Lecture Notes on Kinetics - Mechanism | CHEM 6311, Study notes of Mechanics

Download Lecture Notes on Kinetics - Mechanism | CHEM 6311 and more Study notes Mechanics in PDF only on Docsity! 65 6. Kinetics Kinetics is probably the most basic tool for determining reaction mechanisms. An important point is that reaction mechanisms cannot be directly proven. We have only indirect ways to measure how the atoms were arranged at the transi- tion state. Therefore, a reaction mechanism can only be inferred, often from many varied pieces of experimental data. A. Determining Reaction Mechanisms: 1. Product studies indicate how the ratios of products vary with respect to changes in the reagents. 2. Stereochemistry - we saw this with methylene addition to an alkene. 3. Isotopic labelling - following a label (or labels) from reactants to products can tell us unambiguously what bonds are broken. 4. Trapping or observation of intermediates relies on the formation of an intermediate with a definite lifetime; how long depends on the trapping reagent or spectroscopic method. 5. MO calculations can be used to model reactions. 6. Linear free energy relationships; we have seen their use in the Hammett equation. We will see several more in this course. Basically, they are attempts to relate what happens at the transition state to the ground state prop- erties. 7. Kinetics is probably the most important technique. 66 B. There are several “rules” that we should briefly discuss on how chemists propose a reaction mechanism. When propos- ing a reasonable mechanism: 1. Use Occam’s Razor. William of Occam was born ~1280 in the town of Occam, near London. He studied theology at Oxford University ~1319. His philosophical views were controversial with Pope John 22nd and as a result he fled to Germany where he died of the Black Plague in 1349. He said “Plurality is not to be assumed without necessity...” and “What can be done with fewer assumptions is done in vain with more.” 2. Each individual step in a reaction should be either unimolecular or bimolecular. a. The molecularity is the kinetic order for a single reaction step. i ) A rate law can be written: rate = [A]n [B]m [C]o..... ii) The kinetic order = n + m + o ... b. A two-body collision is 1,000 times more probable than a 3-body collision, solely on statistical grounds. 3. Each step should be energetically and chemically feasible (this requires “experience.”) C. Rate laws of typical reactions 1. General cases A+ B → C+2D Rxn rate = -d[A]/dt = -d[B]/dt = d[C]/dt = (1/2)d[D]/dt The elementary reaction steps of a mechanistic hypothesis allows a rate equation to be written. This would be compared experimen- tally with the kinetics of the reaction of interest. a. Unimolecular A → B i) d[B]/dt = k[A] = -d[A]/dt or kdt = - d[A] / [A] where k = the rate constant for this A→B reaction ii ) In integrated form: kt = In ([A0]/ [A]), where [A0] = initial concentra- tion of A iii ) Plotted: 69 the rate determining step approximations are used. ii) Reaction: A + B k1 k-1 C k2 D k3 E + F 6.7 iii) Analysis (a) step 1: d[C]/dt = kl [A][B] - k-1[C] (b) step 2: d[D]/dt = k2[C] (c) step 3: d[E]/dt = d[F]/dt = k3 [D] iv ) Lets suppose that k2 << kl, k-1, k3. (a) Then step 2 is the rate determining step. (i) the overall rate depends only on this step. (ii) It depends indirectly on any one which proceeds step 2. (b) Therefore, the reaction rate = k2 [C] v ) But suppose C is an intermediate whose con- centration cannot be measured (a) C is connected to A and B by means of a reversible step (b) At the middle of reaction tıme (not at the beginning nor at the end) we can assume that A and B and C are connected by a fast (faster than k2) equilibrium step, i.e. (i) K= [C]/[A][B] = kl/k-1 (ii) ∴ [C] = (k1/ k -1) [A][B]. (iii) The overall rate = k2 (k1/ k -1) [A][B] = kobs [A][B] vi ) An example is the acid catalyzed nucleophilic substitution of alcohols: ROH + H+ k1 k-1 k2ROH2+ + Br- R-Br + H2O 6.8 (a) Assuming the 1st step is a rapid equilibrium: d[RBr]/dt = k2 [ROH2+] [Br¯] (b) Now [ROH2 + ] [ROH][H+ ] = k1 k −1 70 (i) Thus [ROH2+] = k1/k -1[ROH][H+] (ii) and rate = k2kl / k -1[ROH][H+ ][Br¯ ] . vıi ) In this case we could evaluate k1/k-1 in a sepa- rate experiment by using a nonnucleophilic coun- teranion, however, we shall see a much more elegant, direct method later. b. Pseudo first order - Another simplification can come into play when one reagent is present in large excess over the others. A+B →C; rate = k [A][B] i ) If [B] >>[A], then rate = kobs [A], where kobs = k [B], and [B] ≈ constant. ii ) The above is said to be a psuedo first order (rather than bimolecular) reaction. iii ) Normally this will only apply if (a) B is the solvent or (b) there is a buffered solution so that if B is an acid or base, there will be a constant [B]. (c) [B] > > [A], usually ≥ 100 fold. c. The steady state approximation i ) This is a more general, and more rigorous, way to deal with intermediates. ii ) It assumes that one has an intermediate whose concentration is small, and approximately constant, over a large portion of the reaction time. iii) As an example consider the following: (a) The rate of product formation, d[E]/dt = k2 [C][D] (b) If we make the steady state approximation: d[C]/dt = 0 i) Then using this we have that kl [A][B] = k -1 [C] + k2 [C] [D], A + B k1 k-1 k2 Step 1: C C + D E + FStep 2: 6.9 71 (This same expression can be found for -d[A]/dt, try it.) (c) We can further simplify this by assigning a rate determining step: (i) Suppose step 2 is rate determining. Then, (a) -d[C]/dt from step 2 < -d[C]/dt from step 1, or (b) k2 [C][D] < k-1 [C] (c) So k2 [D] < k-I, and, therefore, (ii) suppose step 1 is rate determining: (a) so -d[C]/dt from step 2 > -d[C]/dt from step 1. (b) then k2 [C][D] > k-1 [C] ( or, k2 [D] > k -1 ) (c) and the rate ≅ kl [A][B]. d. Steady state kinetics is also used in the analysis of enzyme catalyzed reactions. i ) In the following reaction an enzyme (E) catalyzes the conversion of a substrate (S) to a product (P), by first forming an enzyme-substrate complex (E•S) which rapidly attains a steady state concentration: yyyyy yyyyy yyyyy yyyyy yyyyy yyyyy  + yyyyy yyyyy yyyyy yyyyy yyyyy yyyyy  k2 P (Product) ES (Enzyme-substrate complex) k1 k -1 S (Substrate) E (Enzyme) E ES S P 6.10 [C] = k1[A][B] k −1 + k2[D] ∴ rate = d[E] dt = k1k2[A][B][D] k −1 + k2[D] d[E] dt ≅ k1k2 k −1 [A][B][D] 74 (e) Finally, if kl and k-l >> k2 then [E.S] = kl/k-I [E][S]. (i) This is equivalent to the case of rapid prior equilibrium discussed previously. (ii) In this case only, Km can be treated as equivalent to an equilibrium constant, that is, Km = k l/k -1 e. Oscillatory reactions i) In these interesting multistep reactions the concentrations of intermediates (and sometimes products) rise and fall periodically. ii) Here is a very simple, hypothetical case. (a) The first step is autocatalytic in X: k1 step 1: A + X → 2X (b) The second step is autocatalytic in Y, and de- stroys X: k2 step 2: X +Y → 2Y therefore, d[X]/dt = kl [A][X] -k2 [X][Y] (Equation 1) (c) The third step involves the destruction of Y: k3 step 3: Y → P therefore, d[Y]/dt = k2 [X][Y] - k3 [Y] (Equation 2) and d[P]/dt = k3 [Y] iii)The course of the reaction: (a) Assume that [A] is very large, i.e. its concentra- tion is constant for the entire reaction period. (b) Initially (i.e. at t= t0), [Y]is small, so (i) d[X]/dt ~ kl [A][X]. (ii) [X] increases rapidly. (c) But the 1st term in Eq.2 is also first order in X. (i) Initially, d[Y]/dt ~ k2[X][Y] (ii) As [X] becomes large, [Y] starts to increase rapidly. (iii) As [Y] becomes larger, the 2nd term in Eq.l increases. (d) At some point, t = tl, [X] reaches its maximum 75 value. (i) [Y] continues to increase, but (ii) [X] decreases as the rate of decomposition of X increases, until (iii) at t = t2, d[X]/dt ~ k2 [X][Y]. (iv) [X] continues to decrease to a small value. (e) At t = t3 , [Y] is maximized. (i) At large [Y] and very small [X], d[Y]/dt ~ - k3 [Y] . (ii) Thus, [Y] must start to decrease. (f) Eventually, we have returned back to the initial state where [X] and [Y] are both small so the cycle starts over. [X] [Y] t0 t1 t2 t3 6.13 First step: A + X 2X k1 Second step: X + Y k2 2Y Third step: Y k3 P iv) A plot of [X] and [Y] (and [P]) as a function of time is shown below. C o n ce n tr at io n s Time [X] [Y] [P] t0 t1 t2 t3 t0 t1 t2 t3 6.14 76 v) It can be shown that the solution of the three rate equations is: k2([X] + [Y]) - k3 In [X] - kl [A] ln [Y] = C, (where C is a constant, try it .) vi ) In the “real world”, this can be a model of an ecological system: A = grass X = sheep Y = lions P = humans (kill lions) A+X → 2X X+ Y → 2Y Y → P (a) In the fist stage of the reaction, sheep eat the grass and become very numerous. (b) In the second, the sheep are eaten by lions, which then reproduce until (c) their population becomes an annoyance to the human population, who (d) in turn kill off a large portion of the lion popula- tion, (and probably increase themselves) vii ) An example of an oscillatory reaction which takes place in a cell is the hydrolysis of esters by the enzyme papain. (a) The general mechanism is thought to be: R OR' O + H2O papain (E) (S) R O O - + R'OH + H+ (P) 6.15 yyyyy yyyyy yyyyy yyyyy E’’ H+ E’ H+ E S S + E P + H+ P P H+ H+ kp kH+ kS k cell wall 6.16 (b) The environment can be depicted ( where E’ & E” are ionized, inactive forms of enzyme.) 79 The reaction co-ordinate is indicated by the dashed line b. Actually, our potential energy surface for H. + H2 was idealized. i ) H. really could collide with H2 at any angle. ii ) The potential surface must be recomputed for each H-H-H angle. iii) For an angle of 60˚, for example, the surface is: (a) Note that the Ea is higher in this case. (b) Actually, the activation energy is the lowest for θ = 180˚. 4 3 2 1 2 3 4 r1 r2 (au) H H H r1 r2 θ = 60o 4 3 2 1 0.5 0.3 3 4 2.5 r1 = r2 = 2.0 c. We can compute the rate constant for this reaction using trajectory calculations. i ) The H. + H2 system is assigned a kinetic energy. (a) This also implies a momentum. (b) Naturally, the kinetic energy is related to the temperature. ii ) The process is repeated many times. iii ) One trajectory with enough kinetic energy to overcome the barrier (and with the two particles moving in the right direction) is shown on the next page. (a) A portion of the starting translational energy is turned into vibrational energy after crossing the barrier. (b) A portion of the starting vibrational energy is 80 turned into translational energy after crossing the barrier. (c) The trajectories shown above are calculations from the Cla- + CH3-Clb → Cla-CH3 + Clb- reaction. The functional form of the reaction is identical to the H• + H2 reaction we have just considered. For each of the trajectories, the particles start at the lower right portion of the potential energy surface and finish on the upper left side. Notice that in trajectory a there is just enough potential energy to cross the barrier. In trajectory b notice the loop that the particles make in the vicinity of the transi- tion state - an event we would never consider. 81 Trajectory c shows what happens when there is not enough kinetic energy to get over the barrier. Finally trajectory d shows what happens when there is more than enough kinetic energy. The excess translational kinetic energy is released as vibrational energy in the product. d. In general, we will need to evaluate 3N-6 vibrational coordinates (where N= number of atoms) to find the path of least energy and the transition state. (i) Clearly this is an extremely difficult task, and we cannot hope to display the results graphically. (ii) What is most commonly done is to simply plot the path of least energy in one dimension. A + B-C [A--B--C] A-B + C activation energy energy of reaction minima transition state minimium energy reaction path P o te n ti al E n er g y Reaction Coordinate 6.21 transition state: energy rises in all directions, except along the reaction coordinate, where it decreases (see H2 + H. surface) e. Characteristics of the reaction energy diagram: (i) The reaction coordinate (a) This is a collection of vibrational co-ordinates. (b) It corresponds to the dashed line in the H2 + H. example. (ii)The energy minima (a) These are defined by their energy gradients - the derivatives of the energy with respect to all 84 c. As another example consider the nucleophilic addition to carbonyl compounds: This shows the relative positions of the atom from the nucleophile which attacks the carbonyl carbon, substitu- ents connected to the carbonyl carbon and the position of the carbonyl oxygen atom for structures labelled A - L. The calculated surface is shown below: 3. The Arrhenius equation a. For a one step reaction: kobs = A e-Ea/RT b. This is called the Arrhenius equation (i) A = pre-exponential factor, related to ∆S‡ (ii) Ea = activation energy (iii) R = gas law constant (iv) T = temperature in Kelvin c.Plotting ln ( kobs ) vs 1/T: (i) The slope = - Ea/R (ii) The intercept = ln A . d. Arrhenius estimates of half-life for the reaction C O R RN + O R R N + - 6.26c 85 A→B: (i) Using A = e kT/ h (assumes ∆S‡ = 0, see below), and (ii) defining: t1/2 = time when [A] = [B], and t1/99 = time when 99[A] = [B]. _________________________________________________ Ea T=50°C T= 100°C_______ t1/2 tl/99 t1/2 t1/99 20 1.2 sec 8.0 sec 1.7 x 10 -3 sec 0.11sec 25 49.1 min. 5.42 hrs. 13.9 sec 1.54 min 30 81.4 days 1.46 yrs 3.26 hrs 21.6 hrs 35 533 yrs 3,534 yrs 114 days 2.08 yrs 40 1.27 x106 yrs 8.44 x106 yrs 264 yrs 1,748 yrs 4. Transition state theory a. Calculations such as those above give a good idea of the relative rates of reactions as a function of the activation energy. (i) But the rates are not particularly accurate, and (ii) the theoretical basis for the relationship is also not clear. b. A more rigorous way to relate rate constants to activation energies can be given if we assume an equilibrium between reactants and the transition state: (i) For the reaction A → B where A‡ is the transi- tion state, (a) K‡ = [A‡]/[A], and (b) kl = κkT K‡ h k = Boltzmann’s const. h = Planck’s const. T = temp ˚K κ= the transmission factor(commonly assumed to be 1.00) = ratio of molecules that will cross over when they get to A‡ . In extremely large molecules, i.e. enzymes, for example, the top of the transition state may be ill-defined, so that, κ < 1. 86 (c) Go over this in Appendix 1, Chapter 2, in your book (ii) Recalling that ∆G‡ = -RTln K‡ (a) where ∆G‡ = ∆H‡ - T∆S‡ (b) then k = κkT h e−∆G ‡ /RT = κkT h e−∆H ‡ /RTe+∆S ‡ /R (c) Where: (i) ∆G‡ is the free energy of activation. (ii) ∆H‡ is the enthalpy of activation (the potential energy difference between the reactants and the transition state). (iii) ∆S‡ is the entropy of activation (related to the rigidity and structure of the transition state versus the reactants). c. Comparing this to the Arrhenius Equation (for the unimolecular reaction): (i) A = eκkT h e∆S ‡ /R (ii) Ea = ∆H‡ + RT (iii) ∆G‡, ∆H‡ & ∆S‡ can be computed by plotting ln(k/ T) vs. 1/T . A + B-C [A--B--C] A-B + C P o te n ti al E n er g y Reaction Coordinate 6.27 ln k T T ( ) 1 slope = - ∆H R ∆S R intercept = + ln κk h observable range 6.27b 89 A compound, R, was decomposed by a reaction that is very well-known to produce cyclobutadiene with shape A2 (if cyclobutadiene is a rectangle) and A2 then can rearrange with a rate constant k1 to isomer A1. Both A1 and A2 are intercepted by an olefin which undergoes another very well defined reaction to produce three isomers, B, C, and D. The only difference between these products is the placement of the deuterium labels. Now if cyclo- butadiene was a square, then A1 and A2 are really only resonance structures and [B] = [C]+[D]. On the other hand, if cyclobutadiene was a rectangle and provided that k2≥k1, then [B] < [C]+[D]. In fact this was found to be the case so cyclobutadi- ene is indeed rectangular in shape. It can easily be seen that k1 ∝ [B] [C] + [D] Carpenter repeated these reactions at different temperatures and as shown below there is curva- ture instead of being a straight line. At lower temperatures the value of ln(k/T) is larger than that D D A1 D D A2 D D P o te n ti al E n er g y Reaction Coordinate 10.8 kcal/mol ∆E ln k T ( ) T 1 quantum mechanical tunneling 6.31 90 extrapolated by the dashed line at higher tempera- ture. Furthermore, a very good quantum mechani- cal calculation puts the activation energy for the conversion from A1 to A2 to be 10.8 kcal/mol. However, Carpenter’s estimate is that ∆H‡ = 4.6 kcal/mol and ∆S‡ = -15 eu. This is a very big differ- ence between the experimental and calculated value and, furthermore, the large negative value of ∆S‡ is difficult to explain. One idea that has come forward (and has been reproduced by calculations) is that this compound undergoes quantum me- chanical tunnelling. A fraction of the molecules cross the classical barrier and a fraction tunnels from one side to the other. The probability to cross the classical barrier is very strongly related to the temperature, but quantum mechanical tunnel- ling does not. Consequently the apparent rate constant becomes too large at lower temperatures as predicted by the classical activation energies. Another way to put this is recall that k = κkT h e−∆G ‡ /RT so one could express the amount of tunnelling by how much κ was greater than 1. Using this it was found that κ = 100 at -10 °C and 800 at -50 °C. A second example is given by a simple bond-breaking reaction. H H3C H H3C H H3C slow fast 6.32 ∆S = 2 eu The activation energy for this reaction is small because the compound is strained. The important point is that ∆S‡ is small in magnitude unlike the 91 CH3-CH3 → 2CH3. example that was presented previously. Here the two radicals are “held to- gether”. However, the reaction shown below has been studied and ∆S‡ was found to be very differ- ent. An explanation that has been advanced for the difference can be outlined as follows. slow fast ∆S = -16 eu 6.33 H H3C H H3C H H3C triplet singlet reaction coordinate (rC-C) p o te n ti al e n er g y 6.34 The key to understanding this problem is that the diradical that is formed has two electronic states, singlet and triplet. As shown above, the first mol- ecule starts out with the two electrons paired in the reactant (a singlet state) and ends up with the singlet diradical. At some energy above this is the triplet state where the two electrons are parallel. Why the energy ordering is this way is beyond what we need to know, however, the important point is that in the second example, this energy 94 f. Volume of activation i ) This is another noteworthy experimental, rate- based tool to investigate the structure of the transition state. (a) For the reaction: (i) Define: ∆V‡ = V3 - (V1 + V2) ∆V = V4 - (Vl + V2) (ii) The closer ∆V‡ is to ∆V, the more the transition state resembles the product, C. (iii) The closer ∆V‡ is to zero, the more the transi- tion state resembles the reactants, A + B. (b) ∆V‡ can be measured by measuring reaction rates at different pressures (keeping the tempera- ture constant at each pressure). (i) [d(ln k)/ dP]T = - ∆V‡/RT, where P = pressure (atmospheres). (ii) Plotting In k versus P gives a straight line with slope of ∆V‡ / RT. (iii) A negative value of ∆V‡ implies a tight, structur- ally rigid transition state where the molar volume of the transition state is less than that of the reac- tants: + 6.39a A + B [A------B] C V1 V2 V3 V4 6.38 ln(k) P ∆V RT slope = 95 ∆V‡ = - 33 cm3/mol ∆V = - 37 cm3/mol iv) A positive value of DV‡ implies a loose, structur- ally flexible transition state: ∆V‡ = + 15 cm3/mol ∆V = + 60 cm3/mol v) The interpretation of the magnitudes of ∆V‡ are not always so straight-forward. For example in the reaction below ∆V‡ = - 44.7 cm3/mol ∆V = - 33.3 cm3/mol It is not possible that the transition state occupies less volume than the product! Obviously the way that the solvent solvates the transition state rela- tive to the product is of importance here. A more straight-forward example of this is given in the following hypothetical reaction: Since two molecules are formed one might think that ∆V (and ∆V‡) should be positive. However, in a polar solvent the formation of the ions causes the solvent shell to contract. Therefore, ∆V (and ∆V‡) is expected to be much smaller. This is called electrostriction and it is a big effect. For the reac- tion shown above ∆V‡ might be close to zero and might even be negative. vi) The drawbacks in measuring ∆V‡ are basically twofold: (a) The temperature must be kept very constant throughout the pressure range studied. (b) It is a small effect. (Suppose ∆V‡ = -15 cm3/mol.; R N N R R N N R N N2R. + 6.39b O+ CH3 O O O O OCH3 6.39c [Aδ+------Bδ-] A+ + B-A-B 6.40 96 then going from 1 to 1000 atms at 25˚C increases the rate constant only 1.8 times.) E. Experimental Techniques for Measuring Rates of Reactions method time scale Ea (kcal/mol) classical isolation 1 hr - weeks ~25 - ~45 Stop-flow technique 10-4 sec - min 10 - 20 NMR 10-6 - 100 sec 5 - 25 ESR 10-10 -10-5 sec 2 - 10 Flash photolysis (picosecond)10-12 - 1 sec 1 -15 (femtosecond) 10-15 sec molecular vibrations Relaxation techniques temperature jump 10-8 - 1 sec 3 - 15 pressure jump 10-6 - 1 sec 5 - 15 1. For relaxation techniques consider a typical reaction sequence: A + E B (fast) k1 k-1 B + C D (slow)k2 6.41 Often times the first step is a proton transfer. Using the steady state approximation for [B] we have: rate = k1k2[A][E][C] k −1 + k2[C] Then using the fact that the first step is fast k-1 > k2[C] and rate ≅ k1k2 k −1 [A][E][C]. a. Now let K = kl/k-1 = [B]/[A][E], and b. [d(lnk)/dT]P = - ∆H/RT2, where ∆H = enthalpy change for the 1st step. c. Suddenly changing T (when ∆H ≠ 0) causes Ink to change. d. This in turn causes the relative concentrations of A, E, and B to change. 2. What is done in a temperature jump experiment is to discharge a capacitor in a small reaction cell. a. This occurs roughly in 10-8 sec. and the temperature 99 rA-B rBC .3 .5 .5 1.0 1.02.0 2.0 6.44 A + B-C → A-B + C ii ) The principle of least motion correctly predicts that a motion like: will not be favorable, and it is not. g. Spectacular failures (in qualitative detail) of this principle are very interesting. i ) They imply that something extremely unusual is occuring to the electronic structure along the least motion path. ii ) Two examples that we shall examine in detail at the end of the semester are: A + C B A B C A B + C 6.45 Both reactions take place by paths which lie very far from the least motion paths. 2. The Hammond Postulate: a highly exothermic reaction will have a T.S. geometrically close to the reactants a. This can be illustrated as follows: P o te n ti al E n er g y Reaction Coordinate A B Ea E’a 6.47 TS1TS2 H H D D + DH DH + +or 6.46 100 T.S.(2) is geometrically closer to the reactants b. By extension, the more endothermic a reaction is, the more the geometry of the T.S. resembles the products. c. Three general solutions are pictured below: P o te n ti al E n er g y Reaction Coordinate A B thermoneutral P o te n ti al E n er g y Reaction Coordinate A B exothermic P o te n ti al E n er g y Reaction Coordinate A B endothermic 6.48 3.Thornton’s rules: a more detailed analysis of Hammond’s concepts. For a review and more examples than are covered here or in your book see W. P. Jencks, Chem. Rev. 85, 511 (1985) a Energy changes along the reacton path: i ) Suppose we model the region in the immediate vicinity of the transition state by a parabola. The diagram for the A → B reaction in the thermoneutral example above is shown on the left side of the drawings below: ii ) Now consider a perturbation in which the energy of B is raised relative to A, by an amount δ∆E0 (a fraction of ∆E0). (a)The change will be transmitted in a linear way along the reaction coordinate. (b) This can be expressed as m• X = δ∆E0, where m is the ratio of the energy ıncrease (δ∆E0) to the distance between A and B along the reaction coordinate. The transition state is initially defined as being at X=0 for the thermoneutral case. (c) This can be illustrated as indicated in the draw- ing on the right side: P o te n ti al E n er g y Reaction Coordinate, X to A to B 0.0 P o te n ti al E n er g y Reaction Coordinate, X to A to B 0.0 m = slope 6.49 101 (i) The line of slope m represents the incremental increase in energy due to δ∆E0, as the reaction moves from A to B. Here the reaction is made endothermic so m is a positive number. (ii) We raised the energy of B relative to A (δ∆E0 is positive), so m is positive, and X (as defined in the diagram) is positive. The arrows indicate the value of δ∆E0 (when X is a negative number then δ∆E0 is negative, i.e. stabilizing and when X is a positive number then δ∆E0 is positive, i.e. destabilizing). Notice that the transition state has moved in the positive X direction, towards B which is in agree- ment with the Hammond principle. (iii) The same analysis holds if the energy of A is raised relative to B. but then m is a negative num- ber and the T.S. is shifted towards A (i e. X is nega- tive). b. Energy changes normal to the reaction path (or dimensions other than the reaction coordinate: i) Consider the simple case of an atom reacting with a diatom: (a) The enegy surface may be represented as: rA-B rBC .3 .5 .5 1.0 1.02.0 2.0 6.50 A + B-C → A-B + C A---B---C A---B---C - or - 104 The structure on the lower left corresponds to com- plete C-X bond breaking to form a carbonium ion and that on the upper right side has complete C-H bond breaking to form a carbanion. Now depending upon R, B and X a reaction path that passes through either a carbonium ion, carbanion, etc could be present, how- ever, for our reference reaction the reaction path follows a diagonal from upper left to lower right and there is only one transition state (with no intermedi- ate) located symmetrically between the reactant and product. (This is called the E2 mechanism). (i) Suppose then one changes the R group so that the carbonium ion, R+ on the lower left side of the idealized potential energy surface, now becomes stabilized (look back at Chapter 1 where we talked about carbonium ions being stabilized by reso- nance). The result of this perturbation is shown in the PE surface below on the left side: rC-X rB-H X H X B H X B - + X B-H+ - - B-H+ X rC-X rB-H X H X B H X B - + X B-H+ - - B-H+ X stabilize carbonium ion stabilize X- 6.54 The net consequence is to stabilize the lower left corner and since this is perpendicular to the reac- tion path direction, the transition state will move towards the direction of energy lowering (the arrow in the drawing) and the new reaction path will be curved. The new transition state will occur at larger values of rC-X (the C-X bond will be more broken) and larger values of rB-H (the C-H bond is less broken). The formation of a stable carbonium ion intermediate is called the E1 mechanism. (ii) Suppose that we make X become a better 105 leaving group. That means that X- is stabilized. The resultant surface is displayed on the right side of the drawing above. Now X- occurs two places on the two-dimensional potential energy surface: It appears as the product (lower right side) and since this is along the reaction path, the transition state moves away from the energy lowering. X- also occurs on the lower left diagonal which is perpend- icular to the reaction path and, therefore, the transition state is moved towards it. The result, as shown, is the addition of two vectors which again curves the reaction path. Here, however, rC-X does not change while rB-H occurs at a longer distance, as before. 4. The Pross-Shaik model of reactivity. This was developed by Addy Pross and Shason Shaik to view many reactions in terms of ground state changes. The idea here can be demonstrated by the first step in the SN1 reaction: R-X → R+ + X-. R..X R+:X- R. + X. R+ + :X- P o te n ti al E n er g y Reaction Path 6.55 β a. On the left side are the energies of the ground and first excited state of R-X. In the ground state the two 106 electrons between C and X are shared covalently. However, the first excited state is one where the two electrons are both associated with X. Now the first step in the SN1 reaction is one where the C-X bond is broken and so the covalent ground state evolves in the R. + .X diradical state. The R-X excited ionic state evolves into R+ + X- ionic state. The problem here is that at the product side the ionic solution is more stable than the diradical state. But the two states, given by the lines do not cross (they both have the same symmetry in terms of their wavefunctions). Instead the two states mix with each other so that the lower state becomes stabilized and the upper state becomes destabilized. They undergo an avoided cross- ing, shown by the dashed line. That amount of inter- mixing between the two wavefunctions can be viewed as resonance, where the amount of stabilization and destabilization is represented by β. b. Suppose that R+ is stabilized: R..X R+:X- R. + X. R+ + :X- P o te n ti al E n er g y Reaction Path 6.56 TS1 TS2 109 e. The same two parabolas are then drawn for the averaged case where the intrisnic reaction barrier is defined by ∆Gin ‡ = ∆GA ‡ + ∆GB ‡ 2 The product parabola is lowered in energy by an amount ∆G, given by the dashed line. The activation energy is expressed as ∆G‡ and the position of the transition state by X‡. f. One can show (see problem 17, p. 247 in your text) for this treatment that ∆G‡ = w + ∆Gin ‡ 1+ ∆G 4∆Gin ‡     2 = w + ∆Gin ‡ + ∆G 2 + ∆G 2 16∆Gin ‡ where w = the work term - the standard free energy required to bring the reactants together in the right configuration for the reaction. Most importantly this includes changes in solvation and X‡ = 1 2 + ∆G 8∆Gin ‡ (normally this last term is small) 0.0 X 0.5 1.0 X ∆G ∆Gin ∆G ∆G 6.61 110 For the present reaction: AH + B- A- + HB ∆G < 0 6.62 k1 k-1 it is easy to see that X‡ < 0.5 and how ∆G‡ is affected by ∆G. g. We shall see in the next chapter that there is an other linear free energy relationship, the Bronsted catalysis law which states for the reaction above that: lnk1 = αlnKeq + C where C = a constant α = the extent of proton transfer at the transition state and this runs from 0 to 1.0 It then suggests that somehow α is related to X‡ in Marcus theory. We will see that this is true later. h. Let us return to the Hammett equation: log k k0     = σρ which leads to δ∆G‡ = δ∆Gρ or δ∆G‡ δ∆G = ρ using the Marcus equation: ρ = ∂∆G ‡ ∂∆G = ∂w ∂∆G + 1 2 1+ ∆G 4∆Gin ‡     and neglecting changes in the work term ρ ≈ 1 2 + ∆G 8∆Gin ‡ We used a very similar treatment before when we looked at the physical basis for the Hammett equation: ρ = sr sp + sr where sr = slope of the reactant parabola at the transition state 111 and sp = the slope of the product parabola at the transition state. Both approaches use equivalent approximations and starting points for their derivation. Both are useful ways to view linear free energy relationships. For example, suppose that the stretching force constant in HA for the proton transfer problem we just finished looking at, then since |sr| is larger, and |sr|/|sp|+|sr| is larger, so ρ becomes larger. 0.0 0.5 1.0 X ∆G 6.63 Notice that the activation energy becomes larger and since ρ is larger, the reaction becomes more sensitive to changes in substituents. Suppose we make ∆G larger. It is clear then that ρ should become larger. 0.0 0.5 1.0 X ∆G 6.64 In this situation notice that the transition state moves in the opposite direction as in the previous case but again the activation energy becomes larger. This points out a general concept called the reactivity - selectivity principle. The less reactive a substance is the more selective it is and vice versa. 114 c. However, sometimes one can have the following situation: A + B C2 C1 ∆E 2 ∆H2 ∆H ∆H1 ∆E 1 6.69 Reaction Coordinate i) Here |∆H1| < |∆H2| as before, ii) but now ∆E1 < ∆E2, so kl > k2 and k-1>>k-2 iii) There will be a difference in the product distri- bution, depending on whether the reaction is carried out under kinetically controlled or equilib- rium controlled conditions. d. To exert kinetic control over a reaction, supply just enough energy for A + B to get over the barrier in the forward direction. i) i.e. use the lowest possible temperature, a short reaction time, mild reagents, a noninteracting (nonpolar) solvent, etc. ii) In the second example above this will favor [C1] vs. [C2]. e. Equilibrium control of this same reaction will favor [C2] vs. [C1]. i) This is because k-2 is very small compared to k-1 ii) That is, by using long reaction times, high tem- peratures, very reactive reagents, polar solvents, etc., one can cause C2 to build up at the expense of C1. G. Isotope Effects 1. Isotope effect arise from the differences in the masses of atomic isotopes. a. They are most evident in the replacement of protium 115 (H) by deuterium (D), because D has a mass twice that of H b. For the most part we will discuss only such replace- ments. 2. Primary isotope effects a. Recall the energy curve for a C-H stretch: P o te n ti al E n er g y rC-H n = 0 n = 1 n = 2 n = 3 vibrational ground state zero point energy correction 6.70 i) For each vibrational level, the energy, εn, = (n + 1/2)hυ, where: (a) h = Planck’s constant, (b) n = 0,1,2,3...are vibrational levels, and (c) υ = frequency of C-H stretch. ii) The frequency of any bond may be expressed as: ν = 1 2π k µ where: (a) k = stretching force constant,and (b) µ = reduced mass = mlm2 / ml + m2. b. Comparing a C- H vs C- D bond being broken: i) µH = l2/13 ii) µD = 24/14 iii) ∴ υD < υH (since µD > µH ), and, as shown below, P o te n ti al E n er g y rC-H 6.71 R. + H. (D.) R-H R-D εo(D) εo(H) ∆GD∆GH ∆νo 116 iv) ε0 (D) < ε0(H) v) It is also clear that ∆GH < ∆GD ∴ (a) kH > kD, or (b) kH / kD > 1 Note: This is a purely vibrational effect. The elec- tronic situation, i.e. the bonding, is identical with the same force constant for H and D. c. A maximum value for kH / kD is approximately kH / kD = e h∆ν0 /2RT (∆υ0 is related to reduced mass difference) i) Considering various bond types: X - H max kH / kD at 25° C C - H 7.0 O - H 11.0 N+- H 6.0 F - H 14.9 M - H 2.7 - 4.2 (where M = a transition metal) These values are related to the value of the vibra- tional frequency itself: if υ is small then ∆υ0 will also be small, etc. ii) The maximum value is reduced somewhat under normal circumstances- see Appendix 2 in your book. In the general situation the H-A bond is not broken by itself but rather abstracted by another reagent, i.e. A_H + B → [A---H---B] → A + H_B 6.72 If the hydrogen atom is transferred from A to B early of late along the reaction path (i.e. X‡ is greater or less than 0.5 using the Marcus approach)then the kinetic isotope effect is reduced. Likewise if the transition state has a nonlinear hydrogen transfer then it is also reduced. 119 i) Not only do ∆υ0 (and ∆υ0‡) depend on the reduced mass difference. they also depend on the frequency of vibration itself. Recall that εn = (n+1/2)hν ii) As the frequency decreases, so do ∆υ0 and ∆υ0‡ . (a) Consider a change in hybridization: C H C H C H νC-H bend = 1340 cm-1 νC-H bend = 1350 cm-1 νC-H bend = 800 cm-1 6.77 (b) On the reaction coordinate this can be repre- sented by: (c) Since ∆υ0‡ < ∆υ0 ; ∆GH‡ < ∆GD‡ and kH / kD > 1.00. (d) Just how much larger kH / kD is than 1, depends on the degree of hybridization change in reaching the T.S. d. General rules for secondary kinetic isotope effects: i) If the % s character increases on going from reactants to T.S, (a) then kH / kD > 1 00. (b) This is a normal secondary isotope effect. ii) If the % s character decreases. (a) then kH / kD < 1.00. P o te n ti al E n er g y Reaction Coordinate 6.78 R-H R-D εo(D) εo(D) εo(H) εo(H) ∆GD ∆GH ∆νo ∆νo 120 (b) This is an inverse secondary isotope effect. e. Examples: Ph C Cl (D) H H C OH (D) H Ph H2O + Ph C Cl H (D) H H2O δ+ δ- H2O C Ph H (D) H + + Cl - kH/kD = 1.30 + + CN - C OH (D) H Ph δ+ NC δ- C OH (D) H Ph NC kH/kD = 0.73 6.79 5. Other isotope effects. a. For example H → T. Now reduced mass is three times as large. However, T is radioactive so this reac- tion is not carried out on a molar basis!The techniques are different. Radioactivity/scintillation counters are used primarily for biochemical reactions where the concentrations are very low. b. 13C → 12C, 34S → 32S, etc. The reduced mass ratio is 4. JACS, 121, 3933 (1999) - a nice example of 1° and 2° isotope effects. The calculated reaction of CCl2 and propene is shown below: C H C H CH3 H (0.909) 0.947 (1.027) 1.026 (0.972) 0.913 (1.005) 1.004 (0.986) 0.989 (theory) experiment 6.79b The transition state is not well defined. ∆G†=10.1 kcal/mol, however ∆S†=-36 eu so ∆H† = -5 kcal/mol. A negative activation barrier! As can be seen there is pretty good agreement between experiment and theory - the two C-C bonds do not form at the same rate. Why does the one bond form before the other? Notice that the carbene lies down on its side when it approaches the olefin. We will discuss this feature in Chapter 8. 121 now much closer to unity. However, one can use them for primary isotope effects which are easier to analyse (and understand) since the bond being broken be- comes the dominate portion of the reaction path. H. An Example of a Very Small Barrier - J. Am. Chem. Soc., 120, 10423 (1998). 1. The IR spectrum of norbornadiene-Fe(CO)3 in the CO stretching region shows a very interesting temperature behavior. At low temerature there are three peaks which correspond to that expected for a molecule with Cs symmetry. 2. At higher temperatures there are two peaks in the ratio of 2:1. This is consistent with an Fe(CO)3 group having C3v symmetry. 3. A mechanism for this behavior is shown below on the right side. 4. The IR spectrum as a function of temperature is shown below on the left side. This can be simulated (just like NMR spectra) so that there is a rate constant for the rotation process illustrated at each tempera- ture. This is shown below on the left. The rate con- stants then can be used to make an Eyring plot on the right side. The ∆H‡ of 0.7 kcal/mol is extraordinarily small to have been measured.