10 WHY DO REACTIONS HAVE ACTIVATION BARRIERS? I'RECIS \ knowledge of activation barriers plays a critical role in chemical kinetics. In Chaptei 5 «i- showed that if one knows how to estimate activation barriers for a series of elemental v reactions, one can predict the mechanisms of the overall reaction. In Chapters 7-9, we showed that if one knows the activation barriers and potential energy surfaces, one can predict the rate constants for each of the elementary steps in the mechanism. So far, we have assumed that we knew the potential, and have used that knowledge to make predictions. In the next two chapters, we will concentrate on finding ways to predict activation barriers for reaction. The main focus of this chapter (10) will be to provide an overview i>l why chemical reactions have activation barriers. We will focus on the general models thai people have proposed to estimate activation barriers for elementary chemical reactions. Some of the ideas are old. However, some of the ideas are more modern and nol completely established. Most of the discussion in Chapter 10 will be qualitative. We will quantify the ideas in Chapter 11. 10.1 INTRODUCTION The idea that reactions are activated dates back to Faraday's work in 1834. Faraday proposed that chemical reactions are not instantaneous because there was an electrical barrier to reaction. Faraday postulated that molecules were held apart by an electrical force and that one needs to overcome that force in order for reaction to happen. Faraday's ideas were proposed 75 years before people could measure an activation barrier. Consequently, Faraday never quantified his ideas. However, Faraday's ideas still were very important in the development of kinetics. Not much progress was made in quantifying the activation barriers until Arrhenius' work appeared in 1889. Arrhenius (1889) proposed that reactions would follow 571 vim. VVI IT I II I Ml Al . 1 II IINM I IAVI /M II V A 111 »N 11/M M 111 I I' ' HAMIIII M'./v.M ii IA II 11 Will I IK INI 11 X II N'ilnNM B73 \ 111 n 11111 v. * law Kalo ■ rule,, x e 1 ...... where rate is the rale of reaction, Li, is the activation barrier, klt is Boltzmann's constant! T is the temperature, and rate,, is a constant. Arrhenius also derived equation (10.11 l>v assuming that only hot (i.e., excited) molecules could react. According lo Arrhenius, ihc activation barrier was associated with the energy the molecules needed to gel them I it x enough to react. Later Bodenstein expanded on these ideas. Bodenstein noted that overall read ions occur via a series of elementary steps where bonds break and form. Bodenstein showed that Arrhenius' law is applicable only to elementary reactions. Overall reactions often show deviations from Arrhenius' law. 10.1.1 Models for Activation Barriers for Elementary Reactions Over the years many models were proposed to understand why elementary chemical reactions are activated. In 1935, Polanyi and Evans discussed the idea that bonds stretch during elementary reactions. The stretching causes a barrier. See Evans and Polanyi (1935, 1936, 1937, 1938). Bonds also break. The bond scission also causes a barrier. More recent studies indicate that activation barriers are subtle. There are many different reasons why activation barriers arise during elementary chemical reactions, the main causes of these effects are listed in Table 10.1. Also • Bonds need to stretch or distort during reaction. It costs energy to stretch or distort bonds. Bond stretching and distortion is one of the major causes of barriers to reaction. • In order to get molecules close enough to react, the molecules need to overcome Pauli repulsions (i.e., electron-electron repulsions) and other steric effects. The Pauli repulsions are another major cause of barriers to reaction. • In certain special reactions, there are quantum effects that prevent the bonds in the reactants from converting smoothly from the reactants to the products. Quantum effects can produce extra barriers to reaction. • There are also a few special cases where the reactants need to be promoted into an excited state before a reaction can occur. The excitation energy provides an additional barrier to reaction. For example, consider the following elementary reaction: H + CH,CH2-> H2 + C2H4 (10.2) Table 10.1 Principal causes of barriers to chemical reactions Bond stretching and distortion Orbital distortion due to Pauli repulsions Quantum effects Special reactivity of excited states Figure 10.1 The changes in the geometry during reaction (10.2). Figure 10.1 shows the changes in the geometry during the reaction. I'irsi ihe hydrogen needs to come in and get close enough to react. Then the hydrogen is transferred I a < >m a carbon atom to the hydrogen. Simultaneously, the other hydrogens Nip up, the i itrbon carbon bond shortens, and an orbital on each carbon rchybridi/.es to form .i I (pi) bond. There are barriers to getting the incoming hydrogen close enough lo the methyl gioup lo react. There are barriers to the stretching of the carbon-hydrogen bond, Then i .i Pauli repulsion associated with the shortening of the carbon-carbon bond. Then i also a quantum effect associated with the formation of the double bond. All of ilusc i iiriis occur simultaneously, and so it is difficult to estimate a barrier for a reaction like rcuction (10.21. One can use quantum-mechanical methods, as described in Chapter II. lo call LllatC Ihc strength of the barrier. However, you do not necessarily know why the barriei arl n from such calculations. Therefore, in this chapter, we will break up the problem and look ill one effect at a time. I want to say at the beginning that, while there are many theories about why ban nr. arise in chemical reactions, the models have not yet been completely quantified excepl In special cases. Consequently, while we do have many models, one still does nol always know which model applies in each situation. 10.2 BARRIERS ASSOCIATED WITH BOND EXTENSIONS In the remainder of this chapter we will discuss each of the effects in Table 10.I. To start off, we will discuss the barriers that arise because bonds need to be distorted during an elementary reaction. The idea that bond stretching could cause barriers to reaction was first discussed by Evans and Polanyi (1935, 1936, 1937, 1938). Polanyi argued that bonds are extended during reaction, and that this phenomenon caused barriers. Polanyi had previously found empirically that barriers to reaction for similar reaction could be fit to what is now called the Polanyi relationship: Ea=E°+yPAHr (10.3) HI MIVAIIIMMMI IMI I'll! ANYI Ml I Allc IM'lllll' B70 where lIs oiled the Intrlnsh at tivation barriei and yP is called the transfer coeffit ient Polanyi ami I Ivans ilcrivcd equation ( 10.3). I lie derivation is given in Section 10.3. Later Marcus (1955, 1964, 1968, 1969) showed that a better approximation is given by AHr 4E° K" (10.4) Equations (10.3) and (10.4) are the key equations that people use to determine how bond stretching affects barriers to reaction. In the next several sections we will show how equations (10.3) and (10.4) arise, and describe when the equations work and when they fail. 10.3 DERIVATION OF THE POLANYI RELATIONSHIP First, let us derive the Polanyi relationship, equation (10.3). The Polanyi relationship was first derived by Evans and Polanyi in 1935. At the time, there was a lot of interest in acid-catalyzed reactions. During an acid-catalyzed reaction an acid reacts with a stable species to yield products. For example, consider the acid-catalyzed reaction CH3CH2OH => C2H4 + HzO (10.5) and assume that the reaction is taking place in aqueous solution, so that the reaction is catalyzed by hydronium ions, [H3Ol+. Reaction (10.5) has been examined at great depth in the literature. The first step is a proton transfer from the hydronium to the ethanol: CH3CH2OH + [H30]+ Then the protonated ethanol loses water: [CH3CH2OH2]+ + H20 (10.6) [CH3CH2OH2]+ + X-► [CH3CH2]+ + H20 + X (10.7) Then another water reacts with the ethyl group to regenerate the hydronium: H20+ [CH3CH2r [H3Oj+ +CH2CH2 (10.8) Evans and Polanyi assumed that in the rate-determining step of an acid-catalyzed reaction, is proton transfer from the acid to the reactant. Next, wc will derive an equation for the activation energy for the proton transfer process. Wc will consider the reaction B — H + R-*B"+ HR+, where R is an proton acceptor, B is a conjugate base, and H is a hydrogen atom. For the purposes of derivation we will assume that one can separate the proton transfer process into two parts: the scission of the bond between the proton and its conjugate base (i.e., the negative ion produced when the proton dissociates), and the formation of a bond between the proton and the reactant to derive equation (10.3). Figure 10.2 will be used in the derivation. Consider the transfer of a proton from an acid BH to a reactant R. We can define two quantities: rbh. the length of the B-H bond: and rhr, the length of the R-H bond as indicated in Figure 10.2. Following Evans and rbh" I luuro 10.2 The energy changes that occur when a proton H is transferred between a con|u(|ntn Imnn II hi ii I it nwctant R. The solid line is the energy of the B-H bond; the dotted line is the energy ot th« H-R bond. l'..l\.ini. we will assume that during the reaction the distance between I) ami R dOU m>i i Itunge. This assumption will be relaxed later in this chapter. During the reaction, the proton-conjugate base bond breaks: run increases while il(ll decreases. Figure 10.2 shows how the energy of the B-H and R-H bonds chance as Ihc reaction proceeds. The energy of a typical B-H bond looks like the solid line In i Igure 10.2, where the energy of the proton increases as the proton moves from its equilibrium position. We have drawn a Morse potential, but the same arguments could be made for other potentials. The energy of the R-H bond looks similar to the potential for ihc li H level. However, when we plot the data onto Figure 10.2, we need to be eaielul hci ause the x axis in Figure 10.2 is the length of the B-H bond. When rMi increases, rl(ll itci leases. As a result, when we plot the potential energy function of the R-ll bond as a function of Tish, the curve will be backwards as indicated by the dotted line in figure II) ' Now consider what happens during the reaction. Initially, the B-H bond breaks, so Ihc potential will ride up the solid contour shown in Figure 10.2. However, once the l< II bond begins to form, the potential can go back down the dotted line in Figure 10.2. During the reaction, the scission of the B-H bond occurs simultaneously with the formation ol ihc R-H bond. Thus, in principle, the total energy should be the sum of the two curve! However, Evans and Polanyi suggested that one can obtain a reasonable approximation In assuming that the potential energy of the system went up the B-H curve as the reaction began, and then went down the R-H curve as the reaction went to completion. Evans ami Polanyi also identified Ea in Figure 10.2 as the activation energy of the reaction. Evans and Polanyi's model, although not quantitative, was quite important. It shows i hat one can consider a reaction as a sum of two processes: one involving bond scission and the other involving bond formation. One can then define the potential energy surface for each of the bond scission processes, and by considering the curve crossing between the two potential energy surfaces, one can get some information about rates. The Evans-Polanyi model has had wide applicability in the physical-organic chemistry literature. As an example of the utility of the Evans-Polanyi model, consider how Figure 10.2 changes when one changes the strength of the acid by replacing one of the ligands on the acid with a different ligand. For the moment, assume that the acid dissociates more easily. By definition, when the acid strength increases, the B-H bond becomes easier to break in solution. This corresponds to an increase in the free energy of the B-H bond. The free energy could increase if either the position or the shape of the U II 1111 vi- m ľi|Miiľ K).' ili.inj'i". lliiwi'vci. in ílu- examples considered by liviins .iii.I Polanyi, the changes mi the shape ol the potential did nol make enough ol b different i Ifl ihc free energy to make a signilieant change in the acid strength. Therefore, Polanyi and Evans proposed that one can get a qualitative picture ol' how a change in acid Strength affected the reaction by simply displacing the solid curve in Figure 10.2 up or down. Figure 10.2 shows the result of an upward displacement. Notice that as the solid curve is displaced upward, the activation energy for proton transfer goes down. For small displacements, the activation energy varies linearly with the change in the energy of the reaction. On a more fundamental level, the Evans-Polanyi model provides an explanation ol why activation energies arise in reactions. The idea is that when a reaction occurs, one bond breaks and another one forms. One can distinguish between two cases: the case in Figure 10.3, where the process is activated; and the case in Figure 10.4, where the curves cross at the minimum in the BH potential so the activation energy is zero. Fundamentally, the two cases are similar. However, when the reaction starts in Figure 10.3, the energy of the system goes up as the B-H bond is stretched, and that increase is larger than the lowering in energy due to formation of the R-H bond. As a result, the total energy of the system goes up as the reaction proceeds. That causes the reaction to be activated. In contrast, in the case in Figure 10.4, the lowering of the total energy of the system due to bond formation is larger than the energy increase due to bond scission. As a result, there is no activation barrier to reaction. It happens that most real cases look closer to the case in Figure 10.3 than the case in Figure 10.4. Consider reaction A + BC -» AB + C. In the initial part of the reaction, A is moving in toward B-C. At long range, the A-BC potential is attractive, but as A gets closer, the A-BC potential is usually repulsive. The electron cloud of A overlaps with the electron cloud of B-C. That produces an electron-electron repulsion (i.e., a Pauli repulsion) of the type discussed in Chapter 9. The Pauli repulsions tend to keep the reactants from getting closer than their van der Waals radii. Now, as the reaction proceeds, the B-C bond begins to break and the A-B bond begins to form. However, the van der Waals radii of the molecules are usually larger than the covalent radii of the atoms. As a result, when atom B begins to be transferred to A, atom A is still pretty far away from B, as illustrated in Figure 10.5. As a result, the B-C bond needs to stretch before it breaks. Generally one observes about 25% bond extensions. The B-C bond loses energy as the , rbh bHo> rrh -® rbh ' Figure 10.3 A diagram illustrating how an upward displacement of the B-H curve affects the activation energy when the B-R distance is fixed. Products Figure 10.4 A diagram illustrating a case where the activation energy is zero. O— 0^D (b) 0 (c) i Igure 10.5 An illustration of the A-BC geometry: (a) A approaching B; (b) the transition state, (c) near Urn tnd of the reaction. H-C bond is stretched. However, the A-B bond is still very long, so the net bonding between A and B is relatively small. This produces a situation that is analogous to the situation in Figure 10.4, where the B-C bond starts to break but the A-B bond hard) starts to form. Figure 10.3 shows that in such a case, there is a finite barrier to reaction Figures 10.3 and 10.5 apply to most reactions. As a result, most reactions are activated A more careful analysis indicates that activation energies can also arise because ol bond distortions. However, the basic implication of the Evans-Polanyi model is thai activation barriers arise because of the Pauli repulsions and because of the bond scissions and bond distortions that occur during reaction. The activation is zero only when no bonds break or distort during reaction, or when the reaction is so exothermic that bond formation dominates over bond scission. Both cases are rare. As a result, most reactions are activated. 10.4 THE POLANYI RELATIONSHIP The analysis in the previous section was all qualitative. However, Polanyi and Evans showed that one can use the model to derive equation (10.3). In the material that follows, we will reproduce Polanyi and Evans' derivation to see how the ideas arise. 11 ii i-i h any i in i aim irr ii in1 nrw 0) c LU Actual potential Line 1 Line 2 Figure 10.6 A linear approximation to the Polanyi diagram used to derive equation (10.11). Consider the reactions A + BC -> AB + C. One can model the reaction as motion along the potential energy in Figure 10.12. The reactants come together at a constant value of rBC, then atom B is transferred, and the reactants move away along a line of constant, rAB. It is useful to consider a slice of the potential energy along the dashed line in Figure 10.12. Figure 10.6 shows how the energy varies along the slice. The horizontal axis (abscissa; x axis) in the figure is the distance along the dashed line, measured from the upper left corner of Figure 10.12. The vertical axis (ordinate; y axis) is the energy measured relative to the reactants. Notice that the potential goes up, reaches a maximum and then goes down. Polanyi suggested that one can view the potential as being composed of a reactant potential E, and a product potential E2. The two potentials are assumed to cross at the transition state. For the purposes of derivation, it is useful to fit lines to the potential energy contour for the reaction as shown in Figure 10.6, where the lines are chosen to fit the tangent of the AB and BC potentials in the transition state. For the purposes of the derivation, it will be useful to assume that E, and E2 are given by Ei = Ereaclanl + |Sli |(rABc - r,) (reactants) E2 = Eproducl + |Sl2|(r2 - rABC) (products) (10.9) (10.10) where Ercaclan, is the reactant energy; Eproduct is the product energy; E, is the line moving up from the reactants; E2 is the line moving from the products; rABC is the distance along a coordinate, moving from left to right along the dashed line as in Figure 10.12; and Slj and Sl2 are the slopes of the two potential energy curves at the transition state. Nulla' Iiiimi Figure Kid I 1 1 i'qiiill In llir energy ill which line I iiilrisci Is line .' Holvlny equation! (10.9) and (!*• 10) il.....Ilaneously for B, ■ E| «= and noting thai • ..... I',........,1 All, yields ISMIShl \ |SI,| E„ = is l| ) + - All, \|SI,| I \Sh\J |SI,| I |SI,| (III III NuhMltuling *•=[ ÄLl(r2_ri) Vp |SI,| + |S12| ISIil |SI,| + |S12| I Ml 12) (I'll I) .......equation (10.1 1) yields Ea = E'a)+yPAHr (10.14) I quution (10.11) is called the Polanyi relationship. Equations (10.11) and (10.14) are the i, \ I \ ans Polanyi results. The implication of equation (10.11) is that if one had some wiiy id change either the enthalpy or the free energy of a reaction, the activation energ) loi the reaction would also change. The activation barrier would also change if |S1|| 01 |SI | were to change, or if (r2 — ri) changed. 10.4.1 Key Predictions of the Polanyi Relationship Physically, one can change the enthalpy of a reaction by adding a substiliieni • ■ i«>i11> lo one of the reactants. For example, consider the reaction between acetic acid and ethanol lo yield ethylacetate: CH,COOH + CH3OH -»• CH3COOCH3 + H20 (10.13) (>ne can vary the acid strength of the acetic acid by fluorinating the methyl group. Thai makes the reaction more exothermic, which in turn leads to a lowering of the activation harrier for the reaction. Hxperimentally, one often finds that the activation barrier for a series of closely related reactions varies linearly with the heat of reaction. Figure 10.7 shows how the activation harriers for a number of hydrogen transfer reactions vary with the heat of reaction. Notice that in the case of hydrocarbons, there is an approximately linear relationship between the heat of reaction and the activation barrier for the reaction. Consequently, the Polanyi relationship has proved to be quite a useful way to correlate data. Another key prediction of the Polanyi relationship is that in general strong bonds will he harder to break than weaker bonds. To see that, it is useful to compare the potentials for forming a strong bond and a weaker one. Figure 10.8 shows the plot of the potential lor breaking a strong bond and breaking a weaker bond. Notice that as you increase I In-strength of a bond, you increase SI,. According to equation (10.12), if everything else is equal, that should increase the barrier to reaction. 11 ii i•< u Ann iu i /\in in'iiin' -.111 n-20 -18 -10 -6 0 5 10 15 20 25 i 11 i i | 11 i i | i i i i 11 11i!iiii|ii, , ,, i i i | , , i 20 0 1 15 c LU ? 10 0 i....i ■ ■■■ i ■■■■ i .... i .... i ■ -20 -15 -10 -5 0 5 AHr, kcal/mol 10 15 20 Figure 10.7 A plot of the activation barriers for the reaction R + HFT -> RH + R' with R, R = H, CH3, OH plotted as a function of the heat of reaction AHr. Weak bond Figure 10.8 A schematic of the curve crossing during the destruction of a weak bond and a strong one for the reaction AB + C -> A + BC. Another important effect is that, if everything else is equal, as you increase (r2 - r,), the barriers to reaction increase. One calls reactions with small values of (r2 - r,) reactions with tight transition states. If (r2 - r,) is small, one does not have to stretch the bonds very much for reaction to happen. Consequently, if everything else is equal and there are no other effects, the barriers to reaction will be small. One calls reactions with large values of (r2 - r,), reactions with loose transition states. If (r2 -n) is large, one has to stretch the bonds a lot in order to get reaction to happen. Consequently, if everything else is equal and there are no other effects, the barriers to reaction will be large. In actual practice, everything is rarely equal. Stronger bonds are not necessarily harder to break than weaker bonds. Reactions with looser transition states do not necessarily have 11 E 1 LU 70 80 90 100 AHr, kcal/mol t:>o i........ 10.9 The activation barrier for the reaction X + CH3X-> XCH3 + X. The nunihi.n. nm I...... Ilm .I. ill,ill.>ns of Glukhoutsev, et al. (1995). Inrgci barriers to reaction. For example, Figure 10.9 shows some data for the reaction X +CH3X->XCH3 + X- (10.16) i..i X = F~, CP, Br , I". Notice that the activation barrier hardly changes as the Strength ..I itie CH3X bond increases. At first, one might think that this result violates the Polanyi relationship, but actually it does not. What is happening physically is that as you change ilie- strength of the CH3X bond, you increase SI| and Sl2 but you decrease (r> r{). Thine! effect is that the barrier to reaction is hardly changed. Another limitation is that one needs to be careful when using the Polanyi relationship with polar and nonpolar molecules. For example, Figure 10.10 shows a plot of lln-activation barrier for a number of hydrogen transfer reactions of the form RH + R' R + HR' (10.17) with R, R' = f-butyl, phenyl, CC13, CF3, CH3CH2, H, H2N, OH, CH3, CH30. Notice that although there is some correlation to AH, the correlation is poor. Roberts and Steele report a correlation coefficient of 0.544. This example shows the difficulty with using die Polanyi relationship for molecules with widely varying properties. In particular, polar molecules and nonpolar molecules do not belong on the same plot. This brings up an important point — activation barriers are subtle. If everything else is equal, increasing the heat of reaction (i.e., making the reaction less exothermic) will increase the barriers to reaction. If everything else is equal, increasing the strength ol Ihe bonds that are being formed and destroyed will increase the barriers to reaction. II everything else is equal, increasing the amount of bond distortion in the transition stale will increase the barrier to reaction. If everything else is equal, pulling electrons out of the transition state, which lowers the Pauli repulsions, will lower the barriers to reaction. However, everything else is rarely equal. A substituent may increase the bond strenglh .....*»"........JI Iři ti' 1(1 25 o E 20 S 15 10 0 -30 -20 -10 0 10 20 AHr, kcal/mol Figure 10.10 A Polanyi relationship for a series of reactions of the form RH + R' -* R + HR'. [Data from Roberts and Steele (1994).] and decrease the amount of bond distortion. It is very difficult to know how the activation barrier will change in such a situation. 10.4.2 Limitations of the Polanyi Relationship Another difficulty is that the Polanyi relationship does not work all of the time. According to the Polanyi relationship, the activation barrier for a reaction should vary linearly with the heat of reaction. Well, the linearity works, only over a limited range of AHr. If one goes to a large range of AHr, one finds that the Polanyi plots are generally nonlinear. Figure 10.11 shows a typical plot that one might observe in the literature. In this case one plots log(k) versus AHr. Note that log(k) is proportional to —Ea, so according to the Polanyi relationship, the plot should be linear. However, the plot is nonlinear. The nonlinearity is typical. Consequently, the Polanyi relationship is limited to small ranges of AH. Another difficulty is that the Polanyi relationships gives unphysical results for very exothermic reactions. Consider a system following the Polanyi equation: Ea = 12 kcal/mol + 0.5(AHr) (10.18) Notice that according to equation (10.18), a reaction with a AHr more negative than —24 kcal/mol will have a negative activation barrier. That is physically impossible. Thus, the Polanyi relationship fails in the limit of very exothermic reactions. Another physical impossibility occurs when AHr > 24 kcal/mol. For example, if we had a reaction with a AHr of +30 kcal/mol, equation (10.18) would predict a barrier of 27 kcal/mol. Yet, the system must go up 30 kcal/mol to get to the products. Consequently, the activation barrier to reaction must be more than 30 kcal/mol. One can generalize this result. In Chapter 4 we found that the activation barrier of an elementary reaction must be greater than the heat of reaction. Equation (10.18) predicts that when AHr is greater that 4-24 kcal/mol, the activation barrier will be less than the heat O 'A a/á. Marcus equation A / / V' -2.4 ^4.8 -7.2 -9-6 AHr, kcal/mol „„ure 10.11 APo,anyiP,ot,ortheeno,izationofN02(CeH4)0(CH2)2COCH3.Notetha«ln(k,l^»................. i,, I .,. [Data of Hupke and Wu (1977).] of reaction reactions. Consequently, the Polanyi relat.onship fails in the limits of very endothermK "rally the Polanyi relationship is a useful approximation for reactions that are „e£S endothermic nor too exothermic. However, the Po.any, re.at,ons.uP ft.......... reactions are very endothermic or very exothermic. 10.5 THE MARCUS EQUATION M-,rcus (1955 1968) proposed an extension of the Polanyi relationship in ordei to Income these d.fficulL with very endothermic and very exothermic reactions. Ml..... considered a general reaction of the form A + BC AB + C (10.19) and derived an equation for Ea using a modification of Polanyi's methods. In the remainder of this section, we will derive a formula for Ea using Marcus' work as a guide. Our derivation will assume that the reaction follows the solid trajectory ol Figure 10.12, where the reactants come together, a reaction occurs, and then the productl fly apart. We will divide the trajectory into three parts: a part where the reactants come together without being significantly distorted, a part where atom B is transferred, and I part where the reactants fly away. In this approximation, the free energy of activation is the work it takes to bring the reactants to point X, w', plus the free energy it takes to 'BC Figure 10.12 An approximation of the minimum-energy trajectory for the reaction A + BC -» AB + C. transfer atom B, E Ea = E' + w' (10.20) Marcus postulated with w' would be nearly constant for a group of closely related reactions, and that E}d could be calculated using a modification of Polanyi's derivation described earlier in this section. For the purposes of derivation, we will define a quantity, rx, as the distance along the dashed line in Figure 10.12 with rx = 0 at point X. We will then assume that the energy of the A-B and B-C bonds follows the dotted lines in Figure 10.13 as a function of rx where the energies show a Lcnnard-Jones dependence of rx. To simplify the analysis, we will fit each of the potential contours in Figure 10.3 with a parabolic function near the transition state as indicated in Figure 10.13. In the next few pages, we will derive Marcus' equation. For the purpose of derivation, we will assume that Eler,(rx) and Erjghi(rx), the energies of the left and right parabolas in Figure 10.13, are given by Eief,(rx) = SS,(rx -r,)- + E, Erigh((rx) = SS2(rx - r2)2 + AHr + E2 (10.21) (10.22) where SS|, SS2, r1( and r2 are fitting parameters. In equations (10.21) and (10.22) we have noted that E|eft and Erighl are functions of rx. We will note in the material to follow that SSi and SS2 are related to the vibrational frequency of the atom B as it is being transferred. Note that since we are measuring energies from the horizontal line at the bottom of the A-B curve in Figure 10.13, E^ is equal to the energy where E|efl equals Eright- For the purpose of derivation, it is useful to define rf as the value of rx where E|eft = Erjght. From equations (10.21) and (10.22) I Igure 10.13 An approximation , lunges. of the change in the potential energy surface that occurs wlu.n All, One can simplify equation (10.23) by assuming ss, = ss2 E2 = Ei (10.24) (1(1 'M Substituting SS2 from equauon substituting E2 from equation (10.25) yields »„n nfiMI solving for r1, and then (10.24) into equation (10.Z3), soiv.. 6 AHr ,10 .'(.I Substituting inar^ into equation (10.21) yields 'r2 + ri Ei =Elcfl(r*)= I Y ^1_^ SS.+E, (H).27) Rearranging equation (10.27), we obtain Ei = 1 + AHr V) SS,(r* - r,)2 + E, = SS2(r* - r2)2 + E2 + AHr (10.23) nil MAIII ll'i I E°. Figure 10.15 illustrates why the maximum occurs. Most reactions show Polany diagrams like those in Figure 10.15a, where a reaction needs to go up a potential......) (a) (b) Ficure 10 15 A Polanyi diagram illustrating how changes in AH for the reaction change the barriers to tha action AB + C - AB + C: (a) normal case; (b) saturation; (c) Marcus inverted reglon. MIM WHY III l III A( III INMIAVI AI I IVA I li ill I i (Míli ni i .nul back down .i)'.iiu lii in (ill I lowcvci. 11 I III' ICIK'I ion is sill In ii ni I \ ( \(illici mih i oiiL' can gci to a situation where the energy curves foi the reactantH and producli..... at the equilibrium point for the ivactanis as illustrated in Figure 10.15b, Notice lhal ,n this particular potential, the activation barrier is zero. Now consider whal happens when the reaction becomes more exothermic as illustrated in Figure 10.15c. Note thai us wi continue to increase the potential, the intersection of the energy curves for the rca< tantl and products moves to the left, causing the activation energy for the reaction lo increasi We call the region where the rate decreases with increasing driving force the Marcus inverted region. Up until 1980, no one had observed Marcus inverted behavior. Howevei a number of electron transfer reactions that show inverted behavior have been discovered since 1980. See W. H. Miller (1991) for details. There is a complication in electron transfer reactions in that the rate of the reaction is moderated by something called | Frank-Condon factor, which can decrease with increasing — AHr. We will discuss this effect later in this chapter. At present, the best available information is that Frank-Condon factors are quite important in the Marcus inverted region. As a result, one cannot explain the data with the Marcus equation alone. Still, the Marcus equation gives the correct qualitative behavior even in the inverted region, and so it is quite powerful. 10.5.2 Strengths of the Marcus Formulation One of the major strengths of the Marcus formulation is that it eliminates the difficulties of the Polanyi relationship for very exothermic and very endothermic reactions. Figure 10.16 is a plot of the height of the barrier as a function of the heat of reaction. A plot of Ea = AH, is also shown in the figure. Notice that the barriers are never negative. Further, the barriers are always greater than the heat of reaction. Consequently, unlike the Polanyi relationship, the Marcus equation never gives a physically impossible result. Another strength of the Marcus equation is that it allows the activation barrier to vary nonlinearly with the heat of reaction. For example, Figure 10.11 showed some data where the activation barrier varies nonlinearly with the heat of reaction. Figure 10.11 shows a fit of the data to the Marcus equation. Notice that the equation fits very well. 250 200 E 150 I uf 100 50 0 ........11" '".............'1......1,1 iT T TIT111 1 ITTITTT7 7 j1 I I 11 I I I I \ Marcus \ equation : \ / - Ea=AHr \ 40 30 20 10 0 -10 -20 -30 AHr, kcal/mol nil M/\i ii 11-( i ( ii i/m ii inj mm One silhllcty in all ol llns is ilini • uIhii .mil Milieus (I'MiK) have shown lli.il I" is mil......I.mi Rather, I" can var) o\....... ol read.....n dial can produce complex iii llllV IOI . Fortunately, Marcus has shown Ihul one can often estimate Ii" from soinellini)' called nil niiiv reactions'. A simple i all ulatlon la shown in Example lo.A. (lenerally the Man uo 111.11h hi allows one to estimate activation barriers from firsl principles. Again an example . iii ulution is shown in Example IDA. It).5.3 Weaknesses of the Marcus Equation i unfortunately, there are many examples where the Marcus equation does uoi In as W( H i cc figures 10.17 and 10.18). Generally failures occur when you considei data taken ■ mi a wide range of AHr. Figure 10.18 shows some data for some electron transits! reactions R*+CH,CN->R++CH3CN_ (10.33) where R*, an excited species, is deexcited via electron transfer lo accionvinlt I'igure 10.17 shows data for intramolecular electron transfer processes across a spacei molecule: R-Sp-R+->R+Sp-R" (10.36) In the first case the activation barrier reaches a maximum and then levels off. In the second i use Ihc activation barrier reaches a maximum and then decreases again. However, the decrease is much smaller than that predicted from the Marcus equation. Both of theiC i ases show that while the Marcus equation eliminates the difficulties with the Polanyi 101 101 10'» ■a I 108 if) I 107 10b 105 i-1-1-1-r Marcus • equation 50 100 150 -AHr, kcal/mol 200 250 Figure 10.16 A plot of the activation barrier calculated from the Marcus equation as a function of the heat of reaction for a reaction with an intrinsic barrier of 12 kcal/mol. Figure 10.17 The rate of intramolecular electron transfer across a spacer molecule plotted as a function ol the heat of reaction. [Data of J. R. Miller et at., (1984)]. Mil MAIM IIM IJUA1IIIN n»ll 25 50 75 -AH, kcal/mol 100 125 Figure 10.18 The rate of florescence quenching of a series of molecules in acetonitrite plotted as a function of the heat of reaction. [Data of Rehm and Weller (1970).] relationship for very exothermic reactions, the Marcus equation does not entirely eliminate the difficulties. Another weakness of the Marcus equation is that one cannot easily explain the trends in Figures 10.9 and 10.10 with the Marcus equation unless one assumes that E" varies over the series of reactions. The Marcus equation is basically a curve-crossing model. As one increases the strength of the bonds, one increases the slopes of the all of the lines, so that the barriers to reaction should increase. However, in Figure 10.9 we find that the activation barrier is not increasing substantially when one makes some substantial changes in the bond strength. People have fit the data in Figures 10.9 and 10.10 by assuming that E" varies over the series of reactions. However, if E° is treated as a variable, the Marcus equation loses its predictive power. Further, the variations in E" have not been physically reasonable. The Marcus equation does not give the correct result for the data in Figures 10.17 and 10.18 because those figures assume that E" is constant in the series of reactions. According to the derivation of the Marcus equation in Section 10.5, E" is constant only when (r2 — n) in equation (10.29) is constant. In practice, (r2 — r,) can change during reaction. Notice that according to equation (10.29), if we change (r2 — n), we change the barrier to reaction; when (r2 — n) increases, the barriers to reaction increase. The Marcus equation gives a different result. According to the Marcus equation, when we decrease (r2 — n), we first decrease the barriers to reaction, and then increase them again. The reason why the data in Figure 10.9 do not show an increase in the barrier to reaction is that (r2 — r,) is increasing as slopes Sli and Sl2 are decreasing. The net result is that the barriers do not change. The data in Figure 10.17 is more complicated. The Marcus equation works in the normal region for the data in Figure 10.17; however, the method fails in the Marcus inverted region. Physically, the situation is as illustrated in Figure 10.19. If one assumes Mudium distance E. = 0 Smaller distance ic'c|ii iii 'AB 'AB ngura 10.19 Changes in the curve-crossing model as (r2-r,) decreases during the reaction Ml i C ■ A + BC. ih.it ihe (r2 -ri) is constant during the reaction, then the situation will be as seen mi iln right of Figure 10.19. The curves will cross in the inverted region, so the barriers in reaction should increase as AH becomes more negative. Note, however, that during I I.....olecular reaction, (r2 — rj) decreases as the reaclants come together. Before the system ri i. to the situation on the right of Figure 10.19, the system will get to the situatii...... ill. middle of Figure 10.19, where the barrier to reaction is zero. Thus the barriet to i. in lion will be zero even though there will be a barrier at smaller values of 0' i, I I lie derivation of the Marcus equation ignores this possibility, so it does not lit the data in Figure 10.17. The situation in the data in Figure 10.18 is even more complex. I Ins is a itnccial reaction where one is transferring an electron across a molecule, (i. \,\ i. constrained, so one cannot reach the situation in the center of Figure 10.19 \t a result, one does observe an inverted region. One does not follow the MorCUl equation exactly, however, because (r2 — rt) still varies to compensate for the inveited region. Generally one finds that the Marcus equation works well when (r2 — ri) is constrained II one considers a unimolecular reaction such as: H H R + R—C—C—C—C—H H H H H H H + R R—C—C—C—C-H H H H H (10.37) (i. — ri) represents the carbon-carbon bond length. The carbon-carbon bond length docs not change significantly during reaction, so it is reasonable to assume that (r2 — ri) is constant in such a situation. However, if one looks at a bimolecular reaction such as: H + CH3CH2R->CH4 + CH2R (10.38) (r2 — ri) changes as one changes the heat of reaction. The Marcus equation does less well in this situation. I'AI II I III I'lll Ml ..... I'AI II I III I 'I M '.KIN'. in ordei to develop .1 bettei model foi the energy .11 the transition state, one needa he able to predict (r2 - i,) al the transition state. Recall from Chaptei 8 thai duri bimolccuiar reaction there are repulsions thai push the reactants apart. These 1 Pauli repulsions. As the Pauli repulsions get stronger, (he reactants are pushed apafl so that (r2 -ri) increases. If the Pauli repulsions weaken, the reacianis can gel close! together so that (r2 - r,) decreases. As a result, the Pauli repulsions play a key role in determining (r2 - r,). The bond energies also play a key role. If you are forming a mi our bond during the reaction, the strong bond tends to pull the reactants together. As a result the transition state gets tighter; specifically, (r2 - r,) decreases. In contrast, weaker bonds increase (r2 - r,). In general, (r2 - r,) is determined by a balance between the bond energies and the Pauli repulsions. in HimoliiiiK Iimii-.iIi.hi i>liit" ( i)(H H) ®- I'niduclH Reaction coordinate Figure 10.20 A diagram of the key molecular orbitals for the reaction D + H2 ■-» DH | H. 10.6.1 The Origin of Pauli Repulsions In order to model (r2 - r,), one needs to understand Pauli repulsions. Pauli repulsions arc basically electron-electron repulsions. Recall from freshman physics that electrons repel each others via a Coulomb force. The Coulomb force occurs with electrons on molecules in the same way that the Coulomb force occurs when there are two isolated electrons. As a result, when two reactant molecules come together, the electrons on one reactanl molecule repel the electrons on the other reactant molecule. That can produce a barrier to reaction. The actual interaction is more subtle than it would appear from the previous paragraph. Recall that when two hydrogen atoms come together, the electrons do not repel. Instead, the electrons on the two hydrogen atoms pair up to form a bond. In Chapter 11 we will show that, because of a quantum effect called exchange, the electron-electron repulsion will go to zero when two electrons pair up into a bond. As a result, the electron-electron repulsion is subtler that it would first appear. Now consider the reaction D + H2 -» DH + H (10.39) The incoming deuterium can form a bond with the hydrogen, so there is an attractive interaction. Yet, the H3 complex is not stable; that is, there is no stable H3 molecule. Consequently, there must also be a repulsive interaction to drive the H3 apart. This repulsion is called a Pauli repulsion. Note that the Pauli repulsion is more complex than a simple Coulomb repulsion because it goes to zero when the electrons in a molecule pair up to form a bond. 10.6.2 A Physical Picture of the Pauli Repulsions One needs to use quantum mechanics to derive an exact expression for the Pauli repulsions. However, it is easy to get a graphical picture of the interactions. In the text that follows, we will describe a qualitative method that one can use to understand the interactions. The ideas will come out of molecular orbital (MO) theory. When I teach this material, I usually review MO theory. The notes from my review are in Chapter 11. Consider the reaction D + H2->DH + H (10.40) I hi deuterium slarts out with an electron in a Is orbital while the II > shirts oul with tWO I |i. nons in a a orbital. The Is orbital is spherical as indicated in Figure 10 20 I hi 0 nibltal Marts out almost spherical, but I have drawn it more elliptical lo make ihc ili.ip.mi i Inner. When the Is orbital interacts with the a orbital, there are two possible moleculai urbilals MOs of the system: a bonding MO where the wavefunction on the deuterium hi. the same sign as the wavefunction on the H2, and an antibonding MO where the wavefunction on the deuterium has the opposite sign as the wavefunction on the II In I ii'iue 10.20 I have arbitrarily assigned a positive sign to the orbital on the deuterium in.l have then let the orbital on the H2 have either a positive or a negative sign. In OtdfH to make the figure easy to see, I have made the positive orbitals light-colored ami the negative orbitals dark-colored. A key thing to remember from diagrams like this is that the light-colored orbitalj Btlract other light-colored orbitals, dark-colored orbitals attract dark-colored orbitall, but dark-colored orbitals repel light-colored orbitals provided the orbitals arc occupied Now consider what happens as the deuterium approaches the H2. In Ihc bonding State nl the transition state (i.e., the bottom state in Figure 10.20), the orbitals are all Ol thi •..ime sign. Orbitals of the same sign attract. Therefore there is an attractive inieiai lion between the deuterium and the H2. In contrast, in the top state, the orbitals are of dilie..... niis. Orbitals of different signs repel. The repulsion causes the orbitals of differenl Igtl lo be distorted. The distortion produces a barrier to reaction. Another way to view this is that there is a bonding (attractive) interaction when Iwo orbitals come together with the same sign and an antibonding interaction when two orbitall i nine together with different signs. For future reference we will call the former case the bonding state of the system, and the latter case the antibonding state of the system, A subtlety is that there is always an antibonding state in the system even when there is no barrier to reaction. Physically, one gets a barrier only when the antibonding Man-is occupied. If all of the antibonding states are empty, there will be no electron-electron repulsions, and therefore no barriers. We need to keep track of the electron count to see if the antibonding state is occupied There are three electrons in the transition state for reaction (10.40). We can put two electrons in the bonding molecular orbital, so we have a two-electron attraction. Howevei. quantum-mechanically, one can put at most two electrons into a molecular orbital The third electron must go into the antibonding orbital. That produces a one-eleclion repulsion. I'Al II I Ml I'lll Mil Reactants come together, H—H bond distorts Hydrogen moves out of H—H bond and into D—H bond C—H bond Separated reactants Figure 10.21 A diagram of the orbital distortions during the reaction D + H2 -> DH + H. Detailed calculations show thai the attraction is 93 kcal/mol while the repulsion is 103 kcal/mol. That produces a net barrier of 10 kcal/mol. Figure 10.21 shows a detailed picture of the orbital distortions. Notice that the orbitals distort during reaction. I like to think of the orbitals as balloons. When the deuterium comes in, it pushes the hydrogen-hydrogen bond out of the way. That allows the hydrogen atom on the left of the H2 to be transferred to the deuterium. In this particular case, the left hydrogen is at the node between the two orbitals at the transition state for the reaction. In other reactions, we have found that the transition state can be slightly earlier or later than the point at which the hydrogen is transferred. However, one usually sees orbitals pushing into one another during reactions. Quantum-mechanically, exothermic reactions have finite activation barriers only when filled orbitals of different signs push into one another, causing orbital distortions. If there are no orbital distortions, an exothermic reaction will not be activated. An endothcrmic reaction will still be activated because the system has to go uphill to get to products. Next, let's consider a slightly different reaction: D + [H2r [DH2 (10.41) The MO diagram for reaction (10.41) is the same as that for reaction (10.40). However, there is an important difference. There are only two electrons in the transition state for reaction (10.41). One can put both electrons into the bonding state. The antibonding state is empty. Consequently, there is no barrier to reaction (10.41). There is a barrier to the following reaction, however D + [H2]+ [DH2 [DH]+ + H (10.42) There is no barrier to forming [DH2]^, but [DH2]+ is a stable molecule. It costs energy to break the [DH2]+ apart. Consequently, there is a barrier to reaction (10.42). ihn i nu geneinli/e these icsulls to any lenclion I'auli repulsions cause Immer. tO ii mi linns The barrier! hum- when orbitals ol different signs push up against one iinolhei II all nl the atomic oibilals have the same sign, there will be little in no barrier. Molecules mlil i.Hhei form MOs where all ol the individual atomic orbitals have the same sign I iiiniiuiiaiely. the MO with all ol the atomic orbitals of the same sign can hold only two . I......ns II there are three or more electrons in the system, there will be an anlibonduig ......m nun Consequently, there will be a barrier to reaction. III O :» Qualitative Picture of the Role of Pauli Repulsions on the Barnors in llftnction WC want to discuss how to quantify these effects. To start off, il is inipoilanl in thai at present it is very difficult to predict the strength of the Pauli repulsion* nils (ienerally, one needs to do quantum-mechanical calculations, and Iben do mil\as to estimate the strength of the Pauli repulsions. So far such calculation* hive been done for only a few special cases, so there is not much to say. Mill. I do have pictures. I think that the pictures are important enough that I wain to in. hide them. figure 10.22 shows a diagram of the orbital distortions during the reaction H + CH,CH, -> H2 + CH2CH, t III 111 ii,,. ase is very much like the case in Figure 10.21. The hydrogen comes in ami pushes il., orbital off one of the methyl hydrogens. Again, the orbitals look like balloons pushing ....., one another. Lee and Masel (1996,1997) showed that the orbital distortion causes |hi n io be a barrier to reaction. Hoactants come logether, (; -H bond distorts Methyl hydrogon moves out of C-bond and into H—H bond C—H bond Separated reactants I igure 10.22 A diagram of the orbital distortions during the reaction H + CH3CH3 -» H2 + CH2CH3. The diagram shows only the interaction with the E state of ethane (the C-H bond). Other MOs of the ethane also distort. I'AIII I III IMII'.ll ill'. !((»/ Koni I: i liin.ik Nonbonding lobe pushs into C—C bond Reactants come together, nonbonding lobe distorts H Separated reactants C—C bond Nonbonding lobes New bonds form Reactants begin to separate CH4 ^ CH3 i ■ w Products Figure 10.23 A diagram of the orbital distortions during the reaction H + CH3CH3 -> CH4 + CH3. Only the 3A1g state of ethane (the state for the carbon-carbon bond) is shown. Figure 10.23 shows a diagram of the orbital distortions during the reaction H + CH3CH3 -► CH4 + CH3 (10.44) This case is more complicated. The hydrogen starts out with a spherical orbital, while the ethane starts out with a carbon-carbon bond, and two nonbonding lobes. When the hydrogen collides with the ethane, the hydrogen orbital pushes into the nonbonding orbital on the methyl group. The nonbonding orbital on the methyl group then pushes into the carbon-carbon bond. The net effect is that the carbon-carbon bond breaks. Again, one can view the process as balloons pushing into one another, causing bonds to break. Notice that the orbital distortions are much larger for the reaction in Figure 10.23 than for the reaction in Figure 10.22. The result is that the barriers to reaction are larger for the reaction in Figure 10.23 than for the reaction in Figure 10.22. Lee and Masel estimate a barrier of 10 kcal/mol for the reaction in Figure 10.22 and a barrier of 42 kcal/mol for the reaction in Figure 10.23. One of the questions in the literature is whether the large barrier to the reaction in Figure 10.23 is caused by the Pauli repulsions or by the fact that bonds need to be distorted during reaction. Notice that you are moving hydrogens out of the way during the reaction in Figure 10.23. Those displacements could be responsible for the barriers to reaction. In order to separate the effects of bond distortion and orbital displacements, Blowers and Masel (1999a,b) calculated the activation barriers for a number of reactions of the form CH3OH + H-> products (10.45) Figure 10.24 compares the activation barriers to the bond distortion energies and the orbital distortion energies. Notice that the activation barriers increase as the orbital 10 20 30 Bond distortion energy 10 20 30 40 50 60 /() Orbital distortion energy m,ii iic. 10.24 A plot of the activation energy for a number of reactions versus the bond ilr.ii n in m.ii.lil. 11 ill il 1 li.tuition energy results of Blowers and Masel (1999a,b). i.....lion energies increase. There is an excellent correlation between the barrier! Mid tin orbital distortion energies. In contrast, there is only a poor correlation between thi lillvution energies and the bond distortion energies. Therefore, it seems Ihal orhilul ill im 1 ions are more important than bond distortions in determining the barriers to n 11 lion. 1 »ne should not infer, from the last paragraph, that bond displacements do not produce hurricrs. That is not true. Instead, bond distortions and orbital distortions are intimately iinnccted. You get orbital distortions whenever you get bond distortions so thai the orbital distortion energy contains a component due to the bond distortion energy. However, the orbital distortion energy has other components; one can get orbital distortions without 111 nding or stretching any bonds. Those orbital distortions are important: more important, m fact, than the bond distortions. Consequently, the orbital distortion energy has a largei Inlluence on the barriers than does the bond distortion energy. In your organic chemistry class, you learned about orbital distortions in terms ol si«i i< 1 Meets. If you have a large group blocking a site, the reaction rate at that sile will he reduced. Steric effects are important to a wide range of reactions, as you learned in organli 1 hemislry. Still, it is important to realize that steric effects are really manifestations of Paull iepulsions and apply only to situations where Pauli repulsions are important. Consider the lollowing reaction: [(CH3)2CH(N02)] + H-> [(CH3)2C(N02)] + H2 (10.46) Hie carbon atom in reaction (10.46) is blocked by a methyl group, and so the reaction rate is slow. Kresge (1974) estimates a barrier of 19 kcal/mol. However, if you change 1 lie reaction slightly [(CH3)2CH(N02)1 + H4 [(CH3)2C(N02)]+ + H2 (10.47) Ihe reaction is unactivated. Physically, during reaction (10.46) the incoming hydrogen needs to overcome a large Pauli repulsion (i.e., electron-electron repulsion) in order in get the reaction to occur. In contrast, in reaction (10.47), a proton, rather than a hydrogen. A Ml III! I 11 III llll III II I i II I'AIII I III IM II Ml IN', I IN IIII IIAIIIIII lľ. H i III Ai I M Hi MM) limit's in The proton does nol conliuii liny electrons, it i it I ytui need elMtKMII In gel I'.mli repulsions, Consequently, there me mi Pauli repulsions in reaction (10.47), The absence of Pauli repulsitms makes reaction (10.47) (inactivated even though reaction (10 16) lias a barrier of 19 keal/niol. There is one other case where Pauli repulsions disappear: a ease where an electron Ii transferred at long distances. Consider the reaction H + F2 -> HF + F (10.48) One might think that there is a Pauli repulsion in reaction (10.48). After all, then are electrons on the hydrogen and the F2 that can repel. However, in practice, the hydrogen transfers an electron to the fluorine at long distances, producing H+ and [F2] . The electron transfer eliminates the Pauli repulsions. Consequently, reaction (10.48) is unactivated. 10.6.4 Empirical Extensions of the Polanyi Relation In the literature, it has been common to account for the effects discussed in Sections 10.6.2-10.6.4 by empirically extending the Polanyi relationship. For example, Roberts and Steele examined a series of reaction of the form BH + R -> B + HR (10.49) where B is a conjugate base and R is a different conjugate base. Roberts and Steele found that the activation barriers for exothermic reactions of the form in equation (10.49) could be fit by the equation E, = E^ HrhHr + VpAH + yx(XB - Xr) + Ys(sB + sr) (10.50) where HBH and HRH are the strengths of the BH and RH bonds, Hhh is a normalization factor (the strength of an HH bond), yp is the transfer coefficient, Xb and Xp are the Mullikin electronegativities of B and R, sB and sr are structure factors fit to the data, and E°, yx and ys are constants. Physically, the first term on the right of equation (10.50) is the intrinsic barrier multiplied by a correction factor to account for the fact that stronger bonds are harder to break than are weaker bonds. The second term is the standard transfer coefficient times heat of reactions as discussed in Section 10.3. The third term is a constant times the difference in electronegativity of the two species. Physically, when the two species have very different electronegativities, there is partial charge transfer, which lowers the Pauli repulsions. The last term in equation (10.50) accounts for the fact that larger groups can have larger Pauli repulsions. At this point, equations like (10.50) have not been derived theoretically. In fact, up until the late 1990s, people did not know why they worked. As a result, the equations are not highly respected in the literature. Still, they work very well. One other unexplained feature at present is that the yp values are different for exothermic and endothermic reactions. yp is usually about 0.3 for exothermic reactions and about 0.7 for endothermic reactions. That has not yet been explained. in / A Mill II I IOII llll It....... I'AIII I III I'lll '.IONS ON llll IIAIIIIII IIS I o Ml ACTION .......in gel some insight iulo eqiiiilmn ( 10.50). by developing a model In eslunale u i,l anil then plugging thai icsull into the Marcus equation. Mowers anil Masel ll#flvcil an equation lo eslunale (r. i'i). The idea in the Ulowers Masel model is In il hi.ne E, by explicitly considering how Pauli repulsions al'leci (r.> hi I el's consitlei tin reaction H + CH3CH3 -► CH4 + CH, 1 I (1 > I 1 Wi dliseussed (he qualitative features of potential energy surfaces lor reactions m Sei llonn I ' 8.4.3, and 8.5. During reaction (10.51), a carbon-carbon bond breaks and I irhon hydrogen bond forms. According to the analysis in Sections 8.4.J. 8.4 1 and í respectively, one can approximate the potential energy surface for the lyitem 11 ílu nm til three terms: the energy to stretch the C-C bond, the energy In lonu .1 new i 11 bond, and the energy associated with the orbital distortions when the reactants come together V(Pcc. Pch) = Ecc(Pcc) + Ech(Pch) + VPauii (10.52) where V(pCc, Pch) is the potential energy surface for the system, Pch and 1',, m the lengths of the C-H and C-C bonds, Ecc is the energy lo sireich ihc ■ iil'nn carbon bond as a function of the bond length, Ech is the energy thai one run m Itinning a carbon-hydrogen bond, and VPauii is the energy associated with the orbital Ii n n 1 ions. Ulowers and Masel approximated ECc and Ech by what are called Morse potentials', Ecc(Pf) = wCc([exp(-aCc(Pcc - Pcc)) _ '] I I Ech(Pch) = wCH([exp(-ach(PcH - PCH)) - 1] - 1) 1 Id .ii 1 in 1 11 where pCc and pch are the lengths of the C-C and C-H bonds; (>;.,. and p|'.„ arc iIh equilibrium bond lengths; wcc and wch are the bond dissociation energies l.n tin . urbon-carbon and carbon-hydrogen bonds in kcal/mol, which are available limn the 1 K(' (the CRC calls them bond strengths); and acc and o the width of the minimum 1 hinges but the depth does nol change. Real atomic potentials look very similar to ilus. which is why the Morse potential has proved so useful. For the purposes of derivation, we will also assume that the Pauli repulsion can be approximated via a simple exponential: Vpauii = V()exp(-PccPcc - PchPch) (10.55) where p"cc. Pch. and V(> arc constants. Figure 10.26 shows a plot of a Pauli potential calculated from equation (10.55). The polential is small unless both the carbon-carbon and carbon-hydrogen bond lengths are small, which corresponds to the case where the reactants are close enough together that significant orbital distortions are seen. The potential stiffens as P increases. A Ml II i| I I i ill H || III il I i il I'AIM I III I'I II Mi iN'm IN llll IIAIII III II'. In III Ai .III IN 001 c lu -4 -3 -2 -1 0 1 2 3 4 5 Distance, angstroms Figure 10.25 A Morse potential for different values of acc ■ 100 2 3 4 5 Distance, angstroms Figure 10.26 A plot of VPauN as a function of the interatomic distance for some typical values of the parameters. Figure 10.27 compares some data for some helium-helium interactions to equation (10.55). Notice the good fit. Therefore, it seems that equation (10.55) is a good approximation. Next we will derive an equation that allows us to calculate Ea from the Blowers-Masel model. Combining equations (10.52)-( 10.55) yields what looks like a complex expression I...... 100 I > t 1 Distance, angstroms tluiiin 10.27 A comparison of Pauli repulsion determined experimentally for helium-helium collision* to .......Ill ulnted from equation (10.55). [Data from Bernstein and Muckerman (1967)]. Iiu the potential energy: V(Pcc, Pch) = wCc([exp(-aCc(Pcc - PCc» - I]2 - 0 - wCc + wCH([exp(-o!cH(PcH - Pch» ~ lf + V0 exp(-|3ccPcc - PchPch) iin 56) figure 10.28 is a plot of the potential energy surface calculated from equation < Hi v'i ini reaction (10.51). Blowers and Masel showed that the potential energy suilaiv is \ niually identical to the ab initio potential energy surface for the same reaction Therefore ihc model does reproduce the qualitative trends in the potential energy surfaces. Next, we will simplify equation (10.56). First it is useful to define nCc and n< n. the I'auling bond order of the carbon-carbon and carbon-hydrogen bond, via ncc = exp(-ofcc(Pcc - Pec)) nCH = exp(-aCH(PcH - PeCH)) (10.57) (10.58) I'auling showed that with this definition, ncc is 1.0 for a first-order bond and 2.0 for a second-order bond. During reaction (10.51), the nCc = 10 and nch = 0.0 at the start Ol I he reaction. As the reaction proceeds, nCc decreases while nch increases, until at the end "l the reaction nCc = 0.0 while nch = 1-0. Substituting equations (10.57) and (10.58) into equation (10.56) yields V(ncc, nCH) = wccflncc - 112 - 1) - Wcc + wCH([ncH - l]2 - 1) + Vp(ncc)qcc(nCH)q™ (10.59) A Ml 11)1 I I I II It III III II I l U I'Al II I III fill Ml in' i I in llll I IAI II III IT. Ill III Al III HI MM C-C distance, angstroms Figure 10.28 A potential energy surface calculated from equation (10.59) with wcc = 95 kcal/mol wCH = 104 kcal/mol, VP = 300 kcal/mol, qcc = 0.7, qCH = 0.5. where Pec qcc = - «cc qcH = fVii «ch V0 exp(+PccPCc + PchPch) At the transition state 3V 3ncc = 0, 3V 3nch 0 (10.60) (10.61) (10.62) Substituting equation (10.59) into equation (10.62), solving for ncc and nCH, and then substituting back into equation (10.59) allows one to derive an expression for the energy of the transition state. For special case that qcc = qcH = 1, Blowers and Masel find 0 (w0 + 0.5AHr)(VP - 2w0 + AHr)2 (VP)2 - 4(w0)2 + (AHr)2 AHr AHr Whe" 4E? <-] when <-- < 1 " 4E? - when — > 1 (10.63) where Ea is the activation barrier and w0 = wCc + wCH It works out that VP is related to the intrinsic barrier, E°, by 'w„ + E"N 2w0 w0 - E? (10.64) (10.65) i......iii in use equation iiomi. om Aral uhooieii E? and wn. calculate! vP ami then |thi(i>i Into equation (10.63), The multi ere virtually Independent oi w„ ('onseqiu-uilv. Ill i in use an average value ol wn lie 10(1 kcal/mol) in all ol Ihc calculations wild iinii error, in / I Qualitative Features of the Model llii lllowcrs Masel model makes four key predictions: . Ihc activation barrier varies nonlinearly with the heal of reaction, approaching zero loi very exothermic reactions and AHr for very endothermic reactions, . I line is no inverted region. . I he results are similar to the Marcus equation when —1 < AH,/4li" I . In. leases in the bond energies can increase or decrease the intrinsic barriers to reaction. | ■in.- 10.29 compares the activation barriers computed from the Blowers M.r.cl ......Ii I lo the activation barrier to barriers computed from the Marcus equation II.'ili models give virtually identical results when —36 kcal/mol < AH < 36 kcal/ntol However, there are derivations at higher energies. The barriers vary nonlinearly with the III n ol reaction in qualitative agreement with Figure 10.11. Still, the Marcus equation predicts and inverted region where E'a' grows lo inflnit) foi m i\ exothermal reactions; no inverted behavior is seen with the Blowcrs-Masel model I In lllowers-Masel model also predicts that Ea approaches AHr at large AH,. While the i in us equation predicts that Ea diverges from AHr at large AHr. I igiire 10.29 compares both models to data. Notice that the data seem lo In dull I. hums-Masel model much better than does the Marcus equation. Physically, the o E 100 80 60 40 '5 20 -20 I i / j -Marcus \ / Jbff Blowers-W'; Masel i i -100 -50 0 50 Heat of reaction, kcal/mol 100 Figure 10.29 A comparison of the barriers computed from the Blowers-Masel model to barriers computed from the Marcus equation and to data for a series of reactions of the form R + HR1 -» RH + R1 with Wo = 100 kcal/mol and E° = 10 kcal/mol. i i tr,• ii im i i Medium distance Ea = 0 Smaller distance Inverted riMjion Even smaller distance Figure 10.30 A plot of the changes in the curve crossing energy as two molecules come together during the reaction AB + C -> A + BC. Blowers-Masel model allows the size of the transition state to vary. Generally the reactants come together; specifically, (r2 — n) changes as the reaction occurs. One can imagine the potential energy surface as being constructed as a series of curve-crossing (see Figure 10.30) models. When the reactants are far away, there is a large barrier to reaction. However, as the distance decreases, the barrier decreases. You always get to the situation that the barrier is zero before you get into the inverted region. The reactants react when the barrier is zero, so the system never reaches the inverted region. As a result, the Blowers-Masel model does not allow Marcus inverted behavior to occur. The other major difference between the Blowers-Masel model and the Marcus equation is that the Marcus equation predicts that increases in the bond energy in the reactants will always increase the intrinsic activation barrier. However, the Blowers-Masel model allows for nonmonotonic behavior. For example, Figure 10.31 shows the variation in the intrinsic barriers with changing bond strength, w0, computed from the Blowers-Masel model with VP = 400 kcal/mol. Notice that the intrinsic activation barrier increases and then decreases with increasing bond strength. Figure 10.31 also shows some ab initio results. The data show some similar trends, although the data do not fit equation (10.63) exactly. Physically, equation (10.63) does not fit exactly because of the assumption qcc = qcH = U if you adjust qCc and qcH- you can fit the data in Figure 10.31 exactly. The final equation qCc ^ 1 is more complex than equation (10.63). Physically, increases in the bond strength in the reactants make the bonds harder to break, which raises the intrinsic barrier to reaction. However, the increases in the bond strength also pull the reactants together. That decreases the size of the transition state, (r2 — rj), which, in turn, lowers the intrinsic barrier to reaction. As a result, the intrinsic barriers to reaction can increase or decrease with increasing w(), as shown in Figure 10.31. One of the main advantages of the Blowers-Masel model is that it allows one to fit a wider data set than docs the Marcus equation. For example, the data in Figure 10.17 are easily fit to the Blowers-Masel model, whereas they do not fit the Marcus equation. The Blowers-Masel model also allows one to fit data as shown in Figure 10.9. Thus, the model is useful. A Ml Hill IOH I.......II "I I'Al II I Ml I'I II UM INI I ON IUI IIAMMIIIIfl Ii I Ml AC I ION T in i.ii'. r I 6 20 III,,wit. Masnl Willi q I II t a. 2 15 y 10 Glukhoutsev 40 60 80 100 Bond energy, kcal/mol 120 rimno 10 31 The variation in the intrinsic barrier with changing bond strength, w0, computed Irom ll.n ....., Ma8el model with VP = 400 kcal/mol. GlukhoutseVs (1995) ab initio calculations ol the activation ......., i, h the reaction X + CH3X -♦ XCH3 + X are included for comparison. I0.7.2 Comparison of the models I i| is useful to compare the three models described so far in this chaplei Figure H) <•' ., plot of the Marcus equation, the Polanyi relationship, and the Blowers M.r.cl ,,,,„,.Minalion for identical sets of parameters. Notice that all three models are vm miil.ii The data in Figure 10.29 follow the Blowers-Masel approximation mosi closely. 100 50 < -100 y Marcus \" Blowers-Masel -50 0 50 Heat of reaction, kcal/mol rigure 10.32 A comparison of the Marcus equation, the Polanyi relationship, and the Blowers-Masel approximation for E° = 9 kcal/mol and w0 = 120 kcal/mol. 000 WHY III ) III A( I M INI I I I/Wl Al IIVA III IN IIAIII III Ml IV 11II I i! INI H II IMA IK IN MIXINi I Mi II il I 11(1/ i»ii the othei two modelu i<»>k very s.....lai to the Blowen Mattel approxlmai.......ft) the range of conditions where most data are measured (i.e., 50 kcal/mol mi, 50 kcal/mol). The Marcus equation deviates from the Blowers Masel approximation onl) for very endothermic and very exothermic reactions. In order to III dala wiih the Polanyi relationship, one needs to draw several lines. However, if one does that, one siill fits datl quite well. In practice, there are some small differences between the models; in partii ulal the Blowers-Masel model fits the data over the widest range. However, in practice, all three models give very similar predictions except for very endothermic or very exothermic reactions. 10.8 LIMITATIONS OF THE MODELS: QUANTUM EFFECTS The big limitations of the models that we have discussed so far is that the models ignore quantum effects. One might initially suppose that quantum effects are not important M reactions. After all, bonds break, and most bonds look fairly classical. Still, there is one important class of reactions where the quantum effects are quite important: four-center reactions. Consider the following reaction: H2 + I2 2HI (10.66) Years ago, people thought that reaction (10.66) proceeded by a four-centered reaction. Two bonds form during the reaction, and two are destroyed. Therefore, one could draw a stick diagram and show two bonds forming and two bonds breaking. However, if you work out the wavefunctions, it turns out that during the reaction, you end up either trying to put four electrons in a single molecular orbital, or having an electron with a spinup and a spindown simultaneously. Quantum-mechanically, one can put at most two electrons into an orbital, one with spinup and another with spindown. It is impossible for an electron to have two different spins at the same time. Therefore, it ends up that the bonding in the four-centered transition state is not possible. In a more general way, reaction (10.66) cannot occur via a four-centered transition state because it is impossible to continuously transform the reactant wavefunctions into product wavefunctions. Instead, one must break some bonds before the reaction can proceed. This effect is not limited to four-centered reactions. There is a whole series of reactions called symmetry-forbidden reactions that have high barriers due to quantum effects. 10.9 THE CONFIGURATION MIXING MODEL We are going to need to do a considerable amount of algebra before we can derive the key equations that we want to use to see whether the bonds are transformed smoothly during the course of a reaction. Pross (1985) summarized the key ideas from several groups in what he calls the configuration mixing model. The idea is to treat a reaction as a process where the reactant wavefunctions are continuously transformed into product wavefunctions. The analysis builds on some of the concepts of MO theory, which we will describe in detail in Chapter 11. Recall that in MO theory, one can write the wavefunction, for any MO of a system as an antisymmetrized product of the wavefunctions, of all of iii, Moms m tin- system I'm example, molecule AH. is given by tin' Wiiveltiiulioii tni tlie u Ihiii,! m n ,|i.ii......, I /2" i'K 1 i|»n)((|)|i I «I'a > (10.67) III re a is the wavefunction for an electron in the bonding orbital ol A with the pin pointing down, <|>„ is the wavefunction for an electron in the bonding orbital ol II u Hli the spin pointing up, <|)|, is the wavefunction for an electron 111 the bonding ,,il,n.il ,,l u with the spin pointing down, and i|f is the antisymmetrizer. There are lluee iinportanl Mi In for excited stales of this system: i|r*, the wavefunction for the <>' orbital ol the A H molecule: V2 (A -B)(<ťA — ďB) llie wavefunctions for both electrons on A: 1 in 6H 1 *a-b+ = <ťAa mid u- , the wavefunctions for both electrons on B: ( 1(1 (,'h A+b:- = fB^B ( II) /(II According to the configuration interaction (CI) model described in Section I I .'. one , .111 write the wavefunction for any state of the system as a sum of the Slater determinate. I'm these MOs, plus the Slater determinates for additional excited states. For example, il \ and B form a bond with a significant dipole moment, then the wavefunction foi the bonding state of the system, *A_B, can be written as C<$o + CA:-B+f A-b+ + Ca+B:-^A+B:- + ^ „' (Ill / I I >v here we used * rather than f in equation (10.71) to indicate that *A_B is a wave! unci ion dial has been computed with the CI model. Similarly, *A.B., the wavefunction foi oni electron being on A and one being on B with no interaction between the two, is a«B» (10.72) For the discussion that follows, we will designate the wavefunction in equation (10.72) .is the [A.B.] state of the system, and the wavefunction in equation (10.71) as the |A lt| state of the system. In some cases, it will also be useful to discuss the [A+B-] state of the system. The [A+B-] state of the system is just the [A-B] state in the limit |cA+B:-1 » |c„|. |ca-b+I. |Co*l- In the material that follows, we will consider how the electronic configuration of the A-B molecule changes as a reaction proceeds and use that information to predict the activation barrier for the reaction between A and B. ...... „,v . rv r , ■■ i i- || . i i | The derivation will start with u simple exiunple where .1 sodium alum approaches 11 chlorine atom and reacts to form sodium chloride: Na. + CI. Na'U (10.73) When the two atoms are far apart, the [Na.Cl.] configuration is the ground stale of ihc system while [Na+CP] configuration is the first excited state. In contrast, in sodium chloride, [Na+CP] state is the ground state and the [Na.Cl.] state is the first excited state. Now consider doing a thought experiment where a sodium atom approaches a chlorine atom but the sodium atoms and the chlorine atom are prevented from exchanging electrons According to calculations, when the sodium atom approaches the chlorine atom, the energy of both the [Na.Cl.] and [Na+CP] configurations go down as indicated in Figure 10.33. However, the [Na+CP] configuration is stabilized much more than is the [Na.Cl.| configuration. That is why the equilibrium structure is [Na+CP]. Now, let's do the same process but allow the sodium atom and chlorine atom to exchange electrons. When the sodium atom and the chlorine atom initially approach one another, they start out in the [Na.Cl.] configuration. However, during the reaction, there-is an exchange of electrons between the sodium and the chlorine to yield the [Na+CP] species. The exchange of electrons occurs gradually during the course of the reaction. At any point along the reaction coordinate, one can represent the electronic structure of the system as a mixture of the [Na.Cl.] and [Na+CP] configurations. For future reference, we will indicate the reaction pathway as the lower dashed line in Figure 10.33. The reactants start out in the [Na.Cl.] configuration, but, as the reaction 11II (;< INI II II IIIAI li IN MIXINi 1 Mi H il I ...... c LU 'Na-CI Figure 10.33 A schematic showing how the energy of the [Na.Cl.] and [Na+CT] configurations of NaCI change as a sodium atom approaches a chlorine atom. ,........Is, I here is a 1 onliniious 1 Imngr in ll.......Iigiii.ilioii ol ih, |) Item liom Ihc |Nn.( 'l«| ......Ho Ihc [Na+CI I slalc Nun 111.11 ii vim did not gel mixing ol the configurations, die sodium ami the chlorine would »tay in the [Na«CU] configuration the atoms could nol get Into ihc [Na+CI I .....Ii,......lion Hence, one would nol gel any reaction. As a result, il is Ihc mixing ol .....durations that allows ihc sodium and chlorine reaction 10 form (metastable) sodium . Monde 1 Mm example illustrates the idea that reactions can be represented as a conversion ol .....I,, Home structure of ihc reactants from one configuration lo another. In die follow Ing I. 1 imNion, we will model the changes in electronic structures, and use that information 10 make prediction about reactions. We will start by noting that the conversion of one slalc into another can he modeled M ......thing called an avoided crossing. Notice that if there were no interactions between iIh |N,i.('l.| and |Na+CP] configurations of sodium chloride, the curves for the energies .1 |he iwo slates in Figure 10.33 would cross. In reality, however, there are strong .......11110ns between the two configurations. In Section 10.9.1, we will show thai when n 1.lies interact, they mix to form an upper state whose energy is greater than thai ol 1 11I111 ol die two individual states, and a lower state whose energy is lower than thai ol ■ nlici of the two individual states. The energy of the two states are represented In the I 1 In id line in Figure 10.33. Notice that because of the interactions between the |Na.('l.| mil |Na' CI I configurations, the energies of the eigenstates of the system never cross. 111 in, the name avoided crossing. One does not always get avoided crossings when in rgy curves for reactions cross. Later in this chapter, we will provide several examples >v here die curves actually cross. However, the significance of the avoided crossing is thai 11 piovides a mechanism lo allow the system to change from one configuration lo anoihci w Inn avoided crossings happen, one can continuously transfer an electron from one stale ol the system to another. That allows bonds to rearrange so that a reaction can lake plat e II there is no avoided crossing, there is no change in the configuration of the reactants ami hence no reaction. As a result, avoided crossings are very important to the litem \ ol reactions. Woodward and Hoffmann (1970) show that one can model most reactions as a change In Ihe electronic configuration of the reactants and an avoided crossing. (The main exception Is in photochemical reactions where the change in configuration can be driven by a photon rather than an avoided crossing.) Hence, the idea of describing a reaction as a change In Ihe electronic configuration of the reactants with an avoided crossing is quite useful. In the materials to follow, it will be useful to draw what is called a configuration mixing diagram for a reaction, where the configuration mixing diagram is a plot showing Ihc energy of each of the configurations of the system change as a function of the reaction coordinate (e.g., bond order). The diagram also shows how the states mix. Figure 10.34 shows a configuration mixing diagram for the reaction of a sodium atom with a chlorine atom. The configuration mixing diagram for the reaction between a sodium atom and a chlorine atom contains the same information as was given in Figure 10.33 There are two solid lines in the coordination mixing diagram: one starting at the [Na.Cl.| state of the reactants and going to the [Na.Cl.] state in the products, and the oihci starting at the [Na+CP] state in the reactants and going to the [Na+CP] state in the products. There are also dashed lines indicating the avoided crossing. The implication 1 As noted in Chapter 8, when an isolated sodium atom reacts with an isolated chlorine atom, a consido.ihli amount of energy is released. The energy needs to be dissipated before a stable NaCI crystal can form. As a ivmiIi. initially reaction (10.73) produces a hot metastable complex not crystalline sodium chloride. wiu why ni i Ml/M IK INU IIAVI ACIIVAIKIN IIAIIIIII II, • CD C lu (a) [Na+Cf]_ l("TI x . (')) / !(«')• 1 [Na • CL] V [Na-CI-] [singlet kk*] •—\ [triplet jut*]---«/ \ ^ — [singlet 7t7t*] — [triplet Ttn*] ^—[Na+Cf] [7t2]-^ Reactants Reaction Products - Reactants Reaction Products coordinate coordinate -► Figure 10.34 Configuration mixing diagram for (a) Na. řrans-2-butene. CI. NaCI, (b) conversion of c/s-butene to of this configuration mixing diagram is that during the reaction, the [Na.Cl.] and the [Na+Cr] states mix continuously to convert the reactants to products. Figure 10.34 also shows the configuration mixing diagram for the conversion of c/s-butene, into trans-butene. H3C CH3 ■ \ / C=C / \ H H H3C H C=C (10.74) The system starts outs with two p electrons in a tt2 orbital. For future reference, we note that there are several excited states of the system: a triplet txtz* state, a singlet Tin* state, and a singlet (;r*)2 state. Now consider what happens when one rotates the :CHCH3 group in the right half of ri.s-2-butene around the carbon-carbon bond, leaving the CH3HC: group on the left fixed. Note that the rotation reverses the sign of all of the p orbitals on the right :CHCH3 group. We start out with an electron in a n orbital with a wavefunction, given by = J=(V-le« + Vright) (10.75) where p...righ, is converted into -(bp...right- As a result, the n orbital is converted into a n* orbital with = j= (<ťp ■left - <ťp.. right) (10.76) Similarly, the jr* orbitals are converted to bonding it orbitals. As a result, the original (jr*)2 state is converted into the ground state of the system while the original n2 state is converted into a (jt*)2 state. Illl CIINI li IIIIIA11( IN MIKINU Mi Mil I III I ..... i.i.i »ii.il 11.■ i >i......In ii ........ proceeds Wc • within the i ululi I|ii»imi. iis we begin lo Iwisl I lit- inoliviilr. wc begin lo mix in some ol Ihe excited I Irsl, the triple! wr" Itttl begins lo internet with the n' Male l.alei Ihe ongiiiul ..... plays a role. Thin diagram Illustrates two important conclusions from the ..i..'million mixing model: I 11 Ihe mixing of excited stales wilh Ihe ground slale plays .......poiiiml role in determining the potential energy surface for a reaction, and (.') often nu. i. in excited states play a role in the reaction, iii. Idea thai the mixing of the ground slate with the excited stales ol the nyati m ....... potential energy surfaces provides an alternative view of why reaction! an hi in iili'il i ill dial when people first started to apply reaction rate theory, il was not obviOUl i. ii livation barriers arose during reactions. The reaction of hydrogen atoms with I i uii i mm tritium molecules illustrates the difficulty: H + DT -> HD + T i in ' i that as this reaction proceeds, a deuterium-tritium bond breaks .mil ,i hydrogen deuterium bond forms. Initially, there is a single D-T bond. Half way iliumi'li in. reaction, there is half a H-D bond and half a D-T bond. At the end of Ihe reaction, ii. n is a single H-D bond. Hence, the total bond order is conserved during reaction Hp lo ihe 1990s, people thought that reactions were activated because bonds broke iiu in), reaction. We now know that the barrier for reaction (10.77) occurs because orbltall 111 distorted. In fact, when one does a quantum calculation of the barriers, one writes the . function for the system i|/ as a sum of the wavefunctions for all of the states in the i' in 'I',,,: in equation (10.78) the Cm terms are a series of coefficients and cbm is the wavefum tion i"i the mth state of the system. When orbitals distort, one finds that one gets contribution! in i|' from the excited states of the system. As a result, one can say that one is mixing . m iied states into the ground state as the reaction proceeds. The configuration mixing model explains in part why activation barriers arise during i. in tion. The general idea is that as a reaction proceeds, excited states are mixed into the uround-state wavefunction. The mixing of excited states raises the energy of the system .mil hence produces an activation barrier. Physically, when excited states gel mixetl Into ili. ground-state wavefunction, the wavefunctions are distorted. Hence, one might waul in ihink about a barrier as being associated with distortion of orbitals in the reactanis ai i reaction proceeds. It is useful to consider how the Marcus-Polanyi relationship arises from Ihe i onliguration mixing model. In the next few pages, we will show that the Marcus equation .uises naturally from the configuration mixing model. Our approach will be to start wilh .1 iwo-state model where one of the states represents the reactant configuration while the niher state represents the product configuration. We will then consider how a substiluenl affects the energy of each states. An analysis similar to that in Section 11.5 will be used lo derive an equation analogous to the Marcus equation. Consider the following reaction: A. + B. -+ A+ +B" (10.79) 'IYMMI IIIY I < lllllll II H N III Ai IIiiN'. OKI UJ |A'U 1 --- ---IA-H-1 [A+FT]--J [A • B •] —' \ [A+B-] [A*B] Reactants Products Reaction _ coordinate Figure 10.35 Configuration mixing model for the reaction A. energy of the [AB] configuration affect the reaction. - B. -> A+ + B showing how changes in the A configuration mixing diagram for the reaction is shown as the solid line in Figure 10.35. The [A.B.] configuration is the ground state of the reactants and an excited state of the products, while the [A+B~l configuration is an excited state of the reactants and the ground state of the products. During the reaction, the configuration of the system moves up the [A.B.] curve and then down the [A+B~] curve. Now consider making a change that displaces the curve for the [A+B~] state up or down. Notice that such a displacement will cause the activation barrier to change. One can model the change by examining how the interaction between the curves shift up or down as the reaction proceeds. Notice the close correspondence between the configuration mixing diagram in Figure 10.35 and the Polanyi diagram in Figure 10.13. In both cases, the activation barrier is given as a curve crossing on a reaction coordinate diagram. Hence, the analysis in Section 11.5 also applies to the configuration mixing diagram Figure 10.35. In particular, if we fit the curves in Figure 10.35 to lines, we can derive the Polanyi relationship. If we fit the curves to parabolas, we will derive the Marcus equation. Hence, the Polanyi relationship and Marcus equation apply as well to the situation depicted in Figure 10.35 as to the situation described in Section 10.5. Pross and Shalk showed that one could use the configuration mixing model to calculate the intrinsic barrier for reaction. The idea is that in the configuration mixing model, intrinsic barriers are associated with the energy to move electrons from one MO to another. If one models the interaction between the system as a curve crossing, then one can get some information about the barriers from Figure 10.36. Figure 10.36 shows four cases. The first two cases show how changes in the gap between the reactions affects the barriers to reaction. Notice that as you increase the gap, I iimcllon coordinate Reaction coordinate Reaction coordinate Reaction i M III l ll' I in'h" 1» 30 A diagram showing how changes in the configuration mixing model affects the intrinsic b»rri«r i......ii Hon. .......e energy to put an electron into an excited state. Consequently, il costs molt .1, n \ lo mix in the excited state. That raises the barriers to reaction. another way to change the barriers is to change the slopes in Figure 10.36. Notice thai I ih. .lopes decrease, the barriers decrease. Physically, the slopes are a measure "I hovi huh Ii ol the excited state is mixed into the ground state during the initial stages "I teai lion II ||ic orbitals in the reactions are distorted more, the initial contribution from the exi Iti d will grow. That will raise the barriers to reaction. Generally, the configuration mi mii)' model makes predictions that are similar to those of the Blowers Masel modi I i .mini, the configuration mixing model allows you to quantify the barriers to reaction; iln details will be discussed in Chapter 11. Unfortunately, the configuration mixing model does not work in detail. The kej ......pi ion in the configuration mixing model is that one can write the wavefunction Ini the transition state as a sum of terms such as that in equation 10.71, where all the i, i ms in the sum are wavefunctions for states of the reactants. Quantum-mechanically, ih.it is not correct. Physically, when a reaction occurs, bonds are extended. The extended bonds contain contributions from orbitals that are bigger than the orbitals in the reai lanl In particular, big orbitals called "diffuse functions" play a key role. The big orbitals art missing from the sum in equation (10.71). As a result, the configuration mixing modi I .lues not work in detail. 10.10 SYMMETRY-FORBIDDEN REACTIONS Another source of derivations from the configuration mixing model occurs dl....... something called a symmetry-forbidden reaction. Throughout the lasi section, we assumed that when the energy contours for two states of the system cross, the stales will mix to provide a pathway for a reaction. However, in 1965, Woodward and Hoffmann i I965a,b,c) noted that sometimes states cannot mix. When states do not mix, one does nol have a convenient pathway to convert the reactants into products. As a result, the reai tiotl late is negligible. Hoffmann showed that the states will not mix when the symmetries oi the states are wrong. Hence, he called reactions that show negligible rates, because Ihc reactant and product configurations do not mix, symmetry-forbidden reactions. Woodward and Hoffmann (1970) wrote a famous book, The Conservation of Orbital Symmetry, which describes, in detail, the role of symmetry in determining rales ol reactions. In the materials that follow, we will summarize the key ideas from Woodward and Hoffmann's analysis. We will also review some other interpretations of the ideas due to Pearson (1976) and Fukui (1975). One should refer to Woodward/Hoffmann (1970), Pearson (1976), or Fukui (1952,1957,1975) for further details. ,,,, i . ... i.i /m in it i . m/wi Ml MVA Ml in 11 Al II III I If i ' 'iyMMI I my II IlllliniH N III Al III INM 818 10.10.1 Forbidden Crossings In the nexl several sections, we will be discussing symmetry Forbidden reai lions namelyi reactions thai cannot occur because symmetry prevents the key slates from mixing, III this section, we will show that one can determine when states can nn\ by examining I quantity p\o(30 defined by ßio(3)= *;(«)3í,(a)*o(ar)dr ( 10.S0) where 9n(yl) is the wavefunction for the initial state of the system, ty\(!X) is the wavefunction for the final state of the system, and K is called the transition HamiltoniM Slates can mix whenever ß|0 is nonzero. No mixing occurs when ß|(l is zero. The derivation is complex and could be skipped without difficulty. It is useful to consider when states can mix. Consider a reaction where molecules A and B come together and react. When the two molecules are far apart, we can describe the electronic structure of the system by a Hamiltonian, 3f[j, with a ground-state wavefunction «Po, ar>d a lirsi excited-state wavefunction where *j] and satisfy HgY? = E°*° (10.81) (10.82) with E° < El>. Now consider doing a thought experiment where we move the two molecules together but do something to prevent the two molecules from reacting. Let's define a Hamiltonian for the nonreactive case as q{X) and *i>\(%) to be the eigenstates of the Hamiltonian that need to mix for the reaction to occur, and we will define E0(?t) and E| (T) to be the corresponding eigenvalues of lKa(%). Figure 10.37 shows a configuration mixing diagram for the reaction between A and B. The solid lines in the figure are plots of EU(X) and E| (X) as a function of the reaction coordinate. We have set up the system so that during the reaction *0 changes from the ground state to the first excited state while changes from the first excited state to the ground state. Now consider moving the molecules together and trying to get the reaction to occur. When reaction occurs, the Hamiltonian of the system will be different from 3üm-E\my + 4MX)ßu>m (l()K/) e+{x) — 2 2 uid wavefunctions given by *-(5D *a(^) when Eo(2) < Ei(3) *b(30 otherwise ¥a(3T) when Eo(3) < EiCD (X) otherwise (I0.HK) (10.89) *o(JT) + *.(2) = *b(Jn Pto(Jt) E'_(a)-E',(ao *i(Jt) i + ßio(«) *l(2) + E'_(T)-E\(T, ßoiPt) E'(X)-E'0(T) *o(*) 1 + (s ßoiW m - e;,(ít) (10.90) (10.91 I ,,,,,,, i, . i i,wi «i i iva i ii in mai ii ill i r 10.11 QUALITATIVE RESULTS Now, let's consider what happens when a reaction occurs. Ai the star) of the reaction f)m(30 = O, so = *0(2). Now, lei's assume thai when (he molecules conic together, PiO(30 is nonzero. Note that according to equation (10.90), when puiiZ) is nonzero, we get mixing of the states. If Pio(Jt) is nonzero, the final wavefunction will approach ^i(T). Therefore, a reaction can occur whenever $WCX) is nonzero when the molecules collide. The dashed line in Figure 10.37 is a plot of E as a function of "X for a typical value of Pio(#). Note that when plo(30 is nonzero, the energy of the lower eigenstate goes up over a hill and eventually ends up at E0. This is typical for a reaction. 10.11.1 Reactions with Negligible Coupling Note, however, that there are some cases where $W(T) is zero when two molecules collide. When $\0(T) is zero throughout the collision process, *_(#) is equal to *0 as the reactants come together. Hence, there would be no coupling of the state. Therefore, we conclude that the motion of the nuclei will convert one configuration of a system into another only when |3io(2) is nonzero during the collision. For future reference, we will call PioW the coupling constant, and note that reactions occur only when the coupling constant is nonzero. In Nobel Prize-winning work, Woodward and Hoffmann (1970) provided several examples where Pio(30 was zero during a reaction. In this section, we will consider a simple example that shows how |3io(30 can be zero. One should refer to Woodward and Hoffmann (1970) or Pearson (1976) for many other examples. Consider the following simple reaction: H2 +D2 > 2HD (10.92) One can imagine that the reaction can go by either the chain propagation mechanism shown in Figure 10.38 or the four-centered reaction shown in Figure 10.39. Wright (1970, 1975) has also proposed a concerted six-centered reaction (e.g., 2H2 + D2 -> H2 + 2HD). Figure 10.38 The chain propagation mechanism for H2/D exchange. i IIIAI II AIM III 'HII I'l l.l / ii I ii ♦ i ♦ i) ii : ii t D ! D 2 -* 11 1 1 ii > , | 1 1 ' ■ i d • I luiiin 10.39 A hypothetical four-centered mechanism for H2/D2 exchange. The dotted lines in the llguio i. 'in 'i.. n iii i or planes that are preserved during the reaction (see the text). This reaction is symmetry lorblddon Al first sight one might think that the four-centered reaction would dominate I he entered reaction requires three molecules to come together simultaneously In the gas i i.r. which is highly improbable. The first step in the chain propagation mechanl m (t'lgure 10.38) is hydrogen-hydrogen bond scission. Hydrogen-hydrogen bond si Isslon . highly activated. In contrast, superficially, during the four-centered reaction 011C ll Imply exchanging bonds within the molecule. As we have drawn the picture, there ll 110 , hunge in bond order anywhere in the reaction. Hence, from a superficial analysis, one mlghl conclude that the four-centered reaction is possible. In fact, however, the four-centered reaction has never been observed. Al high Ii mperatures, reaction (10.92) goes via the chain propagation mechanism. There hai I., en some discussion about the mechanism at low temperatures where the six-centered reaction may predominate. However, there is no evidence that the reaction ever occur. .....ncasurable rate via a four-centered reaction. Calculations of Conroy and Malli i l%9) indicate that the four-centered reaction has an activation energy of 515 kj/moll H\ comparison, the H-H bond strength is only 435 kJ/mol. In the next few paragraphs, we will show that the reason why the reaction docs not occur via the four-centered reactions is that the Pio values for the transformation ol nine of the reactant configurations in reaction (10.92) into production configurations .nc zero. As a result, the reaction in Figure 10.39 cannot occur via a simple gas phasi collision. In order to derive the key results, it will be useful to view the reaction by sitting on a point half way between the hydrogen and the deuterium as indicated by the dol In figures 10.39 and 10.40. One can imagine a transition state where there arc both II II and H-D bonds. If we model this transition state as one big molecule, then we can use molecular orbital (MO) theory to tell us how the configuration of the molecule changes as the reaction proceeds. There are four key MOs in the system. They are depicted In 2 2 I 2 i 2 ® ® ® ® ® i • ® i • 1 1 1 1 ______i______1 t ® ® ® ® i • • i ® CO* O'O" Figure 10.40 A schematic of the key molecular orbitals for the transition state of reaction (10.92). Positive atomic orbitals are depicted as open circles; negative orbitals are depicted as shaded circles. i 111 WHY III I III Al III IN'. IIAVI Al I IVA I H Hl IIAIIIIII ll'l,' Figure I ii in i in i in iin-1 reference wc will label the Ml * In in nr. ol the bonding ol on of ilic hydrogen atoms. The aa Mo will be an mo where one <>i the hydrogens is bound to the other hydrogen via a sigina bond between llie hydrogens, and a a" bond bctwei | the hydrogens and the deuteriums. Figure 10.40 shows a schematic of the wavefunctloni for the various states. All of the orbitals in the wavefunction for the <><> state, <|'..... hav| the same sign. In ^aa*, the orbitals on both hydrogens have the same sign, but the orbital on the two deuteriums have opposite signs. In i|v„, the orbitals on eaeh adjacent II I) have the same sign but the orbitals on the H-H and D-D have opposite signs. In <|'„.„., the wavefunction for the orbitals has alternating signs. Now consider what happens when we move the H2 and D2 together. Wc start out with no net interaction between the H2 and D2, so the wavefunction for the system <4v;ul...... can be approximated by ^reactants = *l a&> (10.93) At the end of the reaction, there are H-D bonds, but there are no net interactions between the two HD. Therefore, the wavefunction for the products *prodiicts can be approximated by ^products = ^(w^lVff. As a result, during the reaction electrons are moved from the an* to the ao* MO. Woodward and Hoffmann (1970) show that it is convenient to represent the transfei of electrons by a correlation diagram where the correlation diagram is very similar to a configuration mixing diagram. We keep track of what happens to each individual MO (i.e., the ip"s) while in the configuration mixing diagram, and we keep track of what happens to the total wavefunction for the system (i.e., the ^s). Figure 10.41 shows a correction diagram for the four-centered reaction depicted in Figure 10.39. The reaction starts out with aa and aa* MOs occupied and the a*a and a*a MOs empty. However, during the reaction, electrons are moved from the aa* to the a*a MO. Now the questions is, whether the reaction can occur as depicted in Figure 10.39. According to the results in Section 10.11, the motion of the H2 and D2 toward one Reactants Products Reaction coordinate T I Jl IAI I I A IIV! Ill Mil I'. II I" ......lint in llie I >' 1111 111 111.111 show..... I ii'iiif 10 l'» will allow 11 if cleclions in I In- "<'' Ml I In be moved into the «*<> MO whenevei |l is noii/eio. Therefore, we enn lell il .....inn can occur by Computing ß. I'oi linlliei discussion, we will note thai there art tWO ......... planes in Figure 10 19 one thai lies hall way between the ll and the !>•. and, one lim goes between the midpoints ol the H H and I) I) bonds. We will designate thou .........in planes as planes I and 2. according to equation (10.80), the coupling constant for conversion ol the aa* itati .......tie a*a state ß......,(»'..1 is tiivL'n by (10.94) lotc that 'K is completely symmetric. Assume that the interaction of the two mirroi plani I Mi, origin for Figure 10.39. If we move up, we move in the y direction. II we move tO ih. right, wc move in the positive x direction. Note that from symmetry !K(x, y) = TO-x, y) = J£(x, -y) = M(-x, -y) It Is useful to rewrite equation (10.94) as (10.93) ß(, an* )(a*a) roc r0 f ^ Wo*o dx dy + t;„. dx dy 0 JO Jo J-00 + \ ^W0.„dxdy+ ^^„.„dxdy (10.96) Jo J-co J-ooJ-oo or the purposes of derivation, it is useful to define a quantity "INT" by roc roo INT= ^.ttf^dxdy Jo Jo I lll.')/i Now consider a quantity INT' given by roo /*0 INT = ^Wa*a dxdy Jo J-00 i in og) INT is related to INT' by a transformation that switches y with -y. Notice thai when we switch y to -y, ^a>„ does not change while •<>> = 0 ( Ml lll.'i Therefore, if we move an H2 molecule toward a D2 in the configuration shown in Figure 10.39, there will be no coupling between the aa" and the a*a MOs. As a result, no reaction will occur. One can get some small amount of coupling if one distorts Ihc geometry (e.g., by twisting the H2 relative to the D2). However, p((I(I. ,<„.„, is always small. As a result, one does not get a significant reaction. Now it is useful to go back and consider why the picture of the transition slanin Figure 10.37 was wrong. In Figure 10.37, we assumed that it will be possible lo simultaneously break a H-H bond and form a H-D bond. However, if we multiply out all of the wavefunctions, we will find that during the reaction, we need to transform wavefunctions that look like i24 into wavefunctions that look like <|> i4, where 4>i, <ť2, 3. and 4 are the atomic orbitals on atoms 1, 2, 3 and 4 (e.g., the two hydrogen atoms and the two deuterium atoms, respectively) and the bars represent spins. Physically, when we try to share an electron between the two states, as we would do if there were partial H-H bond and a partial H-D bond, we would find that we need a single electron to spinup and down at the same time. The electron cannot be both spinup and spindown simultaneously. Therefore, it is impossible to simultaneously break the H-H bond and form a H-D bond. That explains why Conroy and Malli (1969) found that the four-centered reaction has an activation energy of 515 kj/mol. Another way to get the same result is to go back to the stick model of the four-centered reaction in Figure 10.41. In the stick model, one is saying that in the transition state for the reaction, there are four a bonds in the system with four total electrons. According to MO theory, the only way for the four-centered bonding to be stable would be for the four-centered bond to be one of eigenstates of the system. However, the eigenstates of the system are the four states shown in Figure 10.41. Of those, only the aa state is bonding in both directions. One can put two electrons in the aa state, but the other two electrons must be put into other molecular orbitals. These other MO are antibonding orbitals. As a result, the four-centered bonding in Figure 10.41 is not an eigenstate of the system. Therefore, quantum-mechanically, the four-centered bonding cannot occur. A more subtle analysis shows that the reaction H2++ D2 + H—H I I D—D 2HD+ (10.103) where H2+ and D2+ are positively charged molecules, is allowed. However, in reaction (10.103) there are only two electrons in the transition state. One can put both electrons in the SS state so that the bonding is possible. This example illustrates an important feature of Woodward and Hoffmann's analysis: There are many simple reactions that look plausable on paper but cannot occur because they involve bonding schemes that are not possible quantum-mechanically. We can draw a stick diagram like that in Figure 10.39 for many proposed transition states. However, there is no guarantee that the bonding shown in the stick diagram is possible. If it is not, the proposed reaction will have a very high activation barrier. Hence, one has to do some analysis before one assumes that one can get a proposed reaction to occur. A generalization of the analysis above shows that four-center reactions of species with a i.....i|n ail- usually symmetry forbidden ( I hen- are exceptions when- a iionhondine oibilul Id ehangC the syninu-liy ol the system I I liiu-loic, one would lately expect lo see I........uteteil reactions expenmeul.ills 10.12 CONSERVATION OF ORBITAL SYMMETRY ..... , .m imagine repeating the analysis in the last section for a series ol reactioni 1 Wl mid simply work out all of the integrals. However, Woodward and Hoffmann Found .....k that simplifies the analysis. The trick comes from group theory. Recall from i Itnpici .' that we can classify surface structures based on the symmetry element! m ii lace. In the same way, we can classify reaction paths in terms of their synmi. u\ Ii mi nis. For example, if we assume that hydrogen and deuterium arc equivale.....I the . Ii lion in figure 10.39 then the reaction coordinate has three mirror planes: the two line'. Indti iied in the figure and the plane of the paper. There also are three 2-fold axel tWO >ii the plane of the paper at the mirror planes and one perpendicular to the plane ol the i ipi i at the dot. The dot is also an inversion center. In a similar way, one can i lusslf) di. .\ iiimetry of all of the MO of the system. Woodward and Hoffmann (1970) show that ii.. symmetry of the MO determines whether a reaction can occur. If orbital symmein r, ......eived, the coupling constants will be nonzero as the reaction can occur. However, il ii.....hital symmetry is not conserved, no reaction can occur. 10.12.1 Some Results from GroupTheory Wc need to review some group theory to see how this idea arises. Recall thai in group theory, one classifies the symmetry of molecules and the reaction pathways in terms ol their point group. In Chapter 2, we noted that there are only 13 three-dimensional lii.i\ is lattices and three space groups. In the same way, there are only 47 important polnl groups. There are more point groups than space groups because one can have 5-, 7-, X . told axis in molecules, but not in a repeating three-dimensional (3D) structure. Cotton 11971) provides an excellent overview of the application of group theory to mold ulei and reactions, and there is no room to discuss all of the important ideas here Howevei the key feature of the analysis is that if one knows the symmetry elements in a molei lllc ..i reaction path, one can use the tables in Cotton (1971) to find the point group. The most important tables in Cotton's book are the character tables near the end ol i he book. The character tables include a listing of all of the important point groups and their symmetry elements. For example, Table 10.2 is a character table for the D2h point croup. The D2h point group has three perpendicular 2-fold axes, three mirror planes, an inversion center, and an identity element (i.e., an element that does not change anything) Note that the transition state in Figure 10.39 has all of these symmetry elements and no cxlras. Therefore, we say that the reaction in Figure 10.39 follows the D2h point group The character table also lists something called the irreducible representations of the point group. The idea of a group representation is more subtle than there is room lo discuss here. However, basically, if we solve the Schroedinger equation for a molecule (or transition state) that has a given symmetry, then we can classify the independent solutions as being a member of one of several irreducible representations. (The irreducible representations are just a way to classify the solution of the Schroedinger equation in terms of their symmetry properties.) Each independent MO of a system will belong to a single irreducible representation. Hence, one can classify the symmetry of MO in terms of then irreducible representations. 622 WHY III I III Al.tll INUHAVI AI.IIVAIIi ill IIAIIIIII lr •iiimmaiiy DXil Tuble 10.2 A chnriicter table lor the D^, point (iioiip Symmetry Elemeni Irreducible Representation 2-Fold Identity Axis 2-Fold 2-Fold Axis Axis Inversion (lentei xy Minor Plane K2 Mini ii Plane y/. Mirror Plane Source: Adapted from Cotton (19 Rotltlod .mil i Irbitall The character table makes it easy to find the irreducible representation for each MO. Notice all the ones and minus ones in the character table (Table 10.2). The ones and minus ones tell what happens when each of the symmetry operations of the point group acts on a given MO. For example, the xy mirror plane converts y to —y. If we look down the xy mirror plane column in Table 10.2, we find that B2u irreducible representation has a character of —1. Therefore, vJ,b2u> the wavefunction of an MO of B2u symmetry in the D2h point group, will be converted to — *b2u when y is converted to y. One can therefore determine the irreducible representation of a given MO by examining how the symmetry elements in the system affect the given MO. For example, consider 4v„ in Figure 10.40. does not change when it is reflected around the xy or yz mirror plane (i.e., the plane of the paper and plane 2 in Figure 10.40, respectively). However, it switches signs when it is reflected around the xz. mirror plane (i.e., plane 1 in Figure 10.40). Therefore, the irreducible representation of should have a 1 in the column for the xy or yz mirror planes and a —1 in the column for the xz mirror plane. Table 10.2 shows that only the B2ll irreducible representation has the correct symmetry. Therefore, we conclude that has B2u symmetry. The reader may want to show that *a(7 has Ag symmetry, has B3u symmetry, and has B|g symmetry. Now it is useful to go back to our analysis of equation (10.96). Note that we used the symmetry of the various wavefunctions to show that fi(aa*)(,a) is zero. Note that we could have done the analysis directly from the character table. For example, if we start with INT from equation (10.97) and let y go to —y (i.e., reflect around the xz mirror plane), i|v„ does not change while tyaa* is converted into — ^aa», which is completely symmetric (i.e., a member of the Ag irreducible representation), so ^aa* does not change either. Therefore INT' = (-l)(l)(l)INT (10.104) As a result, one could have arrived at the results in Section lO.ll.l directly from the character table without doing detailed analysis. One can generalize this result to show that the coupling coefficients are always zero unless orbital symmetry is conserved. Consider transferring an electron from \|/mr» to some other wavefunction i|/x. The transfer coefficient for the transfer, |3X, is given by (10.105) II >t i si nil with a wnveluut linn nl I * ,. svinmeliy such as t|/„„. ami multiply by 'II, we ill nil cud up Willi a Illinium nl It.„ syinineliy, since '!( is totally symmetric, Now, ll w......1111 > I y by t|', we will gel ii new Function thai we will call 3. One tan show limn .......Iini)' tailed the ,;icnt oi lliiijmnnlil \ llitoiim llial il change! si)'n untlci any nl ii mop operations when we inlegiale ovei all space, hall Ihc integral will be positive iiinl hall will be negative. As a result, ft, will be zero. Notice thai we can tell wliilhci . luingcs sum directly under all oi the group operations from the charactei table h ii- .....iiu ilile representation of i|\ lias ones in any of the places where the irreducible i.....i.iiu>n of >|'„„. has minus ones, or if the irreducible representation ol 'l\ has minus hi any of the places where the irreducible representation of \|/„„. has ones, the lull ill in equation (10.97) will switch signs under at least one of the symmetry operations i iiu 11.1, space group. As a result, p\ will be zero. The only instance where |l, « ill nol in is when the irreducible representations ol'\|/x and i|/„„* have ones and minus oni III in i the same places in the character table. That happens only when i|»x ami I,,,,. havfl ii- inn orbital symmetry (i.e., are members of the same irreducible represenlalion) I he conclusion for the analysis in the last paragraph is that during the reaction hi convert one molecular orbital into another only when both orhilals are ol the urn symmetry (i.e., members of the same irreducible representation). As a result, tnhii.il ■ 111111-11 y is conserved during concerted chemical reactions. in ll EXAMPLES OF SYMMETRY-ALLOWED AND SYMMETRY-FORBIDDEN III ACTIONS Ii Is useful to use these ideas to make some predictions about chemical reactions, (!learl] iiu hisi example is the four-centered reaction in Figure 10.39. Notice that we stall wilh ii i nous in Ag and B2u molecular orbitals and end up with electrons in the A,, ami Bju molecular orbitals. The orbital symmetry changes during the reaction. Hence, the real lion i forbidden. Now consider the reaction H + D2 -> HD + D I III im.I in ie a hydrogen atom reacts with D2. For the purpose of analysis, let's assume thai the n in lion occurs with the hydrogen atoms and the two deuterium atoms lying in a line Figures 10.20 and 10.21 show MO diagrams for the reaction. One starts wilh a Miinictric orbital and an antisymmetric orbital at the start of the reaction, and ends up with orbitals of the same symmetry. Consequently, orbital symmetry is conserved during reliction (10.22). Calculations indicate that the reaction in Figure 10.12 has an activation energy ol I1 k.l/mol while the reaction in Figure 10.39 has an activation energy of 515 kj/niol This example illustrates an important point: We need to look carefully at a reaction pathway before we decide whether a reaction is allowed or forbidden. Many reactions ihai look feasible on paper do not occur because of symmetry constraints. In particulai. ymmetric four-centered reactions are usually forbidden, so they are rarely seen Physically, they have large barriers to reaction. 10.14 SUMMARY In summary, then, the results in this chapter showed that activation barriers are associated wilh • mum villi, lini), • ()iini,ii diitottlon • Quantum effects We have provided some qualitative models to explain the variations. We will quantify (he effects in Chapter 11. 10.15 SOLVED EXAMPLES Example 10.A Using the Marcus Equation to Estimate Barriers for Reactions One of the more intriguing applications of the Marcus equation is in predicting the barriers for reactions. The idea is to estimate E" for an identity reaction, and then use that information to predict barriers. An identity reaction is a reaction where a group is exchanged on a carbon center: X + R-X -> X-R + X OO.A.l) Jzs^ss: MrrN0"^ "v* * - " barrier S qUdt'0" (10-23>- Consequently, E°„, the intrinsic E° - F a. xx E'a,xx where Ea,xx is the activation barrier for reaction (10 A 1) Marcus supposed that when you have a nonidentity reaction X + R -> X - R + Y (10.A.2) 10.A.3) (10.A.4) One can estimate E°xy, the intrinsic barrier for reaction (10.A.3), from a.xy — 2V a,xx ~ a.yy' where E^' yy is the intrinsic barrier for reaction (10.A.I) when X is replaced by Y. To see how this works, let us try to estimate the activation barrier for the reaction H' 4-CH2 = CHCH2OH-► CH2 = CHCH2H' + OH (10.A.5) Equation (10.A.5) is equivalent to equation (10.A.3) with X = H' and Y = OH". Lee et al. (1995a) calculated the activation energy for the reactions OH' + CH2CHCH2OH H' + CH2 = CHCH, • CH2CHCH2OH' + OH CH2 = CHCH2H' + H (10.A.6) (10.A.7) 55.1 kcal/mol. reaction (10.A.7) has an activation barrier of According to equation (10.A.4) E"xy = |(27.9 + 55.1) = 41.5 kcal/mol ,.........., ..............-,„..„■.........«.»A.....1/^«I-A"' l 11 V.oidmg 1» >«|iiiilioii (HI >>l i, «{> - £)' •-(" -A1—\ =21 \ kcal/mol (10.A.9) (41(41.5)/ (10.A.8) | ......parison the actual value is 17.3 kcal/mol. hi k. SUGGESTIONS FOR FURTHER READING I In , onnections between thermodynamics and kinetics are discussed in: \ Promt, Theoretical and Physical Principles of Organic Reactivity. Wiley, New York. IW . '. Sliiiik. H. B. Schlegel, and S. Wolfe, Theoretical Aspects of Physical Organic ( hnniMi i \\ id \ Nrw York. 1992. i In , onnection between orbital pictures and molecular forces is discussed in: I' I W. Bader, Atoms in Molecules: A Quantum Theory, Oxford University Press. Oxford, UK. 1994. 10.17 PROBLEMS I (I. I Define the following terms: (a) Polanyi relationship (h) Configuration mixing model (b) Marcus equation (i) Correlation diagram (c) Intrinsic activation barrier (j) Symmetry-forbidden reaction (d) Transfer coefficient (k) Avoided crossing (e) Marcus inverted region (1) Tight transition state (f) Pauli repulsion (m) Loose transition stale (g) Configuration mixing diagram 10.2 Why do activation barriers arise during chemical reactions? What arc the k<\ factors in determining whether a reaction is activated? Why are activation energiei reduced for fairly exothermic reactions? What happens with very exothermil reactions? Why? 10.3 How do Pauli repulsions affect barriers to reactions? Would there be a barriei III the absence of Pauli repulsions? Explain. 10.4 Describe in your own words how orbitals change as reactions proceed. How do atom transfer processes occur? What extra forces occur when a methyl group is transferred? 10.5 Describe the Polanyi equation. How does it arise? What are the key assumptions in the derivation? When is the equation useful? When does it fail? 10.6 Describe the Marcus equation. How does it arise? What are the key assumptions in the derivation of the Marcus equation? When is the equation useful? When nu mi i M1 does il I n 11 ? wiiii .ne ill. ,nl\.iiii.i)i. ni ilic Mint us t't|iiiil inn ovil ihť I'nl.iiiyi i. i. 111.111 '< 1111 *1 When is ihť M.in us equation appropriate ' When should )......■•■ thi Blowers Masel approximation instead7 10.7 Compare and contrast the Marcus equation, the Polanyi relationship, and thi Blowers-Masel approximation. He sure to consider the key features ol thl derivation and the predictions. Why docs the Marcus equation predict parabol il behavior for very exothermic and endothermie reactions while the Mowers Masel approximation predicts linear behavior? What are the key different assumptions of the two models? 10.8 The data in Figure 10.11 show nonlinear behavior with AH,. (a) The plot actually shows ln(k) versus AHr. Show that if dala in the figure followed the Polanyi relationship and the preexponential were constant, I plot of ln(k) versus AHr should be linear. (b) Use the Marcus equation to explain why the curve is nonlinear. (c) Does the Blowers-Masel approximation predict any different trends? 10.9 What are the key physical forces that are represented by the intrinsic barrier? (a) According to the Marcus equation, what variables affect the intrinsic barrier? (b) According to the Blowers-Masel equation, what variables affect the intrinsic barrier? (c) How would charge transfer prior to reaction affect the intrinsic barrier? (d) How would steric repulsions affect the intrinsic barrier? 10.10 Explain the key trends you see in the data in Figure 10.29. Why is there so much variability in the points? (Hint: look at the Blowers-Masel approximation and equation (10.11) — what varies as you change reactions?) 10.11 Compare the behaviors in Figures 10.17 and 10.18. Why do you see inverted behavior in Figure 10.17 but not in Figure 10.18? 10.12 In your organic chemistry class you learned about steric effects in reactions. (a) How do the steric interactions affect the Blowers-Masel approximation? (b) How does the tightness of the transition state change as the steric interactions change? (c) How will your results in (b) affect the predictions of the Polanyi relationship? (d) How will your results in (b) affect the predictions of the Marcus equation? 10.13 The objective of this problem is to use what you learned in organic chemistry class to estimate the intrinsic barriers for a series of reactions of the form HO + CHR, H20 + CR3 (a) How would the C-H bond energy change as you change the R group from H to CH3 to C(CH3)3. (b) How will the Pauli repulsions change? (Hint: How does the size of the electron clouds change?) (c) Will the barrier go up or down? (d) How would an electron-withdrawing group affect the intrinsic barrier? i ........10 14 l'i.m«|it»ni.i.llnln tin.l.....v...........nil..i«fiii II... ...... I«.n Nil I II II K i ii. E„ k„, I'., ku k. al/mol 1 in \ inu 1 .i . "ii.11 R Iccal/mol tin (moI'MOond) .'ii •I III1 ' CH(CH3)2 14.5 6 x 10" 5 ■ 10" II. 16.5 / ■ Ill1 ' (CII,)..('() 11.5 ..... Dim ol Rhorig and Wagner, Berichte Hunscn Gcsdlschali Physik. Chcm. 9«, 858 I I'W-li hi I I Rhorig and Wagner, Berichte Bunsen Gesellschaft Physik. Chein. 98. 858 i 199 11 examined a number of hydrogen abstraction reactions of the form NH + H-R NH2 + R Their data are given in Table P10.14. (a) Use data in the CRC and the NIST Webbook to estimate the heal >>l reai tion for each of the reactions in the table. (b) Fit the data to a Polanyi plot. How well do they lit? (c) Try fitting the data to the Marcus equation. Assume that the intrinsic bam. i is constant. How well do the data fit? (d) Does the assumption that the intrinsic barrier is constant make sense, oi should the intrinsic barrier vary? (Hint: Look at what determines an iniiinsn barrier.) (e) Examine the variations in the preexponential. According to transition state theory, is the tightness of the transition stale changing? (f) Use the data in Table P10.14 to estimate the role of the tightening ol the transition state in raising or lowering the intrinsic barrier. Should ihe inn him. barrier go up or down? (g) Try fitting the data to the Blowers-Masel equation. Assume (hat Ihe Vr is constant. How well do the data fit? (h) Does the assumption that the VP is constant make sense, or should Vp varj and instead the intrinsic barrier be constant? 10.15 Maslack, Vallombroso, Chapman, and Narvacz, Angew. Chcm. 33 (1994) examined the rate constant for a number of bond cleavage reactions. Then .I.m.i are given in Table P10.15. (a) Maslack et al. showed that ln(k) versus AHr is nonlinear. Show thai if data followed the Polanyi relationship and the preexponential were constant, a plot of ln(k) versus AHr should be linear. (b) Use the Marcus equation to explain why the curve is nonlinear. (c) How well does the Marcus equation fit the data? (d) Does the Blowers-Masel approximation predict any different trends? (e) How well does the Blowers-Masel approximation fit the data? (f) Try adjusting the temperature in the fit; do you get a better fit? (g) What do you conclude from the fact that you need to adjust the temperature'' '"'»'"..... i!> it.u.......m.imi-. ,,i um k mil li.-.il-. ,,l lion l,,, ||„, Reaction ln(k), II.. number cm3/(molsecond) kcal/mol (lc) -7.4 24.8 (Id) -7.0 23.9 (le) -4.7 18.7 (If) -3.2 18.2 dg) -2.5 16.3 (lh) -2.7 15.9 (2a) -0.2 15.2 (2b) 0.7 14.5 (2c) 1.3 13.2 (2d) 1.6 13.2 (2e) 1.8 10.7 (2f) 5.2 7.8 (2g) 5.7 6.6 (2h) 5.6 6.2 Reaction NiiiiiIhi (3a) (3b) (3c) (3d) (3e) (4a) (4b) (4c) (4d) (4e) (5) ln(k), cm3/(molsecond) 5.9 7.0 N I, 9.6 10.5 8.8 9.1 9.6 6.4 7.4 0.2 II.. kcal/inol 5.7 2.8 -1 -12.3 I 7.6 4.6 4 3 I 3.9 4.8 15.6 10.16 Lee, Kim and Lee, J. Computational Chcm. 16 (1995b) examined the activation energy for a number of allyl transfer reactions: X + CH2=CHCH2Y-► CH2=CHCH2X + Y" Table PI0.16 shows a selection of their results (a) Fit the data to a Polanyi plot. How well do they fit? (b) Try fitting the data to the Marcus equation. Assume that the intrinsic barriet is constant. How well do the data fit? (c) Does the assumption that the intrinsic barrier is constant make sense, or should the intrinsic barrier vary? {Hint: Use the additivity assumption to estimate the barriers.) Table P10.16 Activation energy for a number of allyl transfer reactions H Nll: OH F PH2 SH CI OH NH2 SH PH2 H NH2 OH F PH2 SH CI CI CI CI CI AHr, kcal/mol 0 0 0 0 0 0 0 -38.9 -60.1 -19.8 -41.1 Ea, kcal/mol 55.1 43.3 27.9 15.0 33.3 24.9 18.8 5.6 3.5 13.45 7.28 AHr, X Y kcal/mol kcal/mol F H +67.2 74.4 OH II +47.1 64.3 NH2 II +26.1 59.2 CI II +87.5 92.2 SH H +67.3 77.6 PH2 II +47.2 67.4 OH F -16.1 13.1 NH2 F -40.7 5.8 CI F + 16.9 32.7 SH F +0.4 22.0 PH2 F -20.4 15.0 Source: Results of Lee, Kim and Lee, j. Computational Chem. 16 (1995) 1045. i 'i ľ M ii i ivii iii) 11 v fitting tbc dulu tu iln IIIowi'In Mnsel eqtililion Assume lli.il the Vľ is i onstanl 11, iw well do I In ■ i .i. in' le) Dues the asMimplHiii lliiil the Vľ is constant make sense, oi should Vľ vary and instead ľ" be i uii.I.iiiI ' ď) Use the methods In Example I0.A i>> estimate the barriers to .ill "I the reactions. How well does Ihe method wink'.' •" i ' Hcrnasconi and Ni (1994) describe two different methods to estimate the intrinsii barriers lor a series of reactions of the form X + HX -» XH + X (I'll). I /. I ) They were not able to measure the intrinsic barriers directly, hut they urn able io gel an indirect measure by examining relative rates of reactions. The reiultl .in- given in Table PI0.17. The question is how you can do an experiment to see which measure of the intrinsic barrier is accurate. By way of background) nonidentity proton exchange experiments X"+HY->XH + Y~ (P10.17.2) can easily be done, so the object is to find a way to use an exchange reaction to distinguish between the two measures of intrinsic barriers. (a) How many nonidentity reactions can you form from the X groups In Table P10.17? (b) Use Marcus' additivity postulate to estimate the intrinsic barriers for eat h ol these nonidentity reactions. (c) Which nonidentity reaction will show the largest differences? (d) Look in Bernasconi and Ni's paper and find data for that reaction. (e) Which measure of the intrinsic barrier fits best? iii. IK Consider the following series of reactions: HO + CH4-► H20 + CH3 HO + CH3CH3-> H20 + CH2CH3 I nhle P10.17 Intrinsic barriers for a number of reactions of the form X f HX —► XH + X~ K", kcal/mol Estimated Using an Amine Reference, kcal/mol E(a', kcal/mol Estimated Using a 9-Cyanofluorene Reference, kcal/mol |(CH3)NH2]+ llen/ylmaloniumnitrite 9 ('yanofluorene I, t-lndandione I Nitrophenylacetonitrite i t Nitrophenyl)nitromethane (IN itrophenyl)nitromethane I'licnylnitromethane 6.80 11.52 15.40 16.72 17.62 21.74 21.96 22.64 9.37 13.87 13.05 14.37 15.17 19.39 19.61 20.17 IK) I MUCH,), no i čineli,). h........n i II o I C(CH3)i (a) Use Benson's approximation from Chaptei <> to estimate the heal ol reaction for each of the reactions and the energies of the bonds thai form and break (b) Estimate the strength of the Pauli repulsions, assuming that the hydrogen-methyl potential is given by 4.5 A - R0 „ Vr = (1.5 kcal/mol)exp ( ) Be sure to consider that in the reaction H + CH2(CH3)2 ->■ H2 + CH(C1I,), the incoming hydrogen is in close proximity to three methyl groups. (c) Use the Blowers-Masel approximation to estimate the barriers for the reaction. (d) Make a Polanyi plot of the data and estimate the intrinsic activation barrier and transfer coefficient. (e) Why is the transfer coefficient negative? (f) Does the derivation of the Polanyi relationship allow the transfer coefficient to be negative? (g) Does the Marcus equation allow the transfer coefficient to be negative? 10.19 Why are there orbital symmetry barriers to reaction? Where do they arise, and where are they important? How does the orbital symmetry barrier affect the reaction H2 + D2 -» 2HD? What would the orbital diagrams look like for a symmetry-forbidden reaction? 10.20 (a) Show mathematically that the four-centered reaction A2 + B2 -+ 2AB is symmetry-forbidden. (b) What is the physical significance of this result? {Hint: Show that in the four-centered configuration, it is impossible to simultaneously have an A-A, B-B, and two A-B bonds.) 10.21 Lopes et al. (1999) examined the role of acyloxymethyl as a way to protect antibiotics from degradation (hydrolysis) by bacteria. (a) Read the paper and report on the findings (b) The Br0nsted plot is curved. How do Lopes et al explain the curvature? (c) Could the curvature be due the a shift in the position of the transition state as suggested from the Marcus formulation? 10.22 Consider the reaction D + H2 ->- HD + H. Figures 10.20 and 10.21 show diagrams of the key orbitals during the reaction. (a) The pictures only show states that would be expected to be occupied. What will the empty states look like? (Hint: In the empty states, bonding orbitals are converted to antibonding orbitals.) (b) Draw a configuration mixing diagram during the reaction. (Hint: The configuration mixing diagram links up MOs in the ground state of reactants with MOs with the same pattern of plusses and minuses in the excited states of the products and vice versa.) n) ito the orbltali really follow the oonflguruiin......niny dianiam1 Spotliltuli> , .hi you see the exi Ited stati being mixed......the ground stati (dl Aie there .my extra Interactions not considered in Ihe configuration mixing diagram? (e) Draw a correlation diagram foi the system, (Hint'. The correlation diagram links up MOs m the ground state of the reactants with MOs with the same pattern of plusses and minuses m the products.) (0 Do the orbitals really follow the configuration mixing diagram? In othei words, can you see a continuous transition from reactants to products? (g) Are there any extra interactions not considered in the conligiiialion mixing diagram? More Advanced Problems ll).2.t Roberts and Steele J. Chem. Soc, Perkin Trans. 2, (1994), fit the activation h.uncr. for a number of atom transfer reactions to a complex expression. (a) Read Roberts and Steele's paper and explain why all of the factors arift (b) Relate the findings to the Blowers-Masel model. How are the variationi In the Pauli repulsions considered in the Roberts-Steele model.' (c) Are there any factors in the Roberts-Steele model that are ignored In Blowers-Masel's model? 10.24 Zavitsas J. Chem. Soc, Perkin Trans. 3, (1998), refit the activation barriers foi I number of atom transfer reactions. (a) Read Zavitsas paper and explain why all of the factors arise. (b) How does Zavitsas model compare to that of Roberts and Steele J. Chem Soc, Perkin Trans. 2, (1994)? (c) Relate the findings to the Blowers-Masel model. How are the variations in the Pauli repulsions considered in Zavitsas model? (d) Are there any factors in Zavitsas model that are ignored in the lilowci s M.n I model?