Thermodynamics & Kinetics of Chemical Reactions: Entropy, Free Energy & Transition States, Study notes of Chemistry

Download Thermodynamics & Kinetics of Chemical Reactions: Entropy, Free Energy & Transition States and more Study notes Chemistry in PDF only on Docsity! CHAPTER 5 / page 1 CHAPTER 5: STUDY AND DESCRIPTION OF ORGANIC REACTION MECHANISMS THERMODYNAMIC ASPECTS For reaction to proceed spontaneously, the free energy of the products must be lower than free energy of the reactants (ΔG must be negative). Free energy is made up of two components: enthalpy (H) and entropy (S). ΔG = ΔH - T ΔS ΔH – difference in bond energies between the substrate and product (including resonance increments, strain, differences in solvation). ΔH is calculated as a sum of bond energies of all bonds broken, subtracting from this the total of the bond energies of all bonds formed, and adding increments corresponding to changes in resonance, strain and solvation. ΔS – entropy differences are related to disorder and randomness of the system (less ordered systems, the greater entropy). Usually less important compared to ΔH. Preferred conditions: low enthalpy & high entropy Several aspects: 1. Reaction in gasses: entropy factor is more important (in gaseous phase is higher degree of freedom and randomness. In liquids, the entropy factor is less important and in solid state even less important. 2. Stoichiometry of the reaction matters: A + B C + D reactions = low entropic factors a) Increasing number of molecules formed in a reaction (A B + C) corresponds to large gain in entropy. Such reactions are therefore thermodynamically favored. b) Decreasing number of molecules formed during the reaction (A + B C) corresponds to loss in entropy which needs to be compensated by enthalpy factor. 3. Despite the fact that cleavage reactions (A B + C) are entropically favored, many possible cleavages are difficult to perform because of the large increase in the enthalpy. Remember the pyrolytic reactions ! Example: cleavage of ethane to two methyl radicals. The C-C bond (ca 80 kcal/mol) is broken and no new bond is formed to compensate for the enthalpy increase. With increase in temperature, the entropic factor (T ΔS ) takes over (the enthalpy is independent on temperature). 4. Acyclic molecules have more entropy than corresponding cyclic molecules because of the decreasing number of conformations. Ring opening reactions proceed with gain in entropy (ring closing reactions with loss of entropy). CHAPTER 5 / page 2 KINETIC ASPECTS The negative ΔG is a necessary, but not sufficient, condition for a reaction to proceed spontaneously. Free energy of activation (ΔG‡) is another important factor. ΔG ΔG ΔGf ++ ΔGr ++ transition state G++ Reversible reactions: Gr‡ = Gf‡ (energy of TS‡ is the same for forward and reverse rxns) Transition state (‡) – possesses definite geometry and charge distribution, but no finite concentration accumulates, since it is at the top of an energy barrier (system simply passes through the transition state). The free energy is related to the equilibrium constants by the following equation: ΔG‡ = -R T lnK‡ Therefore higher value of ΔG‡ is associated with smaller rate constant. The rates of almost all reactions increase with temperature (Arrhenius equation). Comes from: A + B A---B C K = [A---B]++ [A] [B] The rate of decomposition of the transition state: v [A---B] = κ k T h (κ = transition coeficient, k = Boltzman’s constant, h = Plank’s constant) The rate of reaction is defined: Rate of Reaction = κ k T h [A---B] and [A---B]‡ = K‡ [A] [B] CHAPTER 5 / page 5 Example: in the following case, the transition state #1 will resemble the intermediate B. In addition, the transition state for the second step, transition state #2, also will resemble the intermediate B more closely than the product (C). A C B ΔG ΔGB ++ overall ΔG1 ++ ΔG2 ++ CURTIN-HAMMETT PRINCIPLE The reaction rate of a molecule that can exist in several different conformations in its ground state is defined as: k = xn knΣ n k = overall (observed) reaction rate constant, xn = molar ratio of the nth conformer, kn = specific reaction rate of the nth conformer. E ne rg y Equilibrium Coordinate A B ΔGc ΔGc ΔGbΔGa Product from A Product from B ΔGc ΔGaΔGb = Gb Ga The product ratio (product of A/product of B) = e-(ΔGa‡ - ΔGb‡ +ΔGc)/RT Thus, the least favorable conformation can give rise to the major product, if that conformation has a more favorable transition state to its product than does the more favorable conformation to its product. The difference in energies between the two conformations (their populations) does not matter! CHAPTER 5 / page 6 Example: base-catalyzed trans-elimination of 1,2-diphenyl-1ethyl 1,3,5-triethylbenzoate. This reaction is known to give exclusively trans-stilbene. B Ph H H Ph H OCOR t-BuO t-BuOK Et Et Et O O C H H2 C H Ph H Ph A Ph Ph H H H OCOR Ph H H Ph t-BuO It may seem that the prevalence of trans-stilbene is due to the preference of the conformation B. In fact the composition of the stilbene isomers in the reaction mixture depends solely on the free energy of the activation of the respective transition states A and B. = CURTIN-HAMMETT PRINCIPLE ΔΔG ΔGB ++ conformation A B AButO BButO ΔGA ++ transition state ++ According to Hammond’s postulate, the difference in the energies of transitions states might be inferred from the different energies of the products (23.37 kJ/mol), but not the starting materials. In the cis-isomer, there are two bulky phenyl isomers at the same side of the C=C. Likewise, in the transition state, the phenyl substituents are already shifted toward the spatial positions they would eventually adopt. This is a destabilizing factor that renders such a transition state of higher energy compared to the one of the transition state corresponding to the trans-product. PRINCIPLE OF MICROSCOPIC REVERSIBILITY During the course of reaction, nuclei and electrons assume certain configurations that correspond to the lowest possible energy. If the reaction is reversible, these positions must be the same for the forward and the reverse process as well. If the reaction A C goes through intermediate B, then the reverse reaction must have the same mechanism and go through the intermediate B too. CHAPTER 5 / page 7 SUBSTITUENT EFFECTS – HAMMETT EQUATION Substituents have an effect on the compound reactivity. On the qualitative level, we know how to implement the inductive, field and resonance effects. When we need to quantify these effects, we need to use quantitative methods (equations). Hammett equation: log k / k0 = σ ρ or log k – log k0 = σ ρ ko = the rate constant or equilibrium constant recorded for the compound bearing a substituent =H (the standard rate constant); k = the constant for a compound bearing a substituent X; ρ = the reaction constant (arbitrarily set as 1 for ionization of substituted benzoic acids in water at 25o C); σ = the substituent constant describes the effect of the substituent on free energy of activation. σ sums up the total electronic effects (inductive, field, resonance). Works fine for para- and meta-substituents. Often fails for ortho-substituents, since they conformationally distorted due to crowding from the adjacent group. The Hammett equation is a free-energy relationship. We will demonstrate it for equilibrium constants (it is also valid for rate constants as well). Standard reaction Studied reaction ΔG0 = -RT lnK0 ΔG = -RT lnK Since the Hammett equation can be written as: log K – log K0 = σ ρ and ΔG0 - ΔG = σ ρ 2.3 RT 2.3RT For any given reaction: ρ, R, T and ΔG0 are constants. That means that σ depends linearly on ΔG. CHAPTER 5 / page 10 KINETIC STUDIES - The study of the reaction order The reaction rate does not always correspond to the concentration of all reaction components. To predict any viable mechanism we must know the rate/concentration dependence for all potential reagents (even those that do not appear in the stoichiometric equation). Example 1. In a kinetic study carried out at wide range of concentrations, it was determined that the transmethylation of compound A is a second order reaction to the reactant A (v=k2[A]2). This means that the reaction proceeds via intermolecular (as opposed to intramolecular) mechanism. S O S O O S O S O O S O S O O + S O S O O sulfoniumsulfonate (product) 2-methylsulfonyl-1-methyl- thiobenzene (reactant) A S O S O O Bimolecular Unimolecular Example 2. Nitration of benzene does not depend on the benzene concentration. That means that the attack of the nitration agent is not the rate-limiting step. In fact, the rate limiting step is the formation of the nitronium ion. 2 HNO3 H2NO3 + NO2 + Ar Ar+ H NO2 NO3 - NO2 + H2NO3 + Ar-NO2 Ar+ H NO2 H2O NO3 - HNO3 + + ++ + fast slow fast fast CHAPTER 5 / page 11 Determination of the presence of an intermediate Intermediates are postulated in most mechanisms, and we must often try to determine their presence and structure. How can that be done?: 1) Isolation of an intermediate can sometimes be realized when the intermediate is especially stable. The isolated intermediate must give the required product when subjected to the appropriate reaction conditions! 2) Independent synthesis of the intermediate can sometimes be realized when the intermediate is especially stable. It is necessary that this synthesis be through an alternative route to that of the reaction under examination; otherwise, the strategy is the same as 1 above. Again, the synthesized intermediate must yield the product of the reaction under examination. 3) Detection of an intermediate spectroscopically (short-lived intermediates by UV/visible, IR, and Raman; long-lived intermediates by NMR, MS, ESR, CIDNP). For example: when we propose a radical mechanism, we might be able to see radicals in the EPR or CIDNP spectra. 4) Trapping of an intermediate is often more feasible when dealing with unstable intermediates. If we propose the presence of an intermediate that cannot be isolated (1 above), but we know that the proposed intermediate (if present) would react with (be trapped by) certain other reagents to give very specific products (trapped intermediate), we can employ this strategy. A B C DB + E A + D C E+ presumed trap trap trapped intermediate regular product trapped intermediate regular product We run the reaction in the presence of the trapping agent and see if we find the trapped intermediate in the product mixture. Well known example: proof of the benzyne intermediate by cyclopentadiene trapping. However, benzyne is a relative long-lived intermediate and lives long enough to trap itself and form biphenylene. + trapping agent Biphenylene CHAPTER 5 / page 12 Another example of self-trapping is that of the xylylene intermediate during the photodecomposition of 1,4-dihydrophthalazine. N N -N2 2 2 hν Et2O , -176 oC If the intermediate is very short-lived or not intrinsically reactive with itself, self-trapping will not be possible. Under these circumstances, high concentrations of reactive trapping agent must be employed. N N O2 (10 atm) O O Exchange experiments might be regarded as a special case of trapping that is usually done with isotopic labeling. The similar substrates (substrates with similar reactivity or labeled and unlabeled substrates) are mixed in the same reaction vessel, and the product distribution is determined usually either by NMR or MS. Example: the Claisen rearrangement: twice 13C-labeled starting compound yielded twice labeled product, while a label-free starting compound yielded a label free product. Products with only a single label were not observed. Therefore, the Claisen rearrangement proceeds via intramolecular mechanism !!! O * * O + OHOH * * OH * OH * + but not or Isotopic labeling experiments Experiments used to determine the fate of specific carbon or hydrogen atoms (isotopes of any type of atom can be used in this type of experiment) during the course of a reaction. The starting compound has an isotopic label (tracer) in or near the reaction center, and the location/distribution of that label determined in the products. CHAPTER 5 / page 15 Secondary isotope effects Effects associated with isotopes that are in the vicinity of the reaction, but not directly involved in the bond breaking or making steps. These may be separated further into α- and β-SIE’s depending on whether the isotope one or two heavy atoms removed from the reaction center. Examples: The case below is a β-secondary isotope effect. If the methinyl hydrogen had been replaced by deuterium the secondary isotope effect would have been α. In the case shown below, kC-H/kC-D = 1.34 and is most likely due to the different degrees of hyperconjugative stabilization of the transition states. H3C CH CH3 Br H3C CH CH3 H3C CH CH3 OH D3C CH CD3 OH D3C CH CD3D3C CH CD3 Br H2O KC-H H2O KC-D Isotope effects can be observed with heavier atoms such as 13C, but these are much smaller than hydrogen isotope effects because the 12C/13C mass difference is relatively small compared to that of 1H/2H. Thus the reaction shown below has k12C/k13C = 1.053, *CH2 Br *CH2 OCH3CH3O k12C/k13C = 1.053 CHAPTER 5 / page 16 The study of the product stereochemistry This approach quite often provides significant insights into reaction mechanisms. Example: reduction of acetylenes by alkaline metals in liquid ammonia might proceed via a radical anion mechanism (singlet electron addition to alkyne) or a dianion mechanism (double electron addition to the alkyne). Since this reaction provides only the trans-olefin, it must proceed through the dianion mechanism. This observation is inconsistent with a radical anion mechanism which would afford both olefin isomers, since these radicals easily isomerizes. R R RR RR R R RRR R 2 H+2 H+ RR H+ RR Low energy barrier between isomers would yield a mixture of olefin isomers Radical Anion Dianion Product H H High energy barrier between isomers The study of the relationship between stereochemistry of the starting material and product By comparing the stereochemistry of the SM vs. products we can make judgments about the intermediate/transition state. Example 1. The Meisenheimer’s rearrangement of the optically active amine-oxide = a well known example of the syn-elimination. Erythro Z isomer / Threo E isomer CH3H3C Ph O N CH3 CH3 H H CH3Ph CH3 O N CH3 CH3 H H CH3Ph CH3 O N CH3 CH3 HPh H3C CH3 Threo-starting material E-Product CHAPTER 5 / page 17 The study of catalysis Much information about the mechanism of the reaction can be obtained from the knowledge of which substances catalyze or inhibit a reaction (or do neither). Note: catalysts do not change ΔG. They lower ΔG‡ by providing an alternative pathway for the reaction to proceed. This pathway generally involves several small steps in place of a single large step. Catalysis Catalyst is an agent able to create a new reaction pathway (new reaction coordinate) for a given reaction. The catalyst is not present in stoichiometry amounts. While it may be consumed in an early step in the reaction sequence, it will be released in a later step of that same sequence. Catalysts are not consumed in reactions. Catalysts change the rates of reactions by lowering ∆G‡, it does not affect ∆G0, and affects both the forward and reverse rates equally. Specific acidobasic catalysis H+ or its solvated form (e.g. H3O+) acts in the rate-determining step as a catalyst. In the specific basic catalysis it is OH-. Non-dissociate forms of acids/based are catalytically inactive. Proton transfer between electronegative atoms (O, N) is fast. Other than electron-transfer reactions, proton-transfer reactions are the fastest reactions. This fast equilibrium between the substrate (R) and the conjugated acid (solvated H+) (step A) precedes the rate-limiting step, which is the reaction of the conjugated acid to products (step B). PR + H3O + RH+ + H2O k1 k -1 k2 Step A = fast Step B = slow RH+ For the reaction rate we can write the following equation: Reaction rate = k1 . k2 . [R] . [H3O +] k -1 k2 . K . [R] . [H3O +] K = k1 / k2 = For a specific basic catalyst the equation is very similar. R - PRH + OH - R - + H2O k1 k -1 k2 Step A = fast Step B = slow CHAPTER 5 / page 20 Solvent effects Determining the effect of the solvent on the reaction rate/course may offer insight into the reaction mechanism. We will talk about this in greater detail in ‘substitution reactions’. The description of solvent polarity: there is no single measure of the solvent polarity. μ – molecular dipole moment = property of the individual molecule; It does not account for intermolecular interactions ε – dielectric constant = effect of the substance on the electric field between two conductor plates with opposite charges; It describes the orientation of molecules between conductors, does not account for orientation of solvent molecules around charged/polar solutes. Empirical equations/factors have been developed. Most of them are based on the observed solvatochromism of certain dyes/chromophores: Kosower Z scale (based on the solvatochromism of A), ET(30) scale (based on the solvatochromism of B). N O OEt NO I A B Kamlet and Taft introduced general polarity/polarizability index π* which describes solvent’s ability to stabilize charged/polar species by its dielectric effect. The ability of solvents to act as hydrogen bond acceptors/donors is described by coefficients α/β introduced by Kamlet & Taft. CHAPTER 5 / page 21 Table of solvent parameters. A theoretical calculation of the ∆G of solvation is complicated. Simple empirical rules based on solvation theory have been developed: 1. Reactions during which the formation of the transition state results in the formation or high concentration (localization) of the charge in a small volume are accelerated by increase solvent polarity. 2. Reactions proceeding with recombination or delocalization of the charge are accelerated by decreasing solvent polarity. Reactions most affected by solvent polarity: Nucleophilic substitutions (solvolyses), polar additions, isotopic exchange, reactions of the mesomeric ions (enolates), etc. Three interesting examples: CHAPTER 5 / page 22 Example # 1: CH3 Et H CH3 Et D ROD / RO- ROH / RO- A) In non-polar solvents, the tight ion pair is formed, and the H/D exchange proceeds via a cyclic mechanism (A). Such reaction proceeds with retention of the configuration. H O R D O R Solv Solv ROD H OR A B C retention of configuration racemization inversion of configuration B) In dipolar aprotic solvents, the anion (either non-solvated or symmetrically solvated) is formed (B). The carbanion racemizes before recombination, and a racemic mixture is formed. C) In protic solvents the formed anion is solvated from the side opposite to the leaving proton (C). Such reaction proceeds with inversion of the configuration. Example # 2: A solvent-induced change in the mechanism of the arylazosulfide decomposition: N N N N SR EtOH N N+ SR + SR Benzene The change between radical/ionic mechanisms is caused by the increase in the probability of the heterolytic cleavage as a result of the effective solvation of the ion pair formed while preventing their recombination. Non-polar solvent (benzene) does not solvate the ion pair and does not prevent their recombination. Example 3: Recation of mesomeric ions and enolates. O O O-alkylation C-alkylation A) In dipolar aprotic solvents, enolates are dissociated, cations are well solvated, anions not so well solvated. In the enolates, negative charge is localized on oxygen, and O- alkylation is favored. B) In nonpolar aprotic solvents, cations form ion pairs with enolate anions. The O-atom is masked by the close proximity of the cation. Attacking electrophiles attack the C=C bond, and C-alkylation is favored.