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 transition 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.

B. There are several "rules" that we should briefly discuss on how chemists propose a reaction mechanism. When proposing 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 22^nd 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 Æ 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 experimentally with the kinetics of the reaction of interest.

a. Unimolecular A Æ B

i) d[B]/dt = k[A] = -d[A]/dt

or kdt = - d[A] / [A]

where k = the rate constant for this AÆB reaction

ii ) In integrated form:

kt = In ([A₀]/ [A]), where [A₀] = initial concentra- tion of A

iii ) Plotted:

half-lives: at t_1/2 2[A]=[A₀]

then k=ln2/t_1/2

this is a crude (and easy) way to get an estimate of the rate constant , k, by measuring t_1/2

b. Bimolecular 2A Æ B

i) d[B]/dt = k[A]², or in integrated form

2kt= l /[A] - 1/[A₀] .

ii ) Plotted

c. Bimolecular: A + B Æ C

i ) d[C] = k [A][B] or

ii ) In many cases we will have much more complex rate equations. There are general solutions for all of them, which we shall not cover here.

d. This is the most simple situation for a two-step reaction:

A Æ B Æ C

k₁ k₂

It can be shown by some tedious math that

2. Simplifying approximations

a Rate determining step

i ) Let's now take a look at a specific, multistep reaction to see how the prior rapid equilibrium and

the rate determining step approximations are used.

ii) Reaction:

iii) Analysis

(a) step 1: d[C]/dt = k_l [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 << k_l, 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 t¹me (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] = k_l/k_-1

(ii) \ [C] = (k₁/ k _-1) [A][B].

(iii) The overall rate = k₂ (k₁/ k _-1) [A][B] = k_obs [A][B]

vi ) An example is the acid catalyzed nucleophilic substitution of alcohols:

(a) Assuming the 1st step is a rapid equilibrium:

d[RBr]/dt = k₂ [ROH₂⁺] [Br]

(b) Now

(i) Thus [ROH₂⁺] = k₁/k ₁[ROH][H⁺]

(ii) and rate = k₂k_l / k ₁[ROH][H⁺ ][Br ] .

v¹i ) 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 ÆC;

rate = k [A][B]

i ) If [B] >>[A], then rate = k_obs [A], where k_obs = 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

k_l [A][B] = k _-1 [C] + k₂ [C] [D],

(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 @ k_l [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 enzymesubstrate complex (E·S) which rapidly attains a steady state concentration:

(a) At steady state, defined as d[E·S]/dt = 0,

(i) the rate = k₂ [E·S] ( = d[P]/dt ) and,

(ii) k_l [E][S] = k ₁ [E·S] + k₂ [E·S].

Therefore, V = rate =

The problem here is that at any given time the [E] will be essentially impossible to measure. We can get around this in the following way.

(b) Knowing that [E₀] (the concentration of enzyme initially added) = [E] + [E·S], we can substitute into the steady state equation.

(i) k_l [E₀][S]-k_l [E·S][S] = (k_-1+ k₂) [E·S]

(ii) [E₀][S] - [E·S][S] =

This can be rearranged to:

let the Michaelis constant (not an equilibrium constant). The substituting back into the initial rate expression leads to:

iii) This is the MichaelisMenten equation. It ex- presses the rate of the reaction in terms of several constants, and only one quantity which varies during the reaction, [S].

iv) Under special circumstances the MichaelisMenten equation reduces to simple

expressions.

(a) If [S] << K_m, then V= = k_obs [S], a pseudo first order process.

(b) If [S] >> K_m, the rate = k₂[E₀] .

(i) This is the maximum possible rate, or V_max

(ii) At V_max the reaction is zero order in substrate concentration, and the enzyme is considered to be "saturated" (hence: saturation kinetics).

(iii) In other words, at V_max there is essentially no free enzyme;

[E·S] ~ [E₀].

(c) If [S] = K_m, then

(i) This gives a practical way of determining K_m; it is simply equal to the concentration of the substrate when the velocity of the reaction is half-maximal.

(ii) K_m is easily determined graphically. (Note: K_m has the dimensions of concentration. Does this make sense?)

(d) Cases (a), (b) and (c) above are indicated on the graph of V vs. [S].

(e) Finally, if k_l and k_-l >> k₂ then [E^.S] = k_l/k_-I [E][S].

(i) This is equivalent to the case of rapid prior equilibrium discussed previously.

(ii) In this case only, K_m can be treated as equivalent to an equilibrium constant, that is,

K_m = 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 Æ 2X

(b) The second step is autocatalytic in Y, and de- stroys X:

k₂

step 2: X +Y Æ 2Y

therefore, d[X]/dt = k_l [A][X] k₂ [X][Y] (Equation 1)

(c) The third step involves the destruction of Y:

k₃

step 3: Y Æ 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 ~ k_l [A][X].

(ii) [X] increases rapidly.

(c) But the 1^st 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 = t_l, [X] reaches its maximum

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.

iv) A plot of [X] and [Y] (and [P]) as a function of time is shown below.

v) It can be shown that the solution of the three rate equations is:

k₂([X] + [Y]) - k₃ In [X] - k_l [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 Æ 2X

X+ Y Æ 2Y

Y Æ 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:

(b) The environment can be depicted ( where E' & E" are ionized, inactive forms of enzyme.)

(c) Analysis:

(i) k_p, k_s, k_H+, are diffusion constants through the cell wall.

(ii) Initially, there is no S inside the cell and most of the enzyme is in an ionized form, i.e.

[E] << [E'], [E"].

(iii) As S migrates into the cell, its concentration builds up.

(a) S migrates into cell rather slowly; a realistic situation would be: k >> k_s > k_p ~ k_H+

(b) Initially, however, there is little chance for S to encounter E

(iv) As molecules of S do encounter E, there is a generation of H⁺.

(v) The H⁺ does not diffuse out of the cell, but rather converts E" to E' and E' to E, through their respective equilibra.

(a) The rate of consumption of S increases.

(b) [S] decreases, until there is virtually no more.

(vi) With no more S available, the enzymatic gen- eration of H⁺ ceases.

(a) H⁺ and P diffuse from the cell.

(b) The majority of enzyme returns to the E' or E" forms

(c) The initial state is once again reached.

(d) Graphically, this behavior looks much like a heartbeat.

(i) Notice that there is again an autocatalytic cycle

(ii) The cell wall acts as an inhibitor in a feedback loop.

D. Absolute Rate Theory

There are three ways, in general to display a potential energy surface which is a plot of the potential energy for the molecule(s) as a function of geometrical variables that change during the course of a chemical reaction. These are shown below for an arbitrary reaction where a reactant, R, goes first to an intermediate, I, and then to a product, P, by means of two transition states, TS.

The potential energy surfaces shown above show what happens when thiophene is oxidized.

1. Let us consider the most simple reaction:

H^. + H-H Æ H-H + H^.

a.The reaction potential energy surface is:

An expanded view:

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 E_a is higher in this case.

(b) Actually, the activation energy is the lowest for q = 180°.

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

turned into translational energy after crossing the barrier.

(c) The trajectories shown above are calculations from the Cl_a^- + CH₃-Cl_b Æ Cl_a-CH₃ + Cl_b^- 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.

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.

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 coordinates.

(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

coordinates.

(b) The energy gradients are uniformly positive.

(iii) The transition state

(a) Uniquely defined by the energy gradients.

(b) The energy gradients are positive in every set of directions except one.

(c) The energy gradients are negative along one line (the reaction path), in both directions.

(iv) Both the transition state and the energy minima are called stationary states.

(v) The reaction path

(a) This interconnects the stationary states.

(b) It is much less well defined, in general, for a polyatomic system.

(c) We won't go through all of the technical prob- lems associated with this.

2. The BürgiDunnitz approach

a. The BürgiDunnitz approach is an experimental way to get a rough idea about what happens along the reaction path.

(i) This uses crystal structures to map that path.

(ii) For any reaction, the least energy path must be unique.

(a) In moving away from the minimum, the energy rises more steeply in any direction other than the reaction path.

(b) An external or internal force will cause a mol- ecule to readjust its geometry in the energetically least costly direction, i.e. along the reaction path.

b. Example: ring whizzing in cyclopropeniumML₂+ complexes. Consider the following molecules and the reaction:

i ) The full structures of three complexes are shown below.

(a) Note that the relative orientations of the phenyl rings on the triphenylphosphine rings change.

(b) This causes differences in the intramolecular contacts, which cause the basic structure to change.

ii ) The structures below show just the three cyclopropenium ring carbons, the metal and the two phosphines, from the top. We see a gradual progression from a ground state, where the metal is bonded to two carbons, to just about the transition state where the metal is bonded to all three car- bons.

iii) These results agree nicely with the computations of the potential energy surface for the reaction (next page left), and the projection of the P₂M plane on the cyclopropenium ring, as the species travels along the reaction coordinate (next page right).

iv ) While this experimental method offers direct evidence for what structural changes occur, it does not, of course, tell us anything about the energies or rate constants associated with the reaction.

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, substituents 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: k_obs = A e^-Ea^/RT

b. This is called the Arrhenius equation

(i) A = preexponential factor, related to DS

(ii) E_a = activation energy

(iii) R = gas law constant

(iv) T = temperature in Kelvin

c.Plotting ln ( k_obs ) vs 1/T:

(i) The slope = - E_a/R

(ii) The intercept = ln A .

d. Arrhenius estimates of half-life for the reaction

AÆB:

(i) Using A = e ^{kT/ h} (assumes DS = 0, see below), and

(ii) defining: t_1/2 = time when [A] = [B], and t_1/99 = time when 99[A] = [B].

_________________________________________________

E_a T=50°C T= 100°C_______

t_1/2 t_l/99 t_1/2 t_1/99

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 Æ B where A is the transi- tion state,

(a) K = [A]/[A], and

(b) k_l = kkT K

h

k = Boltzmann's const.

h = Planck's const.

T = temp °K

k= 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, k < 1.

(c) Go over this in Appendix 1, Chapter 2, in your book

(ii) Recalling that DG = RTln K

(a) where DG = DH - TDS

(b) then

(c) Where:

(i) DG is the free energy of activation.

(ii) DH is the enthalpy of activation (the potential energy difference between the reactants and the transition state).

(iii) DS 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)

(ii) E_a = DH + RT

(iii) DG, DH & DS can be computed by plotting ln(k/ T) vs. 1/T .

(a) slope Æ DH (small error)

(b) intercept Æ DS (large error extrapolation goes far outside the experimental region of tempera- ture)

(c) DH and DS errors tend to compensate, giving DG a smaller error than DS.

iv ) Note: the value of DG depends on the tem- perature, whereas DH and DS are temperature independent quantities.

d. Rate dependence on DH and DS

i ) Lets say we have two competing reaction paths:

A ¨ C Æ B

k_a k_b

and that DS is the same for both reactions.

ii ) Then for the following rate ratios:

k_a/k_b DH_a DH_b

2 0.41 kcal/mole

10 1.37 ¨Very large rate differences for

10⁴ 5.49 very small changes in DH.

(a) Variations of DH, by themselves, do not imply much about the structure of the transition state.

(b) Perhaps one can get clues as to what is happen- ing electronically when close comparisons are made.

iii ) On the other hand, if DH is held constant then:

k_a / k_b DS_b - DS_a

2 1.4 e.u. [cal/(mole °K)]

10 14.5

10⁴ 18.3

iv ) DS does provide clues about the structre of the transition state.

(a ) Exp.1: a typical SN₂ reaction, DS = - 42 eu

(i) DS is negative, i.e. the T.S. is highly ordered.

(ii) There is a lower degree of vibrational and rotational freedom than in the reactants.

(b) Exp. 2: a typical bond breaking reaction, DS = +10 e.u.

CH₃-CH₃ Æ 2CH₃· DS = +17 e.u.

(i) The positive value of DS indicates a less ordered T.S.

(ii) Here, the more positive this number is the more closely the T.S. resembles the products. One particle becomes two. The number of rotational and vibrational degrees of freedom increases.

(c) Example three: two special cases where DS has atypical values. We talked before about whether cyclobutadiene was a square structure where there are two different resonance structures, or is it a rectangle which undergoes a conversion from one rectangular form to another. Berry Carpenter has devised a very clever experiment to probe this issue.

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 k₂k₁, 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

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

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 DH = 4.6 kcal/mol and DS = -15 eu. This is a very big differ- ence between the experimental and calculated value and, furthermore, the large negative value of DS 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

so one could express the amount of tunnelling by how much k was greater than 1. Using this it was found that k = 100 at -10 °C and 800 at -50 °C. A second example is given by a simple bond-breaking reaction.

The activation energy for this reaction is small because the compound is strained. The important point is that DS is small in magnitude unlike the

CH₃-CH₃ Æ 2CH₃^. example that was presented previously. Here the two radicals are "held to- gether". However, the reaction shown below has been studied and DS was found to be very differ- ent. An explanation that has been advanced for the difference can be outlined as follows.

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

ordering is reversed. Therefore, the reaction path needs to undergo spin-forbidden crossing:

The molecule in this case is a substituted trimethylenemethane. As shown below, the p orbitals are very straight-forward. With a total of four p electrons, two go into the lowest level and that leaves two for the degenerate pair. In this case it is energetically much more favorable (~10 kcal/ mol for trimethylenemethane itself) for the triplet state where the two electrons are unpaired.

In order for the molecule to undergo a state crossing there must be a coupling of vibrational and electronic states.

(i) The number of possible nuclear configurations in which this is allowed is quite small.

(ii) Thus DS becomes much more negative.

(iii) Since DS becomes more negative, DG be- comes larger, that is, the apparent bond energy is

larger.

In fact in the example shown above the strain is apparently so large that the bond energy is nega- tive(!), i.e. DH is negative but so too is DS so that DG becomes positive.

e. Diffusion controlled reactions

i) These are bimolecular reactions where every encounter is successful.

(a) E_a < kinetic energy of molecules in their first vibrational state (0 to ~3 kcal/mol).

(b) The rate constant is the collisional rate con- stant.

ii) In solution, the rate constant depends only on the viscosity of the solvent:

h = viscosity of solvent

d_A-B = collision distance at the activated complex

s = mean effective radius -related to diffusion

iii) For typical solvents at 25°C:

k~ 10¹³ if E_a ~ 0

k~ 10⁹ - 10^l0 if E_a~ 2-3 kcal/mol

iv) Examples:

CH₃· + CH₃· Æ CH₃ - CH₃ (gas) k=2x 10¹⁰

CH₃· + CH₃· Æ CH₃ CH₃ (H₂O) k=3.2 x 10⁹ (E_a~ 3 kcal/mol)

H⁺ + ^-OH Æ H₂O (H₂O solvent) k= 1.4 x 10¹¹

I· + I· Æ I₂ (H₂O solvent) k= 8.2 x 10⁹

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:

DV = V₃ - (V₁ + V₂)

DV = V₄ - (V_l + V₂)

(ii) The closer DV is to DV, the more the transition state resembles the product, C.

(iii) The closer DV is to zero, the more the transi- tion state resembles the reactants, A + B.

(b) DV can be measured by measuring reaction rates at different pressures (keeping the tempera- ture constant at each pressure).

(i) [d(ln k)/ dP]_T = - DV/RT, where P = pressure (atmospheres).

(ii) Plotting In k versus P gives a straight line with slope of DV / RT.

(iii) A negative value of DV implies a tight, structur- ally rigid transition state where the molar volume of the transition state is less than that of the reac- tants:

DV = - 33 cm³/mol

DV = - 37 cm³/mol

iv) A positive value of DV implies a loose, structur- ally flexible transition state:

DV = + 15 cm³/mol

DV = + 60 cm³/mol

v) The interpretation of the magnitudes of DV are not always so straight-forward. For example in the reaction below

DV = - 44.7 cm³/mol

DV = - 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 DV (and DV) should be positive. However, in a polar solvent the formation of the ions causes the solvent shell to contract. Therefore, DV (and DV) is expected to be much smaller. This is called electrostriction and it is a big effect. For the reac- tion shown above DV might be close to zero and might even be negative.

vi) The drawbacks in measuring DV are basically twofold:

(a) The temperature must be kept very constant throughout the pressure range studied.

(b) It is a small effect. (Suppose DV = -15 cm³/mol.;

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 E_a (kcal/mol)

classical isolation 1 hr - weeks ~25 - ~45

Stopflow technique 10^-4 sec - min 10 - 20

NMR 10⁶ - 100 sec 5 - 25

ESR 10^-10 -10⁵ 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:

Often times the first step is a proton transfer. Using the steady state approximation for [B] we have:

Then using the fact that the first step is fast k_-1 > k₂[C] and .

a. Now let K = k_l/k_-1 = [B]/[A][E], and

b. [d(lnk)/dT]_P = - DH/RT^2, where DH = enthalpy change for the 1^st step.

c. Suddenly changing T (when DH 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

changes by 1 to 10°C.

b. Concentration changes can be measured by normal optical methods, i.e. UV vis spectroscopy, etc.

3. Suppose DH is negative for the above reaction.

a. A temperature increase will shift the equilibrium in the first step to the left.

b. Suppose we can measure [A]:

4. Likewise a sudden pressure change can perturb the equilibrium:

[dlnK / dP]_T = DV / RT

F. Rough Descriptions of Transition State / Reaction Path Changes

1. Principle of Least Motion

a. Stated first by Müller and Peytral in 1924 and then later, more precisely by Rice and Teller.

b. Basically this says an activation energy will be lowest when the atoms change their positions as little as possible, or, given two possible reaction paths, the one requiring the least nuclear motion will be energetically favored.

i) The basic idea is quite simple - the form of any reaction path is approximately given by two parabolas.

ii) The closer they are to each other (i.e. smaller span that the reaction coordinate must take), the lower the value will be of their intersection point (the activation energy).

iii) This is illustrated below:

Clearly, E_a for A' Æ B' is less than E_a for A Æ B.

c. This is obviously a very crude idea, but it works quite well for most cases.

d. There is a lot of common sense in this idea:

i) It costs energy to stretch, bend and rotate around bonds from their equilibrium - ground state - geometries to approach the transition state.

ii ) The fewer changes in these coordinates that need to be made, the smaller will be the energetic cost to deform them to a productive, i.e. reacting, geometry.

e. If one really examines what lies in back of the prin- ciple of least motion, it implies that the transition state structure can be viewed as a resonance combination of the electronic stucture of A and B in the previous case.

i ) The closer that A structurally resembles B, the more resonance should occur.

ii ) Consequently, the transition state becomes stabilized more.

f. There are instances when the principle of least motion breaks down; the most obvious is in a reaction of an atom with a diatomic molecule: A+BC Æ AB + C. From the potential energy surface on the next page, it would appear that the B-C bond should stretch before A comes close to atom B. Clearly this is in error. Actually the principle of least motion is only a qualita- tive indicator of what the reaction path should look like.

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:

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:

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:

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 Æ 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 dDE⁰ (a fraction of DE⁰).

(a)The change will be transmitted in a linear way along the reaction coordinate.

(b) This can be expressed as m· X = dDE⁰, where m is the ratio of the energy ¹ncrease (dDE⁰) 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:

(i) The line of slope m represents the incremental increase in energy due to dDE⁰, 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 (dDE⁰ is positive), so m is positive, and X (as defined in the diagram) is positive. The arrows indicate the value of dDE⁰ (when X is a negative number then dDE⁰ is negative, i.e. stabilizing and when X is a positive number then dDE⁰ 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:

In other words, the dashed line perpendicular to the reaction path at the transition state in this case corresponds to the symmetric vibrational mode.

(b) The cross-section of the energy surface along the dashed line (along the one dimensional vibra- tional coordinate, Z again defined at Z=0 for the thermoneutral case at the transition state), also gives a parabolic shape:

ii) Now apply a perturbation (dDE⁰) which makes the vibration more difficult in the stretching direc- tion.

(a) dDE⁰ = m·Z (where m = a positive number),

(b) We can again apply a line of slope m to the transition state to see what happens.

(i) Raising the energy in the direction towards the right side of the vibration coordinate moves the transition state to the left.

(ii) The values of r_AB and r_BC will be smaller in the transition state.

(iii) Decide for yourself the effect of raising the energy to the left.

c. The net result is that a perturbation which lowers the energy in a direction along the reaction coordi- nate shifts the transition state structure away from the energy lowering, but for directions perpendicular to the reaction coordinate, the transition state will shift towards the energy lowering.

d. Let us consider the elimination of H-X from an alkyl halide as an example that illustrates the utility of this way of looking at energy changes and their conse- quences on reaction paths. Below is an idealized repre- sentation of the potential energy surface for a base, B:^-, reacting with an alkyl halide, R-X to form the proto- nated base, B-H⁺, an olefin and X^-. This is a two-dimen- sional contour surface where the two major variables are plotted: the distance between the base and the proton being pulled off from the alkyl halide, r_B-H, and the C-X bond distance, r_C-X. There are therefore, two minima corresponding to the reactants and products on the upper left and lower right of the diagram.

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:

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 r_C-X (the C-X bond will be more broken) and larger values of r_B-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

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, r_C-X does not change while r_B-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 S_N1 reaction:

R-X Æ R⁺ + X^-.

a. On the left side are the energies of the ground and first excited state of R-X. In the ground state the two

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 S_N1 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 b.

b. Suppose that R⁺ is stabilized:

The first excited state of the reactant and the ground state of the product is stabilized. If we make the assumption that the configuration interaction/reso- nance term, b, is not changed, then the resultant transition state (TS₂) is moved closer to the reactant and occurs with a lower activation energy. It is clear that the stabilization of either the product ground state or the reactant excited state will be linearly related to a stabilization of the transition state.

c. Suppose we move now to a more "complicated" reaction, the S_N2 process. This is modelled by Y:^- + R-X Æ Y-R + ^-:X. The first excited state is one where an electron has been transferred from the nucleophile, Y, to the R-X antibonding orbital.

That energy difference between the ground and ex- cited state is a function of the ionization potential of Y, I_Y, and the electron affinity of R-X, A_R-X. Raising the electron affinity of R-X and R-Y by modifying R will as shown lower the activation energy. Making only the

electron affinity of R-X to change (by modifying X) will do the same, however, the transition state will shift to the left. It can also be seen that in any case

DE = f(I_Y - A_R-X) - b.

5. Marcus theory. This was originally developed in the 1960's by Rudy Marcus to explain electron transfer. It has been more recently used to investigated proton transfer, atom transfer reactions and S_N2 reactions.

a. Let us consider a proton transfer (acid-base) reac- tion which is exothermic:

b. Two "identity reactions" are used:

c. The primary idea behind Marcus theory is that the activation energy and the position of the transition state is a function of the intrinsic reaction barrier, which is an average of the two identity reactions, and the overall thermodynamics of the reaction, DG.

d. For this case we construct two sets of parabolas for the identity reactions as follows:

In each case the reaction coordinates have minima at X = 0 and X = 1 and the transition state occurs at x = 0.5. Note that DG for reaction 1 is greater than that for reaction 2. Why?

e. The same two parabolas are then drawn for the averaged case where the intrisnic reaction barrier is defined by

The product parabola is lowered in energy by an amount DG, given by the dashed line. The activation energy is expressed as DG 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

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

(normally this last term is small)

For the present reaction:

it is easy to see that X < 0.5 and how DG is affected by DG.

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 = alnK_eq + C

where C = a constant

a = the extent of proton transfer at the transition state and this runs from 0 to 1.0

It then suggests that somehow a is related to X in Marcus theory. We will see that this is true later.

h. Let us return to the Hammett equation:

using the Marcus equation:

and neglecting changes in the work term

We used a very similar treatment before when we looked at the physical basis for the Hammett equation:

where s_r = slope of the reactant parabola at the transition state

and s_p = 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 |s_r| is larger, and |s_r|/|s_p|+|s_r| is larger, so r becomes larger.

Notice that the activation energy becomes larger and since r is larger, the reaction becomes more sensitive to changes in substituents. Suppose we make DG larger. It is clear then that r should become larger.

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.

h. One interesting concept from Marcus theory is that when reactions become very exothermic (DG be comes smaller) there reaches a point when DG is zero, i.e.

when DG = -4DG_in then DG = 0.0

(i) As shown in the graph on the left when this condition is met not only does DG become zero but also when the reaction becomes even more exothermic the value of DG rises! This is called the inverted region. One can graphically show this by the plot on the right side. Three values of DG are shown. The lowest parabola on the reactant side shows a very small value of DG. The middle parabola is one where DG 0.0. Finally destabiliz- ing the reactant parabola even more causes DG to rise again. Now the intersection of the parabola occurs on the left side.

(ii) This does in fact occur. The experiment shown below relates to electron transfer in a series of organic molecules. The biphenyl group on one side of this steroid is selectively reduced. The electron is then transferred to the acceptor group by a slow enough process that the overall rate of the reaction can experimentally be measured. Once the DG for the reaction is ~ 1.5 eV there is a decided peak in the observed rate constant (~ 5^.10⁹ which, recall from the discussion on diffusion controlled reactions, corresponds to E_a ~

0.0 kcal/mol).

4. Kinetic versus equilibrium control.

a. Consider the following set of generalized reactions:

b. The normal situation for this system is:

i ) Here |DH_I | < |DH₂| and DE_I > DE₂,

ii ) so k_l < k₂ and [C_l] < [C₂] under all condition,.

c. However, sometimes one can have the following situation:

i) Here |DH₁| < |DH₂| as before,

ii) but now DE₁ < DE₂, so k_l > 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

(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:

i) For each vibrational level, the energy,

e_n, = (n + 1/2)hu, where:

(a) h = Planck's constant,

(b) n = 0,1,2,3...are vibrational levels, and

(c) u = frequency of C-H stretch.

ii) The frequency of any bond may be expressed as: where:

(a) k = stretching force constant,and

(b) m = reduced mass = m_lm₂ / m_l + m₂.

b. Comparing a C- H vs C- D bond being broken:

i) m_H = l2/13

ii) m_D = 24/14

iii) \ u_D < u_H (since m_D > m_H ), and, as shown below,

iv) e_{0 (D)} < e_0(H)

v) It is also clear that DG_H < DG_D \

(a) k_H > k_D, or

(b) k_H / k_D > 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 k_H / k_D is approximately

k_H / k_D =

(Du₀ is related to reduced mass difference)

i) Considering various bond types:

X - H max k_H / k_D 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 u is small then Du₀ 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.

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.

iii) Examples:

iv) In rare instances k_H/k_D can be less than 1.0, e.g.

Here the W-H stretching frequency is only 1845 cm^-1 whereas, the C-H stretching frequency is ~2900 cm^-1. Therefore, the WHC stretching frequency at the transition state which becomes later as k_H/k_D becomes smaller, may well be greater than that for W-H in the reactant. Using the equation previously given, the max k_H/k_D = 3.5. In the reaction: CpW(CO)₃H + CpW(CO)₃^- Æ CpW(CO)₃^- + CpW(CO)₃H, the k_H/k_D = 3.7.

v) Also in extremely rare instances tunnelling may increase k_H/k_D greatly since the probability of tunnelling increses with mass in an exponential manner.

(a) Here protons tunnel but D's do not.

(b) Values of k_H / k_D as high as 50 at room tempera- ture have been observed.

(c) Tunnelling will occur only if there is a narrow barrier (not much configurational space must be transversed).

3. Secondary kinetic isotope effects

a. These may occur when the bond involv¹ng the substitution (e.g. D for H) is not broken.

i) 1.0 < k_H / k_D < 1.5 Æ normal secondary isotope effect

ii) 0.7 < k_H / k_D < 1 Æ inverse secondary isotope effect

b. Consider a reaction which does not involve any change in the zero point energy on going from reac- tants to T.S.:

i) If Du₀ = Du₀ then DG_H = DG_D and k_H / k_D = 1.00.

ii) However, this is usually not the case for reac- tions which occur at the carbon upon which the isotopic replacement is made.

c. A secondary isotope effect results when Du₀ and Du₀ are not equal because of changes on going from ground state to transition state.

i) Not only do Du₀ (and Du₀) depend on the reduced mass difference. they also depend on the frequency of vibration itself. Recall that

e_n = (n+1/2)hn

ii) As the frequency decreases, so do Du₀ and Du₀ .

(a) Consider a change in hybridization:

(b) On the reaction coordinate this can be repre- sented by:

(c) Since Du₀ < Du₀ ; DG_H < DG_D and k_H / k_D > 1.00.

(d) Just how much larger k_H / k_D 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 k_H / k_D > 1 00.

(b) This is a normal secondary isotope effect.

ii) If the % s character decreases.

(a) then k_H / k_D < 1.00.

(b) This is an inverse secondary isotope effect.

e. Examples:

4. JACS, 121, 3933 (1999) - a nice example of 1° and 2° isotope effects. The calculated reaction of CCl₂ and propene is shown below:

The transition state is not well defined. DG=10.1 kcal/mol, however DS=-36 eu so DH = -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.

5. Other isotope effects.

a. For example H Æ 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 Æ ¹²C, ³⁴S Æ ³²S, etc. The reduced mass ratio is

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 C_s 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 C_3v 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 DH of 0.7 kcal/mol is extraordinarily small to have been measured.