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Química Orgânica Física · Kinetics and energy surfaces

A cinética como teste do mecanismo

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O que você vai aprender

  • Derive a rate law from a proposed mechanism with the rate-determining-step, steady-state and pre-equilibrium approximations, and read the composition of the rate-determining transition state from an observed rate law.
  • Use the Eyring equation to convert rate constants into ΔG‡, ΔH‡ and ΔS‡, and interpret the sign and size of ΔS‡ in terms of molecularity and ordering.
  • Apply the Hammond postulate and the energy-span idea to decide which transition state controls the rate and how strongly substituents affect it.
  • Predict product ratios under kinetic control, thermodynamic control and Curtin–Hammett conditions from Gibbs energy differences.
  • Use microscopic reversibility and specific versus general acid catalysis to judge whether a proposed mechanism is consistent with the kinetics.

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Rate laws, order and molecularity

A rate law is measured, never read from the balanced equation. Vary one concentration at a time (initial rates) or flood the reaction with all but one reactant (pseudo-first-order conditions, k_obs = k₂[excess reagent]) and find the exponent of each species. Molecularity belongs to one elementary step and is 1, 2 or, rarely, 3; order belongs to the overall rate law and can be zero, fractional or negative. The rate law tells you the composition of the rate-determining transition state relative to the reactants, not the sequence of events: rate = k[A][B]²/[H⁺] means the highest transition state contains one A, two B and one proton fewer. Species that react only after the rate-determining step never appear. In the SN1 reaction of 2-bromo-2-methylpropane with azide in aqueous acetone, the alkyl bromide disappears at a rate independent of [N₃⁻]: ionization is slow, and azide and water only compete for the cation afterwards.

Steady state and pre-equilibrium

For A + B ⇌ I (k₁, k₋₁) followed by I → P (k₂), setting d[I]/dt ≈ 0 for a reactive intermediate gives rate = k₁k₂[A][B]/(k₋₁ + k₂). Two limits follow. If k₂ ≫ k₋₁, every I goes on to product and rate = k₁[A][B]: the first step is rate-determining. If k₋₁ ≫ k₂, I is in equilibrium with A and B (pre-equilibrium) and rate = Kk₂[A][B] with K = k₁/k₋₁. The steady-state approximation is valid only for intermediates that never build up. The SN1 hydrolysis of chlorodiphenylmethane shows both ideas: the diphenylmethyl cation can return to the chloride or be captured by water, so k_obs = k₁/(1 + α[Cl⁻]) with α = k₋₁/k₂′, and adding chloride slows the reaction (the common-ion, or mass-law, effect). A pre-equilibrium also changes the temperature dependence: the apparent activation energy of Kk₂ is ΔH° + E_a(k₂), which can be negative when the pre-equilibrium is exothermic.

Eyring analysis: ΔG‡, ΔH‡ and ΔS‡

Transition-state theory gives k = (k_B T/h) exp(−ΔG‡/RT), with k_B T/h = 6.21 × 10¹² s⁻¹ at 298.15 K and ΔG‡ = ΔH‡ − TΔS‡. At 25 °C a change of 5.7 kJ/mol in ΔG‡ changes k tenfold; ΔG‡ ≈ 90 kJ/mol corresponds to k ≈ 10⁻³ s⁻¹, and 100 kJ/mol to a half-life of about 10 h. A plot of ln(k/T) against 1/T has slope −ΔH‡/R and gives ΔS‡ from the intercept (for a bimolecular reaction ΔS‡ refers to a 1 M standard state). ΔS‡ reports on order in the transition state: bimolecular associations such as Diels–Alder reactions have ΔS‡ around −100 to −150 J mol⁻¹ K⁻¹; unimolecular cyclic rearrangements such as the Cope (about −58 J mol⁻¹ K⁻¹) lose internal rotations; simple bond homolysis loosens the molecule and has ΔS‡ > 0. In acid-catalysed hydrolyses, an A1 path (unimolecular loss of the leaving group from the protonated substrate) has ΔS‡ near zero, whereas an A2 path (water attacks in the slow step) has ΔS‡ near −100 J mol⁻¹ K⁻¹. Because ΔS‡ is multiplied by T, a path with a higher ΔH‡ but more positive ΔS‡ takes over above the crossover temperature ΔΔH‡/ΔΔS‡.

Hammond postulate and the rate-determining transition state

Two species that lie close in energy and are joined directly have similar structures. An endergonic step therefore has a late, product-like transition state: SN1 ionization resembles the carbocation, so cation-stabilising groups (benzyl < diphenylmethyl < triphenylmethyl; a para-methoxy group) speed it enormously. A strongly exergonic step, such as capture of a carbocation by water, has an early, reactant-like transition state that feels little of the product's stability. In a multistep profile the rate is controlled by the highest transition state measured from the reactant (more exactly, from the lowest preceding state), not by the largest single barrier. If an intermediate returns to the reactant faster than it goes on, it is in pre-equilibrium and the later transition state is rate-determining.

Kinetic control, thermodynamic control and Curtin–Hammett

When products form irreversibly, their ratio equals the ratio of rate constants: [P₁]/[P₂] = exp(ΔΔG‡/RT), 7.5 : 1 for 5.0 kJ/mol at 25 °C. When products interconvert (higher temperature, longer time, reversible steps), the ratio approaches exp(−ΔG°/RT) for the products. Naphthalene sulfonates fastest at C1, because the C1 arenium ion has more contributors that keep a whole benzene ring; naphthalene-1-sulfonic acid is the main product near 80 °C. Sulfonation is reversible, and at 160 °C the less crowded 2-isomer, free of the clash between the sulfonic acid group and the peri hydrogen at C8, accumulates. Low-temperature HCl addition to isoprene and LDA deprotonation of 2-methylcyclohexanone at −78 °C are kinetically controlled in the same sense. When two reacting conformers or isomers A and B interconvert much faster than they react (Curtin–Hammett conditions), the product ratio is [P_B]/[P_A] = K·k_B/k_A, with K = [B]/[A]. This equals exp(ΔΔG‡/RT), where ΔΔG‡ is the Gibbs energy difference between the two product-forming transition states, so the ratio is not set by the conformer populations alone.

Microscopic reversibility and acid catalysis

Under the same conditions a reaction and its reverse follow the same lowest-energy path in opposite directions, with the same intermediates and transition states. A mechanism that hydrates an alkene through a carbocation and dehydrates the alcohol by a concerted E2 in the same medium is impossible, and the slow step of the reverse reaction is the reverse of the slow step of the forward reaction. Acid catalysis comes in two kinetic types. Specific acid catalysis (rate depends only on [H₃O⁺], not on buffer concentration at constant pH) means a fast, reversible protonation precedes the slow step: hydrolysis of benzaldehyde dimethyl acetal is limited by loss of methanol from the O-protonated acetal. General acid catalysis (rate rises with buffer acid at constant pH) means proton transfer itself is in the slow step: hydrolysis of an enol ether such as 1-methoxycyclohexene is limited by proton transfer to carbon. The acid-catalysed bromination of acetone, zero order in Br₂, is limited by enol formation.

What to carry forward

Kinetics never proves a mechanism, but it can rule mechanisms out. Write each candidate mechanism, derive its rate law with the steady-state or pre-equilibrium approximation, and compare it with the measured orders, salt and buffer effects and activation parameters. The rate law gives the composition of the highest transition state, ΔS‡ tells you how ordered it is, the Hammond postulate tells you what it looks like, and product ratios tell you whether transition states (kinetic control, Curtin–Hammett) or products (thermodynamic control) decide the outcome.

Problemas de prática (40)

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Desenhe o mecanismo 6

Steady state and pre-equilibrium

  • Acid-catalysed bromination of acetone: enol formation is rate-determiningDraw the acid-catalysed bromination of acetone through its enol and use the rate law, first order in acetone and H₃O⁺ and zero order in Br₂, to identify enol formation as the rate-determining step.
  • Specific-acid-catalysed hydrolysis of benzaldehyde dimethyl acetalDraw the acid-catalysed hydrolysis of benzaldehyde dimethyl acetal to benzaldehyde and methanol, and derive from the pre-equilibrium protonation why the rate law is k[acetal][H₃O⁺] with no general acid catalysis.
  • General-acid-catalysed hydrolysis of 1-methoxycyclohexeneDraw the hydrolysis of 1-methoxycyclohexene to cyclohexanone and methanol, starting with rate-determining proton transfer to carbon, and explain why this reaction shows general acid catalysis while acetal hydrolysis does not.

Rate laws and the rate-determining step

  • Aminolysis of phenyl acetate with general base catalysis by a second amineDraw the addition of butan-1-amine to phenyl acetate, the removal of the N–H proton of the zwitterionic tetrahedral intermediate by a second amine, and the loss of phenoxide, and connect the second amine to the term second order in amine in the rate law.
  • SN1 hydrolysis of chlorodiphenylmethane and the common-ion effectDraw the SN1 hydrolysis of chlorodiphenylmethane: rate-determining ionization, capture of the diphenylmethyl cation by water and deprotonation of the oxonium ion, and use the reversibility of ionization to explain why added chloride slows the reaction.

Selectivity: control, Curtin–Hammett, reversibility

  • Kinetic sulfonation of naphthalene at C1Draw the sulfonation of naphthalene at C1 through its arenium ion, show the contributor that keeps a benzene ring, and explain why the 1-sulfonate is the kinetic product while the 2-sulfonate is the thermodynamic product at 160 °C.

Preveja o produto 8

Selectivity: control, Curtin–Hammett, reversibility

  • The slow step of acetal formationApply microscopic reversibility to identify the intermediate that is attacked in the rate-determining step of acetal formation.
  • The slow step of the reverse reactionUse microscopic reversibility to identify the rate-determining step of a reverse reaction and predict an isotope-exchange consequence.
  • Quenching a dienolate quicklyPredict the product of kinetically controlled protonation of an extended enolate and contrast it with the thermodynamic product.
  • HCl and isoprene at low temperaturePredict the kinetically controlled adduct of HCl with an unsymmetrical conjugated diene from the more stable allylic cation and the site of fastest capture.
  • Trap the kinetic enolate of 2-methylcyclohexanonePredict which silyl enol ether forms when an unsymmetrical ketone is deprotonated under irreversible, kinetically controlled conditions.
  • Sulfonating naphthalene at 160 °CPredict the thermodynamically controlled product of a reversible aromatic substitution and explain its stability.

Transition states: Eyring and Hammond

  • Which pathway wins on heating?Predict how the balance between two competing pathways changes with temperature from their activation enthalpies and entropies.

Steady state and pre-equilibrium

  • Steady-state rate law for a two-step mechanismDerive the steady-state rate law for a reversible first step followed by an irreversible second step.

Classificar 3

Rate laws and the rate-determining step

  • Order versus molecularityDistinguish the molecularity of an elementary step from the experimental order of an overall rate law.
  • Rate law of an SN1 reaction with azideDeduce a rate law from the effect of each concentration on the initial rate and relate zero order in nucleophile to rate-determining ionization.

Steady state and pre-equilibrium

  • When is the steady-state approximation valid?Recognise the kind of species for which the steady-state approximation is justified.

Comparar 3

Transition states: Eyring and Hammond

  • Rank by entropy of activationRank reactions by ΔS‡ using molecularity and the loss or gain of freedom in the transition state.
  • Which benzylic chloride ionizes fastest?Predict the fastest SN1 substrate among substituted benzyl chlorides from resonance donation to the developing cation.
  • Rank benzylic chlorides by SN1 rateRank phenyl-substituted methyl chlorides by SN1 solvolysis rate from the stability of the carbocations they form.

Calcular 10

Steady state and pre-equilibrium

  • Common-ion retardation of an SN1 solvolysisApply the steady-state rate law for SN1 with ion return to calculate the rate constant in the presence of added chloride.
  • Apparent activation energy after a pre-equilibriumCombine the enthalpy of a pre-equilibrium with the activation energy of the following step to obtain an apparent activation energy, which can be negative.
  • Effective rate constant when neither limit appliesCalculate the effective second-order rate constant from the full steady-state expression when k₋₁ and k₂ are comparable.

Selectivity: control, Curtin–Hammett, reversibility

  • Curtin–Hammett product ratioApply the Curtin–Hammett principle to calculate a product ratio from a conformer equilibrium constant and the ratio of rate constants.
  • Product ratio from ΔΔG‡Calculate a product ratio under kinetic control from the difference in activation Gibbs energies.

Transition states: Eyring and Hammond

  • ΔS‡ of a Diels–Alder reactionCalculate an entropy of activation from a rate constant and ΔH‡ with the Eyring equation and relate its sign to molecularity.
  • ΔG‡ from a rate constantUse the Eyring equation to convert a first-order rate constant into a Gibbs energy of activation.
  • Half-life from ΔG‡Convert a Gibbs energy of activation into a first-order half-life at room temperature with the Eyring equation.

Rate laws and the rate-determining step

  • Second-order rate constant from a pseudo-first-order half-lifeConvert a pseudo-first-order half-life measured with one reagent in large excess into the true second-order rate constant.
  • Rate constant from initial ratesDetermine the orders and the second-order rate constant of an SN2 reaction from initial-rate data.

Encontre o erro 3

Rate laws and the rate-determining step

  • What does a second-order acetal hydrolysis prove?Diagnose an over-interpretation of a rate law that confuses transition-state composition with a collision in the slow step.
  • Which transition state controls the rate?Identify the rate-determining transition state of a two-step profile as the highest point relative to the reactant, not the largest individual barrier.

Selectivity: control, Curtin–Hammett, reversibility

  • Find the flaw in a hydration–dehydration pairDiagnose a pair of forward and reverse mechanisms that violates microscopic reversibility.

Interprete as evidências 4

Transition states: Eyring and Hammond

  • A1 or A2? Read the activation entropyUse the sign and size of ΔS‡ to distinguish unimolecular (A1) from bimolecular (A2) rate-determining steps in acid-catalysed hydrolyses.

Selectivity: control, Curtin–Hammett, reversibility

  • Read a Curtin–Hammett energy profileUse a Gibbs energy profile with rapidly interconverting conformers to predict the major product and its approximate ratio.

Steady state and pre-equilibrium

  • A rate that levels offInterpret saturation of a pseudo-first-order rate constant as a change in rate-determining step or a saturable pre-equilibrium.

Rate laws and the rate-determining step

  • Read the transition-state composition from a rate lawTranslate the exponents of a rate law, including a negative one, into the composition of the rate-determining transition state.

Trabalhe de trás para frente 1

  • A mechanism for third-order brominationWork backwards from a rate law second order in bromine to a mechanism with a pre-equilibrium and bromine-assisted ionization.

Planeje uma síntese 1

  • Design a pseudo-first-order experimentPlan a pseudo-first-order kinetic experiment that yields a second-order rate constant and tests the order in the excess reagent.

Explique por quê 1

  • Why SN1 rates track carbocation stabilityExplain with the Hammond postulate why substituents that stabilise a carbocation speed its formation.

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