iii Ai in in'.......u :'<»'. Still. 1111*1' I Willi SI 11 I'> l(f/l IS Inn simple II VUII It il ik .11 ir.II hull 1*1 ID.') Ill detail, villi ini.i Ilini several excited slulcs oi cycloproptne Contribute to the reikiion. Winn the cyclopropane collides with a collision purlner, the cyclopropane goes into one ul the excited slates. Then lhal excited slate reacts to form a second excited slate. Iwenlually, the excited state decomposes into products. It is not unusual for 20 excited slates to be important to a unimolecular isomerization reaction. The result is a complex reaction 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 sunn-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: X + CH,CH2CH2CH3 -> 2CH3CH2 + X (5.108) However, a competing process is a concerted hydrogen elimination. X + C4Hln-> CH3CHCHCH3 + H2 + X (5.109) 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 considci 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 dichlorosilane and suggested that the primary initiation step is X + SiH2Cl2-> HC1 + SiHCI + 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. 1 ilili '• '.iiimiiiiiy 11t Ilm Inltliitluii |ni>|i.ii|nlli>n ini'i li.iiiKiiin ul Inilli nlri '..........II Slop WC'llkl'Sl In Hill III I i-.li I Hill'. luillk Iii VU'lit llllllCIlls Mllst Hill Iii II tl'tll|H'l .Ii III i I kcal Miiinlyinu Ii, S 0.15—— T mol K 1 Ii. ul iriKts via 11 culiilytic eyele Aluiiis tiiiiislriTi-il iinc iiiuni iil a linio Mll i In- cyclo kcal In. In.Ir ;ill icaclioils salislyini; - 0.07-T m„|.K kcal Mundil havi- a cyclo satislying E„ < 0.05-T mol'K l.........illon Slop lailicals iccombinc ' ••• l>llon.\ im catalytic cyclo where atoms are iransferred one atom at a Ihne '. h SUMMARY OF INITIATION-PROPAGATION MECHANISMS i ibli •> 5 summarizes the key requirements for initiation-propagation mechanisms. First, ii. i. is an initiation step where the weakest bonds in the reactants break to yield radii als Hi. n ihc radicals react via a catalytic cycle where atoms are transferred one at a lime, .mil Ihc cycle leads back to the original radicals. Then there are termination steps. There in i lew reactions that do not follow these general rules. Still, one can use the inks in i 'iir s.5 to predict mechanisms of a wide variety of gas-phase reactions, and be right IU "l the time. I I■ iul il pretty easy to guess a mechanism of a gas-phase reaction, if I know what kinds ■ ■I iiiicimediates are stable. In my experience, the hard part is deciding what intermediates mil u hal initial group of reactions to consider. For example, is a species such as CI I ('(>l I i iblc enough to form? If one can find a way to determine which species are stable, and wlui lypes of reactions to consider, one can predict the mechanism of the reaction using iii. methods in this chapter. Still, 1 have to admit that students find the prediction of mechanisms rather difficult I dunk that the issue is partly information overload. In order to predict a mechanism, you .....I iu know what the relative strengths of bonds are like, and what types of radicals can Ii>i in. There is the trial-and-error procedure, where one guesses at a mechanism and then plugs into equation (5.46) to see if the mechanism is feasible. The only thing that I have lo i\ about this is to stick with it. First, examine the solved examples, and convince yourself iii ii the rules work for these reactions. Next, go back to all of the reactions mentioned n I.ir in this chapter. First consider all of the reactions in Table 2.1. Label each step. Is n .in initiation step or a propagation step? Is the step an association reaction, a ^-hydride ■ llmination...? Estimate the barrier for each step to see if the reaction follows equation • W> Then consider possible side reactions, and ask which side reactions are feasible, finally try to predict the main reaction pathway. In my experience, most students learn in 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, the discussion in this chapter has focused on the reactions of neutral species. However, next, we will discuss the reactions of ions. Iiililn '■> " '.......■ i>«.iim|.I.h, <>l ..... i.......on-, Simplified Mi' hunl in (11,(11.Cll (II. Ml1 • II 1(11,(11.(11 —-CHtf CH3CHC=CHCH3 + H 1 Initio H / Key illttoiiiniim I.. i|w, . ii lonli iniu Ilium imil tiulli ill miiilloim Rci........ [somerization CH3CH2CH=H2 CH3CH=CHCH3 Cracking C10H20 : 2C,H„ Alkylation CH3OH + C6H6 —— CH3C6H5 + H20 Ksterfication CH3COOH + CH3OH : CH3COOCH3 + H20 C10H2() + H+ -> [C,„H2I|H -> C5Hlo + C5H11 + C5H„ + -> C5HI0 + H+ CH3OH + H+ -»• CH3+ + H20 CH3+ +C6H6 -> CH3C6H,,+ CH3C6H6+ -»■ ch3q5h5 + H+ CH3COOH -> CH3COO~ + H+ CH3OH + H+ CH3+ + H20 CH3+ +CH3COO" -+ CH3COOCH3 11 in. ul Vpplii ....... ( K Iniu- rnli.111, rim 11I monomei productlol ( 1 ml,- nil conversion, biological conversion! Pharmaceutical production, niononu'i production, fine chemicals Soap production, fragrance production ' III ill Hew linn 1 1 Inflow nun,in,mi propagation ivaclions .....liny lik,' thai in stable molecules; .........edition have structures similar to tin\ the following mechanism in the presence of HC1. First, the acid dissociates, Ii Mm)' protons (i.e., H+) and chloride ions: Table 5.6 gives some examples of important ionic reactions, including isomerization cracking, alkylation, and esterfication. Isomerization is a process where one takes ti molecule and rearranges it. Cracking is a process where you take a big molecule ami break it into two smaller molecules. Alkylation is the opposite of cracking, where you take two molecules and combine them into one bigger molecule. Esterfication is a special case of alkylation where you combine acids and alcohols to form soaps or detergenls These four reactions represent more than $400 billion of yearly production, so they an certainly quite important. Most ionic reactions occur in solution or on solid catalysts. However, ionic reactions are also seen in certain other systems such as plasma reactors and mass spectrometers. In this section, we will discuss the reactions of ions in the gas phase. Reactions in solution will be briefly mentioned in Section 5.9. First, it is important to note that the reactions of ions are not fundamentally different from reactions of neutral species. The reactions often proceed via an inilia tion-propagation mechanism, where first the ions are formed, then the ions react through a catalytic cycle, and finally ions are neutralized. However, there are some important differences, as outlined in Table 5.7. In practice, many ionic reactions are run in solution, and in solution reactions do nol necessarily follow a catalytic cycle. For example, in an SnI reaction, you form an ion and it reacts, but there is no catalytic cycle reforming the initial ion. Instead you feed the ions into the solution. Industrially, most ionic reactions do follow catalytic cycles, as do most ionic reactions in biology. Thus, although there are some exceptions, the majority of important ionic reactions follow catalytic cycles. For example, the dehydration of ethanol HC1 -> h+ + cr lb. 11 1 he protons react with the ethanol to yield a charged complex: H+ + C2H5OH-► [C2H5OH2]+ 1 hi 11 the charged complex decomposes, yielding a water and an ethyl cation: [C2H5OH2]+-► [C2H51+ + H20 1 in n the ethyl cation decomposes, yielding ethylene and the proton: -> C2H4 + H+ [C2H5]+ I uially, the proton can recombine with chlorine: H+ + CP -> HC1 (5.112) (5.113) (5.1 II) (5.115) (5.1 16) I lure is one subtle point—it is hard to form ions in the gas phase. One often uses electrons to produce ions. So, for example, one could replace reaction (5.112) by H2 + e" H+ + H + 2e" (5.117) CH3CH2OH H2C = CH2 + H20 (5.111) where e~ is an electron. Reaction (5.112) is a classic initiation reaction. Reactions (5.113)-(5.115) are classic piopagation reactions. Reaction (5.116) is a classic recombination reaction. Therefore, 11 are are some similarities between reactions of ions and reactions of radicals. Still, one finds that the actual individual steps are quite different with ions than with ladicals. Consequently, even though both radicals and ions follow initiation-propagation ii .ii I.....v lín prodlil II "I ion inolei lili' líni Hums hi t- i|inlc ililleienl limn llir (iiihIiu Is ii| i.iiIii ill molecule ii,ii lion Foi example, in Section 5.4.4 we noted thai there are two key reactions when i deuterium reacts with ciluine, an exchange reaction D + CH3CH3 -► II I IK II.(II, and a dehydrogenation: D + CH3CH3 -> DH + CH2CH3 In principle, one might also see bond scission: D + CH3CH3-> DCH3 + CH3 However, reaction (5.120) has never been observed experimentally. If one instead runs the reaction with D+, the main reactions are D+ + CH3CH3 D / \ H3C —CH, D / \ H3C—CH3 DCH3 + CH3 (VI IS) (5.119) (5.120) (5.121 (5.122) In reaction (5.121), one forms a complex with a three-centered two-electron bond, while in reaction (5.122), the species decomposes to yield deuteraled methane and methyl radicals. Li et al. (1989) examined these reactions with a molecular beam and found that reaction (5.121) had an activation barrier of zero while reaction (5.122) had an activation barrier of less than 2 kcal/mol. Computations by Carnero (1994) suggest that both reactions are unactivated as is loss of H2 from the complex. By comparison, Lee and Masel (1996) calculated that reaction (5.120) has an activation barrier of 45 kcal/mol. This example illustrates two key differences between the reactions of ions and those of neutral species: (1) species with unusual bonding configurations are stable and (2) the activation barriers for reactions of ions are quite different from the activation barriers of neutral species. The fact that the activation barriers change greatly can be easily understood in light of ideas presented in Section 5.4.5. In Section 5.4.5, we found that the activation barrier for reaction (5.120) was associated with a repulsion between the electrons in the incoming deuterium and the electrons in the ethane, causing the balloonlike orbitals to flatten out. In fact, if one replaces the deuterium with a D+, there will be no electrons on the D+, so there will be no repulsions. As a result, the reaction is not activated. Consequently, positive ions are much more reactive than neutral radicals. The absence of electrons also allows the unusual bonding in the intermediates to occur. As noted above, when a D+ reacts with an ethane, the deuterium ends up in a three-centered two-electron bond. At first, this may seem to be a very strange bonding configuration. However, if you think about it a little bit, you will realize that there is no place else that the deuterium can go. A bare deuterium has a huge electron affinity, 13 electronvolts (eV) (325 kcal/mol), so there is a huge driving force for the deuterium to I 1 onto mi I'lcriinn An ciluine linn ill ni lis iiiiiiiinl bonding orbital* Idled llowcvei, ■ ill 1 HrV diivitie lone. I In- nu.......i >■ liydinpeu will form il bond somewhere. I lie ......I' deuterium craves electrons, nud 11 it- neliesi source ol cleclrons 111 elluuie is die I 1 bond One could imagine the deuterium Instead bonding to a single methyl group II or, that is .1 much less electron rich environment than die carbon carbon bond 11.. i, lore, die most likely place foi the l>' to react is along die carbon carbon bond qui Ml) when a deuterium reads with an ethane, one would expect die I)1 to form I iIm. , < entered two-electron bond as is observed. i lín , .in extend these arguments to the reaction of olefins, for example, considei die II lion ol :i I)1 wilh a substituted ethylene. The carbon-carbon double bond 011 die 1'1 i' ne is ii very rich course of electrons, so that initially the D+ will react to form 1 entered iwo-electron complex: D+ + R2C=CR2 D / \ R2C—CR2 (3.123) in iilhei words D / \ R2C—CR2 (5.124) ii 1 mil obvious that the species in equation (5.124) is stable. Other possibilities include .......1 with charge on one center: D R2C—CR2 (5.125) "i .1 species where the R groups have migrated: R / \ R2C—CRD (5.l2ř.) I experimentally, when R = H, species (5.124) is the most stable; when /? = CH,, pecies (5.125) is the most stable; however, when R = C5H6 the species in (5.126) is die most stable. For the analysis here, it is not really important to know which of these ipecies is the most stable. However, the key thing is that the ions can exist in several isomers, including an isomer where one species bridges between two carbon center! further, there are hardly any barriers to convert one isomer into another because there ne no Pauli repulsions. Now those two factors lead to interesting chemistry. There are two key types of reaction that usually happen only with ions: cracking and isomerization. Cracking is a reaction where one starts with a heavy hydrocarbon and breaks up the hydrocarbon into smaller fragments. The actual reaction involves a series of carbon-carbon bond scission processes hke those in equations (5.120) and (5.122). Notice that such reactions will occur with low barriers if the species are positively charged. In contrast, high barriers will be seen with neutral species. Therefore, if one Ill M III iN'l i U K >N'l 271 wanted to (link ii 11 s 1111 n .i 111« x i one would inn 11 u reaction inula conditions where M positive Ion would form Similarly, Isomerization is a process where one Lakes a ligand (i.e., un H group) imd moves it around a molecule. Willi ;ui ion. one can lonu ;i bridging intermediate like L lit..... equation (5.126), so isomeri/.ation is easy. On the other hand, with a radical, die hridgln| species is unstable. Consequently, isomeri/.ation reactions have much lower barriers wldj ions than with radicals. 5.7.1 The Mechanisms of Reactions of Carbocations Next, 1 want to talk about the detailed mechanism of the reactions of carbocal..... A carbocation is a positively charged hydrocarbon ion. Examples include C3H7 and C3H<)+. Carbocation chemistry has been studied in great detail in solutions and in solid catalysts. The reactions have been less heavily studied in the gas phase. Generally, tin-reactions of carbocations are fairly complex. There are many intermediates and very manj interconnecting reactions. Many of the species have unusual bonding. 1 generally find 11 difficult to predict the stability of the various species. Consequently, it is slightly hardfl to make detailed predictions of the reaction pathways of the ions than to predict reaction pathways of radicals. Still, as we will see, the analysis is easy provided you know wliai intermediates can form. To start off, it is useful to review some of the literature. Martens and Jacobs (1990) and Vogel (1985) point out that there are two key types of carbocations that one CM consider: carbonium ions and carbenium ions. Carbonium ions are formed by adding a H+ or a to a paraffin. Examples of carbonium ions include R / \ H,C—CH, (5.127) R I H3C — CH3 (5.128) H3C—OH (5.129) R /\ H3C — OH (5.130) Years ago, people perceived carbonium ions as having five coordinated carbon atoms, three coordinated oxygen atoms, and fairly localized charges. However, people now know thai there are bridging species and that the charge is distributed throughout the molecule. Carbenium ions are formed by adding a H+ or a R+ to a double bond or the removal of a proton from a paraffin. Examples of carbenium ions include H,C+ (5.131) (5.132) in (3.133) r. nil / \ R2C— CH2 (5.135) I hi I'i muni ions often have three coordinated carbon atoms. However, there are also 1 m, species and cyclic structures. 1 in i.ill\. i arhonium ions react via two pathways: cracking and isomeri/.ation. I)i.....y 1 'mi' ,.1 .1 carbonium ion. a proton reacts with a paraffin to form a bridging species H+ + /?CH2— CH2R H / \ RH2C— CHiR (5 I 16) in Imdging species decomposes to yield a smaller hydrocarbon fragment anil a 1 iti I >c 11111111 ion: H / \ «H2C— CH2R A>CH3 + CH2R+ (5.137) li/alion of carbonium ions occurs when a linear carbonium ion rearranges to form , hildgcd species and then goes back to a linear species: R I H3C- CH3 R I \ H3C —CH3 R I H3C—CH3 (5.138) ■ m 11, iiiuni ions can also crack and isomerize. In the gas phase, reaction (5.138) lias an ......un harrier of less than 4 kcal/mol, which makes the reaction very rapid. Cracking his via ^-scission, where the bond between the a carbon and the f3 carbon breaks to Ii Id an olefin and a smaller carbenium ion: H H R + R—C—C—C—C—H H H H H H H R H R—C—C+ + C=C H H H H (5.139i 1 in 11 are two pathways for isomerization: a standard 1,2 shift, where an R group migrates Id in adjacent site; and a reaction where a cyclic intermediate forms. Reaction (5.140) is mi example of a 1,2 shift: H H R + R —C—C—C—C—R" H H H H H H + R R—C—C—C—C—R" H H H H (5.140) Notice that the R group shifts to the right. These 1,2 shifts are the primary isomerization route with a small hydrocarbon. With a modest-sized hydrocarbon, one also sees Ml Al IK >NM II II IN'. I'/l isuinci i/iilinii vlll i s > lii mill ini'ili,ill'. In.I llir 11 it llt-t ii If ii'iK'ls In liHIII n i villi! intermediate as shown In equation i '< I 111 (5.141) Then the cyclic intermediate undergoes hond scission as shown in reaction (5.142): (5.1-1.') One can also stop reaction (5.142) at the first step to yield a cyclic product. In gas-pha* reactions (5.140)—(5.142) are extremely rapid and often reach equilibrium in a very slum time. In a solvent, the reactions are a little slower because you need to move sol vert molecules out of the way. In the latter case, the intrinsic barriers to reaction arc associated with motions of the solvent molecules, and not the intrinsic barriers to rearrangement o) the reactants. Carbenium ions can also participate in alkylation reactions, where a carbenium ion It' reacts with a paraffin or olefin to yield a carbonium or carbenium ion: B+ + /?3C—C/?2H B / \ «,C—C/?2H (5.143) The carbonium or carbenium ion can then lose a proton to yield a higher-molecular-weigln species: B / \ /?3C— C/?2H /?3C—C/?2B + H+ (5.14-1) There are many combinations here. One can form a cyclic product, and then isomeri/i-the product to form a larger ring. One can then undergo bond scission to yield a myriad of products. For example, Figure 5.7 shows some of the reactions of tridecylcation (Ci3H25 + ). Only sixteen reactions are shown in the figure. However, Martens and Jacobs actually found 70 total reactions. In the top reaction in the figure, the molecule starts out with the charge centered on the carbon atom labeled zero in the figure. During the reaction, the molecule pivots around the 2 carbon, bringing the 0 carbon and the 3 carbon in close proximity. Then a bond forms between the 0 carbon and the 3 carbon, forming a four centered ring. A proton migrates, moving the charge. Then the ring opens again. The other reactions on the page are similar, except that different size rings form. This example illustrates the complexity of ion reactions. Table 5.8 shows the products of the reaction. Note that there are 12 main products and al least 70 reactions in the reaction network. Consequently, the mechanism is rather complex. One thing that people do know is that the stability of the ions varies with the degree of branching. For example, Table 5.9 shows the heat of formation of several C|0H2|^ i .,,.■■«> 5.7 Some of the reactions of a tridecylcation. [Adapted from Martens and Jacobs (1990)|. i .ii. :> 8 The products of tridecylcation (C13H25+) isomerization in different ■Hivlionments Production in the Gas Phase Production in Platinum/CAY Zeolite Catalyst Production in Platinum/USY Zeolite Catalysl ' M. ilivlilodecane 9.3 7.0 II i M. ilivlilodecane 17.7 15.1 17.9 1 Mi ilivlilodecane 19.0 18.4 18.0 1 Mrlhyldodecane 19.1 22.8 19.6 | Mi-iliylilodecane 19.1 23.4 19.4 l 1 Ihylundecane 2.7 1.9 2.7 i 1 ihylundecane 3.4 3.4 3.4 - 1 ihylundecane 3.9 3.6 3.9 (. 1 lliylundecane 2.2 2.0 2.2 i Pmpyldecane 1.0 0.7 1.0 i Pmpyldecane 1.8 1.4 1.8 i Pmpyldecane 0.7 0.3 0.7 '.. .... t Results of Martens and Jacobs (1990]. Ions Notice that the branched ions are more stable than the linear ions. Consequently, ii n|iiilibrium, one would expect to form highly branched species. You remember from organic chemistry that in small molecules, a primary ion is less stable than a secondary inn. which is less stable than a tertiary ion. That rule also works from a simple ion. With ii more complex ion, it is easier to think about the ion stability in terms of Brauman and Blair's (1971) model, where you view the stabilization of the ion, as an interaction between the charged site and the polar stable alkyl groups. The more alkyl groups you get near the . harged site, the more the charge will be stabilized. One can get alkyl groups next to the filhli'M) I lift mlnllvx ntiililllly ..I miiiii' I mil..,1 Intnimmlliilnti Intermediate I K-iii ol i.....tatlon, kcal/niol Intermediute Ileal ol 11iiin........ kc .ll/nml 1- Butyl cation +138 2- Methyl-l -propyl cation +130 2-Butyl-cation +67 2-Methyl-2-propyl cation 0 2-Methyl-2-butyl cation 2- Methyl-2-pentyl cation 3- Methyl-3-pcntyl cation 2,3-Dimethyl-2-pcntyl cation 12 I') 1«) -21 Source: After Martens and Jacobs (1990). charged site by moving from a primary to a secondary to a carbocation. However, one can also stabilize charge by making the alkyl groups more polarizable, or by adding lonjj chains that can coil up to bring the ends of the chains in close proximity to the charged center. In my experience, it is not so easy to predict the relative stability of carbocalions because of chain coiling. However, Martens and Jacobs (1990) show that if one knows the stability of the ion, one can sometimes predict the products of the carbocation reaction from equilibrium. See Martens and Jacobs for details. 5.8 PREDICTION OF THE MECHANISM OF IONIC REACTIONS Next, we will be discussing the prediction of the mechanism of gas-phase ionic reactions Generally, it is pretty easy to guess the mechanism of a ionic reaction in the gas phase, provided the key intermediates are known. For example, let's start with the reaction CH3-CH=CH-CH3 + H+ products (5.145) We would expect that the first step would be the formation of a carbenium ion: CH3-CH=CH-CH3 + H-► [CH3-CH-CH2-CH3]+ (5.146) Then the species could undergo a series of isomerization and cracking reactions as indicated in Figure 5.8. As with radicals, the first step in predicting a mechanism is to write down all of the reactions that could occur. One then does calculation using equation (5.41) to see which reactions can occur. Generally, with a carbenium ion, the intrinsic barriers are all small; usually 1-5 kcal/mol. This means that in the gas phase all of the reactions in Figure 5.8 will occur. As a result, the system will quickly reach equilibrium. The cracking reaction [reaction (5.137)] is slow because it is thermodynamically unfavorable. However, the situation in a solvent is different because there are significant barriers to rearrange the solvent molecules. Unfortunately, we cannot predict the barriers in a solvent, so mechanisms are difficult to guess. As the butene pressure in the system goes up, one also gets alkylation reactions C4H9 + C4H8 + X C8H17++X and transfer reactions C4H9+ + C4*Hg C4H9* + C4*Hg -* C4H8 + C4*H9 + -> C3H6+(CH3C4H8) (5.147) (5.148) (5.149) III Al IK IN'I IN I'I )| All tl< )l 1 Mn )N 27S ::<" * .../>' „ v „ v <• v •H .11' C H'(<'—C H' II H H V H HVC' C—H +H* H'CwC-C^H / \ H H / II H H HvVH -f c~c II II H 1 i|iuu) 5.8 The mechanism of the reaction CH3CH=CH2CH3 + H' -> products at low prosauti) 11 Hon (5.147), the C«H17+ is a transient species, not a true intermediate. Tin- ic: nil on' very complex chemistry. Now, with a carbenium ion, all of the reactions occur with low barriers. Ilowcvn II 11 iln ihe same reactions with a carbonium ion by replacing the butene in rem tlon 1 1 1 u nh butane, there are significant barriers. Unfortunately, at present, such reactions 1 mi been sludied in enough detail to permit useful predictions. n U REACTIONS IN POLAR SOLUTION 1.11 111 this chapter we have been discussing the mechanisms of reactions in Ihe gas However, now we are going to change topics and start to consider reactions In Mtltllion. Prom a practical standpoint, reactions in solution are quite important. Virluall\ •ill organic and inorganic syntheses are run in solution. All biological reactions are run .....lution. Most of the reactions you ran in freshman chemistry and organic elienn h\ .....irred in solution. Most of the chemical reactions in the human body and in oilier In In] 11....... occur in solution. Unfortunately, there is insufficient room in this book to discuss solution chemistT) 1)1 molecules in any detail. You should consult your organic or biochemistry books foi ii 1 ussions of these reaction systems. Still, we will need a few concepts for the discussion later in this book. You probably m member SN1 and SN2 reactions from your organic chemistry class. During an SN I 1 11 lion, a molecule containing a neucleophile breaks apart in solution to yield a carbenium ..... and a neucleophile: H H R N R'—C—C—C—C—H H H H H H H R + R'—C—C—C—C—H + N-H H H H (5.150) in-ie N is a neucleophile. The carbenium ion can then rearrange or crack to yield a new ion: 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 (5.151) Ill ACIIONÍUIN Ml IAI '.lllll All •; 277 1-ľ" -■ > •.....-.u i » Hi. .i .....I nciKIcophile in yield n stable in...I II II i R N i R'— C—C—C—C—H II II II II i'i.'.in. i II II H R R—C—C—C—C—II (5.I.1JI II II II II In SN2 reactions the neucleophile reacts with a stable molecule and exchanges: [NT + R}CN- N'C R3 + AT (5.131 One neucleophile replaces another without the appendant groups undergoing signilii mil rearrangements. The SN2 reactions are very similar to the reactions of radicals in the gas phase (>m observes the same basic chemistry — single atom transfers arc fast; ligand transfers an slower. Everything is more complex because all of the species are charged. The similarities are not as close with SN1 reactions. SN1 reactions look something like ion reactions in the gas phase. First, there is a step where an ion forms. Then the iol rearranges. One observes isomerization cyclizations and cracking. Then a stable specie! forms again. The reaction is not a complete cyclic process, so the kinetics are different However, there are some qualitative similarities between an SN 1 reaction in solution and an ion reaction in the gas phase. The one major difference between reactions in the gas phase and in solution is tlui ionic species are stabilized by the presence of the solution. For example, as noted abovfl in the gas phase, carbonium ions are hard to form. Well, in water or other polar solvent! carbonium and carbenium ions are much more stable than in the gas phase. Consequcnily, one can form carbocations without using electrons to ionize the species. The presence of the solvent also changes the intrinsic barriers to ionic reactions, Generally, intrinsic barriers are higher in solution than in the gas phase. Recall thai m the gas phase, ion reactions like those in equation (5.121) are often virtually unactivated, because there are no Pauli repulsions between the reactants. When the reactants dissolve in solution, everything changes. It is unusual to get a bare ion in solution. Instead, the ion is generally surrounded by a solvent cage as shown in Figure 5.9. In order for a reaction to occur, one needs to first distort the ion cage so the reactants can get close enough to react. The distortion of the ion cage generally produces a repulsion similar to the Pauli repulsions seen in radical reactions. Unfortunately, at present, it is very hard to predict how much the intrinsic barriei changes as a result of the solvent. For example, if one runs a reaction in water, the water Figure 5.9 A schematic of the solvent cage around an ion in solution. i liwlmi'i'ii bunded i .iff mound cmh ion I In- ciiyc needs lo lie disloiled when .......minis, inn t in- exact tmíme <>i ihc distortion is iinclow Consequently, il is nil in iiiiike any predictions. I oiliiiiiili'lv. the changes in the intrinsic barriers do not really change the mechanism ii...... < hie sees nil the same elementary reactions in solution that one sees m the , i |'li i . Ii is |iisi thai the relative rales of reactions change in the solution. The changes n will be discussed in Chapter 13. ' o ili it ul. (IW7) show that one can often, but not always, predict mechanisms ol it.** lion without considering the changes in the intrinsic barriers because of the solvent || In iioi obvious, but if one species can be in two different reactions, then the intrinsic i i" both reactions will change in the presence of the solvent. Often, the intrinsic i*1 .ill ol ihc reactions of a species are changed by a similar amount. In su< h BH .....m .nil predict accurate mechanisms by ignoring the changes that occui in tin llliuli barriers in the presence of the solvent. Id |i i. inc. however, you want the solvent lo affect the rate. Alter all. in organii i \ou want lo only make a few products and not make others. People use Ihc m io direct reactions. For example, if you wanted to get a specific reaction you nl i n i ,i solvent that surrounds the part of the molecule that you want lo protect, so i......sir y could happen there. You can also use a countcrion or an en/yine for a sinulai • In practice, the solvent should be an active participant in reactions because thai on opportunities to optimize selectivity. ..... substantial difference between carbocation reactions in the gas phase and in m is that in solution one can get direct reactions with the solvent. For example, mi lake up protons. The reactions with the solvent limit the stability of the nl.....it ions. That can have important implications for mechanisms. This will be discussed n, i Imptcr 13. In my experience, it is much harder to predict the reaction mechanisms in a solvent ......... the gas phase. One needs much more background. Consequently, the discussion ■ n i" deferred to Chapter 13. '. io REACTIONS ON METAL SURFACES wc want to discuss the mechanisms of another important class of reactions: read ions .......rial surfaces. Table 5.10 lists some important reactions that occur on the surfaces III metal. Metal surfaces are used extensively as catalysts to promote hydrogenation, Ii hydration, and oxidations. Reactions on metal surfaces are also important lo corrosion in.I the production of integrated circuits. Reactions on metal surfaces look similar to radical reactions except that the reactants .....ot radicals. Instead, the reactants are species that have a bond to the surface. Ihc milt v acts like a solvent to stabilize radicals in the same way that water acts as a solvent in stabilize ions. lor example, the reaction 02 + 2H2-> 2H20 can occur in the gas phase. \. 1 ording to Hinshelwood (1946), some of the key steps in the mechanism are X + H2-► 2H + X X + o2-> 20 + X O + H2-> OH + H . . It '».. * II trilllr UVIr I'M i 1| fy\ i i II un' ' i IN Ml I Al 'IIIIII Al I ') 27B I'M'- '• H> sou.....H.lll.....-. Ill ii-. ii I.....•. iii! im.I.ll -.1111.1. .-. Hydrogenation N2 + 3H2 > 2NH.i C2H4 + H2 =>C2H6 Dehydrogenation C2H4 Pt C2H2 + H2 Oxidation 2CO + CO = C2H4 + ^02 2NH3 + 402 : Ag 2C02 + N2 C2H40 N205 + 3H20 Chemical vapor deposition =^=s A1 + 3C2H4 + ^H2 AI(C2H5)3 Fe (s) : Fe 3+ (Aq) I In tin- .11 |.....1111 him Crude oil upgrade Essential oil upgrade Octane enhancemenl Monomei production Catalytic conveners Monomer production Chc.nicals production Connections on integrated circuitl Corrosion H2 + OH H + 02 X + 2H -> H20 + H ^ OH + O -> H2 + X (5.154) Anton and Cadogan (1991) examined the mechanism of hydrogen oxidation on a Pt(l 11) surface. They found that the main reactions were H2 + 2S-> H(a OH(ad) + H(ad) 20(ad) + H2--> 20H(ad) O(ad) + H(ad) -► 0H(ad) 20H(ad)-> H20 + 0(ad) H(ad) + OH(ad)-> H20 (5.155) H2 + 2S 2H I ad I where the notation (ad) is used to designate that the species are attached to the platinum surface. Notice there is a close similarity between the mechanism of hydrogen oxidation in the gas phase [reaction (5.154)] and the mechanism on the surface [reaction (5.155)]. The elementary steps and intermediates are almost the same. Consequently, one can often think about a reaction on a surface as being a gas-phase reaction except that the intermediates of the reaction are bound to the surface. The fact that the reacting species are bound to the surface during the surface reaction has important implications. The surface acts like a solvent to stabilize the radicals. Unpaired electrons in the radicals pair up with the free electrons in the metal to produce i ilili 11. i us I lie icsiill is lluil llic Iiili'iini'illiiles ul Ihc iruclum air slahlc species lalhei lli.in luilit ills Null. I lie species adsoilicd mi niclal suilaces ulten follow ic.u lion pathways ' quite similar io reaction pathway* ol radicals In the gas phase, although the) do not always do so. |'i ople often picture surface reactions via a catalytic cycle, where surface intermediates n tiled .mil then destroyed, lor example, figures 5.10a and 5.10b show the I atalytil li loi the production of water via the mechanism in equation (5.155). Wain |l i | in. cd in two reactions in equation (5.155): (1) the disproportion of Oil ami l.'l the I ie.h lion between Had and OH(ad). The first reaction dominates when the oxygen titration is low. The second pathway dominates when the oxygen concentration In high In the lust pathway, the surface starts out covered by oxygen. Hydrogen adsorbs on the ii io yield OH groups. Two OHs then couple to yield H20 and a bare sile. (>\\ gi n Hi- ii adsorbs on the bare site to get back to the initial conditions. In the second pathway, the surface starts out empty. Hydrogen and oxygen adsorb I in adsorbed species then reacts to yield water and regenerate the clean surface Ml sieady-state catalytic reactions can be viewed as occurring via a catalytic cycle In ie surface species are formed and destroyed. The cycles are similar to those discussed pi. vlously for gas-phase reactions. However, there is an important difference beiwei n >■ i. nous in the gas phase and on a surface. In the gas phase, radicals are rather unstable i" - us As a result, one has to go to rather high temperatures before one gels a lin-li .......ili concentration of radicals to get a reasonable rate. In contrast, when a radical binds ' i hi face, the dangling bond in the radical forms a bond to the surface, producing a 1»' us that is stable. As a result, one can get a reasonable concentration of adsorbed ' "ii' als al low temperatures. For example, at 300 K, less than one hydrogen molecule in io dissociates in the gas phase, while on a platinum surface, more than 99.9% of the hydrogen dissociates. The result is that hydrogen on platinum is much more reactive than l phase H2. The hydrogen on platinum is not as reactive as a hydrogen radical in the gas phase, however. The hydrogen atoms on the surface are bound to the platinum and arc no longei radicals. As aresult, the hydrogen atoms on the surface are less reactive than free hyd.....- n radicals in the gas phase. However, the effect of the reduced reactivity is smaller than ili. effects of the enhanced concentration of hydrogen atoms. As a result, at moderate .....peratures, the hydrogen/oxygen reaction occurs much faster over a platinum catalysl -0 tH, o oooo WWWWV o o o o WWWWV ^ H H O O O O O wwwwv (H H o olo o o WWWWV b) + 1/2 02 I .....IH O WWWWV o WWWWV ^wwww T * H H O WWWWV Figure 5.10 Catalytic cycles for the production of water (a) via disproportion of OH groups and (b) via the inaction OH,ad| + H(ad) -> H20. líniu in Mu- niis phase Ai muni lempcriilure, n lis tli> >> ■ i ■ 11 ox\ecu mixture remains slnhto im years However, m one puts •> plati......> wire In the mixture, the mixture «di explode In general, melal smlaees slahih/c leaelive mlerinediales. Thai greatly speeds up die initiation steps in die mechanism. The speedup of the initiation reaction is why catalytic reactions have so much higher rales than do gas-phase reactions. Masel (1996) also notes that surfaces can stabilize di- and iriradicals. That allows new chemistry to occur on a surface that would not occur in the gas phase. Details are given in Chapter 12. There is one key difference in the nature of the initiation reaction between the mechanism of the reaction in the gas phase and the mechanism of a reaction on die surface. In the gas phase, the initiation step always involves the scission of a bond. On a surface, the key initiation is the adsorption of one or more of the rcactants onto the surface. During adsorption, a molecule attaches itself to a surface that changes the properties at the molecule so that the reactions change as well. We call the species bond to surfaces adsorbed molecules or adsorbates. Masel (1996) has presented a detailed discussion of adsorption. There is some important notation. We will use a subscript (ad) to designate an adsorbed species. That way we can distinguish between adsorbed species, and species in the gas phase. 5.11 GENERAL CONCEPTS ABOUT ADSORPTION AND REACTION ON SURFACES Before we talk about specific mechanisms of surface reactions, we need to discuss some general concepts for surface reactions. There is a lot of information here, and students often find it overwhelming. My best advice is to stick with it. The material will get easier after you have read it a couple of times. The first general concept is that surface reactions occur in what is called an adsorbed phase. The adsorbed phase is a layer of gas that is bound to the surface. Figure 5.11 shows what the adsorbed phase is like when ethylene adsorbs on platinum. The surface consists of an ordered layer of platinum atoms indicated as circles in the figure. Platinum, is a metal, and metals have many lots of free electrons. When ethylene adsorbs on platinum, some of the ethylene adsorbs intact. Other ethylene molecules react to form a variety of hydrocarbon fragments. Some of the hydrocarbon fragments form strong bonds to the platinum atoms, while some of the other fragments are free to move around. The result is a complex mixture that has properties somewhere between those of a liquid and those of a solid. i n wi nm i i iní i ľ i'. ai IN AND III AI III IN i IN '.I IIII AI ;i ri .•Ml ľ. n|ile spend then entue liletinii . siinl\me .id.oiplion There is n tremendous amount >il Information kbOUl die topic I'in now. the key dung lo remember is that a surface ........usually occurs in a dun Inyei ol molecules attached to the surface. The Inyei can .....deied like a liquid oi ordered like a solid. Still, the key feature is that there is a dun Inyei ol adsorbed molecules and reaction occurs in that layer. in. next key point is that there are two types of adsorption: molecular adsorption h ...native adsorption. In molecular adsorption, the molecule adsorbs intact, "lid. ... i.iiive adsorption, a bond breaks in the adsorbed molecule. One finds that bonds i., il . i.iIn during a surface reaction, so it is not unusual for a molecule to dissociate on .1.1 ...iplion. i i •! .sample, when 02 adsorbs onto platinum at 300 K, the oxygen dissociates, yielding wren atoms. H2 dissociates into two hydrogen atoms. C2H4 dissociates to yield a I II,( poup and an adsorbed hydrogen. At 100 K, H2 still dissociates. However, < > and i II, adsorb molccularly. Mu fact that bonds can break during adsorption is quite important, because n dial one can produce atoms or other very reactive species in the adsorbed layw i ntmcqucntly, surface reactions are often quite rapid. I In- high reaction rates on metal surfaces make metal surfaces very active catalyitl i i its arc good at producing radicals, so they tend to catalyze radical reactions lonu .......his arc rarely catalyzed by metals, although there are a few exceptions. In order to know whether a surface will catalyze a given reaction, it is important to I iiin\ whether a given molecule dissociates on a given surface. Figure 5.12 gives a chart III h one can use to see if a given molecule dissociates. Note that metals in the middle i......1 the periodic table are the most active for dissociation. \noiher key concept for surface reactions is that the reactions occur on distinct surface ..lies Surface sites are delined as the place on the surface where reactions can on m CO 300 K Dissociated Molecular Sc Ti V Cr Mn Fe Co Ni Cu Y Zr Nb Mo Tc Ru Rh Pd Au. La Hf Ta W Re Os Ii Pt Au N2 300 K Dissociated Activatod Sc Ti V Cr Mn Fe Co Ni Cu Y Zr Nb Mo Tc Ru Fill I'd A(| La Hf Ta W Re Os ľl Au o2 300 K NO 300 K Dissociated Sc Ti V Cr Mn Fo Co Ni C u Y Zr Nb Mo Tc R u Rh Pd Ag La Hf Ta W Ro Os lr Pt 02 100 K Dissociated Molecular Sc Ti V Cr M n Fe H NI Cu Y Zr Nb Mo Tc Ru 4 Pd Ag La Hf Ta W Re Os Pt Au NO 100 K Sc Ti V Cr Mn Fe Co Ni Cu Y Zr Nb Mo Tc Ru Rh Pd Au La Hf Ta W Re Os lr Pt Au Dissociated M olec III.u Sc Ti V Cr Mn F. Á 00 Ni Cu Y Zr Nb Mo Tc Ru Ri 1 Pc Ag La Hf Ta W R 3 Os lr Pt Au Key □Dissociativel I Molecular adsorption |_| adsorption Molecular and dissociative adsorption Activated dissociative adsorption No ^//J adsorption □ Limited data Figure 5.11 An illustration of the adsorbed phase when ethylene adsorbs on platinum. Figure 5.12 The metals that dissociate CO, NO, N2, 02, and CO at various temperatures. Pt(111) Pt(110) Pt(100) Pt(210) Figure 5.13 Pictures of some common surface structures. Sometimes, the surface site consists of a specific arrangement of metal atoms. Olhei times, only a single bare metal atom is sufficient. Experimentally, one often funis ili.u the rates of surface reactions are greatly enhanced if a specific arrangement of atoms is available on the surface of the catalyst. However, that is never an absolute requircmcni Often, there is an ordered arrangement of sites in the surface of a catalyst. Figure 5.1 shows a picture of some surface structures. The surface atoms are indicated as spheres m the photo. Notice that one has an ordered arrangement of surface atoms. If each surface atom is a reactive site, then one will also have an ordered arrangement of surface sites. 5.12 PREDICTION OF THE MECHANISMS OF SURFACE REACTIONS Next, we want to discuss how one can predict the mechanisms of reactions on nieial surfaces. The material will build on the analysis in Sections 5.4.1-5.4.3, 5.4.5, and 5.4.6. where we showed that one can combine a series of empirical rules and the Polanyi relationship to predict the mechanism of reactions. First, I want to note that all reactions on metal surfaces are propagation reactions In Section 5.2, we noted that when we run a reaction in the gas phase, the first step is always an initiation step. Reactions on metal surfaces are a little different, in that there are always dangling bonds and free electrons in a metal surface, so that the surface is, in effect, a radical. Consequently, one does not have to form radicals at the start of the surface reaction. Instead, the bare surface is a radical. Consequently, one does not need an initiation step. In Section 5.4.1, we noted that all propagation reactions go in catalytic cycles where radicals are formed and destroyed. In the same way All surface reactions occur in cycles where bare surface sites are formed and destroyed. (5.156) Figure 5.10 shows a typical catalytic cycle. The surface starts out empty. Hydrogen and oxygen adsorb. The adsorbed species then react to yield water and regenerate the clean surface. Notice that we start with a bare surface site at the beginning of the reaction, and end with a bare surface site at the end of the reaction in agreement with requirement (5.156). ■r- i'in i ne i kin nr mi mimiani'immh miiiiau mi ai imn\ 283 Ai 14 >i 11111 f in rule I 3 68), when nulu iiľ. n .ii l in the PJU phase, llu- inci danism |'iiui,ill\ n.I,is "I a seiies ol clcmciilniy icm lions when' a single atom is linnslciicd I ruin one •|n i |m in the next as described in So i..... s 4.6. i u nuns on metal surfaces arc very similar to radical reactions in the gas ptlMC I n m nilly, molecules first adsorb. Then, they transfer one atom al a lime to yield products I m example. Figure 5.14 shows the mechanism of methanol decomposition CH,OH CO + 2H2 (5.157) illinium surface called Pt(110). The methanol adsorbs molecular!) Then the III iiiuiol loses one hydrogen at a time to yield products. This mechanism differs from in . hanism of methanol decomposition in the gas phase because there is no initiation . H...... Still, the methanol follows the general rule that atoms are transferred one atom .1 i niiie into products. ||i lore we proceed, we need to define some notation. We will designate a surfaci i \ [he symbol S. With this notation, the second reaction in Figure 5.14, where Ml III orbed methanol reacts with the surface site to yield an adsorbed methoxy (CH3O) plus III id .orbed hydrogen atom, can be written CH,OHlad) + S CH30(ad) + H(ad) (5.I5K) hdc the subscript (ad) is used to indicate that we have an adsorbed species. Ncxl we need to note that, just as in the gas phase, one does not always simplj ......., one atom at a time during a surface reaction. For example, Figure 5.15 shows .......cchanism of ethanol decomposition on platinum. The ethanol first sequential > ,,, hydrogenates to yield an acetyl (CHCO) intermediate. Then the C-C bond breaks to , „ kl ( O and an adsorbed methyl group. CH3OH Formyl V////A Z77777A Adsorbed\ methanol H O V////A Methoxy V o=c7 |> V////A " Formaldehyde Figure 5.14 The mechanism of methanol decomposition on Pt(110). I'lll IHI IK IN > M Mil Ml I HANI'.M'. i H MIMIAI.I HI At:HON!I 288 *1>0 xc c V)//////A Methyl + carbon monoxide co 1 7..... Acetyl O V////A Adsorbed \ ethanol c o H X/////X Acetaldehyde V////A Ethoxy Figure 5.15 The mechanism of ethanol decomposition on Pt(111). Notice the acetyl group could still lose hydrogens, so the fact that the C-C bond breaks is an exception to the rule that molecules gain or lose one atom at a time during reactions on solid surfaces. 5.12.1 Intrinsic Barriers to Surface Reactions Next we will discuss how one can predict mechanisms of reactions on solid surfaces. In Section 5.4.6, we noted that one can use the Polanyi relationship to predict the mechanism of the reactions in the gas phase. Lee and Masel (1996) showed that one can use the same ideas to predict the mechanism of reactions of simple hydrocarbons, oxygenates, and amines on metal surfaces. The idea is that when you run a reaction on a surface, the reactants and products of the reaction are stabilized. However, in many cases, the intrinsic barriers to the reaction do not change. In that case, one can use equation (5.41) to predict the barriers to surface reactions. One then follows the procedures in Section 5.4.6 to predict the mechanism of the reaction. There are two major changes when we have reactions on metal surfaces. There is a proximity effect that bonds only in close proximity to the surface can break, and there is a role of d electrons, which can weaken bonds and thereby make them easier to break. The proximity effect is discussed in Section 10.11 of Masel (1996). The basic idea is that bonds can break only when the bonds are in close proximity to the surface. For example, Figure 5.16 shows the bond energies and geometry of methanol adsorbed on platinum. Notice that the CH bond is weaker than the OH bond. Therefore, on the basis of thermodynamics, one would expect the C-H bond to be easier to break than the O-H bond. However, experimentally, the OH bond is easier to break than the CH bond. Lee and Masel [1996] analyzed this case and found that the issue is what Masel [1996] called a proximity effect. Figure 5.17 shows a diagram of the transition state for C-H scission bond. During reaction, the carbon and hydrogen both need to form a bond to the platinum. There is no problem forming a platinum-hydrogen bond, but it is hard to form a platinum-carbon bond because the hydrogen gets in the way. Masel [1996] calls HI ki.,il/mnl 103 kcal/mol 'M ki ul/iniil Figure 5.16 Bond energies in methanol. *s /C H -Extended bond Figure 5.17 The transition state for C-H scission in adsorbed ethanol. till i proximity effect because only bonds close to the surface can break. Generally, the ........\ effect adds about 10-20 kcal/mol to intrinsic barriers, so it is quite important. i |,2 A Specific Example: Ethanol Decomposition of Platinum Hi example of the analysis, in this section, we will try to predict the mechanism I ih inn! decomposition on platinum. We will assume that the intrinsic barrier foi the ii nl a O-H or C-H bond is about 15 kcal/mol. The intrinsic barrier for the scission i hi i c or C-O bond is about 45 kcal/mol as described in Section 5.4.6. In principle, ethanol could decompose by a host of different reaction pathways. The ii. ni.il can sequentially dehydrogenate. The C-O bond could break before any other - i nun occurs. Now, let's ask which reaction is most favored if the intrinsic barrier foi III ' ission of a C-C or O-C bond is 45 kcal/mol while the intrinsic barrier to the scission i C II or O-H bond is 15 kcal/mol. i i mho 5.18 shows the bond energies in ethanol. Notice that the C-C bond is the J rsi bond in the molecule. Therefore, if all of the intrinsic barriers to bond scission i equal, one would expect the C-C bond in ethanol to be easier to break than the i II. O-H and C-O bonds. However, if one considers the difference in intrinsic barrier, ..... would come to a very different conclusion. Notice that the intrinsic barrier for the ion of a C-C or O-C bond is 30 kcal/mol higher than the intrinsic barrier of the i sion of a C-H or O-H bond. Consequently, the C-O bond would be harder to break 84 kcal/mo 110 kcal/mol —H 98 kcal/mol 83 kcal/mol 96 kcal/mol Figure 5.18 Bond energy in ethanol. zoo ri...... hi ii i, h 11 ii mi i in ii i ,i i ii .......n ill III.....>TI '.l U '.Mill Al I III Al III HI'. 'Jill iii.ill i'lllii-i 11 it- () II 111 ( llliunil I liciclote, lii ill) i In- (' IlindO II bond should hicilk at Iowa temperatures than the C-C bond Now consider whether the oil or C-H bond should be easici to break Notice tii.it the C-H bond is weaker than the O-H bond. Therefore, on the basis of bond cnergii alone, the C-H bond should break more easily than the Oil bond. However, if one looks at the reaction in detail, one finds that there is a proximity effect that substantially raises the intrinsic barrier for C-H bond scission. During reaction, carbon-surface and hydrogen-surface bonds form. Figure 5.17 shows a possible transition state loi ihc reaction. Notice that it is hard to form a carbon-surface bond because the carbon hydrogen bonds get in the way. One does not know how much energy this costs withouj doing detailed calculations. However, Masel and Lee 11996] did quantum-mechanical calculations that suggest that these repulsions raise the intrinsic barriers by 24 kcal/mol One can do the same analysis on the O-H bond scission step. In the case of O-Hj bond scission, there are some extra intrinsic barriers due to the repulsions of the enipiv dangling bonds on the ethanol. However, those effects would be expected to be mu< Ii smaller than the extra 24 kcal/mol repulsion if the C-H bond breaks. Therefore, on I In basis of an analysis of the intrinsic barriers, one would expect O-H bond scission to have a lower activation barrier than C-H, C-C, or C-O bond scission, even though the () II bond is the strongest bond in ethanol. Therefore, we conclude that, per an analysis of tin-intrinsic barriers, the first step in ethanol decomposition should be O-H bond scission In yield an cthoxy intermediate, as shown in Figure 5.15. That is what has been observed experimentally on all of the transition face metals that have been examined previously except Pt(110)(l x 2) and Pt(331), where the notation Pt(llO) (2 x 1) and (311) refefll to platinum catalysts with surface arrangements called (110) (2 x 1) or (311) as shown in Figure 5.19. One can continue the analysis to predict a mechanism of ethoxy decomposition on metals. Again, we note that the C-C bond in ethoxy is 10 kcal/mol weaker than the C-ll bond. Therefore, if one considered bond energies only, one would expect the C-C bond in ethoxy to break before the C-H bond. However, the intrinsic barriers go in the opposite directions. The intrinsic barriers for C-H bond scission are much lower than the intrinsic barriers for C-C bond scission. If one plugs numbers into equation (5.41), one finds that Pt(311) Pt(110)(2x1) Figure 5.19 The arrangement of the platinum atoms in Pt(311) and Pt(110) (1 x 2). 1 i mil activation barriei foi ethoxy dehydration should be It) .Ml kcnl/inol lowei In ictlvaiion barriei foi C-C b.....i iclssli......ethoxj Consequently, the ethoxj ii .pl.t -., 1111 < -1111 a 11 v dehyilidgenale lo produce an acctaldehydc and an acetyl inlermediiile I h. iiled in Figure 5.15. i In ni.iiv.i-. changes after the acetyl intermediate forms, in Chapter 5, we will provide methods lo estimate heats ol reactions. If one uses the methods from Chaptei 6, in I mi Is ill.it on platinum, the reaction CH3CO„ ■+ CH3ad + COa, i. .1 by about 44 kcal/mol over the reaction CH3COad -► Had + CH2COad (3.159) (5 .160) lion i 5.160) has a 20-40 kcal/mol lower intrinsic barrier than does reaction (3 I 59) hill lower intrinsic barrier is overcome by the 44-kcal/mol difference in All, I In n lore, according to equation (5.41), reaction (5.159) should have a lower activation i ...... ili.in reaction (5.160). Consequently, one would expect the acetyl intermediate to i .....pose via reaction (5.159) rather than reaction (5.160). Experimentally, one finds thai lliu ., decomposition follows the mechanism in Figure 5.15 in all of the cases examined ..in except in the case of ethoxy decomposition on Rh(lll). ilu point of this exercise is that one can make useful predictions with very lew .......i.. r rough estimates of the heats of adsorption and intrinsic barriers. In other work, i mil Masel (1997) have suggested that one can successfully predict the mechanism ol iii I. nc. methanol, ethanol, acetaldehyde, methyl iodide, methylamine, and ethyl iodide i.....nposition on most of the loose-packed faces of transition metals examined to date ....... the same simple analysis! Now, I would like to say that this procedure will always work, but unfortunately, I .......I For example, Cong et al. (1998) found that on a stepped surface, such as thai nl I'll I 10) (1 x 2) shown in Figure 5.19, ethanol decomposes by one of two pathways: I hi pathway shown in Figure 5.15 or, a pathway where the C-C bond breaks at low Ii mpcrature. This second pathway is not expected from the materials in this section drown and Barteau [1996] also found an exception during ethanol decomposition on i in 100). These exceptions are currently being actively investigated but, are not well Understood at present. In my experience, one can guess at a mechanism of a surface reaction and be righl of the time. However, there are occasional exceptions and the exceptions are noi understood. ft. 13 GENERIC TYPES OF SURFACE REACTIONS l lure is one other detail of importance to the prediction of mechanism of surface reactants. '.nine of the species that participate in the reaction might not be strongly attached to the in lace. People in the literature often discuss the participation of gas-phase species, and weakly bound complexes on surface reactions. There are three generic types of surface reaction: Langmuir-Hinshelwood mech-inisms, Rideal-Eley mechanisms, and precursor mechanisms. Figure 5.20 shows a ii hematic of these three mechanisms for a hypothetical reaction A + B -> AB. In the ' ll li..... I Yl'l 'i > II HI IIII ACl III , "ltd " A A .14 W B A B -BA \ A ^B. \ \ 11 vxw-mxvWxx? x\\\\\v B, b^a R I 4 A B-A (a) Langmuir-Hinshelwood B B-A B ' 4 A a-A B. I! / J A ^\\\\? (b) Rideal-Eley ^A (c) Precursor Langmuir-Hinshelwood mechanism, A and B first adsorb onto the surface of the catl lyst. Next, the adsorbed A and B react to form an adsorbed A-B complex. Finally, ilu A-B complex then desorbs. In what's now called the Rideal-Eley mechanism,1 the real tant A chemisorbs. The A then reacts with an incoming B molecule to form an A II complex. The A-B complex then desorbs. In the precursor mechanism, A adsorbs. Nexl B collides with the surface, and enters a mobile precursor state. The precursor rebounds along the surface until it encounters an adsorbed A molecule. The precursor then react! with the A to form an A-B complex which desorbs. Each of these reactions in Figure 5.20 can also occur in reverse as shown in Figure 5.21. For example, one can run the Langmuir-Hinshelwood reaction in reverM by adsorbing an A-B molecule, heating to allow the A-B to decompose into adsorbed A and B, and then desorbing the A and B. Alternatively, if the adsorbed A-B molecule Figure 5.21 ' In their original papers, Rideal and Eley [1940 194H did nm iti i(i,i, .ii mechanism. Ml o| the surface reactions that have been studied so far can he viewed as following a ......i IImshelwood mechanism, a Rideal-Eley mechanism, a precursoi mechanism, ......niibinalion of these three mechanisms. i ii , (ample, Figure 5.21 shows the mechanism of ethylene hydrogenation seen ni ven I pii in,- on a platinum catalyst. First, the hydrogen adsorbs and dissociates. Then. Ihe III i in adsorbs. Then, one adds one hydrogen at a time to the adsorbed ethylene lo yield i. Notice that in the mechanism in Figure 5.21, the reactants adsorb and then react i In lore, Ihe mechanism in Figure 5.21 is a classic Langmuir-Hinshelwood mechanism iiuii llinshelwood mechanisms are quite common in the literature. Most catalytic ........s go via Langmuir-Hinshelwood mechanisms. i i .„ lions used to deposit films on surfaces, on the other hand, arc thoughl lo often go i , . . ,li via Rideal-Eley mechanism. For example, one grows CVD (chemical vapOl i p. uioiii diamonds by heating methane to produce CH-s groups in the gas phase. The i II, 1'ioups then react with a cool surface to produce a diamond film. Harris (1990, 1993) illllncd the mechanism theoretically and proposed the mechanism shown in Figure 5.22. i Ii I Ihe methyl groups adsorb on the surface, forming a hydrocarbon layer. The (II, groups from the gas phase react with the hydrocarbon layer to form a hydrogen-terminated lil i.......,1 surface. Finally, other CH3 groups react with the layer to remove excess hydrogen |h ilrogen also desorbs). Notice that each of these reactions follow the general reaction , heme CH3 + As-► products (5.161) 11.1, a gas-phase species, CH3 g, reacts with a surface intermediate, As, to form products II, n, e, classically, this is a Rideal-Eley mechanism. In my experience, the key difference between catalytic reactions and film growth i. ii nous is that in catalysis, one usually runs the reaction at low temperatures. When ih, temperatures are low, the reactants do not dissociate until the reactants adsorb onio ii, . surface. In that case, reactions happen only after all of the species adsorb. Only i hi Miiiiir-Hinshelwood reactions are seen. On the other hand, film growth reactions often Figure 5.22 A Rideal-Eley mechanism for diamond deposition. •ItIMMAIIY m Figure 5.23 A precursor mechanism for the reaction CO + 02 -> C02 + OaC|. run at much higher temperatures. Radicals can form in the gas phase. When a radical hits u surface, the radical can react before it adsorbs. One can observe Rideal-Eley mechanisms of radicals. I do not have a good example of a precursor mechanism. People cannot actually observe precursors because the precursors remain on the surface for perhaps only 10 \ seconds. Consequently, one can never tell for sure whether a reaction goes via a precursor. There is some experimental evidence that the reaction 2CO + 02 2CO. (5.162) can go via the precursor mechanism shown in Figure 5.23. First the oxygen adsorbs and dissociates. Then a CO collides with the surface and wanders around until the CO finds an adsorbed oxygen atom. Then the two species react to form a C02 molecule, which desorbs. If one glances at Figure 5.23, it is not immediately obvious that this is a precursor mechanism and not a Langmuir-Hinshelwood mechanism. After all, the CO and 02 both enter the surface phase and then react. However, people say that this is a precursor mechanism because the CO reacts with the oxygen before the CO has had time to form a strong bond with the surface. Of course, at this point, no one has directly observed the precursor, so no one knows for sure whether the CO forms a bond to the surface before it reacts. In my view, the best available evidence is that the CO bonds before the CO reacts. However, there are others who disagree. In my experience, most reactions that were originally thought to proceed via a precursor mechanism are now thought to go by other mechanisms. Still, people discuss precursor mechanisms at length in the literature. Therefore, I thought that it would be important to mention them. 5.14 REACTIONS ON ACID SURFACES So far we have been talking about reactions on metal surfaces. Next, we want to move on and briefly mention reactions on solid acids. Students usually think about acids as being liquids, but there is a whole class of solid acids. For example, one can put protons into an ion exchange resin to create a solid acid. In the analysis and detergent industries, one uses special silica aluminates, specifically mixtures of Si02 and alumina I \l I),I culled ;ct>lilr\. The /collies me hiplilv poious slmcliiics whose siuliiees have | || ,,, , t.tlls lieilcd Io create Lewis oi Hiwisled acid sites It is possible Io in.ike ,i ....hi. wiih ,i |iK„ ol 10. Recall thai by comparison. I N II..SO, has a pK, Ol I and n ah i lias a pK„ of 7. Meat lions on solid acids are a cross between reactions on niclal surfaces and reactions .......I solution. As in a metal, the reactions occur at distinct sites on the acid's .....,, and as in a niclal, the reaction usually follows a l.angiiuiir Hinshelwood rale Mo as in a metal, the reactions go via a catalytic cycle where protons aie .........., ,1 and regenerated. Still, on a solid acid, most of the key intermediates are lh. reaction pathways look more like reactions in solution than reactions on .....11 surfaces. in particular, reactions on solid acids usually follow the mechanisms outlined m m i I and 5.4.1. Olefins, paraffins, and alcohols react with protons in the acid to Iťld i iirhcnium or carbonium ions. The carbenium and carbonium ions read as described i nous 5.4 and 5.4.1. One observes the same cracking, isomerization, ami alkylaiion . n lions m a solid acid that one observes in the gas phase or in a superacid solution I he III difference is that there are mass transfer limitations that limit the products produi ed i.. 11 ihc leading molecules are big. Solid acids are widely used as catalysts. They are also used as ion traps, for example, loti rgents and other cleaners. I In analysis of mechanisms of reactions in solid acids is very similar to ihc analysis i. lions of ions in the gas phase. Martens and Jacobs (1990) suggest that all ol Hli Ideas m Sections 5.4 and 5.4.1 apply to solid acids, and indeed the results in i u 5.8 show that reactions in solid acids behave very similarly to ionic reactions ... ih, i'as phase. Generally, one can predict the qualitative behavior of solid acid a ilysls using the material in Sections 5.4 and 5.4.1 (see Martens and Jacobs (1990) lin details). ri r. SUMMARY I., iniiuiary, then, in this chapter we reviewed the mechanisms of simple reactions in the | . phase and in various environments. We found that there are basically two types ol . Ii non pathways: radical pathways and ionic pathways. The radical pathways usually .. insist of an initiation step where radicals are formed, a series of propagation steps where products are produced in a catalytic cycle, and a series of termination steps where radicals ... oinbine. We found that one can often predict the reaction pathway using the Polanyi i. lationship, some knowledge of intrinsic barriers, and a few empirical rules. In particular, Ihc intrinsic barriers for atom transfer reactions are usually 30 kcal/mol lower than the Intrinsic barriers for transfer of molecular ligands. As a result, atomic transfer reactions .1 .ii,illy dominate in the gas phase. Ionic reactions are more complicated in that there are no empirical rules for prediction of reaction mechanisms. Generally, ionic pathways consist of a series of .iddiiions, isomerizations, and bond scissions. In the gas phase all of the processes have low intrinsic barriers, which means that the reactions quickly reach equilibrium. However, if one runs the reaction in a solvent, the rearrangements of the solvent determines the intrinsic barrier. The forces due to the rearrangements of the solvent cage are still not well understood, which makes prediction of reaction pathways more difficult. •II II VI II I KAMIM I '. W.\ 5.16 SOLVED EXAMPLES Example 5.A The reaction CH3CH3 >IIC ('II. i II. obeys the following mechanism: -»• 2.CH,. + X -> CH4 + .CH2CH, -> CH2CH2 + H. + H CH3CH3 + X •CH3 + CH3CH3 •CH2 + CH3X 2CH3. + X (+ other reactions) a) Make a diagram of the reaction similar to that in Figure 5.3 b) Identify the initiation step, the transfer step, the propagation steps, the termination step! Also indicate whether the propagation steps are associations, ^-hydrogen transfer, etc c) Does the mechanism obey the rules in the Section 5.4.7 d) Estimate the activation barrier for each step. e) Calculate the temperature where equation 5.35 is satisfied. Solution a) CH3CH3 initiation +X CH, . CH3 transfer . X + H2C = CH2 b) Step I —initiation Step 2 — chain transfer Step 3 — propagation (|3-hydrogen elimination) Step 4 — propagation (hydrogen transfer) Step 5 — termination c) Docs follow rules There is an initiation step (step I) There is a catalytic cycle (step 3 and 4) There is a termination step (step 5) We will verify the activation barriers below CH2CH3 ill I inn,ilr 11n• .h malum hmiicis i 11 || consider: (11,(11, I X » 2(11,. 1 II 'Hi |i I 1 ".limale All, I mm NISI Web book (http:/Avebbook.nist.gov) 1 hcrcl'ore \ll,K 11 ,t 11,) 20.0 AHr(CH3.) = +34.8 kcal/mol AH, = 2(34.8) - (-20.0) = 89.6 kcal/mol estimate BA using Table 5.4. This is a simple bond scission reaction, From ruble 5.4 EA = I + AHr = 90.6 kcal/mol CH3. + CH3CH3 CH4 + .CH2CH3 Nlep I estimate AHr From the NIST web book AH,(CH3CH3) = -20.0 kcal/mol AHf(CH3.) = +34.8 kcal/mol AH,(CH4) = -17.9 kcal/mol AH,(.CH2CH3) = +28.4 kcal/mol I herefore AHr = 17.9 + 28.4 - 34.8 - (-20.0) = -4.3 kcal/mol ■ i. |> ' estimate EA using Table 5.4. This is an atom transfer reaction. From Table 1 I EA = 10 kcal/mol + 0.3(-4.3) = 8.7 kcal/mol CH2CH3 + X-► CH3CH2 + H + X Slcp I: estimate AHr From the NIST web book AH,(CH2CH3) = +28.4 kcal/mol AHf(CH2CH2) = +12.5 kcal/mol AHf(H) = +52.1 kcal/mol AH = 52.1 + 12.5 - 28.4 = +36.2 kcal/mol Siep 2: estimate EA using Table 5.4. This is as endothermic ^-scission reaction. From Table 5.4 EA = 15 + 0.7(36.2) = 40.3 kcal/mol "• I «11,(11, . II. I., II, II Step I: estimate All, From the NIST web book AH,(H) = +52.1 kcal/mol AHKCH3CH3) = -20.0 kcal/mol AHf(CH2CH3) = +28.4 kcal/mol AH2 = 0 kcal/mol Therefore AHr = 0 + 28.4 - 52.1 (-20.0) = -3.7 kcal/mol Step 2: estimate Hr using Table 5.4. This is a hydrogen transfer reaction. From Table 5.4 EA= 10 + 0.3(-3.7) = 5.9 kcal/mol •CH3 + .CH3 + X-► CH3CH3 + X Step 1: estimate AHr AHr = —89.6 kcal/mol (reverse reaction 1) Step 2: estimate Hr using Table 5.4. This is a recombination reaction. From Table 5.4 EA = 1 kcal/mol e) Calculate temperature such that EA < 0.15 for initiation mol K kcal EA < 0.07——-— for all propagation mol K kcal moFK for cata]ytic cycle 'mol0 K For initiation, EA = 90.6 kcal/mol T > 90.6/0.15 = 604 K For propagation. T > 40.3/0.05 = 806 K Therefore any temperature above 806 K will satisfy all constraints. M II VI III XAMI'I I I > uii|ili 11 1 < m.1,1,1 ílu-i, ,11, iw h, 1 n.,1, m,, h.im.m for ethylene production from NMHItf (II,(II, 1 X .'( II (II,. I ('11,(11, CI l.i + .CH2CH3 .CH2 + CH3X ——► H2C = CH2 + H. + X H. + CH3CH3 -► CH4 + .CH, 2CH3. + X —-> CH3CH3 + X II 1 'm . iliis mechanism follow all of the rules at 810 K? 1 ' 1 ilii-. mechanism more or less likely than the mechanism in Example 5.A? Suliilioii I in does follow the rules! II There is an initiation step (step 1) ' 1 There is a catalytic cycle (steps 2,3,6) i) There is a termination step (step 5) 1 heck all steps obey constraint in equation 5.36 Steps 1,2,3,5 do (see Example 5.A) I heck step 6 H. + CH3CH3 -»• CH4 + .CH3 Nil pi: estimate AHr From the Nist webbook AHf(H.) = +52.1 kcal/mol AHf(CH3CH3) = -20.0 kcal/mol AHf(CH4) = -17.9 kcal/mol AHf(.CH3) = +34.8 AHr = 34.8 + (-17.9) - 52.1 - (20.0) = -15.2 kcal/mol Step 2: estimate EA using Table 5.4. This is a ligand transfer reaction to hydrogen. Front Table 5.4 EA = 45.0 + 0.5 + (-15.2) = 37.4 kcal/mol This reaction is in the catalytic cycle (k ;" t! \ kccil 0.05- T = (0.05)(810 K) = 40.5- mol / mol Therefore all constraints are satisfied h) Which mechanism is better?