5 PREDICTION OF THE MECHANISMS OF REACTIONS I'HECIS lii i liapter 4, we showed that if one knows the mechanism of a reaction anil the rate .....slants for all of the elementary steps in the mechanism, one can calculate a rate i|liution for the reaction. I he objective of this chapter is to give some insights into the mechanisms of reaction Wi « ill start with gas-phase reactions and present some empirical rules to predict reaction mi. i nanisms. We will then move on to reactions in liquids and on solid surfaces, and ii 11ins how mechanisms can be predicted in those cases. We will find that wc can often '1' i accurate predictions with relatively simple methods. 8.1 INTRODUCTION I In- study of the mechanisms of simple reactions had its start in the latter part oi 11u-nincteenth century. At the time, people were still unsure why the kinetics of simple relictions were so different from what one would expect from the stoichiometry of I i. k lion. In 1898, people first realized that reactions followed complex mechanisms. In Ihe 1920s and 1930s, there were several attempts to predict mechanisms. However, I lie work was largely unsuccessful. That is as far as the theory went up until 1950. In fact, up until 1985, no one seriously discussed the idea that one might be able to predict the mechanisms of simple reactions. However, as computers became faster, people started to talk about the idea that one might be able to predict mechanisms, and not just measure mechanisms. Unfortunately, much of the information has been discussed only al meetings and has not made it into the general literature. Still, I believe that the ideas thai are coming out now are important enough, and simple enough, that students and others might to know about them. In this chapter, we will discuss how one can predict reactions mechanisms. 235 5 PREDICTION OF THE MECHANISMS OF REACTIONS nil cis in i haptci I. we showed that if one knows Ihc mechanism of a reaction ami 11 u- lale ......i.mis foi all of the elementary steps in the mechanism, one can calculate a rate ><|u,iiion lor the reaction. I lie objective of this chapter is to give some insights into the mechanisms of reaction. \\. w ill siaii with gas-phase reactions and present some empirical rules to predict reaction .....luinisms. We will then move on to reactions in liquids and on solid surfaces, and ill i uss how mechanisms can be predicted in those cases. We will find that we can often >•>1 uccurate predictions with relatively simple methods. i. I INTRODUCTION I he study of the mechanisms of simple reactions had its start in the latter part ol the nineteenth century. At the time, people were still unsure why the kinetics of simple reactions were so different from what one would expect from the stoichiomeii y ol i ivuction. In 1898, people first realized that reactions followed complex mechanisms. In Ihc 1920s and 1930s, there were several attempts to predict mechanisms. However, ilic »nik was largely unsuccessful. That is as far as the theory went up until 1950. In fact, up until 1985, no one seriously discussed the idea that one might be able to predict the mechanisms of simple reactions. However, as computers became faster, people i ii led to talk about the idea that one might be able to predict mechanisms, and not just measure mechanisms. Unfortunately, much of the information has been discussed only ai meetings and has not made it into the general literature. Still, 1 believe that the ideas thai ue coming out now are important enough, and simple enough, that students and Others might to know about them. In this chapter, we will discuss how one can predict reaction! mechanisms. 235 Iiim. wt u ill diM uss why niii Itlon fellow complex mechanisms it ml then wo will work mil 10(TIC iiili-s leu prediction <>l the mechanisms ol ^as pliiisi* icaclions, turftCC u-.k linns, and reactions in solution, M you can predict mechanisms, you know all ol chemistry, M clearly wc cannot cover everything, However, we wanted i<> provide some principles ilui readers could use to understand mechanisms. In n- X is ii collision pailnei lie, iiuv olhei spei ics lliul can collide with Hi., us described .....is I I and H.I I I. Notice lliul we lonned a reactive species, namely, a radical in ll |i I I he radical then reacts to form products. Hie formation of a reactive species is a .......hi feature ol most chemical reaction mechanisms. 5.2 WHY DO REACTIONS FOLLOW COMPLEX MECHANISMS? The first question we want to ask is why reactions follow complex mechanisms; why don't the reactants instead combine in one step to produce products? The answer is prclly simple. In most reactions in industrial practice, the reactants are stable molecules. A stable molecule is by definition stable, that is, relatively unreactive. As a resuli. a direct reaction between the reactants is usually slow. There are exceptions, of course. If you react sodium with chlorine, you get a direct reaction between the reactants. However, most organic or inorganic reactions start with stable, unreactive species. If you start with a stable unreactive species, you need to do something to the species before that species becomes reactive. Generally, most reactions follow the general mechanism Reactants ' , reactive species Reactants + reactive species-> products (5.1) As seen in reaction (5.1), you first produce a reactive species, and then the reactive species reacts with more reactants to form products. You can form the reactive species by breaking a bond in the species or by exciting the species with, for example, a laser. The excited species can be a vibrational^ excited molecule or something like a radical or an ion. In the gas phase, the excited species is most often a radical. However, if you start with an unreactive molecule, you almost always need to create a reactive molecule before significant reaction can occur. For example, let's consider the case that was discussed in Chapter 4: the reaction between hydrogen and bromine to yield HBr: H2 + Br2 => 2HBr In Chapter 4, we noted that the mechanism of the reaction is X + Br2 —^ 2Br + X Br+H2 —HBr+H H + Br2 —HBr + Br X + 2Br —i-> Br2 + X (5.2) (5.3) H + HBr ——-> H2 + Br i PREDICTION OF MECHANISMS OF REACTIONS .....A', PHASE SPECIES i ......lainder ol this chapter, we will discuss how one can predict the mechanism nit i\ ni different reactions. We will first do gas-phase reactions, then move on to pi h.....is iii liquids and on surfaces. I.i start with gas-phase reactions. Most gas-phase reactions follow what are called u,m propagation mechanisms. Initiation-propagation mechanisms were mentioned in i Imptci 4, Consider the idealized reaction A + B Pi +P2 (5.4) In n i molecule of A reacts with a molecule of B to produce two products. P, and Pj n ui imn is said to follow an initiation-propagation mechanism when the rea< lion i'(.i \ .is follows: I usi, there is an initiation step where a collision partner X collides with one ol the . I mi molecules, and breaks a bond in the reactant to produce two radicals, which I will call R] and R2: X + B-> Ri + R2 (5.5) I hen there may be a transfer step where one of the radicals reacts to form a more .• m live species. Rt, and R4, and two side products C and D: A, + H Ri + C R2 + A-> Ra + D (5.0) I hen the radicals react with the reactants or the products in a series of propagation .hps to yield a series of new radicals, R5 and R(,: + A-> P, + R5 R4 + B->• P2 + tf6 additional propagation steps occur where /?5 and R6 react to regenerate Ri and R4: (5.7) #6 +A P, + «4 R5 + B -► P2 + «3 I anally, the radicals can recombine in what is called a termination step: X + Rx + R2-> B + X (5.8) (5.9) Iheic mi' 11111II v dllleienl i < ii 1111111 .il 11 >l is Null, till' key Icaluic ol ,111 111111.11 Mill propagation mechanism is thiii lii'si (lii'ii- is .in initiation sup whore i in I n .ils iirc formed Then, there arc a scries nl steps when- radicals rend In form products. I'inally, theie are n series of termination slops where radicals are destroyed. Most gas-phase reactions go via initiation propagation mechanisms, hor example. Rice (1932) examined the decomposition of a wide variety ol organic compounds, including aliphatic and oletinic hydrocarhons. aldehydes, ketones, esters, amines, and carbolic ai idl and found that in all cases, the reaction went by an initiation-propagation mechanism Therefore, if one wanted to guess at a mechanism of a gas-phase reaction, one would guess an initiation-propagation mechanism. Let's look back at mechanism (5.3) and show that it follows an initiation-propagation mechanism. Reaction 1 in mechanism (5.3) is an initiation step. Reactions 2 and 3 ail propagation steps. Reaction 4 is a termination step. Reaction 5 is the reverse of reaction 2. Clearly, mechanism (5.3) is an initiation-propagation mechanism. Another important initiation-propagation reaction is the reaction that happens in the cylinders of your car. During the power cycle of your car, a mixture of air and gasoline is put into the cylinders of the car. Nothing happens until the spark plugs fire, producing radicals. Then the propagation reactions occur, producing power. Termination reactions include reactions with tetraethyl lead (TEL) or methyl fert-butyl ether (MTBE), and reactions that occur on the walls of the cylinders. Table 5.1 Examples of initiation-propagation mechanisms Reaction Example Mechanism Combustion, e.g., CH4 + 02 => C02 + 2H20 + other products Free-radical polymerization, e.g., ethylene => polyethylene with a free-radical catalyst, R2 Ozone depletion Hydrocarbon pyrolysis 02 ^ 20 O + CH4 -> CH3 + OH OH + CH4 -> H20 + CH3 CH3 + 02 -> CH3 + O + O CH3 + OH -» CH2 + H20 + other products OH -> walls CH3 walls R2 -> 2R. R. + C2H4 -> R(C2H4). RC2H4. + C2H4 -» R(C2H4)2. R(C2H4)n. + C2H4 -»■ R(C2H4)n+1. R(C2H4).m + R(C2H4).n ->• R(C2H4)m+nR 02 + hv, -> 20 O + O2 + X -+• O3 03 + hv2 -> 02 + O Cl + o3 -» 02 + CIO CIO + o -> o2 + CI X + CH3COH -> CH3 + COH + X CH3 + CH3OH -* CH3CO + CH4 CH3CO + CH3OH -* CH4 + CH3CO COH + X-.-CO + H + X H + CH3COH -» CH4 + COH H + CH3COH CH3 + CO + H2 2CH3 + X -> C2Hft + X H + CH3 + X -» CH4 + X H + CH3CO + X -> CH3COH + X ....... > i' 'i, ' '. till rvii » mít inn im it iín l 'Mi H'Ai IA IH »N M liiblr "> I gives seveinl olhei ouunples ol imlinlioii piopngnl......em lions Renders I.....Ill convince Iheinselves thai eiich ol these reactions is an initiation piopngnlion .....I.....ism Most y.v, phase reactions of neutral species go via initiation propagation reactions. In the literature, initiation-propagation reactions of gas-phase species arc also ■ ill. J radical mictions, since the reactive species arc radicals. They are also called II, 1.;/<7 2Br + X (5.11) (5.12) I he hydrogen atoms can react with each of the stable species in the system: H + Br2->• HBr + Br (5.13) .............'>•" "" »Ii ' nrmi TO I ■ li Ml /V.....II. II I Ulli II i II II i Iii II I II i . Mi (5.15) (5.16) (5.17) (5.18) (5.19) (5.20) (5.21) I also want to mention that years ago, people thought about the possibility of the reaction occurring via a four-centered complex: The bromine atoms can react wiih each of the stable species in the system: Br + H2-► HBr+H Br + HBr-► Br2 + H Br + Br2 -► Br2 + Br The radicals can react with each other: 2Br + X 2H + X H + Br + X -> Br2 + X -> H2 + X -> HBr + X Br2 + H2 H-H - / 1 -Br-Br -> 211 Br They also considered reactions without collision partners, such as Br2 -► 2Br 2Br-► Br2 (5.22) (5.23) (5.24) Notice that reactions (5.11 )-(5.24) are all of the possible reactions between H2, Br2, HBr, H, and Br. Therefore, the mechanism must involve only these elementary reactions. Next, let's try to eliminate reactions. In Chapter 4, we noted that Every elementary reaction must have at least two reactants and at least two products. (5.25) Consequently, reactions (5.23) and (5.24) cannot occur. In contrast, reactions (5.11)—(5.22) are perfectly good elementary reactions, There are two or more reactants and two or more products in each step, which means that in principle all of these reactions could occur. A more precise analysis shows that they all do occur at some rate. Therefore, in a sense, reactions (5.11)-(5.22) represent the complete mechanism at the reaction between H2 and Br2. The question is: Why did we say that reaction (5.2) follows mechanism (5.3) and not include reactions (5.11), (5.15), (5.17), (5.18), and (5.20)-(5.22)? This brings up a subtle point. When we say that reaction (5.2) follows mechanism (5.3), we are not saying that reactions (5.12)-(5.14), (5.16), and (5.19) are the only reactions that occur. Rather, we are saying that reactions (5.11), (5.15), (5.17), (5.18), and (5.20)-(5.22) occur so slowly that they have a negligible effect on the overall rate. More precisely, we can state that nu nu in in i ;IIANtrlMOI INI ma I ION WHIN MM.IIANI'iM'i .■I I I lir mechanism represents 1 lit* piiiuipul mule hum iviicliiiits in products, hul olhei i. hi lions can, and do occur. low the nexl question we want to address is: Could one have guessed the mechanism iii. nou (5.2) without doing any experiments? The answer is "Yes". There are B seriei i mimical rules that one can use to tell you about the mechanisms of chemical reactions, .......ni often use those empirical rules to guess the preferred mechanism of a chemical it Empirical Rules to Predict Mechanisms of Gas-Phase Radical Itoiictions: Reactions to Include I in I. are two kinds of empirical rules: empirical rules to exclude reactions, and empirit il mies to include reactions. I ri's start with the empirical rules to include reactions. Note that when one lias in ......iiion -propagation reaction, there are certain requirements that one needs to satisfy to " il c the reaction happen. The first empirical rule is that There must be at least one initiation reaction. (5.26) I Include this rule because students often forget that there must be an initiation step hi any initiation-propagation reaction. Typically, the initiation step has a small rale .....sequently, students sometimes assume that the initiation step will not be important In 11 il- reaction. However, in practice, no reaction happens until the initiation step occurs I iirlier in this chapter, we noted that the reactions in the engine of your car follow an miii.iiion-propagation reaction in which the initiation process is activated by the Bpark plugs. II you never fire the spark plugs and start the reaction, no reaction ever on in . I lie same is true for any initiation-propagation reaction. Typically, the initiation slips have a much lower rate than do any other steps in the reaction mechanism. However, il one ignores the initiation step, one will never derive a correct rate equation. You need to include at least one initiation step in every mechanism independent Ol W hether the rate of the initiation reaction is small. The next key requirement is The propagation reactions must occur in a cycle where radicals react with the reactants to form new radicals and then the new radicals react to form the original radicals again. (5.27) acquirement (5.27) is more complicated to understand, so let's go back to mechanism i \3) and see what we mean. Notice that there are two propagation reactions in mechanism (5.3), which we have relabeled reactions (5.13) and (5.16). In reaction (5.16), a bromine nlom reacts with H2 to yield HBr plus a hydrogen atom. In reaction (5.13), a hydrogen atom reacts with a Br2 to yield a second HBr and regenerate the original bromine atom, i onsequently, one can view reactions (5.13) and (5.16) as a cyclic process where hydrogen atoms are created via reaction (5.16) and lost via reaction (5.13). Bromine atoms arc rrcated in reaction (5.13) and then lost via reaction (5.16). I Ml I HI . Ill IIN I II IMI Ml I IIANI'tM I II IIIIIIAIII III I'l HHi HBr — ~" H ^ Br2 Figure 5.1 A cycle for HBr formation via reaction (5.3). Figure 5.1 shows a pictorial representation of this mechanism. Figure 5.1 is a different representation of the mechanism than you are used to. In the figure, we label the key radicals and show how those radicals are formed and destroyed. Reactants and products are indicated by the ends of the arrows. Two arrows come together when a given pair of species react. The products of the reaction are indicated by arrows leading away from the encounter. For example, the curved arrows on the left of the figure indicate that H2 is reacting with a bromine atom to yield a HBr and a hydrogen atom. This should be convinced that Figure 5.1 is a valid representation of the propagation steps for mechanism (5.3) before proceeding with this chapter. Notice the cycle in Figure 5.1. Hydrogen atoms are formed and destroyed in a cyclic process. The cyclic nature of the propagation reaction is a key feature of initiation-propagation reactions. Radicals are very reactive species. Consequently, if one had a significant concentration of radicals in the system, one could produce a considerable amount of product by going around the cycle many times. Later in this book, we will refer to the cycle as a catalytic cycle. Catalytic cycles are key features of most reactions. All kinds of cycles have been observed experimentally. One can have a simple cycle with only two radicals as shown in Figure 5.1 or a more complex cycle with eight intermediate species as shown in Figure 5.2. Still the key feature is that the cycle needs to loop back on itself (i.e., form the original intermediates) in order to produce a significant amount of product. For example, if one produced a chain of radicals and never got back to the original radicals, the reaction would go around the cycle once and stop. Relatively little product would be produced. In contrast, if the cycle loops back on itself, you can go around and around the cycle and produce lots of product. Consequently, cycles that loop back on themselves have a much higher net rate than do cycles that do not loop back on themselves. In actual practice, usually only cycles that loop back on themselves show appreciable steady-state rates. Noncyclic reactions are seen in laser photolysis, however. A related requirement is that All of the steps in the catalytic cycle must have low barriers. (5.28) Again, in order to get much product, each of the steps in the catalytic cycle must have low barriers or else the cycle will be too slow. As a general rule of thumb, if you are Adilyl ( 11A NAI111 - NAD FADH^ FAD CoA-SH 1 in. il. iii-I 11111.11. il. ■ 3 Succinate GTP GDP + P Succinyl-CoA NADH C02 11,......B.2 The tricarboxylic acid cycle. This cycle is the principal way that cells produce NADU, llm NAI )l I 1I1 m |uently used to convert ADP to ATP (adenosine diphosphate to adenosine triphosphate). running the reaction at a temperature T, then the EA, the activation barrier for each ol the 1 should satisfy EA < [0.07 (kcal-mol)/KJT (5.29) 1 .1 reasonable amount of product. Equation (5.29) tells you that if the propagation Il ps have high barriers, you need to run the reaction at high temperature lo fei .1 11 1 unable amount of product. I In- last key requirement for an initiation-propagation mechanism is that: There should be at least one termination reaction where two radicals combine to yield stable species. (5.30) Requirement (5.30) is different from requirements (5.26) and (5.27) because it says thai iluie should be a termination reaction, not that there must be a termination reaction I here are some reactions where the termination reactions have extremely tiny rates. Still, il there were no termination reactions, radicals would build up in the system and could never be destroyed. Consequently, the system would have to get to steady state. Instead I he reactive mixture would explode. One can show that one can never get a steady-stale reliction unless there is at least one termination reaction. Therefore, if one observes I Hleady-state reaction, there must be at least one termination reaction. One often seel i vcral termination reactions. lixplosive reactions are exceptions. One can design a high explosive so thai the iiibstance explodes in the absence of a significant termination reaction. In actual praeliee. such systems are dangerous. However, they can be designed. Industrially, however, one often adds something to a reacting mixture to enhance the termination reactions and thereby prevent unnecessary explosions. For example, in your ear, people add octane enhancers [e.g., methyl t-butyl ether (MTBE) and tetraethyl-lead (TEL)] to better control the combustion process. The MTBE and TEL participate in termination reactions, so the fuel burns more smoothly. If you do not add MTBE, you get lilllc explosions (knock), which tears your engine apart. In practice, one needs termination in (it mi Will INDIA III IN ľlll H'Al lAIH IN Ml I IIANI'.M'i Inlilii '.K«v n«|iiii«iiii>Mi-. i..i Hi.- 11 n * i h.....mu im mni linn I Br, Reliction Requirement Krsiill (5.26) There nuisl be an milialioii reaction (5.27) Musi have an catalytic cycle (5.30) There should be a termination reaelion Must im.hide reaelion (5.1 I) or (5.12) Musi include reaction (5.1) and (5.Id) Should include either (5.1°). (5.20) or (5.21). reactions to prevent explosions. That is why people add MTBE to gasoline. In practice] most reactions have termination steps, which is why we can state (5.28) as a requiremenfl even if it is not an absolute requirement. Before we proceed, it is useful for the student to look back to Table 5.1 and verily that each of the conditions in Table 5.2 is obeyed for the reactions in Table 5.1. Let's apply these rules to the HBr reaction. Requirement (5.26) (see Table 5.2) says thai there must be an initiation reaction. If we apply requirement (5.26) to the HBr case, iltcn we conclude that either reaction (5.11) or reaction (5.12) must occur during the reaction Requirement (5.27) says that we must have a catalytic cycle to have an appreciable rate If we apply requirement (5.27) to the HBr case, then we conclude that reactions (5.13) and (5.16) must be included in the mechanism. Requirement (5.30) says that there should be a termination step. If we apply requirement (5.30) to the HBr case, we conclude that either reaction (5.19), (5.20), or (5.21) should be included in the mechanism. Therefore, we know that the mechanism of HBr production must include reactions (5.13) and (5.16), and either reaction (5.11) or reaction (5.12), and either reaction (5.19) or reaction (5.20) or reaction (5.21). We do not know anything about reactions (5.14), (5.15), (5.17), (5.18), or (5.22). 5.4.2 Empirical Rules to Predict Mechanisms of Gas-Phase Radical Reactions: Reactions to Exclude Next, we will provide some rules to exclude reactions from a mechanism. Generally, these rules are based on the fact that if the rate of a reaction is small enough, the reaction will have a negligible effect on the overall rate. These rules are harder to apply because you must include at least one initiation reaction even though the initiation reaction has a low rate, and you must include at least one termination reaction even though the termination reaction has a low rate. Still, if there is more than one possible initiation reaction or termination reaction, one can use the rules in this section to choose the one that will be the most important. The overall plan will be to estimate rate constants for reactions, and if a reaction has a small enough rate, exclude it. Generally, the methods in this chapter will let us determine rate constants to within only a few orders or magnitude. That is not good enough for detailed kinetics. However, it is good enough to eliminate reactions from mechanisms. Our general approach will be to use various methods to estimate the activation barriers for reactions. We will eliminate all reactions that do not satisfy equation 5.29. For the purpose of discussion we will assume that all of the reactions follow Arrhenius' law: ku = k" cxp (5.31) where ku is the rate constant for the reaction, k" is the preexponential, Ea is the activation energy: kB is Boltzmann's constant, and T is the absolute temperature. Níni iiiiimiIci the inilialiiin steps Ihe Inilliilioti steps, relictions O II) .mil (VI.'). .n. .1 unlmolťculnr rtmlions We have u single Peuclanl colliding with n collision i in yield products. In duplet K, we will lind that unimolcculut leaclions usually . |iice\ponenlial ol about I0m- Ä /(molecule second) independent of the details ......is (5||) and (5.12) are both unimolecular reactions, and their preexponentials i Mini.n Consequently, one could not distinguish between these two reaction* imi. oi the preexponentials, The two reactions do, however, have quite diffcrcnl Hon energies. To see that, we have to introduce our next empirical rule, i In m livation energy for an initiation reaction, E!, can be approximated via El, = AHA„A + 1 kcal/mol (5 12) ..... MIA a is the enthalpy change during the initiation reaction. I it'-, use equation (5.32) to estimate the activation barriers for reactions (5.11) and i i According to data in the Chemical Rubber Company Handbook (CRC) hydrogen ii i i bond energy of 103 kcal/mol while bromine has a bond energy of 45 kcal/mol n. n inn according to equation (5.32), reaction (5.11) will have an activation Daniel i ,1.....i 104 kcal/mol while reaction (5.12) will have an activation energy ol aboul iil/mol. If we plug back into equation (5.31), we find that at 800 K. reaelion (-> I 11 ill in a factor of I015 slower than reaction (5.12). Reaction (5.11) is much slowei than "•n (5.12). Therefore, little error will be created if we ignore reaction (5.1 I) i Inc can generalize the ideas in the previous paragraph to show The most important initiation reaction wil breaks the weakest bond in the reactants. be the reaction thai (5.33) Physically, the weakest bond in the molecule is the easiest to break. Consequently, the ■ ikcsl bond breaks first. There are some special cases where there are (wo bonds in iin reactants that have almost the same energy. In those cases, both bonds break with n i unable rates. Still, such cases are exceptions. There is usually only one initiation step I.......g that step the weakest bond in the molecule breaks. I he implication of the discussion in the last paragraph is that while we have to include i. a. lion (5.12) in the mechanism of HBr formation, we can ignore reaction (5.1 I) Now, let's consider reactions (5.15) and (5.18), which are called exchange reactions In an exchange reaction, one atom in a molecule is replaced by another identical atO...... id. same molecule. Exchange reactions often occur at a reasonably high rate. However, hange reactions do not consume any reactants or radicals or produce any product!. ' onsequently, they do not affect the overall rate of reaction. Therefore Exchange reactions can be ignored when one is deriving a rate equation. (5 111 \s a result, there is no reason to include reactions (5.15) and (5.18) in the mechanism ol 11 Br formation. Physically, the reactions do occur at reasonably high rates. However, Ihe ii actions do not contribute to the formation of HBr. Therefore, they are not part of the incclianism of HBr formation. Next, let's consider reactions (5.14) and (5.17), which are called inhibition reactions In inhibition reactions, the products of the reaction react with a radical to produce a I til 1*1« I I II lit ' 11 it mm I in I it in I 'I ii ii 'm in in in Ml 1,1 IANI' IM' i irm lain iniilri ill.- In llir i ase nl i «'ni 11< his l *i III ,iinl (»i I /), III mid* Willi n radical In I >i i»lut <- iilliii II, iii III, II oik- stints with the product anil ends up with the n-.u i.iiii-.. one will definitely slow down I tic- reaction, which is why reactions nl I lux type are tailed "inhibition" reactions. Now the question is whether we need to include reactions (5.14) and (5.17) in thfl mechanism. The answer is "Yes", provided that the reactions satisfy equation (5.2M According to data in the CRC, reaction (5.14) is 16 kcal/mol exothermic. The heal <>l reaction is similar to that of reaction (5.13). Consequently, there is no reason to exclude reaction (5.14) on the basis of energetics, and we will see in the nexl paragraph dial there is no reason to exclude reaction (5.14) on the basis of kinetics, either. Reaction (5.17), however, is 40 kcal/mol endothermic. Consequently, if everything else is equal] one would expect reaction (5.17) to be much less important than reaction (5.14). We can quantify the effect by using another empirical rule, called the Polanyi relationship, to estimate the difference in the rates of the two reactions. The Polanyi relationship was proposed by Polanyi and Evans to try to correlate activation barriers to reactions. According to the Polanyi relationship, Ea, the activation barrier for a given reaction, is related to AHr, the heat of reaction, by Ea = E(a'+yPAHr (5.35) where E(a' is the intrinsic activation barrier and yP is the transfer coefficient. We will discuss the Polanyi relationship in detail in Chapters 10 and 11. Generally, E" and yP vary with the reaction types and the shape of the bonds which form. One needs to consider these effects in detail, if one wants to predict an accurate value of the activation barrier. Fortunately, in Section 5.4.5, we will find it possible to get a rough approximation to the activation barrier if we assume that E° is 12-16 kcal/mol for atom transfer reactions while yP is 0.2-0.5 for exothermic reactions and 0.5-0.8 for endothermic reactions. If we substitute into equation (5.35) with yP = 0.7, we find that reaction (5.17) should have an activation barrier of about 34 kcal/mol. By comparison, the same calculation shows that reaction (5.14) will have an activation barrier of only 6 kcal/mol. Both reactions have similar preexponentials. Clearly, at 400 K, reaction (5.14) will be much more important than reaction (5.17). Still, in the homework set, we will ask the reader to show that if one loads HBr into the reactor and runs reaction (5.2) in reverse, one will need to include reaction (5.17). Actually, the HBr formation reaction is not very reversible. One needs to heat the system to almost 4000 K to get a reasonable conversion. Still, reaction (5.17) is important for the reverse reaction at 4000 K, even though it is not so important for the forward reaction at 400 K. One can quantify this idea more carefully, using the material from Section 2.6. Generally, in order for an elementary reaction to have a significant rate, the activation energy for the reaction needs to be less than a critical value, Ecril, where Ea < E™1 = ( 0.07- 0.15 kcal mol-K kcal mol-K kcal T for propagation reactions -T for initiation reactions (5.36) II.......c different values ol I he initial temperature In.....tuition and propugt....... H ii lions because Initiation it-actions have much sniallei rales than do pnipagalion .. .. nous Equation (5.36) allows you to till whether a given reaction will be iniporliinl ,i , given temperature, according to equation (5.36), at 400 K we should include all reactions where Ii, -mil/ kcal inoi Knioo K) 28 kcal/mol. On the previous page we found thai Reaction (| II) has an activation barrier of 6 kcal/mol while reaction (5.17) has a burriei ol i i , ul/mol. Reaction (5.14) satisfies equation (5.36) so it should be included al 100 l\ i ii iion (5.17) docs not satisfy equation (5.36) at 400 K, so it should not be included II ion K Reaction (5.17) should be included al temperatures above 500 K, howevei \i this point, we have eliminated reactions (5.11), (5.15), (5.17), and (5.18) from thl .....Imnism of HBr formation and shown that we must include reactions (5.12) (3.14) .....1 (3 16). Next, let's consider reactions (5.19)—(5.21), which are all termination reai...... \i ■ niiling to rule, we know that we must have at least one termination reaction The in, Hon is which one, and whether we need to have only one termination reaction 01 io mi hide all three termination reactions. The question of which one is easy. There is mother empirical rule: One should include all termination reactions where the species produced in the initiation reaction combine to yield stable products. If there are transfer steps (defined later), one should include the species produced in the transfer steps, too. (S W) In a action (5.12), we produce bromine atoms, so we must include reaction (5.19) in ill. mechanism. It is not obvious whether to include reaction (5.20) and (5.21). Anoihei . mimical rule is that Radical recombination reactions will usually have an activation energy of less than 2 kcal/mol (5.3X) 0.05——T for reactions in the catalytic cycle mol-K I bus, the rate constant for reactions (5.20) and (5.21) should be significant. One can, In in i. find conditions where reactions (5.20) and (5.21) are quite significant. Consequent!} d I were guessing the mechanism of reaction (5.2), I would include reactions (5.20) and (5.21). However, when Bodenstein and Lund did experiments on HBr in 1906, they picked i onditions where the hydrogen atom concentration was much less than the bromine atom concentration. In those special circumstances, the rate of reaction (5.19) is much larger ili,in the rate of reactions (5.20) and (5.21). Consequently, it was acceptable to ignore ri action (5.20) and (5.21). I want to note, however, that if you are predicting a mechanism, you are not going to know a priori that people are going to only want to take data under conditions where I he hydrogen atom concentration is much less than the bromine atom concentration, i 'onsequently, if one is asked to predict a mechanism, one should include reactions (5.20) .aid (5.21) even though under some conditions reactions (5.20) and (5.21) will not In-important. At this point, we have considered all of the reactions above except reaction (5.22). Reaction (5.22) is different from all of the rest in that it is a four-centered reaction niMlK.IHINUI llll MM IIANIMMIII INI HA Iii lAIH in MM. MM Figure 5.3 A stick diagram of a proposed transition state for reaction (5.22). Many years ago, people used to talk about four-centered reactions. They would assume a transition state such as in Figure 5.3 and calculate properties. However 1965, Woodword Hoffman pointed out that a symmetric four-centered transition state is impossible with neutral molecules. We provide a detailed derivation of the rule in Chapter 10. However, the thing to remember for now is that in a symmetric four-centered transition state, one ends up trying to put four electrons into a single molecular orbital. That is quantum-mechanically forbidden. One can get sort of a four-centered transition state by first breaking two of the bonds. However, that is an energetically unfavorable process. Consequently Four-centered transition states can be ignored when the reactants are neutral nonpolar molecules. (5.39) In summary, according to the rules in the last two sections, the mechanism of HBr formation must include reaction (5.12)-(5.14), (5.16), and (5.19) and probably should include reaction (5.20) and (5.21). If we look back to Bodenstein's mechanism of the reaction in equation (5.3), we see that we got the answer almost right. We predicted that of all the reactions which Bodenstein said occurs do occur. The only issue was that we included reactions (5.20) and (5.21), while in Bodenstein's original work [1906] these two reactions were ignored. Now it is hard to say that Bodenstein was wrong, since reaction (5.20) and (5.21) were not important under the specific conditions he considered. Still, I would include those two reactions in the mechanism since they are important under many conditions. 5.4.3 Additional Rules for More Complex Reactions Next, we want to extend the ideas in the last two sections to more complex reactions. To keep this effort in perspective, note that we have spent several pages analyzing reaction (5.2), the formation of HBr, and believe it or not, reaction (5.2) was a particularly easy case. There were only two reactants. They were both diatomics, and all of the radicals were atoms. Anything else is going to be more difficult. Still, one can make useful predictions. There are two general difficulties with extending these ideas to more complex reactions: (1) there are additional intermediates to consider and (2) there are many more reactions to consider. Let's talk about the issues of the number of intermediates first. One of the problems with calculating the mechanism for a more complex reaction is that there are many more intermediates to consider. For instance, I have found that my chemical intuition is usually not able to predict all of the intermediates that could form. For example, let's consider the mechanism of HI formation: III liiiiiiiilliin is wis s.....I.ii In Illli Iniiiiiiliiiii. mul so one iiuj'ln expeel llie mei li.inr.ni nl teiielion (5.401 lo be very sinnliii in llie mei liamsin nl Illli liumalioii | mcelianism O l|| In lact, howevei, an III inlei ineili.ile hums iliiini)' ie.ii iion (5.40) while no si tl 11 llll inli'iiilfiliale is seen during Illli Immalum i . .Minul tell you a simple way i" predict thai the 114 intermediate will be important ........I in im mat um. even though ihe I Md intermediate is not important to HBi formation .......ni calculate the stability of the intermediates using the quantum mechanical methodN 11 || led m Chapter 1 1. However, that is complicated. For now. my best advice is thai il n unsure whether an intermediate is stable, include il in the mechanism. I he woril ih ii will happen is that you will have some extra reactions to consider. However, il you in wrong, you will miss some key chemistry. ■ I l Intrinsic Barriers for Propagation Reactions i in-. vim know which intermediates are stable, you can start to predict mechanisms t ii • >i iiliire is the same as in Section 5.4: Mu . Start with a guess of all of the intermediates in the system. Use quantum mechanic! (Chapter 11) to verify that you are not missing any intermediates. . Write down those intermediates. . Use various rules to pare down the list of reactions. I In hard part is the last step. The general scheme is to use an empirical rule to estimate in activation barriers for each of the reactions in a mechanism. Earlier in this chapter. WC presented three empirical rules to determine barriers to reactions: equation (5.32) Im initiation reactions; equation (5.38) for termination reactions; and, a more gcneial i• l.uiimship, equation (5.35) for propagation reactions. According to equation (5.35). E„ Mi. activation barrier for a given reaction, is related to AHr, the heat of reaction, by Ea = E(a» + yPAHr (5 ii . H2 + I2 2HI (5.40) where E" is the intrinsic activation barrier and yP is the transfer coefficient. Equation i Y 11) is the most important relationship in this chapter because it allows you to estimate in activation barrier for a given reaction from the heat of reaction. In this section, we will examine equation (5.41) in more detail to see how it can be used to produce mechanisms nl reaction. There are two terms in equation (5.41): an intrinsic barrier, and the transfer coefficient nines the heat of reaction. The second term on the right side of equation (5.41), ypAII, In easy to understand. This term is associated with the energy you gain or lose in < (inverting the reactants into products. If you have an exothermic reaction, you gain energy as the reaction proceeds. The heat of reaction is negative, which means that, ticcording to equation (5.41), the reaction is a little easier. In contrast, if you have an endothermic reaction, you lose energy as the reaction proceeds. The heat of reaction is |iositive, which means that, according to equation (5.41), the reaction is a little harder, there is a coefficient, yP, in propagation reactions. yP is usually between 0.3 and 0.7. therefore, one does not get the full effect of the heat of reaction in equation (5.41). Si ill, there is the qualitative feature that exothermic reactions tend to be favored over 1'iidothermic reactions. I lie mínusu ii.....n in ii|ii.iiii>n i vi 111-, h.Hiii'i in iiiuiiisi,ukI Physically, the iiiiiinsii barrici provides a measure ol ihe energy il lakes lo make llie reaction happen il ihe heal ol reaction were /.cro. In order to get a physical picture ol why intrinsic barriers arise, consider an example, the reaction H + CH3CH3-► CH4 + CH, (5.42) Reaction (5.42) is a simple reaction where a hydrogen atom approaches an ethane molecule and induces the carbon-carbon bond in the ethane to break. If one looks at reaction (5.42) in detail, it is not at all obvious that the reaction would be activated. After all, one can imagine the reaction proceeding as in Figure 5.4. One starts off with a hydrogen and an ethane with a 88-kcal/mol carbon-carbon bond. During the reaction, the hydrogen comes up to the ethane and forms a partial carbon-hydrogen bond. Simultaneously, the carbon-carbon bond breaks. If one imagines that one has a partial carbon-hydrogen and a partial carbon-carbon bond, then one might imagine writing the bond energy of the system, Ehond, at any time during the reaction it: Eb0„d(t) = EH-c(t) + Ec-h3(t) (5.43) where EM-c(t) is the energy of the H-C bond at time t and Ec_cH, (t) is the energy of the carbon methyl bond at time t. If you have half of a carbon-carbon bond and half of a carbon-hydrogen bond, the Ehond comes out to be (0.5X102 kcal/mol) + 0.5(88 kcal) = 95 kcal/mol (5.44) According to this simple model, it seems that you have more total bond energy in the transition state than you do in the reactants, and this conclusion is verified by detailed quantum-mechanical calculations. Therefore, it is not obvious that reaction (5.42) is activated. In fact, however, reaction (5.42) has an activation barrier of 44 kcal/mol. Lee and Masel |I996, 1997] examined the barriers to reaction (5.42) in some detail, and found that the main contribution to the barrier comes from the orbital distortions that occur during the reaction. During reaction (5.42), the Is orbital in the hydrogen interacts with what is called the 3A,g orbital on the ethane to induce the ethane's carbon-carbon bond to break. Diagrams of both orbitals are shown in Figures 5.5 and 5.6. In the figure, lightly colored lobes arc used to designate positive orbitals, while darkly colored lobes are used to designate negative orbitals. When the hydrogen atom is far away from the ethane, the Is orbital in hydrogen looks spherical, while the 3A,g orbital has a central balloon-shaped orbital plus two kidney-shaped non-bonding lobes. The 3A|g orbital is constructed from a p orbital on each carbon plus some s orbitals on the hydrogens as indicated in Figure 5.5. The diagram on the left of Figure 5.5 shows H M H H^C-C^H 111.1111 > ■ i reaction h , tin- lone pairs on 11 u- radical lonu a hoi h I during rcaelion In those cases, 11 it- intrinsic harriers an- also small. Yv is approximately 1.0 foi both initiation ami termination reactions, in contrast, during .1 propagation reaction, one bond breaks and another forms. Figure 5.6 shows that one has to push the bonding orbitals out of the way before new bonds ean form. It eosls encrgj to push the orbitals out of the way. Consequently, propagation reactions generally have intrinsic barriers between 8 and 70 kcal/mol. Next, we want to discuss how intrinsic barriers change as one changes the reaction First, we want to note that there is a fundamental difference between transferring an individual hydrogen atom and transferring a molecular ligand. Let's go back to the ethani hydrogenolysis case mentioned earlier. If we ignore exchange, then there are two possible reactions which can occur when ethane reacts with a hydrogen atom. The carbon-carbon bond can break to yield methane: -»• CH4 + CH3 H + CH3CH3 or the incoming hydrogen can pick off a hydrogen atom from the ethane: H + CH3CH3 -> H2 + CH2CH3 (5.45 (5.46) rill 1 ill.Ill IN m HU Ml ii lANI'iMiH INI IIAIII IN I'llOl'AUAIIllNMK.IIANI-.M-. :M..i Lee and Masel [1996] found that reaction (5.45) has an intrinsic barrier of 45 kcal/mol while reaction (5.46) has an intrinsic barrier of 12 kcal/mol. Experimentally, only reaction (5.46) is seen. It is important to consider why reaction (5.46) has a much lower intrinsic barrier than does reaction (5.45). When reaction (5.46) occurs, the hydrogen atom comes up along with the CH bond axis in the ethane. One has to distort the orbitals in the break in the C-H bond. However, the remaining orbitals in the ethane are largely unperturbed. In contrast, when reaction (5.45) occurs, the hydrogen approaches the ethane along iis C-C axis. The incoming hydrogen needs to push the C-H bonds in the ethane out of the way before any reaction can occur. That costs extra energy, so reaction (5.45) has a much higher intrinsic barrier than does reaction (5.46). Lee and Masel 11996] showed that this extra barrier is associated with the energy to distort orbitals and not just the energy to move the C-H bonds. For example, the reaction CH3OH + H CH3 + H20 (5.47) has almost the same intrinsic barrier as reaction (5.45) even though the hydrogen does not have to displace any atoms during reaction (5.47). The hydrogen does need to displace the lone-pair orbitals on the OH, however. This example shows that the intrinsic barriers are associated with orbital displacements and not just motions of individual bonds. A key point is that it does not matter what molecular ligand one is transferring, one always sees intrinsic barriers of a similar magnitude. So for example, if we look at a series of radical transfer reactions such as R + H R, + RH (5.48) where R and R\ are two hydrocarbon ligands one usually finds that the intrinsic barrier is between 40 and 60 kcal/mol independent of R and Rt. In contrast, hydrogen transfer reactions of the form Rt-H+R-► Ri+HR (5.49) I ihlu B 3 Th« Intrinsic barriers for II111 oxc limni" mm Hon X" + CH»X -> XCH3 + X" X tun insii Barrier, kcal/mol F II CI 12.8 Hi 10.8 1 9.7 .Source: Results of Glukhoulsev, Pross and Radom |1«W5|. ........sic barriers between 8 and 15 kcal/mol, independent of R and R\. II one looks ill, one finds that the intrinsic barriers do depend slightly on R and /<', lypieal 111 id..... in activation barriers are ±5 kcal/mol. A 5-kcal/mol change in the barrlei niii. n hi lo change the rate by a factor of 103 at 300 K. However, such a changi ........portanl to the prediction of the reaction mechanism. Therefore, we will i|......1 variations for the discussion in this chapter. These variations will be discussed In 1 li,i|ii. 1 11. 1 in oilier thing that one finds is that if one goes to stiffer orbitals, the intrinsic barriers • > I hi example, Table 5.3 give the intrinsic barriers for a number of isotopic exchange .....is as reported by Glukhoutsev et al. (1995). Notice that as one goes to siillei atoms, ■.........isic barriers rise and then fall. Consequently, it is harder to transfer a still atom ....... spongier one. The differences are only a couple of kilocalories, though. I nblc 5.4 gives a selection of other intrinsic barriers and transfer coefficients foi iiii.niii types of reaction. The reader should refer to this table when predicting mi 1 lianisms of reactions. I umples 5.A and 5.B show several examples to indicate how Table 5.4 is used to 1. 1. inline whether a certain mechanism is feasible. One should study those examples 1» fore proceeding with this chapter. h 4.5 Extra Reactions with Larger Molecules Si 1 1.n. we have been discussing reactions of simple diatomic molecules. Next, we waul In discuss reactions of organic molecules containing several atoms. Ilyser (1970), Fossey, et al. (1995), Perkins (1994), and Alfassi (1999) offer good n \ n-ws of the reactions of organic molecules in the gas phase. Generally, most organil molecules follow mechanisms such as those we have already discussed, but, there ,m .....e extra reactions that can occur: . Chain transfers • Addition reactions . Isomerizations . Disproportionalions • Fragmentations . fl Scissions We will review each of these types of reaction next. cn Table 5.4 Intrinsic barriers and transfer (coefficients for different types of reaction of neutral species). Actual E°, EiJ to Assume yp to Assume when Reaction When Predicting Predicting Example kcal/mol Mechanisms, kcal/mol Actual yp Mechanisms Simple bond scission AB + X^A + B + X 0-1 1 1.0 1.0 (X = a collision partner) Recombination A + B + X^AB + X 0-3 1 0.0 0.0 (X = a collision partner) Atom transfer reaction 0 > AHr > -40 Rx + R[ -* R + x - /?, (x = an atom) 8-16 12 0.2-0.6 0.3 Atom transfer reaction 0 < AHr < 40 R-x+R-*- R + x-(x = an atom) 8-16 12 0.4-0.8 0.7 Atom transfer reaction AHr < —40 R - x + R -> R + x-Ri 0-2 0 0 0 (x = an atom) Atom transfer reaction AHr > 40 R-x+ R~* R + x- Ri (x = an atom) 0-5 2 0.9-1.0 1.0 Ligand transfer reaction to hydrogen H+/?-/?,-> H/?+ /?, 40-50 45 0.4-0.6 0.5 Other ligand transfer reactions x +/?-/?!->. xR+ Ri (x = an atom) >50 50 0.3-0.7 0.5 Endothermic ^-hydrogen elimination RCH2CH2. + X RCH 12-25 15 0.5-1 0.7 = CH2 + H. + H Another common ic.it lion is culled n |l scission Keuclions (5.61) nnd (5.62) aflj examples ol |) scission relictions. R - CH2CH2. + X O R-C-O. + X R. + C2H4 + X R. + C02 + X (5.61) (5.62 In both cases, one breaks the a bond at the p position from the radical, and lorn is 11 double bond with the p carbon. All ^-scission reactions break a bonds at the p posiimn ^-Scission reactions are very common in the decomposition or combustion reactions modest-sized molecules. P Scissions are one example of a larger class of reaction called fragmentulioi reactions. Fragmentation reactions are reactions where a radical breaks apart to form a stable species and a new radical. For example 10.......1 HI ol till Ml 1 IIAIIľ.M i'l till IIA I 11 Hl ľ.....'Ai lAlli ill Ml 1 IIAMf.M'. (CH3)3CO. + X-► (CH3)2C=0 + CH3. + X (5.63) is a fragmentation reaction. Again, one forms a double bond that drives the reaction. The fragmentation reactions can also occur in reverse: CH3. + H2C=CH2-► CH3CH2CH2. (5.64) In this case, the radical adds across the double bond. Reaction (5.64) and the reactions where you add species to radicals are called addition reactions. There is also the possibility of a disproportion reaction. In a disproportion reaction, two radicals react to form a stable species. For example, reaction (5.65) is a disproportion reaction: CH,CH2. + CH3CH2. CH3CH3 + CH2=CH2 (5.65) Reaction (5.65) is similar to the termination reactions discussed earlier in this chapter, except that we get two species as products instead of one. Finally, if you have radicals with double bonds, there is the possibility of the radicals isomerizing as shown in equation (5.66): (5.66) In practice, reaction (5.66) does not occur with small molecules because the ring strain is too high. However, it can occur when there is a chain of five or more carbons. Therefore, it is called five-centered isomerization. This behavior gives a lot more reactions to consider when predicting a mechanism. /\|>|ili< ntion lo tliti I'khIii nllnii <>l lli<- Mix:hiiniHinn ol Honctiomi It 1 . upply the findings in the last two sections to predict mechanisms ol rcuctiOM, .....I M.isel I I'l')/). and Hroadhell el ill ( 1994) have scpaialcly shown that one can in.......(5.41) to predict mechanisms ol reactions. The idea is to follow the method 1 illu .1 111 Set lion 5.2. First, we postulate a series of possible intermediates, then look I .11 ..1 ii.. possible reactions ol the intermediates, and use equation (5.41) to sec which ......us urc favored. A» noted previously I he most important initiation reaction will be the reaction that breaks the weakest bond in the reactants. (5.67) 111 denu- equation (5.67) from equation (5.41) by noting that the intrinsic barriet l"' ......,11.111 reaction is usually 1-2 kcal/mol and y> is about 1.0. AHr for the scission ol ,1 bond is smaller than AHr for scission of a strong bond. Consequently, according .....urn (.1.41), a weak bond will always be easier to break than a strong bond d.....\y ílu initiation step. Ill 1 * m 111 us!, during a propagation reaction, a weak bond may or may not be the easiest to I i.m example, reactions (5.45) and (5.46) are both propagation reactions. Reaction 111 is kcal/mol exothermic, while reaction (5.46) is only 8 kcal/mol exothe.....1 qucntly, on the basis of thermodynamics alone, one would expect reaction (5.45) 1 m.nil over reaction (5.46). Note, however, that the intrinsic barrier to reaction ...... 10 kcal/mol less than the intrinsic barrier reaction (5.45). That difference in .......,1, barriers is more than enough to overcome the difference in AHr for the two lions. Consequently, according to equation (5.41), one would expect reaction (5.46) c 1» favored over reaction (5.45). 1 ■ pcrimentally, reaction (5.46) has an activation barrier of 10 kcal/mol, while reaction 1 i) has never been observed. Lee and Masel calculate an activation barriet Ol mi. til/mol for reaction (5.45). 1 in. can generalize this result to say that During propagation reactions, if everything else is equal, it is usually easier to transfer atoms than to transfer molecular ligands. (5.68) I lute are exceptions to equation (5.68) in cases where the differences in AHr are large. However, equation (5.68) works about 98% of the time. ft.4.7 Example: Acetaldehyde Decomposition Ii \t. let's use the methods we have discussed so far to predict a mechanism for a simple i',is reaction: the decomposition of acetaldehyde: CH3CHO CH4 + CO (5.69) I 11 si, we note that this reaction is likely to go by an initiation-propagation mech-iinsm. One cannot be 100% sure of that, but 99.9% of gas-phase reactions go by 11111 iation-propagation reactions. i-mm ill HOND! III! Ml i IIANI'.M i H INIIIAIInN I'M! II'AUAIl( >N Ml I .1 IANTiM'i 200 Nc\t. Id's 11 v I" picdlcl I lit- im-i linilisill Accouhlig In rule (5 U|, | hi- lllsl slcp in the met danism will be ii stop where the weakest hoiul in I lit- acetaldehyde breaks According to the data in the CRC, the carbon carbon bond is the weakest bond in acelaldcliyilo Therelbre, one would expect the initial step in acetaldehyde decomposition to he scissioa of the carbon-carbon bond CH3COH + X CH3 + HCO + X (5.70) to yield a methyl group and a formyl group. Next, the methyl and formyls need to real 1 via a catalytic cycle to produce the CO and CH4 products, and to regenerate each a the radicals. For example, we might postulate that the methyl radical reacts with the acetaldehyde to produce methane. There are two possible reactions: CH3 + CH3COH CH, + CH3COH -> CH4 + CH3CO -> CH4 + CH2COH (5.71) (5.72) However, the H-C bond in the methyl group has a bond energy of 9 kcal/mol while the C-H bond in the formyl group has a bond energy of 96 kcal/mol. Reactions (5.71) and (5.72) have similar intrinsic barriers because both are atom transfer reactions. Therefore, one would expect reaction (5.71) to be slighthy faster than reaction (5.72). One can quantify these ideas as follows. First note that according to the CRC, the C-H bond in CH4 has a bond strength of 104 kcal/mol. Consequently, reaction (5.71) has a heat of reaction of 96 - 104 = -8 kcal/mol, while reaction (5.72) has a heat of reaction of 99 - 104 = -5 kcal/mol. One can calculate the activation barriers for both reactions from equation (5.35). EJ5.71] = 12 kcal/mol + 0.3*(-8 kcal/mol) = 9.6 kcal/mol. Ea[5.72] = 12 kcal/mol +0.3*(-5 kcal/mol) = 10.5 kcal/mol. Therefore reaction (5.71) should be slightly faster than reaction (5.72), although both would likely occur at room temperature. One might also consider the reaction CH3 + CH3OCH-► CH3CH3 + OCH (5.73) We analyze reaction (5.73) in problem 5.C. It has a high barrier so it is unlikely to occur. In order to complete the cycle, we need the CH3CO fragment to decompose to yield CO and regenerate the methyl group: CH,CO + X CH3 + CO + X Similarly, the formyl (HCO) radical can react via a chain transfer process: HCO + HCOCH-,-► H2CO + CH3CO A second chain transfer process is HCO + X-► H + CO + X H + CH3COH-► H2 + CH3CO Analysis is given in Example 5.C. (5.74) (5.75) (5.76) (5.77) .......nilil also cousiilei n closed evi I o II i CHjCOH I II, i (Oil (3.78) Ntiiiiiuii iV/X) has a high intrinsic barrier, bul it is 19 koal exothermic. Consequently. i in do ti i .Mi ulanou to sec when to include it |we include u ii equation (5 16) ll MlUlledl I ii, . .ill illation is given in Example 5.C. .1 io iircils termination reactions. The following reactions are possible: CH, + COH + X-► CH3COH + X CH3 + HCO-► CH4 + CO X + 2CH3-► C2H6 + X 2COH-► HCOCOH 2COH-► CO + H2CO (5.79) (5.80) (5.81) (5.82) (5.83) in n lias to go back and use equation (5.41) to see which of these reactions ut i i ill. Reaction (5.80) is thermodynamically favored by 3.5 kcal/mol over reaction Reaction (5.79) is a standard termination reaction, so it has an intrinsic barriet ol i i ul/mol. Reaction (5.80), on the other hand, is a hydrogen atom transfer reaction ii ......,n transfer reactions have intrinsic barriers of 10-14 kcal/mol. If one substitute! iii.ition (5.46) one finds that reaction (5.80) is unactivated. Therefore reactions (5,19) > sill should both occur. Note however, that reaction (5.79) is required according I/) while (5.80) is optional. Consequently, one would expect reaction (5.79) to I'., dominate over reaction (5.80). A similar argument applies to reaction (5.83). At this we have considered reactions (5.70)-(5.83). There are no other alternatives oi .m\ II..... reactions. Consequently, reactions (5.70), (5.71), (5.76)-(5.79), (5.81). and (5 s.'i ii litute the mechanism of acetaldehyde decomposition. Hi nson analyzes this case in more detail and notes that at 700 K and 1 aim: [CH3] = 10" 113 mol/liter [CH3CO] = 10" 136 mol/liter [CHO] = 10 146 mol/liter [HJ = 10 17 mol/liter I'hcrcfore, Benson concluded under these conditions, reaction (5.81) is much faster than i i, lion (5.79), (5.80), (5.82), or (5.83). Reaction (5.79) is still required, however. liaison's analysis can be generalized to any reaction. If one radical has a much in-her concentration than the rest, then only that radical will matter to the termination .. actions. In the literature, it has been proposed that one way to identify the most important termination reaction is to look at the catalytic cycle. If one reaction is much -.lower than all the rest, then the intermediates involved in that reaction will have the highest concentrations. Consequently, the most important termination reactions will be the reactions involving the intermediates participating in the slowest step in the catalytic cycle. Unfortunately, that criterion is not easy to use. For example, in the acetaldehyde case, reaction (5.71) is exothermic while reaction (5.74) is endothermic. Reaction (5.71) is an .iimn iiunslci reaction, so ai'iottling In lablc 5 4. tcacliou t *> /1) should have n harrier ul I.' | III * ( 14.8) / (i kcal/mol In contrast, reaction (5.74) should have a harrier nt I I 1.0» III I 2. I kcal.inol ('onscqucnlly. il c\ci Yllnne else were equal, teai lion (5.71) would be quickei than reaction (5.74). In fact, however, reaction (5.74) has an unusually large preexponential, 10" A / (molecule-second). As a result, reaction (5.71) is slower than reaction (5.74) ai hid temperature. Under such conditions, CH3 is the key intermediate so only reaelion (5.81) matters. Still, I am not sure how you would know that the preexponentials were so different. (We discuss the preexponentials in Chapter 9.) Therefore, my advice is that il you have a case where the barriers are close, include all the termination reactions. I 'hat way you will be sure that you are right. One can extend this analysis to many different reactions. More than 95% of the mechanisms of gas-phase reactions can be predicted by these procedures. I find it pretty easy to guess a mechanism of a gas phase reaction, if I know what kinds of intermediales are stable. In my experience, the hard part is deciding what intermediates and what initial group of reactions to consider. For example, is a species such as CH2COH stable enough to form? If one can find a way to determine which species are stable, and what types of reactions to consider, one can predict the mechanism of the reaction using equation (5.46), assuming that the reaction occurs via an initiation-propagation mechanism. 5.5 GAS-PHASE REACTIONS THAT DO NOT FOLLOW INITIATION-PROPAGATION MECHANISMS Next, we will discuss the gas-phase reactions that do not follow initiation-propagation mechanism. As noted previously, in 1932, Rice showed that all of the mechanisms of gas-phase reactions of hydrocarbons that had been studied before 1930 follow initiation-propagation mechanisms. However, since 1930, people have discovered a few gas-phase reactions that do not follow initiation-propagation mechanisms. In the next few sections, we will discuss gas-phase reactions that do not follow initiation-propagation mechanisms. There are three groups of exceptions: • Reactions where there is no feasible initiation-propagation mechanism from reactants to products . Reactions where one of the steps in the catalytic cycle has a high barrier so that no reaction is feasible . Reactions where concerted processes are possible In the next several sections, we will discuss these exceptions. Mi'iiilliui |5H4) is special in Ihal llieic is in.....nation ptopaeatton mechanism leading ....... 11 i.uiis lui example, one can imagine a lusi step such as: Na I CI. NaCI I CI (3.83) 11 ' 1 then what happens? II the chlorine atom reacted willi ihe sodium atom, one table Na( 'I and no radicals. There would be no catalytic cycle. As a result, tins loll cannot follow an initiation propagation mechanism. In fact, no one has Found I a .Is ti* cycle for reaction (5.84). I'ol.uivi (ll)32) showed that if you react Cl2 with sodium dimers, the reaction does go 1 la a catalytic cycle: X + Na2-► 2Na + X Na + Cl2-► NaCI + CI CI + Na2-> NaCI + Na X + 2Na-► Na2 + X (.*>.«(.) (3.87) (3.88) (3.89) However, reaction (5.88) will not occur unless there are sodium dimers in the system I'oI.uim showed that if there are no sodium dimers, the reaction goes by Na + Cl2-► NaCI + CI Nawan + CI ■+ NaCI (5.90) (5.911 III n action (5.91), the chlorine atoms react with sodium atoms that are on the walls ol 11 ic vessel. I his brings up an important point: if there is no catalytic cycle leading from real Uultl in products, then there will be no way for an initiation-propagation mechanism to 0CCU1 in ucli a case, the products must be formed by some other route: either a direct, com ertt d o H lion between the reactants, as in step (5.90); or a wall reaction, as in reaction (5.91) 6.5.2 Association Reactions t »nc key class of reactions where no direct conversion is feasible are called association n 111 lions. In an association reaction, two stable species come together to form products. For example, the reaction NH, + HC1 NHXl (5.92) 5.5.1 Reactions with No Feasible Initiation-Propagation Mechanism First, let's consider reactions where no initiation-propagation reactions is feasible. A key feature of the initiation-propagation reactions is that there is a catalytic cycle where radicals are formed and destroyed. If there is no catalytic cycle, one cannot have an initiation-propagation mechanism. For example, Polanyi (1932) examined the reaction of sodium with chlorine in the gas phase: 2Na + Cl2 => 2NaCl (5.84) is a simple association reaction, because two stable species come together to form products. It is easy to show that with an association reaction, there is no simple iniiiation-propagation mechanism leading from reactants to products. Consequently, the reaction must follow some other mechanism. Usually, an association reaction will occur via a direct concerted process. For example, reaction (5.92) goes via NH3 + HC1 + X NH4CI + X (5.93) ilium iiniislii icmiion, so according in Table 5.4| reacllon t"i /I) should have n hitrrloi of 12 + 0.3 x (-14.8) ■ 7.d kiiil/mol. In COntTUt, rcm'iiuii O Mi should Imvc u hinder ol 1 | 1.0 x 11.3 = 12.3 kcal.inol Consequently, il everything else were equal, reaction (5.71) would be quicker than reaction (5.74). In fact, however, reaction (5.74) has an unusually large preexponential, 10" A / (molecule-second). As a result, reaction (5.71) is slower than reaction (5.74) at high] temperature. Under such conditions, CHi is the key intermediate so only reaction (5.81) matters. Still, I am not sure how you would know that the preexponentials won- m different. (We discuss the preexponentials in Chapter 9.) Therefore, my advice is thai il you have a case where the barriers are close, include all the termination reactions. I li.n way you will be sure that you are right. One can extend this analysis to many different reactions. More than 95% of the mechanisms of gas-phase reactions can be predicted by these procedures. 1 find it prett) easy to guess a mechanism of a gas phase reaction, if I know what kinds of intermediates are stable. In my experience, the hard part is deciding what intermediates and what initial group of reactions to consider. For example, is a species such as CH2COH stable enough to form? If one can find a way to determine which species are stable, and what types of reactions to consider, one can predict the mechanism of the reaction using equation (5.46), assuming that the reaction occurs via an initiation-propagation mechanism. 5.5 GAS-PHASE REACTIONS THAT DO NOT FOLLOW INITIATION-PROPAGATION MECHANISMS Next, we will discuss the gas-phase reactions that do not follow initiation-propagation mechanism. As noted previously, in 1932, Rice showed that all of the mechanisms of gas-phase reactions of hydrocarbons that had been studied before 1930 follow initiation-propagation mechanisms. However, since 1930, people have discovered a few gas-phase reactions that do not follow initiation-propagation mechanisms. In the next few sections, we will discuss gas-phase reactions that do not follow initiation-propagation mechanisms. There are three groups of exceptions: . Reactions where there is no feasible initiation-propagation mechanism from reactants to products . Reactions where one of the steps in the catalytic cycle has a high barrier so that no reaction is feasible • Reactions where concerted processes are possible In the next several sections, we will discuss these exceptions. 5.5.1 Reactions with No Feasible Initiation-Propagation Mechanism First, let's consider reactions where no initiation-propagation reactions is feasible. A key feature of the initiation-propagation reactions is that there is a catalytic cycle where radicals are formed and destroyed. If there is no catalytic cycle, one cannot have an initiation-propagation mechanism. For example, Polanyi (1932) examined the reaction of sodium with chlorine in the gas phase: 2Na + Cl2 => 2NaCl (5.84) i ii Hon i'i.MIi is special in that llirir is no inilialiou propagation inei haiiisui Iraduir I.....i leaclanls I oi example, one can imagine a lusi step such as Na I < I Na< 'I Ml (3.8S) However, then whal happens? If the chlorine atom reacted with the sodium atom, one immiIiI gel stable NaCI and no radicals. There would be no catalytic cycle. As a result. Ibis Il Ii lion cannol follow an initiation propagation mechanism. In fact, no one has found I • iialvtic cycle for reaction (5.84). Polanyi (1932) showed that if you react Cl2 with sodium dimers, the reaction does go ■ i i catalytic cycle: X + Na2 -> 2Na + X Na + Cl2 -► NaCI + CI CI + Na2-► NaCI + Na X + 2Na-► Na2 + X (5.86) (5.87) (5.88) (5.89) However, reaction (5.88) will not occur unless there are sodium dimers in the system. I'olunyi showed that if there are no sodium dimers, the reaction goes by Na + Cl2-> NaCI + CI Nawa„ + CI NaCI (5.90) (5.91) In reaction (5.91), the chlorine atoms react with sodium atoms that are on the walls ol the vessel. This brings up an important point: if there is no catalytic cycle leading from reactants lo products, then there will be no way for an initiation-propagation mechanism to OCCUI in such a case, the products must be formed by some other route: either a direct, concet ted reaction between the reactants, as in step (5.90); or a wall reaction, as in reaction (5.911, '< '> 2 Association Reactions i Ine key class of reactions where no direct conversion is feasible are called association n actions. In an association reaction, two stable species come together to form products. Tor example, the reaction NH, + HCI NH,C1 (5.92) Is a simple association reaction, because two stable species come together to form products. It is easy to show that with an association reaction, there is no simple iiuiiation-propagation mechanism leading from reactants to products. Consequently, the teaction must follow some other mechanism. I Isually, an association reaction will occur via a direct concerted process. For example, o action (5.92) goes via NH3 + HCI + X NH4C1 + X (5.93) No iikIk ills mi" needed In gcnciiil. 11111>11- .i'.'.ih Hilton ii'ilcllons mini vill louicil reactions. Initiation propagation reactions seldom apply Anoilui kiml oi association reaction is a Diels Aldci reaction: + X ,C. V I R V I + x (5.94) Again, there is no initiation-propagation reaction leading from reactants to products. ;nul so the reaction must occur via some other pathway such as a concerted addition. Thai is what occurs. Generally, association reactions occur via concerted processes. Initiation-propagation reactions do not apply. 5.5.3 Reactions with High Barriers to Propagation Steps In all of the reactions in Sections 5.4.1 and 5.4.2, one does not observe an initia tion-propagation reaction because there is no feasible initiation-propagation mechanism leading from reactants to products. There are other groups of reactions, where there is a feasible initiation-propagation mechanism. However, one of the steps in the catalytic cycle has a high barrier. As a result, the catalytic cycle is slow; and most of the product molecules are produced by some other route. A key example is the methane coupling reaction: 2CH4 C2H6 + H2 (5.95) Reaction (5.95) was first studied in 1919, but in 2000 people were still trying to determine a conclusive mechanism. In 1932, Kassel considered whether the reaction could occur via a simple initiation-propagation mechanism with an initiation step CH4 -> CH3 + H with simple propagation cycle -> CH3 + H2 -> C2H6 + H and three termination steps: H + CH4 CH3 + CH4 H + CH3 -► CH4 2H-► H2 2CH3 -> C2H6 (5.96) (5.97) (5.98) (5.99) (5.100) (5.101) I lie 1 yt le in so slow bei niisc their is 11 liuyc him lei to 1 cm lion (Y9H) KcnclIon (5,98) ' iIh iiM'isc oi if, 11 lion <•> I.' 1, which was illiisti.itiil in hpiii' "1 I Figure V-l shows 1I111I II I.....anil with nullum' anil imllivls (I.e., the piodiii Is 111 I ii'iin > I), you need to disloit iIiIiiiIn in the methane before the reaction can happen Consequently, reaction (S.98) "ii ' 11 1 itpei led to have .1 large barriei 1 «e and Masel 11996] calculate an ai tivation ' ol "11 kcal/mol. The analysis 111 Example SC gives 52.5 kcal/mol. Therefore, the ' .1 Hi 1 yi le violates requirement (5.28). The reaction is never going to he last enough 1.....III! c much product at a reasonable temperature. wh.ii happens in practice is that the overall reaction is just really slow. You ct......il lllli ' ii.ilylic cycle to occur al a reasonable rale. No other set of gas phase reactions Ii 1 11 >li■ Consequently, the reaction is slow unless you heal to 800 K. ■ 1 Unimolecular Isomerization Reactions 11......uplii alions discussed in this section are particularly important to what I will 1 ill •/••< ular isomerization reactions. A unimolecular isomerization reaction is a reaction In 11 it single molecule rearranges to form a new product. Examples include Cyclopropane CH3NC : propene CH3CN (5 ..... (5.103) 1 111 isomerization reactions proceed via initiation-propagation reactions. However, in ' 11 nun (5.102) there is no feasible initiation-propagation reaction leading from react.mis In products. There is a possible mechanism for reaction (5.103), with an initiation step 1 propagation step mil 1 termination step: CH3NC CH3NC + CH3 CH3 + NC -» CH3 + NC CH3 + NCCH3 -y CH3CN (5.104) (5.105) (5.106) However, reaction (5.105) includes C-N bond scission, so it has a large intrinsic barrier, I he reaction violates requirement (5.28). Consequently, the catalytic cycle for reaction 1 J 105) is not feasible. Most isomerization reactions follow what is called the Lindemann mechanism. In Ihc lindemann mechanism, the reactants first collide with a collision partner to yield an 1 m ued species, and then the excited species reacts to form products. One can write the mechanism in a simple form as Methylisocyanide + X excited isocyanide excited methylisocyanide + X methylcyanide (5.107) However, Kassel found that this mechanism did not fit his data. There have been many arguments on this point. Hinshelwood showed that methane coupling could occur via this sequence of reactions. However, in practice, one does not produce much product via this route. (Qualitatively, the mechanism in (5.107) can explain much of the data on unimolecular Isomerization reactions. Experimentally, collision partners are needed to get unimolecular reactions to happen. Further, it has been shown that excited species are involved in the reaction process. in ai in nr. i ii ii lir. snii. mil h,um,ni i'i iii/i is ion simple ii von louk ni reaction o 102) In detail, yofl lnul lhal leveral excited Itttei Ol cyclopropane contribute in Ihe reaction Whin the cyclopropane collides with a collision partner, the cyclopropane goes into one ol the excited stales. Then that excited slate reacts to form a second excited stale. Eventually, the excited state decomposes into products. It is not unusual for 20 excited stales in be important to a unimolecular isomerization reaction. Ihe result is a complex react..... network, even though the reaction appears simple. 5.5.5 Concerted Eliminations A larger molecule also shows extra complications in the reaction path; there may be some extra pathways in addition to the standard initiation-propagation steps. For example during butane pyrolysis, the initiation step is the scission of a carbon-carbon bond: -> 2CH3CH2 + X X + CH3CH2CH2CH;, However, a competing process is a concerted hydrogen elimination. X + C4H10-► CH3CHCHCH3 + H2 + X (5.108) (5.1091 Concerted reactions, where you break two or more bonds simultaneously, have high intrinsic barriers. For example, reaction (5.109) has an intrinsic barrier of about 60 kcal/mol. By comparison, reaction (5.108) has an intrinsic barrier of less than 2 kcal/mol. At first, one might think that it is obvious that reaction (5.108) will predominate. However, note that reaction (5.108) is 80.6 kcal/mol exothermic while reaction (5.109) is only 22.4 kcal/mol endothermic. Therefore, one does have to consider whether the difference in the heats of reactions is sufficient to overcome the difference in intrinsic barriers. If you plug into equation (5.41), you find that the answer is "No". Besides, reaction (5.109) does not produce any radicals, so you will not get a catalytic cycle. Still, in the literature, people discuss the concerted elimination processes in great detail, and say that there are cases where the concerted eliminations dominate. In my experience in the gas phase, it is usually okay to ignore the concerted eliminations if one is considering only small molecules with no fluorines or chlorines. Under such circumstances, one can predict mechanisms easily. Large molecules are harder to consider because there are so many reactions and concerted eliminations occur. The result is a very complex reaction pathway. You need a computer to keep track of it. The one place where there is some reasonable evidence for concerned eliminations is the reactions of chlorinated or fluorinated molecules. For example, Swihart and Carr (1998) examined decomposition of dichlorosilanc and suggested that the primary initiation step is X + SiH2Cl2->HCl + SiHCl + X (5.110) At first sight, this reaction would not be expected to occur, because of the large barriers. However, the hydrogen in HC1 is ionic. Later in this chapter, we will find that the reactions of ionic species have much lower intrinsic barriers than do the reactions of neutral species. The ionic transition state allows reaction (5.110) to occur. Generally, one observes concerted eliminations only with chlorine and fluorine, and even then the concerted elimination process occurs only in the initiation step of the mechanisms. • ■•I'l' • '• '......111.iiy nt II,,'.............. I m . M ' I - il.....mi'' li.nii'.nr. nl 1.1.In .il-. 111 ili'p wi'ukisi bond 111 iiiieliiiils lni'iik In vii'lil unheals Musi run ill 11 leinpt'inliiri' I ki nl •iillilvlliK Ii, • O.IS , T null K Hmli' ill ii'iuls viu 11 catalytic cycle m.....■ 11 Hastened one atom al a lime M11,1 he cycle In, 11 iil> all reactions satisfying li„ - 0.07 T .....i Is kcal Miniilil have a cycle salislying h, < 0.05-T molK ImiiiIiiiiIioii Slep radicals lecombine • ••"> no catalytic cycle where atoms are transferred one atom al a lime ti 11 SUMMARY OF INITIATION-PROPAGATION MECHANISMS 1 1I1I1 ' > su111111uriy.es the key requirements for initiation-propagation mechanisms. First, Il..... .in initiation step where the weakest bonds in the reactants break to yield radicals ii- 11 Ihe radicals react via a catalytic cycle where atoms are transferred one al a time, aid 1I1. cycle leads back to the original radicals. Then there are termination steps. There lew reactions that do not follow these general rules. Still, one can use the rules In 1 11. ' 5 to predict mechanisms of a wide variety of gas-phase reactions, and be right tM'l of the time. I 11 nd il pretty easy to guess a mechanism of a gas-phase reaction, if I know what kinds 1 lull 1 mediates are stable. In my experience, the hard part is deciding what intermediates 1.1 whul initial group of reactions to consider. For example, is a species such as ("II.('< )l I II enough to form? If one can find a way to determine which species arc stable, and W/llHl lypes of reactions to consider, one can predict the mechanism of the reaction using iln methods in this chapter. Still, I have to admit that students find the prediction of mechanisms rather difficult, I dunk thai the issue is partly information overload. In order to predict a mechanism, you in • .1 in know what the relative strengths of bonds are like, and what types of radicals can 1...... there is the trial-and-error procedure, where one guesses at a mechanism and then I in ■ into equation (5.46) to see if the mechanism is feasible. The only thing that 1 have lo ' tiboul this is to stick with it. First, examine the solved examples, and convince yourself in a the rules work for these reactions. Next, go back to all of the reactions mentioned ■ 1 11 in this chapter. First consider all of the reactions in Table 2.1. Label each step. Is II an initiation step or a propagation step? Is the step an association reaction, a P-hydridc ' luiiination...? Estimate the barrier for each step to see if the reaction follows equation ■ 16. Then consider possible side reactions, and ask which side reactions are feasible. 1 in.illy try to predict the main reaction pathway. In my experience, most students learn i" predict mechanisms, if they work at it. !»7 REACTIONS OF IONS \i this point, I want to change topics and start to look at other classes of reactions. So far, Ihe discussion in this chapter has focused on the reactions of neutral species. However, next, we will discuss the reactions of ions.