Transition States in Multistep KineticsWhen we determine ΔG‡, what are we actually measuring?RTGhTkk/Brxn‡eΔ−=For simple kinetics, answer is easy.ΔG‡Transition States in Multistep KineticsWhen we determine ΔG‡, what are we actually measuring?RTGhTkk/Brxn‡eΔ−=But what about multistep reactions? Where do we draw ΔG‡?ΔG‡Transition States in Multistep KineticsRTGhTkk/Brxn‡eΔ−=If we assume pre-equilibrium between R and I,]R[]I[]P[IRIIII‡‡→→→==∂∂Kkkt⎟⎟⎠⎞⎜⎜⎝⎛=→]R[]I[IRKKR→IkI→I‡I‡()]R[ee//BIR‡IIRTGRTGhTk→→Δ−Δ−⎟⎠⎞⎜⎝⎛=0{}]R[e/B‡IIIR⎟⎠⎞⎜⎝⎛=→→Δ+Δ− RTGGhTk0Eyring (TST) KeqTransition States in Multistep KineticsRTGhTkk/Brxn‡eΔ−=If we assume pre-equilibrium between R and I,{}⎟⎠⎞⎜⎝⎛=→→Δ+Δ− RTGGhTkk/Bobs‡IIIRe0‡IIIR→→Δ+Δ GG0Conclusions: “Rate-determining”transition state is the one that’s highest in energy.Reaction barrier is between starting material and highest-energy transition state.Transition States in Multistep KineticsWhen we determine ΔG‡, what are we actually measuring?RTGhTkk/Brxn‡eΔ−=Reaction barrier is between starting material and highest-energy transition state.ΔG‡ΔGtot‡ΔGtot‡ΔGtot‡Transition States in Multistep KineticsGood way to think about it: Imagine I is your new starting material.Exception to “Reaction barrier is between starting material and highest-energy transition state.”:Uncoupled forward reactions.Ereaction coordinateRIPk1k2What is the rate of this reaction?Really, this is two reactions.First reaction is complete before second one even gets started.So, kobs≈ k2.ΔGobs‡Seeing Trends in Kinetics:The Hammond PostulateIf ΔGO>> 0,thenΔG‡large,“late” TS.In general, endothermic reactions have transition states that resemble products…Ground-state thermodynamics do correlate with transition-state kinetics.Hammond PostulateIf ΔGO<< 0,thenΔG‡small,“early” TS.Ground-state thermodynamics do correlate with transition-state kinetics.…and exothermic reactions have transition states that resemble reactants.Hammond PostulateVariation in ΔGOshows up in ΔG‡.(So, the less exothermic the reaction is, the later and higher-energy the transition state.)Only holds for series of reactions with same mechanism.Hammond Postulate Example:E2 EliminationPhPhBBHCl+++DMSOHClPhPhB = KOtBu 18.8NaOEt 19.3NaOPh 20.2ΔH‡(kcal/mol)decreasing basicity,decreasing [BH]/[B-],less exothermic,later,higher-energy TS.Hammond Postulate Example:E2 EliminationPhPhEtO EtOH X+++DMSOHXPhPhX = OBz 20.6OAc 19.8Cl 19.3ΔH‡(kcal/mol)increasing leaving-group quality,more exothermic,earlier,lower-energy TS.O’Ferrall Plots: Visualizing HammondCan model reaction coordinate asinverse parabolic potential well.O’Ferrall Plots: Visualizing HammondWhat happens when we make Xa better leaving group?Model by tilting entire plotforward.O’Ferrall Plots: Visualizing HammondWhat happens when we make Xa better leaving group?Model by tilting entire plotforward.O’Ferrall Plots: Visualizing HammondResult:ΔH‡decreases (as reaction becomes more exothermic)ΔH‡O’Ferrall Plots: Visualizing HammondResult: Transition state occurs “earlier”, looks more like starting materials.‡The Curtin-Hammett PrincipleA1A2B1B2Given an equilibrium that favors starting material A2over A1, which product is favored: B1or B2?Question:ΔG21OKeq=[A2][A1]−ΔG21ORT= e()I.e., ifthen what is∂[B2]∂t∂[B1]∂t?rate2rate1=(ΔG21Ois a negative number here)k2k1The Curtin-Hammett PrincipleA1A2B1B2ΔG21OGiven an equilibrium that favors starting material A2over A1, which product is favored: B1or B2?Question:Depends solely on relative transition state energies, noton equilibrium.Answer:ΔG1‡ΔG2‡k2k1The Curtin-Hammett PrincipleA1A2B1B2ΔG21OΔG1‡ΔG2‡∂[B2]∂t∂[B1]∂t=k2[A2]k1[A1]=k2Keqk1k2k1The Curtin-Hammett PrincipleA1A2B1B2ΔG21OΔG1‡ΔG2‡∂[B2]∂t∂[B1]∂t=k2[A2]k1[A1]=k2Keqk1−ΔG21ORTe()−ΔG2‡RTe()h()kT=−ΔG1‡RTe()h()kTEyring (TST)Eyring (TST) Keqk2k1The Curtin-Hammett PrincipleA1A2B1B2ΔG21OΔG1‡ΔG2‡∂[B2]∂t∂[B1]∂t=−ΔG21O− ΔG2‡+ ΔG1‡RTe()k2k1The Curtin-Hammett PrincipleA1A2B1B2ΔG21OΔG1‡ΔG2‡∂[B2]∂t∂[B1]∂t=−ΔG21O− ΔG2‡+ ΔG1‡RTe()GTS2- GTS1GTS2- GTS1= ΔG2‡− ΔG1‡+ ΔG21O(because ΔG21Owritten as negative number)∂[B2]∂t∂[B1]∂t=−(GTS2- GTS1)RTe()k2k1The Curtin-Hammett PrincipleA1A2B1B2ΔG21OΔG1‡ΔG2‡GTS2- GTS1∂[B2]∂t∂[B1]∂t=−(GTS2- GTS1)RTe()So, relative reaction rates depend only on relative transition-state energies, and not on starting-material ground-state