as for example, a transition state and an unstable interme- diate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of the molecular structures.” This stimulated organic chemists to think about transition states and try to generalize plausibly about reaction coordinates. by permission, E. Menger

The more exothermic a reaction - the more similar the transition state to starting material (in both energy and structure) Starting Material Product endo product At least among one-step reactions that are closely analogous, such as X + H-R. . X-H + R…..

The more exothermic a reaction - the more similar the transition state to starting material (in both energy and structure) There is “likely” a continuum between starting material and product with respect to the factors that influence stability. endo product An effect mostly influencing the energy of the product of an endothermic reaction should have a similar (slightly smaller) influence on its (late) transition state. Rates of slower reactions should be more sensitive to overall  G! An effect mostly influencing the energy of the product of an exothermic reaction should have a small influence on its (early) transition state. e.g. resonance stabilization H PhCH 2 H

Reactivity/Selectivity “Principle” More Reactive Less Reactive More Selective Less Selective

‡ GG 1  G  Reactivity ‡ k  e -  G /RT  /4  G /sec ‡ ‡  G  Relative Reactivity ‡ k 1  e -  G /RT ‡ 1 k 2  e -  G /RT ‡ 2 k 1 /k 2 = e -  G /RT  10 -3/4  G ‡ ‡ ‡ GG 2 ‡ GG or  G  -4/3 (log(k) -13) ‡ or  G  -4/3 log(k 1 /k 2 ) ‡ Rates converge with increasing T Increase discrimination by lowering T !

Reactivity & Selectivity + X H 3 C-CH 2 -CH 3 + HX H 3 C-CH 2 CH 2 + HX H 3 C-CH-CH 3 43% X = Cl 57% 8% X = Br 92% ‡  G  4/3 log(4) = 0.8 kcal/mole ‡  G  9/3 log(35) = °C ( N.B. : Br shows greater selectivity despite increased T ) k 1°  6k 2°  2 Note correction for number of primary & secondary H atoms.  k 2° /k 1° = 4  k 2° /k 1° = 35 (  3  14) n-Pr-Xi-Pr-X

° F Cl Br I ” Reactivity & Selectivity 2° ” X + HX H 2 C-CH 2 CH 3 H 3 C-CH 2 -CH 3 H 3 C-CH-CH 3

Reactivity/Selectivity “Principle” More Reactive Less Reactive More Selective Less Selective Cl Br 1° abstraction 2° abstraction k 2 /k 1 ~ 4 k 2 /k 1 ~ 35  G ‡  4.5  G ‡  0.8

Reactivity/Selectivity “Principle” Less Reactive More Selective Br 1° 2° k 2 /k 1 ~ 35  G ‡  4.5 How can transition states be more different than products are? Br H R Br – H R + Maybe polar character helps transition states. (2° cations are much more stable than 1°)  G  2.5!

Sometimes factors involved in stabilizing Transition States can be different from those involved in stabilizing either starting materials or products.

Chain H-X Addition to Alkene C=C H-X X C-C X XH cyclic machinery

X + C=C (+ X-H) X-C-C-H + X F Cl Br I X-C-C (+ X-H)    Chain H-X Addition to Alkene 83 Only HBr works fast enough in both steps. Average Bond Energies = 17 But only HBr works. Why?

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