Enthalpy-Entropy Compensation• Strategy of adding binding groups to increase ligand affinity works.• But, for additive weak interactions, increased binding enthalpy is compensated by decreased entropy.ΔΔHa= -1.5 kcal/molTΔΔSa= -1.0 kcal/mol…but thermal energy within deeper potential well doesn’t sample as many conformations;so, entropy lost.HOHMultiple Interactions (“Multivalency”):Building Better Ligands & CatalystsPossible reasons for designing/understanding multiple interactions:•Stronger binders¾Higher-affinity pharmaceuticals¾Improved organometallic ligands•More selective binders¾Decrease non-specific drug targeting¾Improved response of analytical detectors•Faster catalysts¾Better-designed synthetic enzymes¾Improved polymerization catalysts•More selective catalysts¾Enhanced enantioselectivity, substrate specificityMultivalency (“Chelate Effect”): ConceptsWhat is the advantage of tethering two ligands together?(sequential step 1 & step 2 above)KAB≠ KAKB;difference illustrates advantage/ disadvantage of connecting ligands.Effect is primarily entropic.Degree of Cooperativity (α)What is the advantage of tethering two ligands together?i.e., How much better or worse does the “B”in A-B bind than you would expect?α =KABKAKBExample:Cd2+OH2OH2H2O OH2H2O OH2Cd2+OH2OH2H2O NH3H2O NH32 NH3Cd2+OH2OH2H2O OH2H2O OH2Cd2+OH2OH2H2OH2NH2O NH2H2N NH2KAKB= (KA)2= 2.5 × 105M-2KAB= 2.4 × 106M-1α = 9.5(because A & B are the same)Degree of Cooperativity (α)What is the advantage of tethering two ligands together?α =KABKAKBExample:Cd2+OH2OH2H2O OH2H2O OH2Cd2+OH2OH2H2O NH3H2O NH32 NH3Cd2+OH2OH2H2O OH2H2O OH2Cd2+OH2OH2H2OH2NH2O NH2H2N NH2ΔG°(kcal/mol)ΔH°(kcal/mol)TΔS°(kcal/mol)-6.75-7.96-7.12-7.03-0.37+0.93Degree of Cooperativity (α)Works well for very small ligands with little conformational freedom.Works less well with larger ligands.Example:Influenza hemagglutinin protein. Binds multiple sialic acid molecules on your cells at once.OOHHOHOAcHNHOCO2OMeOOHHOHOAcHNHOCO2OONHHNOOHHOHOAcHNHOCO2OO33Ksialic acid= 400 M-1Ksialic dimer= 40000 M-1α = 0.25therapeuticflu inhibitorDegree of Cooperativity (α)Works well for very small ligands with little conformational freedom.Works less well with larger ligands.Example:Influenza hemagglutinin protein. Binds multiple sialic acid molecules on your cells at once.OOHHOHOAcHNHOCO2OMeOOHHOHOAcHNHOCO2OONHHNOOHHOHOAcHNHOCO2OO33α = 0.25ΔGsialic acid= -3.3 kcal/molΔGsialic dimer= -5.8 kcal/mol(ΔΔG = +0.8 kcal/mol)Enhancement Factor (β)α is a bit unfair for multivalent pharmaceuticals;Even though ΔG values aren’t additive, divalent molecule still a better ligand for influenza than monovalent. OOHHOHOAcHNHOCO2OMeOOHHOHOAcHNHOCO2OONHHNOOHHOHOAcHNHOCO2OO33ΔGsialic acid= -3.3 kcal/molΔGsialic dimer= -5.8 kcal/mol(ΔΔG = +0.8 kcal/mol)Enhancement Factor (β)OOHHOHOAcHNHOCO2OMeOOHHOHOAcHNHOCO2OONHHNOOHHOHOAcHNHOCO2OO33Ksialic acid= 400 M-1Ksialic dimer= 40000 M-1β = 100β =KA2KAJust a measure of how much better a ligand A2is than A. (Says nothing about contribution of entropy.)Phenomenological KineticsHow do we understand, predict rates of reactions from experimental data?What do we even mean by “rate”? Consider:N2O5→ 2 NO2+ ½ O2Could define rate =d[N2O5]dtd[NO2]dtd[O2]dtor orFor this reaction, by definition from stoichiometry,d[N2O5]dtd[NO2]dtd[O2]dt== 2-12(For every molecule of N2O5consumed, 2 of NO2created, etc…)Understanding Reaction RatesN2O5→ 2 NO2+ ½ O2d[N2O5]dtd[NO2]dtd[O2]dt== 2-12True at any time.Don’t need to know anything about reaction mechanism to say this.Measuring Reaction RatesN2O5→ 2 NO2+ ½ O2Δ[N2O5]Δt(0.012 M) – (0.009 M)(0 min) – (10 min)== -3 × 10-4M/minMeasuring Reaction RatesN2O5→ 2 NO2+ ½ O2Δ[N2O5]ΔtΔ[NO2]ΔtΔ[O2]Δt= -3 × 10-4M/minby the same method,= 6 × 10-4M/min= 1.5 × 10-4M/minUnderstanding Reaction RatesN2O5→ 2 NO2+ ½ O2Δ[N2O5]ΔtΔ[NO2]ΔtΔ[O2]Δt== 2-12True over any time period.Again, don’t need to know anything about reaction mechanism to say this.Phenomenological KineticsMost important question:How does rate depend on reactant concentration?For any reaction, can write a rate law:rate = k[reagent]xk: rate constantx: reaction order• k and x are determined experimentally.• k = f(rxn, temp) and is always a positive number.• x has no necessary relationship to the coefficients of the balanced chemical equation. Is usually an integer or fraction. Can be zero.Method of Initial Ratesrate = k[reagent]xProblem: rate changes, over time, as [reagent] changes.So, if we don’t have a good way of monitoring [reagent], how do we determine k and x?Solution: [reagent] is defined at t = 0, and is effectively constant at beginning of reaction (first few %)So measure multiple rates at t = 0 for different starting concentrations [reagent], solve for kand x.Method of Initial RatesConsider: 2 NO + Br2→ 2 NOBrFor rate = k[NO]x[Br2]y, what are k, x, y?Solution: Vary each concentration independently;solve for variables.8.78 × 10-33.001.5032.93 × 10-31.001.5021.3 × 10-31.001.001{d[Br2]/dt}0(M/min)[Br2]0(M)[NO]0(M)Run #(answer on the board)Consider: 2 NO + Br2→ 2 NOBrFor rate = k[NO]x[Br2]y, what are k, x, y?Solution: Vary each concentration independently;solve for variables.8.78 × 10-33.001.5032.93 × 10-31.001.5021.3 × 10-31.001.001{d[Br2]/dt}0(M/min)[Br2]0(M)[NO]0(M)Run #(answer on the board)Method of Initial Ratesd[Br2]/dt= k[NO]x[Br2]yd[Br2]/dt(run 1)= k[NO](run 1)x[Br2](run 1)yd[Br2]/dt(run 2)= k[NO](run 2)x[Br2](run 2)yd[Br2]/dt(run 1)k[NO](run 1)x[Br2](run 1)yd[Br2]/dt(run 2)k[NO](run 2)x[Br2](run 2)y=For runs 1 and 2, we made [Br2] the same. [Br2](run 1)y= [Br2](run 2)y.Method of Initial Ratesd[Br2]/dt(run 1)[NO](run 1)xd[Br2]/dt(run 2)[NO](run 2)x=1.3 × 10-3M/min2.93 × 10-3M/min=(1.00 M)x(1.50 M)x=( )1.00 M1.50 MxSolution: x = 2.Using the same method for runs 2 and 3,([NO](run 2)x= [NO](run 3)x),y = 1;Plug these x and y into any relation, getk = 1.3 × 10-3M-2min-1.Method of Initial RatesPros:Decomposition or other subsequent reactions play no role in initial rate, so data very reliable.Good for enzyme, polymerization kinetics.Cons:Have to perform multiple reaction runs; each run contributes only one data point.Must start data acquisition at t = 0, the only time concentrations are known.There is a better way: integrated rate