( ) Thermodynamic selectivity (reversible reactions)

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1 II. Reactivity A) General Principles Fundamental Equations G o Ea / RT = "RT ln K eq k = Ae 1) Kinetics vs Thermodynamics Kinetic selectivity (irreversible reactions) [ P 1 ] [ ] = ln k 1 P 2 ( ) = " #G 1 " #G 2 k 2 RT Thermodynamic selectivity (reversible reactions) [ P 1 ] [ ] = lnk eq = " #Go RT P 2 Temperature Enhancement of Selectivity Difference in Product Energies with Corresponding K eq Values ΔG K (25 o C) K (-78 o C)* 0.5 kcal 2.33 (70:30) 3.63 (78:22) 1.0 kcal 5.4 (85:15) 13.2 (93:7) 2 kcal 29.3 (97:3) 174 (>99:<1) 3 kcal 134 (99:1) 2300 (>99.9:<0.1) * -78 o C = C 2 (s) / acetone 81

2 2) Rate Limiting Step k A 1 k B 2 k C 3 D The rate of formation of product D will be equal to the rate of the slowest step. In this example, compare k 1 to k 2 to k 3. The rate of formation of D, d[d]/dt is equated to the smallest of those three rate constants multiplied by a concentration term (i.e., if k 2 << k 1, k 3, d[d]/dt = d[c]/dt = k 2 [B]). Interlude: Kinetic vs. Thermodynamic product control Thermodynamic Product Control -occurs when there is an equilibrium between starting material and products -a single irreversible step precludes thermodynamic product control P' S P G o P' S G o P P' P K = [ P] [ P' ] = e "G o P' "GP o ( ) RT depends only on the relative energies of P & P Example: NaMe Me Na (more stable than Na ) 82

3 Kinetic Product Control k 2 k 1 P' S P G 2 G 1 S P' P [ P] [ P' ] = k 1[ S] k 2 [ S] = k 1 k 1 = e ""G RT G = G k 2 k 1 " G2 2 Example: Li Li Li + - Ni-Pr 2 (less stable than ) K eq k 1 P' S P Le Chatelier s Principle -P exclusive product Example: RC 2 Me - Me - RC 2 Me - RC 2-83

4 3) Steady State kinetics A k 1 k -1 k 2 B C assume [B] is very small and d[b] dt 0 (true for reactive B) What is d[c] dt? d[c] dt = k 2 [B] d[b] dt = k 1 [A] k -1 [B]- k 2 [B] = 0 k 1 [A] = k -1 [B] + k 2 [B] k 1 [A] = [B]( k -1 + k 2 ) [B] = k 1 [A] ( ) k -1 + k 2 When k 2 << k -1 d[c] dt d[c] dt " k = k 1 [A]% 2 $ ' = k 2 K eq [A] # & k -1 " k = k 1 [A] % 2 $ ' #( k -1 + k 2 )& When k -1 << k 2 d[c] dt = k 1 [A] 84

5 4) ammond Postulate Me N 2 Me N 2 Me N N 2 Me N 2 minor MAJR Me Me Me Me N 2 N 2 N 2 N 2 loss of Me N 2 If this reaction is based on kinetic control, why base the argument on the stability of the intermediate (a thermodynamic principle)? Because the loss of + is exothermic TS TS SM PDT PDT SM Exothermic Endothermic ammond Postulate: if a reaction is exothermic, the transition state of the reaction will resemble the starting material but if the reaction is endothermic the transition state will resemble the product. 85

6 5) Curtin-ammett Principle consider: Mg Mg PhC Ph Mg Major product Ph Et 2 Et 2 Mg vs Mg Et 2 Ph Et 2 C 3 predict kinetic product distribution based on the energy of competing transition states k 2 K eq slow fast slow P' S' S P ow to pick the product distribution? k 1 [ P] ow to relate [ P' ] to energies? Curtin-ammett Principle [ P] [ P' ] = k 1 [ S] k 2 [ S' ] = k 1 K eq k 2 86

7 a) choose a common reference point b) relate ΔG 1 and ΔG2 to common reference point c) [ P] [ P' ] = e ("G 1 "G2 ) RT [ P] [ P' ] depends only on barrier heights Curtin-ammett principle: competing transition states kinetic product distribution is based on the energy of t.s. 2 t.s. 1 G 1 G 2 P' S' S P Now consider: 2 Why trans-diaxial? 87

8 A B A C-3 attack C-2 attack B C-3 attack C-2 attack Principle of Microscopic Reversibility Forward and reverse reactions share a common transition state 88

9 Experimental Methods For Kinetic Analysis 1) Reaction rders Working out the kinetic expression for product formation can often provide valuable information about the details of the mechanism. Some examples: Et S N 1 Et v = k 1 [A] first order Cl NaCN S N 2 CN v = k 2 [A][B] second order What can we learn from this? 89

10 + CN CN ex: -catalyzed by addition of R 3 N (Lapworth, J. Chem. Soc. 1903, 83, 998) Prelog examined rate vs. [R 3 N] found the rate [R 3 N] 2 Two amines present in the transition state (Prelog & Wilhelm, elv. Chim. Acta. 1954, 37, 1634) NR 3 CN CN R 3 N so: amine salts serve as acid catalyst & as counterion to nucleophilic - CN 2) Kinetic Isotope Effects definition: a difference in the rate of reaction based on replacing with D a) Primary Isotope Effect E C- stretch = (n + 1/2 )hν n = 0, 1, 2, 3,... = 1 2" k µ " D = " 1 2 hc # "# D # " " % D k $ kt = e k D ( ) = " 1 2 hc $ 1 " 1 & ( ' % # µ = m 1 m 2 $ & " m 1 + m 2 % & 1.35' # * hc # % k 2kT 1" 1 & () - # , $ 1.35'./ % e = e T ) & $ ( ' k D " k ( 3000) % # $ e 298 &' = 6.5 k D note that 6.5 is the upper limit for a primary isotope effect 90

11 E C C D Vibration Coordinate reaction k /k D ( o C) ( 3 C) 2 C C(C 3 ) 2 * + ( 3 C) 2 C * C(C 3 ) 2 + *- 6.1 (25 o C) PhC 2 * + Ph C 2 + * 4.0 (191 o C) b) Secondary Isotope Effect change in C- bond stretching frequency as a consequence of rehybridization at carbon " k # $ = e k D [ ] 1 2 u u D ( u r u rd ) % & ' where u = h kt using ν D = ν / 1.35 ( ) # k = e T " $ % D k D for sp 3 sp 2 (normal) maximum value is 1.41 (typical values are 1.15 to 1.25) for sp 2 sp 3 (inverse) values are less than 1 with a minimum of 0.71 (typical values are 0.8 to 0.9) & '( 91

12 reaction k /k D ( o C) * Me * + CN Me CN 0.73 (25) * 2 CC 2 * + * 2 C C 2 * 1.37 (50) Example of mechanistic implications: C(D) (D) k /k D = 12.2 => concerted (symmetric) TS 3) Van t off Plots ΔG = Δ - TΔS from transition state theory recall that k 1 = kt "G h e RT G # = "RTln hk 1% $ kt & so, plot ln(rate) vs. (temperature) -1 for a reaction: ln(k 1 /T) (T) -1 ln k 1# " T $ = %& 1 # R " T$ + C' 92

13 S = T + Rln " hk 1$ # kt % " R = slope of the line Significance: large and positive: bond breaking (and little bond formation) in TS, formation of reactive intermediate small: equal amounts of bond breaking, formation (often concerted) S large and positive: cyclization reactions, cycloadditions, addition reactions S negative: eliminations and fragmentations (forming 2 molecules from 1) Example: S 2 + S 2 = 24.7 kcal mol -1 S = -5 kcal mol -1 K -1 concerted reaction 93

14 4) Substituent Effects & Linear Free Energy Relationships ex: C 2 K a C 2 + electron donating K a decreases electron withdrawing K a increases ammett quantified this effect with the parameters σ meta and σ para : " K = log$ $ K # a o a % ' ' & ""G o ammett, JACS 1937, 59, 96 C 2 K a 1) C 2 + C 2 K a C 2 + 2) In each case, use a series of substituents (Cl, Me, N 2, C 3 ) and plot log K a $ # & vs σ " K o a % 1) log K a $ # & = 0.489σ " K o a % 2) log K a $ # & = 0.212σ " K o a % generalize this as: log K # " K o $ = %& 94

15 ammett used this to study rates: let s examine: log# " k $ & = '( k o % Et 2 slow Et -Et fast Et very fast plot log# " k $ & vs σ ρ = 2.61 k o % electron-withdrawing substituents increase rate by stabilizing the anionic transition state of the rate-determining first step; also, EWG s stabilize the transition state more strongly than benzoate (ρ > 1) CAVEAT - no guarantee that log# " k $ & is linear k o % ex: K + Why is plot non-linear? N N - important resonance interaction (unaccounted for by σ) 95

16 New parameters: σ- : K a + " $ # log K a ' & ) o % K a ( σ + : 3 C C C % Cl 2 C 3 90% 2 C3 C 3 + # " log% k & ( $ k o ' application: ex: 2 Fe 3 slow fast Plot log# " k $ & vs. σ, σ + k o %.C. own & Y. kamoto, JACS 1958, 80,

17 Deviations from linearity may give mechanistic information: 2 N N N 2 N N2 N N N N 2 semicarbazone log(k/k o ) = 3.5 change in the rate determining step = " Noyce, JC 1958, 23,