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orbital vacant \ T 4 p he Study of Chem ical Reactions carbocation alleyl group The most interesting and useful aspect of organic chemistry i s the study of reactions. We cannot remember thousands of specific organic reactions, but we can organize the reactions into logical groups based on how the reactions take place and what interme diates are involved. We begin our study by considering the halogenation of alkanes, a relatively simple reaction that takes place in the gas phase, without a solvent to com plicate the reaction. In practice, alkanes are so unreactive that they are rarely used as starting materials for most organic syntheses. We start with them because we have already studied their structure and properties, and their reactions are relatively uncom plicated. Once we have used alkanes to introduce the tools for studying reactions, we will apply those tools to a variety of more useful reactions. The overall reaction, with the reactants on the left and the products on the right, is only the first step in our study of a reaction. If we truly want to understand a reaction, we must also know the mechanism, the step-by-step pathway from reactants to products. To know how well the reaction goes to products, we study its thermo dynamics, the energetics of the reaction at equilibrium. The amounts of reactants and products present at equilibrium depend on their relative stabilities. Even though the equilibrium may favor the formation of a product, the reaction may not take place at a useful rate. To use a reaction in a realistic time period (and to keep the reaction from becoming violent), we study its kinetics, the variation of reac tion rates with different conditions and concentrations of reagents. Understanding the reaction ' s kinetics helps us to propose reaction mechanisms that are consistent with the properties we observe. 4-1 I ntroduction The reaction of methane with chlorine produces a mixture of chlorinated products, whose composition depends on the amount of chlorine added and also on the reaction conditions. Either light or heat is needed for the reaction to take place at a useful rate. When chlorine is added to methane, the first reaction is 4-2 Ch lori nation of M eth ane H H-C-H H methane chlorine I I H + C ] - Cl heat or light ) H - C - Cl H chloromethane (methyl chloride) I I + H - Cl hydrogen chloride 1 25 1 26 Chapter 4: The Study of Chemical Reactions T his reaction may continue; heat or light is needed for each step: I H-C-Cl I H H I H-C-Cl I H + Cl CI CI-C-Cl H + I CI C]-C-Cl CI + I I I HCl HC] HCl This sequence raises several questions about the chlorination of methane. Why is heat or light needed for the reaction to go? Why do we get a mixture of products? Is there any way to modify the reaction to get just one pure product? Are the observed products formed because they are the most stable products possible? Or are they fa vored because they are formed faster than any other products? The answers to these questions involve three aspects of the reaction: the mecha nism, the thermodynamics, and the kinetics. 1 . T he mechanism i s the complete, step-by-step description of exactly which bonds break and which bonds form in what order to give the observed products. 2. Thermodynamics i s the study of the energy changes that accompany chemical and physical transformations. It allows us to compare the stability of reactants and products and predict which compounds are favored by the equilibrium. 3 . Kinetics i s the study of reaction rates, determining which products are formed fastest. Kinetics also helps to predict how the rate will change if we change the reaction conditions. We will use the chlorination of methane to show how we study a reaction. Before we can propose a detailed mechanism for the chlorination, we must learn everything we can about how the reaction works and what factors affect the reaction rate and the product distribution. A careful study of the chlorination of methane has established three important characteristics: 1 . The chlorination does not occur at room temperature in the absence of light. The reaction begins when light falls on the mixture or when it is heated. Thus, we know this reaction requires some form of energy to initiate i t. 2. The most effective wavelength of light is a blue color that is strongly absorbed by chlorine gas. This finding implies that light is absorbed by the chlorine mole cule, activating chlorine so that it initiates the reaction with methane. 3 . The light-initiated reaction has a high quantum yield. This means that many molecules of the product are formed for every photon of light absorbed. Our mechanism must explain how hundreds of individual reactions of methane with chlorine result from the absorption of a single photon by a single mole cule of chlorine. 4-3 The Free-Ra d i ca l Cha i n Reaction A c hain reaction mechanism has been proposed to explain the chlorination of methane. A chain reaction consists of three kinds of steps: 2. Propagation steps, i n which the reactive intermediate reacts with a stable mole cule to form another reactive intermediate, allowing the chain to continue until the supply of reactants is exhausted or the reactive intermediate is destroyed. 3. Termination steps, side reactions that destroy reactive intermediates and tend to slow or stop the reaction. 1 . The initiation step, which generates a reactive intermediate. 4-3 The Free-Radical Chain Reaction 1 27 In studying the chlorination of methane, we will consider just the first reaction to form chloromethane (common name methyl chloride). This reaction is a substitution: Chlorine does not add to methane, but a chlorine atom substitutes for one of the hydrogen atoms, which becomes part of the HCl by-product. H I H -C -@ I H methane h eat or light (hv) + @- Cl chlorine ) H I H -C I H --@ + @- Cl chloromethane (methyl chloride) 4 -3A T he I n itiation Step: Generation of Radicals Blue light, absorbed by chlorine but not by methane, promotes this reaction. There fore, initiation probably results from the absorption of light by a molecule of chlorine. Blue light has about the right energy to split a chlorine molecule ( CI2) into two chlorine atoms (242 kj/mol or 58 kcal/mol).* The splitting of a chlorine molecule by absorption of a photon is shown as follows: : CI : CI: .. .. r-- + photon (hv) : CI + CI : . . .. Notice the fishhook-shaped half-arrows used to show the movement of single unpaired electrons. Just as we use curved arrows to represent the movement of electron pairs, we use these curved half-arrows to represent the movement of single electrons. These half-arrows show that the two electrons in the CI - Cl bond separate, and one leaves with each chlorine atom. The splitting of a Cl2 molecule is an initiation step that produces two highly reactive chlorine atoms. A chlorine atom is an example of a reactive intermediate, a short-lived species that is never present in high concentration because it reacts as quickly as it is formed. Each CI' atom has an odd number of valence electrons (seven), one of which is unpaired. The unpaired electron is called the o dd electron or the radical electron. S pecies with unpaired electrons are called radicals or free radicals. R adicals are electron-deficient because they lack an octet. The odd elec tron readily combines with an electron in another atom to complete an octet and form a bond. Figure 4 -1 s hows the Lewis structures of some free radicals. Radicals are often represented by a structure with a single dot representing the unpaired odd electron. Free radicals may play a role in dis eases and accelerate aging. In the course of everyday life, reactive oxygen species are encountered in the environment and produced in the body. These compounds break down body's into short-lived and hydroxyl DNA. The radicals, which can react with the proteins reSUlting damage accumulates and may result in heart disease, cancer, and premature aging. --- --- ---- ----- Lewis structures : I C Wrillen : r B H: O H H: C H CH3 methyl radical HH H: : CC HH CH3 CH2 ethyl radical CI chlorine atom Br bromine atom HO hydroxyl radical Free radicals. Free radicals are reactive species with odd numbers of electrons. The unpaired electron is rep resented by a dot in the formula. ... Figure 4- 1 *The energy of a photon of light is related to its frequency v by the relationship E = hv, where h is Planck's constant. Blue light has an energy of about 50 kJ (60 kcal) per einstein (an einstein is a mole of photons). 2 128 Chapter 4: The Study of Chemical Reactions P RO B L E M 4 - 1 Draw Lewis structures for the following free radicals. (a) The n-propyl radical, CH3-CH2-CH2 (c) The isopropyl radical 4 -3 8 ( b) The t-butyl radical, (CH3hC' (d) The iodine atom Propagation Steps When a chlorine radical collides with a methane molecule, it abstracts (removes) a hydro gen atom from methane. One of the electrons in the C -H bond remains on carbon while the other combines with the odd electron on the chlorine atom to form the H - CI bond. First propagation step yo' + Cl H-C-H I H methane chlorine atom H I H-C ' I H methyl radical + H -Cl hydrogen chloride T his step forms only one of the final products: the molecule of HCl. A later step must form chloromethane. Notice that the first propagation step begins with one free radical (the chlorine atom) and produces another free radical (the methyl radical). The regen eration of a free radical is characteristic of a propagation step of a chain reaction. The reaction can continue because another reactive intermediate is produced. In the second propagation step, the methyl radical reacts with a molecule of chlorine to form chloromethane. The odd electron of the methyl radical combines with one of the two electrons in the CI - CI bond to give the CI- CH 3 bond, and the chlo rine atom is left with the odd electron. Secon.d propagation step Y/'1 + C l-Cl H-C I H methyl radical chlorine molecule H I H-C -Cl I H chloromethane + Cl ' chlorine atom In addition to forming chloromethane, the second propagation step produces anoth er chlorine atom. The chlorine atom can react with another molecule of methane, giving HCI and a methyl radical, which reacts with Cl2 to give chloromethane and regenerate yet another chlorine atom. In this way, the chain reaction continues until the supply of the re actants is exhausted or some other reaction consumes the radical intermediates. The chain reaction explains why many molecules of methyl chloride and HCI are formed by each photon of light that is absorbed. We can summarize the reaction mechanism as follows. KEY MECHANISM 4-1 Free-Radical Halogenation Like many other radical reactions, free-radical halogenation is a chain reaction. Chain reactions usually require one or more initiation steps, to form radicals, followed by propagation steps that produce products and regenerate radicals. Initiation: Radicals are formed. Light supplies the energy to split a chlorine molecule. CI-Cl + hv ( light) 2 Cl 4-3 The Free-Radical Chain Reaction Propagation: A radical reacts to generate another radical. Step 1: 1 29 A c hlorine radical abstracts a hydrogen to generate an alkyl radical. H-C-H H I I H + Cl' H I H -C' I H methyl radical + H-Cl continues the chain Step 2: T he alkyl radical reacts with Cl2 to generate the product and a chlorine radical. + H I H-C ' I H C l - Cl -----7 H I H - C -CI I H chloromethane + CI ' The chlorine radical generated in Step 2 g oes on to react in Step I, continuing the chain . The overall reaction is simply the s um o f the propagation steps: H I H-C-H I H Question: + C ]- C! H I H - C -Cl I H + H - CI W hat factor s characterize the propagation steps of a chain reaction? (a) Write the propagation steps leading to the formation of dichloromethane ( CH2Cl2 ) from chloromethane. (b) Explain why free-radical halogenation usually gives mixtures of products. (c) How could an industrial plant control the proportions of methane and chlorine to favor production of CCI4? To favor CH3Cl? P RO B L E M 4 - 2 4-3C Term i nation Reactions I f anything happens to consume some of the free-radical intermediates without gener ating new ones, the chain reaction will slow or stop. Such a side reaction is called a termination reaction: a step that produces fewer reactive intermediates (free radi cals) than it consumes. The following are some of the possible termination reactions in t he c hlorination of methane: H 1r1 H -C- + C l ' I H C l r1 H I H - C -Cl I H Cl-Cl 0 130 Chapter 4: The Study of Chemical Reactions H H 1r1 I H -C + C-H I I H H H HH I I H-C-C-H I I HH H H- H -C c ollides with wall I I H Cl collides with wall CI -1Mal l J ? -B H Some anti-cancer agents act by generating highly reactive hydrox yl radicals, which damage and degrade the DNA of the rapidly dividing tumor cells. As a result, the cells die and the tumor shrinks. One example of a radical generator is bleomycin, which is used for the treatment of testicular cancer. The combination of any two free radicals is a termination step because it decreases the number of free radicals. Other termination steps involve reactions of free radicals with the walls of the vessel or other contaminants. Although the first of these termination steps gives chloromethane, one of the products, it consumes the free radi cals that are necessary for the reaction to continue, thus breaking the chain. Its contri bution to the amount of product obtained from the reaction is small compared with the contribution of the propagation steps. While a chain reaction is in progress, the concentration of radicals is very low. The probability that two radicals will combine in a termination step is lower than the probability that each will encounter a molecule of reactant and give a propagation step. The termination steps become important toward the end of the reaction, when there are relatively few molecules of reactants available. At this point, the free radicals are less likely to encounter a molecule of reactant than they are to encounter each other (or the wall of the container). The chain reaction quickly stops. P RO B L E M 4 - 3 PROBLEM-SOLVING In a free-radical chain reaction, H?ltp initiation steps generally create new free radicals. Propagation steps usually combine a free radical and a reactant to give a product and another free radical. Termination steps generally decrease the number of free radicals. Each o f the following proposed mechanisms for the free-radical chlorination o f methane is wrong. Explain how the experimental evidence disproves each mechanism. (a) C l2 + hv Cl ; (an "activated" form of C12) Cl ; + C H4 H CI + CH3CI ( b) C H4 + hv CH3 + H CH3 + C l2 C H3Cl + CI CI + H- H CI P R OB L E M 4 - 4 Free-radical chlorination of hexane gives very poor yields of l -chlorohexane, while cyclo hexane can be converted to chlorocyclohexane in good yield. ( a) How do you account for this difference? (b) W hat ratio of reactants (cyclohexane and chlorine) would you use for the synthesis of chlorocyclohexane? 4-4 Equilibri u m Constants and Free Energy N ow that we have determined a mechanism for the chlorination of methane, we can consider the energetics of the individual steps. Let's begin by reviewing some of the principles needed for this discussion. Thermodynamics i s the branch of chemistry that deals with the energy changes accompanying chemical and physical transformations. These energy changes are most u seful for describing the properties of systems at equilibrium. L et's review how ener gy and entropy variables describe an equilibrium. 4-4 Equilibrium Constants and Free Energy The equilibrium concentrations of reactants and products are governed by the equilibrium constant of the reaction. For example, if A and B react to give C and D, then the equilibrium constant Keg i s defined by the following equation: )C+ D [products] [C][D] Ke g [reactants] [A][B] The value of Keg t ells us the position of the equilibrium: whether the products or the reactants are more stable, and therefore energetically favored. If Keg is larger than 1, the reaction is favored as written from left to right. If Keg is less than 1, the reverse re action is favored (from right to left as written). The chlorination of methane has a large equilibrium constant of about 1.1 X 1 0 1 9 . A + B ( = ---- 13 1 CH4 K eq + = [CH3Cl][ HCI] [CH4] [CI2 ] = CI2 ( ) CH3C1 + HCI 1 . 1 X 1 019 The equilibrium constant for chlorination is so large that the remaining amounts of the reactants are close to zero at equilibrium. Such a reaction is said to go to completion, and the value of Keg i s a measure of the reaction's tendency to go to completion. From the value of Keq w e can calculate the change in free energy ( sometimes called Gibbs free energy) that accompanies the reaction. Free energy is represented by G, and the change (Ll) in free energy associated with a reaction is represented by LlG, the difference between the free energy of the products and the free energy of the reactants. LlG is a measure of the amount of energy available to do work. LlG = ( free energy of products ) - (free energy of reactants) If the energy levels of the products are lower than the energy levels of the reactants (a "downhill" reaction), then the reaction i s energetically favored; and this equation gives a negative value of LlG, corresponding to a decrease in the energy of the system. The standard Gibbs free energy change, LlGo, is most commonly used. The symbol designates a reaction involving reactants and products in their standard states (pure substances in their most stable states at 25C and 1 atm pressure). The relation ship between LlGo and Keg i s given by the expression K eq = e -t:J.GOjRT or, conversely, by R = 8 . 3 1 4 Jlkelvin-mol (l.987 cal/kelvin-mol), the gas constant T absolute temperature, in kelvins* e 2 .7 1 8, the base of natural logarithms The value of RT at 25C is about 2.48 kJ/mol (0.592 kcal/mol). = = where The formula shows that a reaction is favored ( Ke q > 1) i f it has a negative value of LlGo (energy is released). A reaction that has a positive value of LlGo (energy must be added) is unfavorable. These predictions agree with our intuition that reac tions should go from higher-energy states to lower-energy states, with a net decrease in free energy. will often include the degree sign, however, to distinguish absolute temperatures (K) from equilibrium constants (K) = A reaction with a negative t:J.G is favored. PROBLEM-SOLVING H?nl/ t:J.G is A reaction with a positive unfavorable. *Absolute temperatures (in kelvins) are correctly given without a degree sign, as in the equation 25C = 298 K. We as in 25C 298K. 132 C hapter 4 : T he Study of Chemical Reactions S O LV E D P RO B L E M 4- 1 C alculate the value of !:!.Go for the chlorination of methane. S O L UTION - 2.303RT(log Keq) Keq for the chlorination is l . 1 X 1 019, and log Keq !:!' Go = = 1 9 .04 At 25C (about 298 Kelvin), the value of RT is RT = ( 8 . 3 1 4 llkelvin-mol) ( 298 kelvins) = = 2478 llmol, or 2.48 kJ/mol Substituting, we have !:!'Go = ( -2.303)(2.478 kllmol) ( 1 9.04) - 108.7 kJ/mo\ ( -25 .9 kcaIjmol) This is a large negative value for !:!. Go, showing that this chlorination has a large driving force that pushes it toward completion. In general, a reaction goes nearly to completion ( >99%) for values of !1Go that are more negative than about 1 2 kJ /mol or 3 kcalj mol. Table 4- 1 shows what per centages of the starting materials are converted to products at equilibrium for reactions with various values of !1Go. - P RO B L E M 4 - 5 The following reaction has a value of !:!'Go = - 2 . 1 kJ/mol (-0.5 kcaIjmol ) . (a) C alculate Keq a t room temperature (25 C ) for this reaction a s written. ( b) Starting with aiM solution of CH3Br and H 2 S, calculate the final concentrations of all four species at equilibrium. P RO B L E M 4-6 A t room temperature (25 C), the reaction of two molecules of acetone to form diacetone alco hol proceeds to an extent of about 5%. Determine the value of !:!'Go for this reaction. II 2CH3-C-CH3 acetone o OH II I CH3-C-CH2-C(CH3)2 o diacetone alcohol TABLE 4-1 Product Composition as a Function of !:!'GO at 25C 100% !:!'GO kJ/mol kcal/mol K Conversion to Products + 4.0 + 2.0 0 .0 -2.0 -4.0 - 8.0 -12.0 -16.0 -20.0 ( + 1 .0) ( +0.5) (0.0) ( -0.5) ( - 1 .0) ( -1.9) ( -2.9) ( -3.8) ( -4.8) 0 .20 0.45 1 .0 2.2 5 .0 25 1 27 638 3200 17% 31% 50% 69% 83% 96% 99.2% 99.8% 99.96% ::l '0 0 t) '" 90% 80% 70% 60% 50% 0.0 -4.0 -8.0 -12.0 !:!'Go, kllmol -1 6.0 -20.0 c... .u;.. .. u (\) >a 0 B .... a 0 4-5 Enthalpy and Entropy Two factors contribute to the change in free energy: the change in enthalpy and the change in entropy multiplied by the temperature. tlGo tlGO tlHo tlso = = = 133 4-5 Entha l py and Entropy tlHo - Ttlso = ( free energy of products) - ( free energy of reactants) ( enthalpy of products) - (enthalpy of reactants ) ( entropy of products ) - (entropy of reactants) At low temperatures, the enthalpy term (tlHO) i s usually much larger than the entropy term (-TtlSO), a nd the entropy term is sometimes ignored. 4 -5A E nthalpy The change in enthalpy (tlHO) is the heat of reaction-the amount of heat evolved or consumed in the course of a reaction, usually given in kilojoules (or kilocalories) per mole. The enthalpy change is a measure of the relative strength of bonding in the prod ucts and reactants. Reactions tend to favor products with the lowest enthalpy (those with the strongest bonds). If weaker bonds are broken and stronger bonds are formed, heat is evolved and the reaction is exothermic ( negative value of tlHO). I n an exothermic reaction, the enthalpy term makes a favorable negative contribution to tlGo. I f stronger bonds are broken and weaker bonds are formed, then energy is consumed in the reaction, and the reaction is endothermic ( positive value of tlHO). I n an endothermic reaction, the enthalpy term makes an unfavorable positive contribution to tlGo. The value of tlHo for the chlorination of methane is about - 105. 1 kJ/mol ( - 25.0 kcaIjmol) . This is a highly exothermic reaction, with the decrease in enthalpy serving as the primary driving force. 4-5 8 E ntropy Entropy i s often described as randomness, or freedom of motion. Reactions tend to favor products with the greatest entropy. Notice the negative sign in the entropy term (-TtlSO) of the free-energy expression. A positive value of the entropy change (tlSO), indicating that the products have more freedom of motion than the reactants, makes a favorable (negative) contribution to tlGo. In many cases, the enthalpy change (tlHO) is much larger than the entropy change (tlSO), and the enthalpy term dominates the equation for tlGO. Thus, a nega tive value of tlso does not necessarily mean that the reaction has an unfavorable value of tlGo. The formation of strong bonds (the change in enthalpy) is usually the most important component in the driving force for a reaction. In the chlorination of methane, the value of tl so i s + 1 2. 1 J/kelvin-mole. The -Ttlso term in the free energy is -TtlSo = - ( 298K) ( 1 2. 1 J/Kelvin-mol) = - 3610 J/mol = - 3. 6 1 kJ/mol ( -0.86 kcaIjmol ) The value of tlGo tlGo = - 1 08.7 kJ/mol i s divided into enthalpy and entropy terms: tlHo - TtlSo = = - 105. 1 kJ/mol - 3.61 kJ/mol = - 108.7 kJ/mol ( -25.9 kcaIjmol) The enthalpy change is the largest factor in the driving force for chlorination. This is the case in most organic reactions: The entropy term is often small in relation to the enthalpy term. When we discuss chemical reactions involving the breaking and forming of bonds, we can often use the values of the enthalpy changes (tlHO), under the assumption that tlGo == tlHo. We must be cautious in making this approximation, however, because some reactions have relatively small changes in enthalpy and larger changes in entropy. 134 C hapter 4 : The Study of C hemical Reactions S O LV E D P RO B L E M 4 - 2 Predict whether the value o f I::!.so for the dissociation o f Cl2 i s positive (favorable) o r nega tive (unfavorable). What effect does the entropy term have on the sign of the value of I:Go :!. for this reaction? SOL UTION Two isolated chlorine atoms have much more freedom of motion than a single chlorine molecule. Therefore, the change in entropy is positive, and the entropy term ( -TI::!.S O) is negative. This negative (favorable) value of ( -TI::!. SO ) i s small, however, compared with the much larger, positive (unfavorable) value of I::!.Ho. The chlorine molecule is much more sta ble than two chlorine atoms, showing that the positive enthalpy term predominates. PROBLEM-SOLVING In general, two smaller molecules (or Hi-ltv P RO B L E M 4 - 7 fragments, such as radicals) have more freedom of motion (greater entropy) than one larger molecule. When ethene i s mixed with hydrogen i n the presence of a platinum catalyst, hydrogen adds across the double bond to form ethane. At room temperature, the reaction goes to completion. Predict the signs of I::!.Ho and I::!.So for this reaction. Explain these signs in terms of bonding and freedom of motion. Pt catal yst ) HH I I H-C-C-H I I HH ethane ethene P RO B L E M 4 - 8 For each reaction, estimate whether I::!. So for the reaction is positive, negative, or impossible to predict. (a) C3H6 + C7HI6 heptane propene n-decane (b) The formation of diacetone alcohol: 0 0 CloH22 catalyst heat II 2 CH3-C-CH3 0 -OH OH II I CH3-C-C-C(CH3)2 H+ II CH3-C-OCH3 0 + II (c) CH3-C-OH + CHpH HzO 4-6 Bond-Dissociation Enthal pies We can put known amounts of methane and chlorine into a bomb calorimeter and use a hot wire to initiate the reaction. The temperature rise in the calorimeter is used to calculate the precise value of the heat of reaction, fl.Ho. This measurement shows that 105 kJ (25 kcal) of heat is evolved (exothermic) for each mole of methane converted to chloromethane. Thus, fl.Ho for the reaction is negative, and the heat of reaction is given as fl.Ho - 1 05 kli mol ( -25 kcaljmol ) I n many cases, we want to predict whether a particular reaction will be endother mic or exothermic, without actually measuring the heat of reaction. We can calculate an approximate heat of reaction by adding and subtracting the energies involved in the breaking and forming of bonds. To do this calculation, we need to know the energies of the affected bonds. The bond-dissociation enthalpy (BDE, a lso called bond-dissociation energy) is the amount of enthalpy required to break a particular bond homolytically, that is, in such a way that each bonded atom retains one of the bond's two electrons. In contrast, when a bond is broken heterolytically, one of the atoms retains both electrons. = 4-7 Enthalpy Changes in Chlorination Homolytic cleavage (free radicals result) 135 A:B : CI : CI: U ' " .. f.. r A' + 2 : Cl ' .. 'B 6.HO 6.HO = bond-dissociation enthalpy 242 kJ/mol (58 kcal/mol) = Heterolytic cleavage (ions result) A:B n (CH3)3C - 1 : r- -----;. -----;. A+ (CH3hC+ + + :B : Cl :(6.H o varies with solvent) Homolytic cleavage (radical cleavage) forms free radicals, while heterolytic cleavage (ionic cleavage) forms ions. Enthalpies for heterolytic (ionic) cleavage de pend strongly on the solvent's ability to solvate the ions that result. Homolytic cleav age is used to define bond-dissociation enthalpies because the values do not vary so much with different solvents or with no solvent. Note that a curved arrow is used to show the movement of the electron pair in an ionic cleavage, and curved half-arrows are used to show the separation of individual electrons in a homolytic cleavage. Energy is released when bonds are formed, and energy is consumed to break bonds. Therefore, bond-dissociation enthalpies are always positive (endothermic). The overall enthalpy change for a reaction is the sum of the dissociation enthalpies of the bonds broken minus the sum of the dissociation en thaI pies of the bonds formed. 6.Ho 2: (BDE of bonds broken ) - 2: ( BDE of bonds formed) By studying the heats of reaction for many different reactions, chemists have developed reliable tables of bond-dissociation enthalpies. Table 4-2 gives the bond-dissociation enthalpies for the homolysis of bonds in a variety of molecules. = We can use values from Table 4-2 t o predict the heat of reaction for the chlorination of methane. This reaction involves the breaking (positive values) of a CH3 - H bond and a CI - CI bond, and the formation (negative values) of a CH3 -CI bond and a H - CI bond. Overall reaction 4-7 Enthal py Changes in Chlorination Bonds broken CI- Cl CH3 - H Total 6.Ho = 6.Ho (per mole) +242 kJ (+58 kcal ) +435 kJ (+ 104 kcal ) +677 kJ (+162 kcal ) +677 kJ + (-782) kJ B onds formed 6.Ho (per mole) H - Cl -431 kJ (-103 kcal ) CH3 - CI -351 kJ (-84 k cal) Total -782 k J (-1 87 kcal ) = - 1 05 kJ/mol (-25 kcal/mol ) The bond-dissociation enthalpies also provide the heat of reaction for each indi vidual step: First propagation step C I + C H4 Breaking a CH3 - H bond Forming an H - CI bond Step total ' CH3 + HCI +435 kJ/mol (+ 1 04 kcaljmol ) -43 1 kJ/mol (-103 kcaljmol ) + 4 kJ/mol (+ 1 kcal/mol ) 136 TABLE 4-2 Chapter 4 : The Study of Chemical Reactions Bond-Dissociation Enthalpies for Homolytic Cleavages A:B A + B Bond-Dissociation Enthalpy Bond-Dissociation Enthalpy Bond Bonds to secondary carbons Bond H - X bonds and X - X bonds kJ/mol kcal/mol kJ/mol kcal/mol Methyl bonds F-F CI-CI Br-Br I-I H-F H-CI H-Br H-I HO-H HO-OH D -D H-H 435 444 159 242 192 151 569 431 368 297 498 213 435 456 351 293 234 381 410 448 339 285 222 381 410 448 339 285 222 381 104 106 38 58 46 36 136 103 88 71 119 51 104 109 84 70 56 91 98 107 81 68 53 91 98 107 81 68 53 91 (CH3hCH-H (CH3hCH-F (CH3hCH-CI (CH3hCH-Br (CH3hCH-I (CH3hCH-OH (CH3hC-H (CH3hC-F (CH3hC-CI (CH3hC-Br (CH3hC-I (CH3)3C-OH 397 444 335 285 222 381 95 106 80 68 53 91 Bonds to tertiary carbons Bonds to primary carbons CH3-H CH3-F CH3-Cl CH3-Br CH3-1 CH3-OH 381 444 331 272 209 381 91 106 79 65 50 91 Other C- H bonds PhCH2-H (benzylic) CH2=CHCH2-H (allylic) CH2=CH-H (vinyl) Ph-H (aromatic) C-C bonds 356 364 464 473 85 87 111 113 CH3CH2-H CH3CH2-F CH3CH2-Cl CH3CH2-Br CH3CH2-1 CH3CH2-OH CH3CH2CH2-H CH3CH2CH2-F CH3CH2CH2-CI CH3CH2CH2-Br CH3CH2CH2-I CH3CH2CH2-OH CH3-CH3 CH3CH2-CH3 CH3CH2-CH2CH3 (CH3hCH-CH3 (CH3)3C-CH3 368 356 343 351 339 88 85 82 84 81 S econd propagation step PROBLEM-SOLVING Bond-dissociation enthalpies are for H ?nv 'CH3 + Cl2 CH3Cl + CI ' +243 klimol (+58 k cal/ mol ) Breaking a Cl - Cl bond -352 klimol ( - 84 kcal/ mol ) Forming a CH3 - Cl bond - 1 09 klimol ( -26 kcal/ mol ) Step total Grand total = +4 kli mol + ( - 1 09 kli mol ) = - 105 klimol ( -25 kcalimol ) calculating values of t'lHo, use positive BDE values for bonds that are broken and negative values for bonds that are formed. breaking bonds, which costs energy. In The sum of the values of I1Ho for the individual propagation steps gives the overall enthalpy change for the reaction. The initiation step, Cl2 - 2 C I ' , is not added to give the overall enthalpy change because it is not necessary for each molecule of prod uct formed. The first splitting of a chlorine molecule simply begins the chain reaction, which generates hundreds or thousands of molecules of chloromethane. P R OB L E M 4 - 9 (a) Propose a mechanism for the free-radical chlorination of ethane, C H3 - CH3 + Cl2 hv C H3 - CH2Cl + HCI 4-8 Kinetics a nd the Rate Equation ( c) C alculate the overall value of 6 Ho for this reaction. 1 37 (b) Calculate 6Ho for each step in this reaction. A lter native M echanis m The mechanism we have used is not the only one that might be proposed to explain the reaction of methane with chlorine. We know that the initiating step must be the splitting of a molecule of C12, but there are other propaga tion steps that would form the correct products: (a) CI' + C H3 - H C H3 - Cl + H- 6.Ho +435 kJ - 351 kJ = (b) H' Total -105 kJ ( -25 kcal) This alternative mechanism seems plausible, but Step (a) is endothermic by 84 kJ/mol (20 kcal/mol). The previous mechanism provides a lower-energy alternative. When a chlorine atom collides with a methane molecule, it will not react to give methyl chlOIide and a hydrogen atom (6.Ho +84 k J + 20 k cal ) ; it will react to give HCI and a methyl radical (6.Ho + 4 kJ +1 k cal) , the first propagation step of the correct mechanism. = = = = + C l - CI H - Cl + C I' 6.Ho = +242 k J - 431 kJ = -189 k J ( -45 k cal) +84 kJ ( +20 k cal) P R OB L E M 4 - 1 0 (a) U sing bond-dissociation enthalpies from Table 4-2 (page 1 36), calculate the heat of reaction for each of the steps in the free-radical bromination of methane. (b) Calculate the overall heat of reaction. B r2 + C H 4 heat o r light ) CH 3 B r + HB r Kinetics i s the study of reaction rates. How fast a reaction goes is just as important as the position of its equilibrium. Just because thermodynamics favors a reaction (nega tive 6.GO) does not necessarily mean the reaction will actually occur. For example, a mixture of gasoline and oxygen does not react without a spark or a catalyst. Similarly, a mixture of methane and chlorine does not react if it is kept cold and dark. The rate of a reaction i s a measure of how fast the products appear and the reac tants disappear. We can determine the rate by measuring the increase in the concentrations of the products with time, or the decrease in the concentrations of the reactants with time. Reaction rates depend on the concentrations of the reactants. The greater the con centrations, the more often the reactants collide and the greater the chance of reaction. A rate equation ( sometimes called a rate law) i s the relationship between the concen trations of the reactants and the observed reaction rate. Each reaction has its own rate equation, determined experimentally by changing the concentrations of the reactants and measuring the change in the rate. For example, consider the general reaction A+BC+D The reaction rate is usually proportional to the concentrations of the reactants ([A] and [B]) raised to some powers, a and b. We can use a general rate expression to represent this relationship as rate kr[AY'[B]b where kr is the rate constant, and the values of the powers (a and b) m ust be deter mined experimentally. We cannot guess or calculate the rate equation from just the stoichiometry of the reaction. The rate equation depends on the mechanism of the reaction and on the rates of the individual steps in the mechanism. In the general rate equation, the power a i s called the order o f the reaction with respect to reactant A, and b i s the order of the reaction with respect to B. The sum of these powers, (a+b), i s called the overall order of the reaction. The following reaction has a simple rate equation: = 4-8 Kinetics and the Rate Equation C H3 - Br + -OH H,O/acetone - ) CH3 - OH + Br- 138 C hapter 4 : The Study of Chemical Reactions E xperiments show that doubling the concentration of methyl bromide, [ CH3Br], dou bles the rate of reaction. Doubling the concentration of hydroxide ion, [-OH], also doubles the rate. Thus, the rate is proportional to both [ CH3Br] and [-OH], so the rate equation has the following form: rate kr[CH3Br][-OH] This rate equation is first order i n each of the two reagents because it is proportional to the first power of their concentrations. The r ate equation is second order overall be cause the sum of the powers of the concentrations in the rate equation is 2; that is, ( first order ) + ( first order ) = second order overall. Reactions of the same overall type do not necessarily have the same form of rate equation. For example, the following similar reaction has a different kinetic order: = ( CH3 h C - Br + -OH - H20/acetone '> ( CH3 h C - OH + B r- Doubling the concentration of t buty l bromide [ ( CH3 h C - Br] causes the rate to double, but doubling the concentration of hydroxide ion [-OH] has no effect on the rate of this particular reaction. The rate equation is rate kr[ ( CH3 ) 3C -Br] This reaction is first order in t buty l bromide, and zeroth order in hydroxide ion (pro portional to [-OH] to the zeroth power). It is first order overall. The most important fact to remember is that the rate equation must be deter mined experimentally. We cannot predict the form of the rate equation from the stoi chiometry of the reaction. We determine the rate equation experimentally, then use that information to propose consistent mechanisms. = - S O LV ED P R OB L E M 4 - 3 Chloromethane reacts with dilute sodium cyanide ( Na+ -C=N) according to the follow ing equation: CH3-CI + T=N CH3 - C - N + C lchloride acetonitrile cyanide chloromethane When the concentration of chloromethane is doubled, the rate is observed to double. When the concentration of cyanide ion is tripled, the rate is observed to triple. (a) What is the kinetic order with respect to chloromethane? (b) What is the kinetic order with respect to cyanide ion? (c) What is the kinetic order overall? (d) Write the rate equation for this reaction. SOLUTIO N (a) When [CH3CI] is doubled, the rate doubles, which is 2 to the first power. The reaction is first order with respect to chloromethane. (b) When [-CN] is tripled, the reaction rate triples, which is 3 to the first power. The reaction is first order with respect to cyanide ion. (c) First order plus first order equals second order overall. (d) rate kr[CH3CI][TN] = P R OB L E M 4 - 1 1 The reaction of t-butyl chlOlide with methanol ( CH3hC - Cl t-butyl chloride + C H3 - OH methanol rate = ( CH3hC - OCH3 methyl t-butyl ether + H CI is found to follow the rate equation kr[ ( CH3hC - CI] (a) W hat is the kinetic order with respect to t-butyl chloride? ( b) What is the kinetic order with respect to methanol? (c) What is the kinetic order overall? 4-9 A ctivation Energy and the Temperature Dependence of Rates P R OB L E M 4 - 1 2 1 39 Under certain conditions, the bromination of cyclohexene follows an interesting rate law: (a) What is the kinetic order with respect to cyclohexene? (b) What is the kinetic order with respect to bromine? (c) What is the overall kinetic order? P R OB L E M 4 - 1 3 G H Br Br H When a small piece of platinum is added to a mixture of ethene and hydrogen, the following reaction occurs: H Pt catalyst ) H-C-C-H H I I H I I ethene hydrogen H eth ane Doubling the concentration of hydrogen has no effect on the reaction rate. Doubling the con centration of ethene also has no effect. (a) What is the kinetic order of this reaction with respect to ethene? With respect to hydro gen? What is the overall order? (b) Write the unusual rate equation for this reaction. (c) Explain this strange rate equation, and suggest what one might do to accelerate the reaction. Each reaction has its own characteristic rate constant, kr. Its value depends on the conditions of the reaction, especially the temperature. This temperature dependence is expressed by the Arrhenius equation, kr Ae -E,,/RT = 4-9 Activation Energy and the Tem perature Dependence of Rates where Ea A = R T = = = a constant (the "frequency factor") activation energy the gas constant, 8 .3 1 4 llkelvin-mole ( 1 .987 callkelvin-mole) the absolute temperature The activation energy, E a, is the minimum kinetic energy the molecules must have to overcome the repulsions between their electron clouds when they collide. The exponential term e -Ea/RT corresponds to the fraction of collisions in which the particles have the minimum energy Ea needed to react. We can calculate Ea for a reaction by measuring how kr varies with temperature, and substituting into the Arrhenius equation. The frequency factor A accounts for the frequency of collisions and the fraction of collisions with the proper orientation for the reaction to occur. In most cases, only a small fraction of collisions occur between molecules with enough speed and with just the right orientation for reaction to occur. Far more collisions occur without enough kinetic energy or without the proper orientation, and the molecules simply bounce off each other. The Arrhenius equation implies that the rate of a reaction depends on the fraction of molecules with kinetic energy of at least E a. F igure 4-2 shows how the distribution of kinetic energies in a sample of a gas depends on the temperature. The black curved line shows the molecular energy distribution at room temperature, and the dashed lines show the energy needed to overcome barriers of 4 kllmol ( 1 kcal/mol), 40 kllmol ( 1 0 kcal/mol), and 80 kl ( 1 9 kcal/mol). The area under the curve to the right of each barri er corresponds to the number of molecules with enough energy to overcome that 140 Chapter 4: T he Study of Chemical Reactions The dependence of kinetic energies on temperature. This graph shows how the number of molecules with a given activation energy decreases as the activation energy increases. At a higher temperature (red curve), more collisions have the needed energy. Figure 4-2 i room temperature (3000K) \ 40 kJ/mol 80 kllmal energy (E) barrier. The red curve shows how the energy distribution is shifted at 1 00C. At 100C, many more molecules have the energy needed to overcome the energy barriers, espe cially the 80 kJ/mol barrier. For smaller temperature changes, chemists often use an approximation: For reactions with typical activation energies of about 40 to 60 kJ/mol ( 1 0 to 15 kcal/mol), the reaction rate approximately doubles when the temperature is raised by loC, as from 27C (near room temperature) to 37C (body temperature). Because the relative rate constant, kreb i ncreases quickly when the temperature is raised, it might seem that raising the temperature would always be a good way to save time by making reactions go faster. The problem with raising the temperature is that all reactions are accelerated, including all the unwanted side reactions. We try to find a temperature that allows the desired reaction to go at a reasonable rate without producing unacceptable rates of side reactions. 4-10 Transition States The activation energy Ea represents the energy difference between the reactants and the transition state, the highest-energy state in a molecular collision that leads to reaction. In effect, the activation energy is the barrier that must be overcome for the reaction to take place. The value of Ea is always positive, and its magnitude depends on the relative energy of the transition state. The term transition state i mplies that this configuration is the transition between the reactants and products, and the molecules can either go on to products or return to reactants. Unlike the reactants or products, a transition state is unstable and cannot be iso lated. It is not an intermediate, because an intermediate i s a species that exists for some finite length of time, even if it is very short. An intermediate has at least some stability, but the transition state is a transient on the path from one intermediate to another. The transition state is often symbolized by a superscript double dagger ( :j: ) , and the changes in variables such as free energy, enthalpy, and entroPl involved in achieving the transition state are symbolized IlG:j:, IlH:j:, and IlS:j: . IlG is similar to Ea, and the symbolllG:j: is often used in speaking of the activation energy. Transition states have high energies because bonds must begin to break before other bonds can form. The following equation shows the reaction of a chlorine radical with methane. The transition state shows the C - H bond partially broken and the H - Cl bond partially formed. Transition states are often enclosed by brackets to emphasize their transient nature. . /H H - C ...... H transition state H I H-C-H I H + 'Cl + H-Cl 4-10 Transition States transition state :j: 141 .... Figure 4-3 t ('t0: ['f{O _________ C+D (products) reaction coordinate __ Reaction-energy diagram for a one-step exothermic reaction. The reactants are toward the left, and the products are toward the right. The vertical axis represents the potential energy. The transition state is the highest point on the graph, and the activation energy is the energy difference between the reactants and the transition state. Reactio n-Energy D i agrams The concepts of transition state and activation ener gy are easier to understand graphically. Figure 4-3 s hows a reaction-energy diagram for a one-step exothennic reaction. The vertical axis of the energy diagram represents the total potential energy of all the species involved in the reaction. The horizontal axis is called the reaction coordinate. The reaction coordinate symbolizes the progress of the reaction, going from the reactants on the left to the products on the right. The tran sition state is the highest point on the graph, and the activation energy is the energy difference between the reactants and the transition state. The heat of reaction (!::..HO) i s the difference in energy between the reactants and the products. If a catalyst were added to the reaction in Figure 4-3, i t would create a transition state of lower energy, thereby lowering the activation energy. Addition of a catalyst would not change the energies of the reactants and products, however, so the heat of reaction and the equilibrium constant would be unaffected. S O LV E D P R OB L E M 4 - 4 Enzymes serve as biological cata lysts. the They speed up reactions substrates) without changing the energies of reactants (called and products. Without enzymes, most of the reactions in our cells would not go fast enough to keep us alive. Consider the following reaction: This reaction has an activation energy (Ea) of + 1 7 kJImol ( +4 kcal/mol ) and a !:!.Ho of +4 kJ/ mol (+ 1 kcal/mol ) . Draw a reaction-energy diagram for this reaction. We draw a diagram that shows the products to be 4 kJ higher i n energy than the reactants. The barrier is made to be 17 kJ higher in energy than the reactants. SOLUTION --- i P R OB L E M 4 - 1 4 E, CH4 + CI +" r - -- ---- ---- --. _ [H3C ---H- -- C1)" transition state kl _ _ _ :k: kJ reaction coordinate __ (a) Draw the reaction-energy diagram for the reverse reaction: (b) What is the activation energy for this reverse reaction? (c) What is the heat of reaction (!:!.HO) for this reverse reaction? CR3 + RCI CR4 + CI 142 Chapter 4: The S tudy of Chemical R e actions PROBLEM 4-1 5 (a) Draw a reaction-energy diagram for the following reaction: The activation energy is 4 kllmol ( 1 kcal/mol), and the overall !J. Ho for the reaction is 1 09 kllmol ( -26 kcal/ mol ) . (b) Give the equation for the reverse reaction. (c) W hat is the activation energy for the reverse reaction? - 4-11 Rates of M u ltiste p Reactions Many reactions proceed by mechanisms involving several steps and several intermedi ates. As we saw in Section 4.7, for example, the reaction of methane with chlorine goes through two propagation steps. The propagation steps are shown here, along with their heats of reaction and their activation energies. Just the propagation steps are shown because the rate of the initiation step is controlled by the amount of light or heat available to split chlorine molecules. CH4 + CI ' 'CH3 + HCI 'CH3 + Cl2 CH3C1 + CI Step - !::. HO (per mole) +4 kJ (+ 1 kcal ) 1 09 kJ ( - 26 kcal ) E a (per mole) l 7 kJ (4 kcal ) 4 kJ ( 1 kcaJ ) I n this reaction, CI and CH3 are reactive intermediates. Unlike transition states, reactive intermediates are stable as long as they do not collide with other atoms or molecules. As free radicals, however, CI and CH3 are quite reactive toward other molecules. Figure 4-4 shows a single reaction-energy profile that includes both propa gation steps of the chlorination. The energy maxima (high points) are the unstable transition states, and the energy minima (low points) are the intermediates. This com plete energy profile provides most of the important information about the energetics of the reaction. I n a multistep reaction, each step has its own characteris tic rate. There can be only one overall reaction rate, however, and it is controlled by the rate-limiting step ( also called the rate-determining step). In general, the highest energy step of a multistep reaction is the "bottleneck," and it determines the overall rate. How can we tell which step is rate limiting? If we have the reaction-energy dia gram, it is simple: The highest point in the energy diagram is the transition state with the highest energy-the transition state for the rate-limiting step. The highest point in the energy diagram of the chlorination of methane (Figure 4-4) is the transition state for the reaction of methane with a chlorine radical. T he Rate- lim iting Step rate-limiting transition state Combined reaction-energy diagram for the chlorination of methane. The energy maxima are transition states, and the energy minima are intermediates. (Units are kllmo\.) Figure 4-4 I . . : ;;:51 . . . .. : C H3 l - ;-w = :a cD C ___ . I .. .. . reaction coordinate -----+- C H3 Cl + Clo W" " - I kl/m o 4- 1 2 Temperature Dependence of Halogenation This step must be rate limiting. If we calculate a rate for this slow step, it will be the rate for the overall reaction. The second, faster step will consume the products of the slow step as fast as they are formed. 143 We now apply what we know about rates to the reaction of methane with halogens. The rate-limiting step for chlorination is the endothermic reaction of the chlorine atom with methane to form a methyl radical and a molecule of HCI. Rate- limiting step 4-1 2 Tem peratu re Dependence of Halogenation C H4 + Cl' ' CH3 + H Cl The activation energy for this step is 17 kJ/mol (4 kcal/mol). At room temperature, the value of e -EJRT is 1 300 X 1 0-6 . This value represents a rate that is fast but controllable. In a free-radical chain reaction, every propagation step must occur quickly, or the free radicals will undergo unproductive collisions and become involved in termination steps. We can predict how quickly the various halogen atoms react with methane given relative rates based on the measured activation energies of the slowest steps: Relative Rate (e-Ea/RT x 1 0 6) Reaction fa (per mole) p. C l' Br' I' + + + + CH4 CH4 CH4 CH4 -- -- --- HF + ' CH3 HCl + 'CH3 HBr + 'CH3 HI + 'CH3 27C(3000K) 227C(5000K) 5 kJ ( 1 .2 kcal) (4 kcal) 7 5 k J ( 1 8 kcal) 1 40 k J ( 34 kcal) 1 7 kJ 1 40,000 1 300 9 X 1 0-8 2 X 1 0- 19 300,000 18,000 0.0 1 5 2 X 1 0-9 These relative rates suggest how easily and quickly methane reacts with the differ ent halogen radicals. The reaction with fluorine should be difficult to control because its rate is very high. Chlorine should react moderately at room temperature, but it may be come difficult to control if the temperature lises much (the rate at SOooK is rather high). The reaction with bromine is very slow, but heating might give an observable rate. Iodi nation is probably out of the question because its rate is exceedingly slow, even at SOOK. Laboratory halogenations show that our predictions are right. In fact, fluorine reacts explosively with methane, and chlorine reacts at a moderate rate. A mixture of bromine and methane must be heated to react, and iodine does not react at all. PRO B L E M 4- 1 6 The bromination of methane proceeds through the following steps: (a) (b) (c) (d) t:.Ho (Ee r mole1 hv Br2 2 B r' + 1 92 k J ( 46 kcal ) C H4 + Br' --3> ' CH3 + HBr 6 7 k J ( 16 kcal ) --3> ' CH3 + Br2 C H3Br + Br' - 1 0 1 k J ( -24 kcal ) Draw a complete reaction-energy diagram for this reaction. Label the rate-limiting step. Draw the structure of each transition state. Compute the overall value of t:.Ho for the bromination. 192 k J Ea (Eer mole1 (46 kcal ) 7 5 kJ ( 1 8 kcal ) 4 k J ( 1 kcal ) iodination of methane. (b) Compute the overall value of t:.Ho for iodination. (c) S uggest two reasons why iodine does not react well with methane. (a) U sing the B DEs in Table 4-2 (p. 1 36), compute the value of t:.Ho for each step in the P R O B L E M 4- 1 7 144 Chapter 4: The Study of Chemical Reactions U p to now, we have limited our discussions to the halogenation of methane. Beginning our study with such a simple compound allowed us to concentrate on the thermody namics and kinetics of the reaction. Now we consider halogenation of the "higher" alkanes, meaning those of higher molecular weight. 4 -13A Chlori nation of Propa ne: Prod uct Ratios 4-1 3 Selectivity in Halogenation Halogenation is a substitution, where a halogen atom replaces a hydrogen. R - H + X2 R-X + H-X In methane, all four hydrogen atoms are identical, and it does not matter which hydro gen is replaced. In the higher alkanes, replacement of different hydrogen atoms leads to different products. In the chlorination of propane, for example, two monochlorinated (just one chlorine atom) products are possible. One has the chlorine atom on a primary carbon atom, and the other has the chlorine atom on the secondary carbon atom. I carbon 2 carbon \ / CH3- CH2-CH3 propane + hv, 2 5C ) Cl I C H2-CH2-CH3 l -chloropropane, 40% (n-propyl chloride) + Cl I CH3-CH - CH3 2-chloropropane, 60% (isopropyl chloride) The product ratio shows that replacement of hydrogen atoms by chlorine is not random. Propane has six primary hydrogens (hydrogens bonded to primary carbons) and only two secondary hydrogens (bonded to the secondary carbon), yet the major product results from substitution of a secondary hydrogen. We can calculate how reac tive each kind of hydrogen is by dividing the amount of product observed by the num ber of hydrogens that can be replaced to give that product. Figure 4-5 shows the definition of primary, secondary, and tertiary hydrogens and the calculation of their relative reactivity. Replacing either of the two secondary hydro gens accounts for 60% of the product, and replacing any of the six primary hydrogens accounts for 40% of the product. We calculate that each secondary hydrogen is 4.5 times primary ( 1 0) hydrogens Six primary ( J O) hydrogens H I R - C -H I H R I R - C -H I H secondary (2) hydrogens R I R - C -H I R tertiary (3) hydrogen relative reactivily ) replacement C H3 -CH2-CH2- Cl primary chloride 40% 6 hydrogens = 6 .67% per H Definitions of primary, secondary, and tertiary hydrogens. There are six primary hydrogens in propane and only two secondary hydrogens, yet the major product results from replacement of a secondary hydrogen. Figure 4-5 Two secondary (2) hydrogens replacemen C12, il v Cl I CH3 -CH- CH 3 secondary chloride 60% 2 hydrogens = 30.0% per H The 2 hydrogens are 6 = 4.5 times as reactive as the 1 hydrogens. 4-13 Selectivity in Halogenation Initiation: S plitting of the chlorine molecule 1 45 Cl2 + hv ------c> 2 C I First propagation step: A bstraction (removal) a of primary or secondary hydrogen pri mary radical secondary radical Second propagation step: Reaction with chlorine to form the alkyl chloride ' CH2- CH2- CH3 p rimary radical + C l2 ------c> C l-CH2- CH2-CH3 primary chloride ( l -chloropropane) + C I or CH3- CH-CH3 s econdary radical + CI2 ------c> CI I CH3-CH- CH3 secondary chloride (2-chl oroprop ane) + C l The mechanism for free-radical chlorination of propane. The r propagation step forms either a primary ust radical or a secondary radical. This radical determines whether the final product will be the primary chloride or the secondary chloride. Figure 4-6 as reactive as each primary hydrogen. To explain this preference for reaction at the sec ondary position, we must look carefully at the reaction mechanism (Figure 4-6). When a chlorine atom reacts with propane, abstraction of a hydrogen atom can give either a primary radical or a secondary radical. The structure of the radical formed in this step determines the structure of the observed product, either l -chloropropane or 2-chloropropane. The product ratio shows that the secondary radical is formed prefer entially. This preference for reaction at the secondary position results from the greater stability of the secondary free radical and the transition state leading to it. P R O B L E M 4- 1 8 What would be the product ratio in the chlorination of propane if all the hydrogens were abstracted at equal rates? P R O B L E M 4- 1 9 Classify each hydrogen atom in the following compounds as primary ( 1 0), secondary (2), or tertiary (30). ( a) butane (c) 2-methylbutane (b) isobutane ( d) cyclohexane (e) norbornane (bicyclo[2.2. 1 Jheptane) 4- 1 3 8 F ree-Rad ical Sta b i lities Figure 4 -7 shows the energy required (the bond-dissociation enthalpy ) to form a free radical by breaking a bond between a hydrogen atom and a carbon atom. This energy is greatest for a methyI carbon, and it decreases for a primary carbon, a secondary carbon, and a tertiary carbon. The more highly substituted the carbon atom, the less energy is required to form the free radical. From the information in Figure 4 -7, w e conclude that free radicals are more sta ble if they are more highly substituted. The following free radicals are listed in de creasing order of stability. 1 46 Chapter 4: The Study of Chemical Reactions Formation of a methyl radical Formation of a primary ( JO) radical /ili0 Bond-dissociation enthalpy = 4 35 kJ ( 1 04 kcal) C H3 - CH2- CH3 H' + CH3 - CH2- CH2 t"W = 4 1 0 kJ (98 kcal) Formation of a secondary (2 ) radical C H3 - CH2 - CH3 Enthalpy required to form a free radical. Bond-dissociation enthalpies show that more highly substituted free radicals are more stable than less highly substituted ones. Figure 4-7 Formation of a tertiary (30) radical H + C H3 - CH-CH3 t"W = 397 kJ (95 kcal) CH3 C H 3 - C -H CH3 I I H- t"W = 38 1 kJ (9 1 kcal) R R-C ' R > I I R > R-C' H I I H > R-C H I I H > H-C' H I I > > Me ' methyl te rti ary > s ec ondary > p rimary > In the chlorination of propane, the secondary hydrogen atom is abstracted more often because the secondary radical and the transition state leading to it are lower in energy than the primary radical and its transition state. Using the bond-dissociation enthalpies in Table 4-2 (page 13 6), w e can calculate D.Ho for each of the possible reaction steps. Abstraction of the secondary hydrogen is 3 kcallmol ( 1 3 kllmol) more exothermic than abstraction of the primary hydrogen. 1 H: CH3- CH2-CH3 + Cl - -;. CH3- CH2- CH2 ' + H -Cl Energy required to break the CH3CH2CH2+ H bond Energy released in forming the H + Cl bond Total energy for reaction at the primary position: + 4 1 0 kllmol ( +98 kcal/mol) -43 1 kllmol ( - 1 03 kcal/mol) - 2 1 kllmol ( - 5 kcal/mol) I Energy required to break the CH3 - CH + H bond Energy released in forming the H + Cl bond Total energy for reaction at the secondary position: C H3 + 3 97 kllmol ( + 95 kcal/mol) -43 1 kllmol ( - 1 03 kcal/mol) - 34 kllmol ( - 8 kcal/mol) A reaction-energy diagram for this rate-limiting first propagation step appears in Figure 4-8. The activation energy to form the secondary radical is slightly lower, so the secondary radical is formed faster than the primary radical. 4- 1 3 S electivity in Halogenation 1 47 1 Ea 2 E-.- -- al >-> CHFH2CH3 + C I - 1----------- I d ifference in activation energies (about 4 kJ) - I radical 1 3 kJ difference CH3<;:HCH3 + HCI 2 radical reaction coordinate - - Reaction-energy diagram for the first propagation step in the chlorination of propane. Formation of the secondary radical has a lower activation energy than does formation of the primary radical. Figure 4-8 S O LV E D P R O B L E M 4 - 5 Tertiary hydrogen atoms react with Cl' about 5 .5 times as fast as primary ones. Predict the product ratios for chlorination of isobutane. S O L UTI O N There are nine primary hydrogens and one tertiary hydrogen in isobutane. n ine primary hydrogens = H3C -I H 1C-C- H ", . o ne tertiary hydrogen I H3C = = (9 primary hydrogens ) X ( reactivity 1 .0) ( 1 tertiary hydrogen) X ( reactivity 5 .5) 9.0 relative amount of reaction 5 .5 relative amount of reaction Even though the primary hydrogens are less reactive, there are so many of them that the primary product is the major product. The product ratio will be 9 .0:5.5, or about 1 .6: 1 . fraction of primary fraction of tertiary = 9 .0 9.0 + 5 .5 5 .5 9.0 + 5 .5 = = 6 2% 3 8% C H2-CI I CH3 -C-H I CH3 m ajor product 62% PRO B L E M 4-20 CH3 I CH3 -C-Cl I CH3 m inor product 38% U se the bond-dissociation enthalpies in Table 4-2 (page 1 36) to calculate the heats of reaction for the two possible first propagation steps in the chlorination of isobutane. Use this informa tion to draw a reaction-energy diagram like Figure 4 -8, comparing the activation energies for formation of the two radicals. PRO B L E M 4-2 1 Predict the ratios of products that result when isopentane (2-methylbutane) is chlorinated. 1 48 Ch apter 4: The Study of Che mic al Reactions ( a) W hen n-heptane burns i n a gasoline engine, the combustion process takes p lace P R O B L E M 4-22 too quickly. The explosive detonation makes a noise called knocking. When 2,2,4-trimethylpentane (isooctane) is burned, combustion takes p lace in a slower, more controlled manner. Combustion is a free-radical chain reaction, and its rate depends on the reactivity of the free-radical intermediates. Explain why isooctane has less tendency to knock than does n-heptane. (b) Alkoxy radicals ( R - 0 ' ) are generally more stable than alkyl ( R' ) r adicals. Write an equation showing an alkyl free radical (from burning gasoline) abstracting a hydrogen atom from t-butyl alcohol, (CH3hCOH. Explain why t-butyl alcohol works as an anti knock additive for gasoline. 4- 1 3C B rom i nation of Propa ne F igure 4-9 shows the free-radical reaction of propane with bromine. Notice that this reaction is both heated to 1 25C and irradiated with light to achieve a moderate rate. The secondary bromide (2-bromopropane) is favored by a 97: 3 product ratio. From this product ratio, we calculate that the two secondary hydrogens are each 97 times as reactive as one of the primary hydrogens. The 97: 1 reactivity ratio for bromination is much larger than the 4.5: 1 ratio for chlorination. We say that bromination is more s elective than chlorination because the maj or reaction is favored by a larger amount. To explain this enhanced selec tivity, we must consider the transition states and activation energies for the rate limiting step. As with chlorination, the rate-limiting step in bromination is the first propagation step: abstraction of a hydrogen atom by a bromine radical. The energetics of the two possible hydrogen abstractions are shown below. Compare these numbers with the energetics of the first propagation step of chlorination shown on page 1 46. The bond dissociation enthalpies are taken from Table 4-2 (page 1 36). 1 H: CH3- CH2-CH3 + Br C H3-CH2-CH2 ' + H - Br Energy required to break the CH3CH2CH2+ H bond Energy released in forming the H + Br bond Total energy for reaction at the primary position: + 4 1 0 kJ/mol ( + 98 kcal/mol) - 368 kJ/mol ( - 88 kcal/mol) +42 kJ/mol (+ 10 kcal/mol) === === === === === === === === === === === === === === ==== -- - -=-- -=-- - - - -= -:: -=-- ::.-: ' :-== - :::: . Br I CH3 - CH2- CH2 primary bromide, 3 % Relative reactivity + Br I CH3 - CH - CH3 secondary bromide, 97% + H Br six primary hydrogens (5 = 0 . 5% per H 3% two secondary hydrogens 97% 2- = 01 48 . 5-0 per H 1 The 2 hydrogens are .... Figure 4-9 ; = 97 times as reactive as the I hydrogens . The free-radical reaction of propane with bromine. This 97 :3 ratio of products shows that bromine abstracts a secondary hydrogen 97 times as rapidly as a primary hydrogen. Bromination (reactivity ratio 97 : 1 ) is much more selective than chlorination (reactivity ratio 4.5: 1 ). 4- 1 4 The H ammond Postulate 1 49 difference in activation energies (over 9 k J) i reaction coordinate ---+ 1 3 kJ di fference ... Figure 4-1 0 Reaction-energy diagram for the first propagation step in the bromination of propane. The energy difference in the transition states is nearly as large as the energy difference in the products. C H3 I Energy required to break the CH3 - CH + H bond Energy released in forming the H + B r bond Total energy for reaction at the secondary position: + 3 97 kJ/mol (+ 9S kcal/mol) - 368 kJ/mol ( - 88 kcallmol) + 29 kJ/mol ( + 7 kcallmol) The energy differences between chlorination and bromination result from the difference in the bond-dissociation en thai pies of H - Cl (43 1 kJ) and H - Br (368 kJ). The HBr bond is weaker, and abstraction of a hydrogen atom by Br' is en dothermic. This endothermic step explains why bromination is much slower than chlorination, but it still does not explain the enhanced selectivity observed with bromination. Consider the reaction-energy diagram for the first propagation step in the bromi nation of propane (Figure 4- 1 0). Although the difference in values of t:.. Ho between abstraction of a primary hydrogen and a secondary hydrogen is still 1 3 kJ/mol (3 kcallmol), the energy diagram for bromination shows a much larger difference in activation energies for abstraction of the primary and secondary hydrogens than we saw for chlorination (Figure 4-8). Figure 4- 1 1 summarizes the energy diagrams for the first propagation steps in the bromination and chlorination of propane. Together, these energy diagrams explain the enhanced selectivity observed in bromination. Two important differences are apparent in the reaction-energy diagrams for the first propagation steps of chlorination and bromination: 2. T he transition states forming the 1 and 2 radicals for the endothermic bromina 4-14 The Ham mond Postulate 1. The first propagation step is endothermic for bromination but exothermic for chlorination. tion have a larger energy difference than those for the exothennic chlorination, 1 50 Chapter 4: The S tu dy of Chemical Reactions a bout 4 kJ about 9 k J CH3CHFH2 + HBr difference difference in Ea 1 1 3 kJ 2 C H3<;:HCH3 + H B r 2 C H3<;:HCH3 + HC1 reaction coordinate reaction coordinate ( b) C HLORINATION exothermic (a) B RO M INATION endothermic TS c lose to products large difference i n Ea The energy diagrams for bromination and chlorination of propane. (a) In the endothermic bromination, the transition states are closer to the products (the radicals) in energy and in structure. The difference in the 1 and 2 activation energies is about 9 kJ (2.2 kcal), nearly the entire energy difference of the radicals. ... Figure 4-1 1 TS close to reactants small difference i n Ea (b) I n the exothermic chlorination, the transition states are closer to the reactants in energy and in structure. The difference in activation energies for chlorination is about 4 kJ (1 kcal), only a third of the energy difference of the radicals. even though the energy difference of the products is the same ( 1 3 kJ, or 3 kcaJ) in both reactions. In general, we will find that these differences are related: In an endothermic reaction, the transition state is closer to the products in energy and in structure. In an exothermic reaction, the transition state is closer to the reactants in energy and in structure. Figure 4- 1 2 compares the transition states for bromination and chlorination. In the product-like transition state for bromination, the C - H bond is nearly broken and the carbon atom has a great deal of radical character. The energy of this transition state reflects most of the energy difference of the radical products. In the reactant-like transition state for chlorination, the C - H bond is just beginning to break, and the Comparison of transition states for bromination and chlorination. In the endothermic bromination, the transition state resembles the products (the free radical and HBr). In the exothermic chlorination, the free radical has just begun to form in the transition state, so the transition state resembles the reactants. Figure 4-1 2 transition state \ close in . :j: I reactants - close in structure '--v-/ - - - - - - }Icr; I products I:> --reactants I energy t transition state :j: -- / close i n structure products '---v-J I 4-14 The H ammond Postulate carbon atom has little radical character. This transition state reflects only a small part (about a third) o f the energy difference of the radical products. Therefore, chlorina tion is less selective. These reactions are examples of a more general principle, called the Hammond postulate. HAMMOND POSTULATE: Related species that are closer in energy are also closer i n struc ture . The stlUcture of a transition state resembles the structure o f t he closest stable species. This general IUle tells us something about the transition states in endothermic and exothermic reactions. The transition state is always the point of highest energy on the energy diagram. Its structure resembles either the reactants or the products, whichever ones are higher in energy. In an endothermic reaction, the products are higher in ener gy, and the transition state is product-like. In an exothermic reaction, the reactants are higher in energy, and the transition state is reactant-like. Thus, the Hammond postulate helps us understand why exothermic processes tend to be less selective than similar endothermic processes. PROBLEM 4-23 1 51 (a) Compute the heats of reaction for abstraction of a primary hydrogen and a secondary hydrogen from propane by a fluorine radical. CH3 - CH 2 - CH3 + CH3 - CH2 -CH 3 + F' F' C H3 - CH 2 - CH2 + HF C H3 - CH - CH3 + HF (b) How selective do you expect free-radical fluorination to be? (c) What product distribution would you expect to obtain from the free-radical fluorination of propane? P R O B L E M - S O LV I N G S T R AT E G Y P ropos i n g React i o n M ech a n isms Throughout this course, we will propose mechanisms to explain reactions. We will discuss methods for dealing with different types of mechanisms as we encounter them. These tech niques for dealing with a variety of mechanisms are collected in Appendix 4 . At this point, however, we focus on free-radical mechanisms like those in this chapter. Free-Radical Reactions General principles: Free-radical reactions generally proceed by chain-reaction mecha nisms, using an initiator with an easily broken bond (such as chlorine, bromine, or a peroxide) to start the chain reaction. In drawing the mechanism, expect free-radical intermediates (especially highly substituted or resonance-stabilized intermediates). Watch for the most stable free radicals, and avoid any high-energy radicals such as hydrogen atoms. 1 . D raw a step that breaks the weak bond in the initiator. 2 . Draw a reaction of the initiator with one of the starting materials. A free-radical reaction usually begins with an initiation step in which the initiator under goes homolytic (free-radical) cleavage to give two radicals. One of the initiator radicals reacts with one of the starting materials to give a free-radical version of the starting material. The initiator might abstract a hydrogen atom or add to a double bond, depending on what reaction leads to formation of the observed product. You might want to consider bond-dissociation enthalpies to see which reaction is energetically favored. ( Continued) 1 52 C hapter 4: The Study of Chemical Reactions 3. Draw a reaction of the free-radical version of the starting material with another 4 . Draw termination step(s). starting-material molecule to form a bond needed in the product and generate a new radical intermediate. Check your intermediates to be sure that you have used the most stable radical intermedi ates. For a realistic chain reaction, no new initiation steps should be required; a radical should be regenerated in each propagation step. The reaction ends with termination steps, which are side reactions rather than part of the product-forming mechanism. The reaction of any two free radicals to give a stable mole cule is a termination step, as is a collision of a free radical with the container. B efore we illustrate this procedure, let's consider a few common mistakes. Avoiding these mistakes will help you to draw correct mechanisms throughout this course. C ommon Mistakes to Avoid the substituents of each carbon atom affected throughout the mechanism. Three-bonded carbon atoms in intermediates are most likely to be radicals in the free-radical reactions we have studied. If you draw condensed formulas or line-angle formulas, you will likely misplace a hydrogen atom and show a reactive species on the wrong carbon. 2 . D o not show more than one step occurring at once, unless they really do occur at once. PROBLEM-SOLVING 1. Do not use condensed or line-angle formulas for reaction sites. Draw all the bonds and all Free-radical bromination is highly HiltZ; S ample Problem selective, chlorination is moderately selective, and fluorination is nearly nonselective. Draw a mechanism for the reaction of methylcyclopentane with bromine under irradiation with light. Predict the major product. CH3 hv In every mechanism problem, we first draw what we know, showing all the bonds and all the substituents of each carbon atom that may be affected throughout the mechanism. H /CH3 H' '..., C H ..., / ...... / C / C\ / '-... H H H-C-C-H H Bf, hv 1 . Draw a step involving cleavage of the weak bond in the initiator. I H I The use of light with bromine suggests a free-radical reaction, with light providing the energy for dissociation of Br2. This homolytic cleavage initiates the chain reaction by generating two Bf" radicals. Initiation step 2 . Draw a reaction of the initiator with one of the starting materials. B r - Br Bf" + 'Br First propagation step One of these initiator radicals should react with methylcyclopentane to give a free-radical version of methylcyclopentane. As we have seen, a bromine or chlorine radical can abstract a hydrogen atom from an alkane to generate an alkyl radical. The bromine radical is highly selective, and the most stable alkyl radical should result. Abstraction of the tertiary hydro gen atom gives a tertiary radical. HBr 3. Draw a reaction of the free-radical version of the starting material with another 4 -15 Rad ical I nhibitors 1 53 starting-material molecule to form a bond needed in the product and to generate a new radical intermediate. The alkyl radical should react with another starting-material molecule, in another propa gation step, to generate a product and another radical. Reaction of the alkyl radical with Br2 gives l -bromo- l -methylcyclopentane (the major product) and another bromine radi cal to continue the chain. Second propagation step C H3 Br H " / C/ ... ,, .... H C / ...... H H H - C- C -H /C \ / + Br 4. Draw termination step(s). H I H I m aior product It is left to you to add some possible termination steps and summarize the mechanism developed here. A s practice in using a systematic approach to proposing mechanisms for free radical reactions, work Problem 4-24 by going through the four steps just outlined. PRO B L E M 4-24 2 ,3-Dimethylbutane reacts with bromine i n the presence of light to give a good yield of a monobrominated product . Further reaction gives a dibrominated product. Predict the structures of these products, and propose a mechanism for the formation of the monobrominated product. PRO B L E M 4-2 5 In the presence of a small amount of bromjne, cyclohexene undergoes the following light promoted reaction: c yclohexene (a) Propose a mechanism for this reaction. (b) Draw the structure of the rate-limjting transition state. (c) U se the Hammond postulate to predict which intermediate most closely resembles this transition state. (d) Explain why cyclohexene reacts with bromine much faster than cyclohexane, which must be heated to react. o + trace Br2 hv rY Br V + H Br 3 -bromocyclohexene We often want to prevent or retard free-radical reactions. For example, oxygen in the air oxidizes and spoils foods, solvents, and other compounds mostly by free-radical chain reactions. Chemical intermediates may decompose or polymerize by free-radi cal chain reactions. Even the cells in living systems are damaged by radical reactions, which can lead to aging, cancerous mutations, or cell death. Radical inhibitors are often added to foods and chemicals to retard spoilage by radical chain reactions. Chain reactions depend on the individual steps being fast, so that each initiation step results in many molecules reacting, as in the 4 -15 Radical Inhibitors 1 54 Chapter 4: The Study of Chemical Reactions r eaction-energy diagram at the left of the following figure. (Only the radicals are shown.) add I B chain reaction (exothermic I with inhibitor I The diagram at right in the figure shows how an inhibitor (I) can stop the chain by reacting with a radical intermediate in a fast, highly exothermic step to form an inter mediate that is relatively stable. The next step in the chain becomes endothermic and very slow. "Butylated hydroxyanisole" (BHA) is often added to foods as an antioxidant. It stops oxidation by reacting with radical intermediates to form a relatively stable free radical intermediate (BHA radical). The B HA radical can react with a second free rad ical to form an even more stable quinone with all its electrons paired. OH RO + or R CH3 HO R H3C CH3 (R = O ROH or RH + r r OCH3 OCH3 BHA BHA radical + or R O R ROCH3 + or RCH3 r 0 a quinone 0 CH3 vitamin E alkyl chain) Radical inhibitors also help to protect the cells of living systems. Like B HA, vitamin E is a phenol (an aromatic ring with an - OH group), and it is thought to react with radicals by losing the OH hydrogen atom as just shown for B HA. Ascorbic acid (vitamin C) is also thought to protect cells from free radicals, possibly by the fol lowing mechanism: RO . or R . + 0 Y H H CH-CH20H or RH ROH OH HO ascorbic acid (vitamin C) stabilized free radical P R O B L E M 4-26 Draw resonance forms to show how the B HA radical i s stabilized b y delocalization of the radical electron over other atoms in the molecule. PRO B L E M 4-27 Write an equation for the reaction of vitamin E with an oxidizing radical ( RO ) to give ROH and a less reactive free radical. 4- 1 6 Reactive Intermediates The free radicals we have studied are one class of reactive intermediates. Reactive inter mediates are short-lived species that are never present in high concentrations because they react as quickly as they are formed. In most cases, reactive intermediates are fragments of molecules (like free radicals), often having atoms with unusual numbers of bonds. Some of the common reactive intermediates contain carbon atoms with only two or three bonds, compared with carbon's four bonds in its stable compounds. Such species react quickly with a variety of compounds to give more stable products with tetravalent carbon atoms. Although reactive intermediates are not stable compounds, they are important to our study of organic chemistry. Most reaction mechanisms involve reactive inter mediates. If you are to understand these mechanisms and propose mechanisms of your own, you need to know how reactive intermediates are formed and how they are likely to react. In this chapter, we consider their structure and stability, and in later chapters, we see how they are formed and ways they react to give stable compounds. Species with trivalent (three-bonded) carbon are classified according to their charge, which depends on the number of nonbonding electrons. The carbocations have no nonbonding electrons and are positively charged. The radicals have one non bonding electron and are neutral. The carbanions h ave a pair of nonbonding electrons and are negatively charged. H H -C+ H carbocation 1 55 4 -16 Reactive Intermed iates I I H H-C ' H radical I I H H-C : H carbanion I I H H '" C: / carbene The most common intermediates with a divalent (two-bonded) carbon atom are the carbenes. A carbene has two non bonding electrons on the divalent carbon atom, making it uncharged. 4- 1 6A Car bo catio ns A carbocation (also called a carbonium ion or a carbenium ion) is a species that contains a carbon atom bearing a positive charge. The positively charged carbon atom is bonded to three other atoms, and it has no nonbonding electrons, so it has only six electrons in its va 2 lence shell. It is sp hybridized, with a planar structure and bond angles of about 1 20. For example, the methyl cation ( CHt) is planar, with bond angles of exactly 1 20. The unhy bridized p orbital is vacant and lies perpendicular to the plane of the C - H bonds (Figure 4- 1 3). The structure of CHt is similar to the structure of BH3, discussed in Chapter 2. With only six electrons in the positive carbon's valence shell, a carbocation is a powerful electrophile (Lewis acid), and it may react with any nucleophile it encoun ters. Carbocations are proposed as intermediates in many organic reactions, some of which we will encounter in Chapter 6 . Like free radicals, carbocations are electron-deficient species: They have fewer than eight electrons in the valence shell. Also like free radicals, carbocations are stabilized by vacant orbital p ... Figure 4-1 3 top view side view Orbital diagram of the methyl cation. The methyl cation is similar to BH3. The carbon atom is (J bonded to three 2 hydrogen atoms by overlap of its sp hybrid orbitals with the s orbitals of hydrogen. A vacant p orbital lies perpendicular to the plane of the three C - H bonds. 1 56 C hapter 4 : The S tudy of Chemical Reactions alkyl substituents. An alkyl group stabilizes an electron-deficient carbocation in two ways: (1) through an inductive effect, and (2) through the partial overlap of filled orbitals with empty ones. The inductive effect is a donation of electron density through the sigma ( (J" ) bonds of the molecule. The positively charged carbon atom withdraws some electron density from the polarizable alkyl groups bonded to it. Alkyl substituents also have filled sp 3 orbitals that can overlap with the empty p orbital on the positively charged carbon atom, further stabilizing the carbocation (Figure 4-14). Even though the attached alkyl group rotates, one of its sigma bonds is always aligned with the empty p orbital on the carbocation. The pair of electrons in this (J" bond spreads out into the empty p orbital, stabilizing the electron-deficient carbon atom. This type of overlap between a p orbital and a sigma bond is called hyperconjugation. In general, more highly substituted carbocations are more stable. R I R -C+ I R most stable Stability of carbocations > p vacant orbital \ R I R-C+ I H > H I R -C+ I H > H I H -C+ I H least stable carbocation A Figure 4-1 4 alkyl group 3 > 2 > 1 > methyl Effect of alkyl substituent on carbocation stability. A carbocation is stabilized by overlap of filled orbitals on an adjacent alkyl group with the vacant p orbital of the carbocation. Overlap between a u bond and a p orbital is called hyperconjugation. Unsaturated carbocations are also stabilized by resonance stabilization. I f a pi ( 1T ) bond is adjacent to a carbocation, the filled p orbitals of the 1T bond will over lap with the empty p orbital of the carbocation. The result is a delocalized ion, with the positive charge shared by two atoms. Resonance delocalization is particularly effective in stabilizing carbocations. Carbocations are common intermediates in organic reactions. Highly substituted alkyl halides can ionize when they are heated in a polar solvent. The strongly elec trophilic carbocation reacts with any available nucleophile, often the solvent. C H3 I .. H3 C - C \ B r: I '-"' " C H3 P R O B L E M 4-28 C H3 I H3 C -C+ :Br: I .. CH3 .. C H3cH?qr-I Y E - ) The triphenylmethyl cation is so stable that some of its salts can be stored for months. Explain why this cation is so stable. triphenylmethyl cation 4- 1 6 Reactive Intermediates PRO B L E M 4-29 1 57 Rank the following carbocations i n decreasing order of stability. Classify each as primary, secondary, or tertiary. + (a) The isopentyl cation, ( CH3 hCHCH 2 - CH2 (b) The 3-methyl-2-butyl cation, C H3 - CH - CH(CH3h + (c) The 2-methyl-2-butyl cation, C H3 - C( CH3)CH2CH3 + 4- 1 68 F ree Rad icals Vitamin B12 is an essential dietary factor and a deficiency results in anemia and neurological damage. The vitamin assists two different enzymes in the production and the stabilization These of methyl radicals. then methyl radicals are Like carbocations, free radicals are sp 2 hybridized and planar (or nearly planar). Unlike carbocations, however, the p orbital perpendicular to the plane of the C - H bonds of the radical is not empty; it contains the odd electron. Figure 4- 1 5 shows the structure of the methyl radical. Both radicals and carbocations are electron deficient because they lack an octet around the carbon atom. Like carbocations, radicals are stabilized by the elec tron-donating effect of alkyl groups, making more highly substituted radicals more stable. This effect is confirmed by the bond-dissociation enthaIpies shown in Figure 4-7: Less energy is required to break a C- H bond to form a more highly substituted radical. R I R-C' I R most stable Stability of radicals 3 > > used for the synthesis of important cellular components. R I R-C' I H > H I R -C ' I H > H I H-C' I H least stable 2 > 1 > methyl Like carbocations, radicals can be stabilized by resonance. Overlap with the p orbitals of a 1T bond allows the odd electron to be delocalized over two carbon atoms. Resonance delocalization is particularly effective in stabilizing a radical. H I H CH3 C " '-.... / C C I I H H ,, / / H H I C CH3 or H 'C I H C I H fj. / """' / . ' ' fj H I C CH3 C I H C I H p 1IIi- +- orbital odd electron ..... Figure 4- 1 5 top view side view Orbital diagram of the methyl radical. The structure of the methyl radical is like that of the methyl cation (Figure 4- 1 3), except there is an additional electron. The odd electron is in the p orbital perpendicular to the plane of the three C-H b onds. 1 58 C hapter 4: The S tudy of Chemical Reactions P R O B L E M 4-30 Rank the following radicals i n decreasing order of stability. Classify each as primary, second ary, or tertiary. (a) The isopentyl radical, ( CH3hCHCH2 - CH2 (b) The 3 -methyl-2-butyl radical, C H3 - CH - CH(CH3h ( c) The 2-methyl-2-butyl radical, C H3 - C(CH3)CH 2 CH3 4 -16C Ca rbanions A carbanion h as a trivalent carbon atom that bears a negative charge. There are eight electrons around the carbon atom (three bonds and one lone pair), so it is not electron deficient; rather, it is electron rich and a strong nucleophile (Lewis base). A carbanion has the same electronic structure as an amine. Compare the structures of a methyl carbanion and ammonia: H I H-C:I H methyl anion H-N: methyl a nion H I I H a mmonia ammonia .. Figure 4- 1 6 The hybridization and bond angles of a simple carbanion also resemble those o f 3 an amjne. The carbon atom is sp hybridized and tetrahedral. One of the tetrahedral positions is occupied by an unshared lone pair of electrons. Figure 4- 1 6 compares the orbital structures and geometry of ammonia and the methyl anion. Like amines, carbanions are nucleophilic and basic. A carbanion has a negative charge on its carbon atom, however, making it a more powerful base and a stronger nucleophile than an amine. For example, a carbanion is sufficiently basic to remove a proton from ammonia. Comparison of orbital structures of methyl anion and ammonia. Both the methyl anion and ammonia have an 3 sp hybridized central atom, with a nonbonding pair of electrons occupying one of the tetrahedral positions. '. 0 Carbanions that occur as intermediates in organic reactions are almost always bonded to stabilizing groups. They can be stabilized either by inductive effects or by resonance. For example, halogen atoms are electron withdrawing, so they stabilize carbanions through the inductive withdrawal of electron density. Resonance also plays an important role in stabilizing carbanions. A carbonyl group (C 0) stabilizes an adj acent carbanion by overlap of its 1T bond with the nonbonding electrons of the carb anion. The negative charge is delocalized onto the electronegative oxygen atom of the carbonyl group. = C - C -H / IV H H H I . + - : OH lo H / C- '" H :0: . - P2 '" / C=C / H H resonance-stabilized carbanion '" H H 1 + Hp 2 This resonance-stabilized carbanion must be sp h ybridized and planar for effective delocalization of the negative charge onto oxygen (Section 2-6). Resonance stabilized carbanions are the most common type of carbanions we will encounter in organic reactions. 4 - 1 6 Reactive Intermediates PRO B L E M 4-3 1 1 59 Acetylacetone (2,4-pentanedione) reacts with sodium hydroxide to give water and the sodium salt of a carbanion. Write a complete structural formula for the carbanion, and use resonance forms to show the stabilization of the carbanion. o H 3C - C -CH2-C - CH3 acety lacetone (2,4-pentanedione) PRO B L E M 4-32 II 0 II Acetonitrile ( CH3 C = N ) i s deprotonated by very strong bases. Write resonance forms to show the stabilization of the carbanion that results. 4 -1 6 D C arbenes Carbenes are uncharged reactive intermediates containing a divalent carbon atom. The simplest carbene has the formula : CH2 and is called methylene, j ust as a - CH 2 group i n a molecule i s called a methylene group. One way of generating carbenes i s to form a carbanion that can expel a halide ion. For example, a strong base can abstract a proton from tribromomethane ( CHBr3 ) to give an inductively stabilized carbanjon. This carbanion expels bromide ion to give dibromocarbene. Br 1 "1 Y\ Br- C - H -OH I Br tribromomethane H2O + Br-C : I Br a carbanion B r/ Br" C: + Be dibromocarbene The electronic structure of dibromocarbene is shown next. The carbon atom has only six electrons in its valence shell. It is sp 2 hybridized, with trigonal geometry. An unshared pair of electrons occupies one of the sp 2 hybrid orbitals, and there is an empty p orbital extending above and below the plane of the atoms. A carbene has both a lone pair of electrons and an empty p orbital, so it can react as a nucleophile or as an electrophile. empty p o rbital Methylene itself is formed when diazomethane (CH2N2) is heated or irradiated with light. The diazomethane molecule splits to form a stable nitrogen molecule and the very reactive carbene. H/ -:\) / G u Br " """C Bf ,, ; n onbonding e l ectrons p ai,;d in this sp- orbItal hybr'd'"d H, '- + ' C N- N : l,( d iazomethane heat or light ) methylene nitrogen The most common synthetic reaction of carbenes is their addition to double bonds to form cyclopropane rings. For example, dibromocarbene adds to cyclohexene to give an interesting bicyclic compound. 1 60 Chapter 4: The Study of Chemical Reactions / '" Br :C cyclohexene -----c'> Br dibromocarbene N o simple carbenes have never been purified or even made in a high concentra tion, because when two carbenes c ol l i de, th ey immediately dimerize (two of them bond together) to give an alkene. R ".. C : / R O C / "' Br Br + : C ".. / R R very fast ) Carbenes and carbenoids (carbene-like reagents) are useful both for the synthe sis of other compounds and for the investigation of reaction mechanisms. The carbene intermediate is generated in the presence of its target compound, so that it can react immediately, and the concentration of the carbene is always low. Reactions using car benes are discussed in Chapter 8. PRO B L E M 4-33 When i t i s strongly heated, ethyl diazoacetate decomposes to give nitrogen gas and a carbene. Draw a Lewis structure of the carbene. + : N =N o - CH - C II - O - C CH 3 ethyl diazoacetate (I S U M MARY Reactive Intermediates carbocations radicals carbanions carbenes Structure Stability Properties - C+ I I I I 3 > 2 > 1 > + CH3 3 > 2 > 1 > ' CH3 electrophilic strong acids electron deficient nucleophilic strong bases both nucleophilic and electrophilic I -C, C . - I- I - : CH3 > 1 > 2 > 3 "" C: ../ I C hapte r 4 G l ossa ry activation energy ( Ea) T he energy difference between the reactants and the transition state; the minimum energy the reactants must have for the reaction to occur. (p. 1 39) bond-dissociation enthalpy (BDE) T he amount of enthalpy required to break a particular bond homolytically, to give radicals. (p. 1 34) A :B ----> A' + B' carbanion A strongly nucleophilic species with a negatively charged carbon atom having only three bonds. The carbon atom has a nonbonding pair of electrons. (p. 1 58) carbene A h ighly reactive species with only two bonds to an u ncharged carbon atom and with a nonbonding pair of electrons. The simplest carbene is methylene, : CH 2 . (p. 1 59) Chapter 4 G lossary carbocation (carbonium ion, carbenium ion) A strongly electrophilic species with a posi tively charged carbon atom having only three bonds. (p. 1 55) catalyst A substance that increases the rate of a reaction (by lowering Ea) without being con sumed in the reaction. (p. 1 4 1 ) chain reaction A m ultistep reaction where a reactive intermediate formed in one step brings about a second step that generates the intermediate needed for the following step. (p. 1 26) initiation step: T he preliminary step in a chain reaction, where the reactive intermediate is first formed. propagation steps: The steps in a chain reaction that are repeated over and over to form the product. The sum of the propagation steps should give the net reaction. termination steps: Any steps where a reactive intermediate is consumed without another one being generated. enthalpy (heat content; H) A measure of the heat energy in a system. In a reaction, the heat absorbed or evolved is called the heat of reaction, t:.Ho. A decrease in enthalpy (negative t:. HO) i s favorable for a reaction. (p. 1 33) endothermic: C onsuming heat (having a positive t:. HO). exothermic: G iving off heat (having a negative t:. HO). entropy (S) A measure of disorder or freedom of motion. An increase in entropy (positive t:. SO) is favorable for a reaction. (p. 1 33) equilibrium A state of a system such that no more change is taking place; the rate of the for ward reaction equals the rate of the reverse reaction. (p. 1 3 1 ) equilibrium constant A quantity calculated from the relative amounts of the products and reactants present at equilibrium. (p. 1 3 1 ) For the reaction aA + bB cC + dD 1 61 the equilibrium constant is free energy (Gibbs free energy; G) A measure of a reaction's tendency to go in the direction written. A decrease in free energy (negative t:. G) is favorable for a reaction. (p. 1 3 1 ) Free-energy change i s defined: t:. G = t:. H - Tt:.S standard Gibbs free energy change: ( t:. GO ) The free-energy change corresponding to reac tants and products in their standard states (pure substances in their most stable states) at 2 5C a nd 1 atm pressure. t:. Go i s related to K eq b y K eq = e -!:J.Go/RT ( p. 1 3 1 ) Hammond postulate Related species (on a reaction-energy diagram) that are closer in energy are also closer in structure. In an exothermic reaction, the transition state is closer to the reac tants in energy and in structure. In an endothermic reaction, the transition state is closer to the products in energy and in structure. (p. 1 49) heterolytic cleavage (ionic cleavage) T he breaking of a bond in such a way that one of the atoms retains both of the bond's electrons. A heterolytic cleavage forms two ions. (p. 1 34) homolytic cleavage (radical cleavage) The breaking of a bond in such a way that each atom retains one of the bond's two electrons. A homolytic cleavage produces two radicals. (p. 1 34) A + B inductive effect A donation (or withdrawal) of electron density through sigma bonds. (p. 1 56) intermediate A molecule or a fragment of a molecule that is formed in a reaction and exists for a finite length of time before it reacts in the next step. An intermediate corresponds to a rel ative minimum (a low point) in the reaction-energy diagram. (p. 1 40) reactive intermediate: A short-lived species that is never present in high concentration because it reacts as quickly as it is formed. (p. 1 55) 1 62 C hapter 4: The Study of Chemical Reactions kinetics T he study of reaction rates. (p. 1 37) mechanism The step-by-step pathway from reactants to products showing which bonds break and which bonds form in what order. The mechanism should include the structures of all inter mediates and arrows to show the movement of electrons. (p. 1 26) potential-energy diagram See reaction-energy diagram. ( p. 1 4 1 ) radical (free radical) A highly reactive species i n which one o f the atoms has an odd number of electrons. Most commonly, a radical contains a carbon atom with three bonds and an "odd" (unpaired) electron. (p. 1 57) radical inhibitor A compound added to prevent the propagation of free-radical chain reactions. In most cases, the inhibitor reacts to form a radical that is too stable to propagate the chain. (p. 1 53) rate equation (rate law) T he relationship between the concentrations of the reagents and the observed reaction rate. (p. 1 37) A general rate law for the reaction A + B C + D is rate = k r[A]O[B] b kinetic order: The power of a concentration term in the rate equation. The preceding rate equation is ath order in [ A], bth order in [ B], and a+bth order overall . rate constant: The proportionality constant k r i n the rate equation. rate-limiting step (rate-determining step) The slowest step in a multistep sequence of reactions. In general, the rate-limiting step is the step with the highest-energy transition state. (p. 1 42) rate of a reaction T he amount of product formed or reactant consumed per unit of time. (p. 1 37) reaction-energy diagram (potential-energy diagram) A plot of potential-energy changes as the reactants are converted to products. The vertical axis is potential energy (usually free ener gy, but occasionally enthalpy). The horizontal axis is the reaction coordinate, a measure of the progress of the reaction. (p. 1 4 1 ) :t:/ transition state products reaction coordinate resonance stabilization S tabilization that takes place by delocalization of electrons in a pi bonded system. Cations, radicals, and anions are often stabilized by resonance delocaliza tion. (p. 1 56) substitution A reaction in which one atom replaces another, usually as a substituent on a car bon atom. (p. 1 27) thermodynamics The study of the energy changes accompanying chemical transformations. Thermodynamics is generally concerned with systems at equilibrium. (p. 1 30) transition state (activated complex) The state of highest energy between reactants and prod ucts. A relative maximum (high point) on the reaction-energy diagram. (p. 1 40) Study Problems 1 63 I 3. 2. 1. 4. E ssential Problem-Solvi ng Skills i n Chapter 4 E xplain the mechanism and energetics of the free-radjcal halogenation of alkanes. Based on the selectivity of halogenation, predict the products of halogenation of an alkane. C alculate free-energy changes from equilibrium constants. C alculate enthalpy changes from bond-dissociation enthalpies. Determine the order of a reaction, and suggest a possible mechanism based on its rate equation. Use energy diagrams to discuss transition states, activation energies, intermediates, and the rate-determining step of a multistep reaction. Use the Hammond postulate to predict whether a transition state will be reactant-like or product-like. Describe the structures of carbocations, carbanions, free radicals, and catbenes and the structural features that stabilize them. Explain which are electrophilic and which are nucleophilic. 5. 6. 7. 8. S tudy Problems 4-34 4 -35 Define and give an example for each term. (a) homolytic cleavage ( b) heterolytic cleavage (d) carbocation (e) carbanion ( g) carbonium ion ( h) i ntermediate (k) rate equation (j) transition state (m) r ate constant ( 0) reaction mechanism (p) substitution reaction (q) activation energy (s) rate-limiting step (t) Hammond postulate Consider the following reaction-energy diagram. (c) free radical carbene (i) catalyst ( I) e quilibrium constant ( 0) chain reaction (r) bond-dissociation enthalpy (u) resonance stabilization (f) 4-36 4 -37 4 -38 Label the reactants and the products. Label the activation energy for the first step and the second step. Is the overall reaction endothermic or exothermic? What is the sign of !:J.HO? W hich points in the curve correspond to intermediates? Which correspond to transition states? Label the transition state of the rate-limiting step. Does its structure resemble the reactants, the products, or an intermediate? Draw a reaction-diagram profile for a one-step exothermic reaction. Label the parts that represent the reactants, products, transition state, activation energy, and heat of reaction. D raw a reaction-energy diagram for a two-step endothermic reaction with a rate-limiting second step. Treatment of t-butyl alcohol with concentrated HCI gives t-butyl chloride. CH3 CH - C - OH 3 (a) ( b) (c) (d) I I C H3 + W+ ClC H3 - C - Cl CH3 I C H3 I + H 20 I-butyl alcohol I-butyl chloride When the concentration of H + is doubled, the reaction rate doubles. When the concentration of t-butyl alcohol is tripled, the reaction rate triples. When the chloride ion concentration is quadrupled, however, the reaction rate is unchanged. Write the rate equation for this reaction. 1 64 4-39 C hapter 4: The Study of Chemical Reactions (a) C H3CH2CH(CH3h Label each hydrogen atom in the following compounds as primary ( 1 0), secondary (2), or tertiary (3). ( b) ( CH3 hCCH2C(CH3h d) (c) 4-40 4 -41 Use bond-dissociation enthalpies (Table 4-2, p. 1 36) to calculate values of !1Ho for the following reactions. (a) C H3 - CH 3 + 1 2 -- CH3CH2I + H I (b) C H3CH2CI + H I -- CH3CH2I + H Cl (c) ( CH3 h C - OH + H Cl -- (CH 3 h C - CI + H 20 ( d) C H3CH2CH3 + H 2 -- CH3CH3 + CH4 (e) C H3CH20H + H Br -- CH3CH2-Br + H 20 Use the information in Table 4-2 (p. l 36) to rank the following radicals in decreasing order of stability. O CH3 ( cb 4-42 4-43 For each of the following alkanes, 1 . D raw all the possible monochlorinated derivatives. 2 . D etermine whether free-radical chlorination would be a good way to make any of these monochlorinated derivatives. 3 . W hich monobrominated derivatives could you form in good yield by free-radical bromination? (a) cyclopentane (b) methylcyclopentane d) 2 ,2,3,3-tetramethylbutane (c) 2 ,3-dimethylbutane ( Write a mechanism for the light-initiated reaction of cyclohexane with chlorine to give chlorocyclohexane. Label the initi ation and propagation steps. hv -----7 c yclohex ane chlorocyclohexane 4-44 O CI + HCl Draw the important resonance structures of the following free radicals. ( a) C H2 = CH - CH 2 (b) < /- 0 CH2 (c) C H3-C-O II (d) * 4-45 In the presence of a small amount of bromine, the following light-promoted reaction has been observed. 0 (e) H 3C -U C H3 + V hv 0 H 3C C H3 Br + (Hint: Notice which H atom has been lost in both products.) ( b) E xplain why only this hydrogen atom has been replaced, in preference to any of the other hydrogen atoms in the starting material. (a) Write a mechanism for this reaction. Your mechanism should explain how both products are formed. S tudy Problems 4-46 1 65 For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed. ( a) cyclohexane ( b) methylcyclopentane (c) 2 ,2,3-trimethylbutane (0 hexane (e) 3 -methyloctane (d) decalin 4-47 4-48 4-49 When exactly I mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorina tion reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane. (a) Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the formation of these compounds from chloromethane. (b) How would you run this reaction to get a good conversion of methane to CH3Cl? Methane to CCI4? The chlorination of pentane gives a mixture of three monochlorinated products. (a) Draw their structures. ( b) Predict the ratios in which these monochlorination products will be formed, remembering that a chlorine atom abstracts a secondary hydrogen about 4.5 times as fast as it abstracts a primary hydrogen. (a) Draw the structure of the transition state for the second propagation step in the chlorination of methane. ethylbenzene 4-50 4-51 " '4-52 'CH3 + C l 2 C H3Cl + C l' Show whether the transition state is product-like or reactant-like, and which of the two partial bonds is stronger. (b) Repeat for the second propagation step in the bromination of methane. Peroxides are often added to free-radical reactions as initiators because the oxygen-oxygen bond is homolytically cleaved rather easily. For example, the bond-dissociation enthalpy of the 0 - 0 bond in hydrogen peroxide ( H - 0 - 0 - H) is only 2 1 2 kllmol (5 1 kcal/mol). Give a mechanism for the hydrogen peroxide-initiated reaction of cyclopentane with chlorine. When dichloromethane is treated with strong NaOH, an intermediate is generated that reacts like a carbene. Draw the structure of this reactive intermediate, and propose a mechanism for its formation. When ethene is treated in a calorimeter with H2 and a Pt catalyst, the heat of reaction is found to be - 1 37 kllmol ( - 32.7 kcal/mol ) , and the reaction goes to completion. When the reaction takes place at 1400 K, the equilibrium is found to be evenly balanced, with Keq = l . Compute the value of D.. S for this reaction. CH 2 - CH 2 + H 2 Pl catalyst +--- " '4-53 l D.. H = - 1 37 kllmol *4-54 When a small amount of iodine is added to a mixture of chlorine and methane, it prevents chlorination from occurring. Therefore, iodine is a free-radical inhibitor for this reaction. Calculate D.. Ho values for the possible reactions of iodine with species present in the chlorination of methane, and use these values to explain why iodine inhibits the reaction. (The I - Cl bond-dissociation enthalpy is 2 1 1 kllmol or 50 kcal/mol.) Tributyltin hydride (Bu3SnH ) is used synthetically to reduce alkyl halides, replacing a halogen atom with hydrogen. Free radical initiators promote this reaction, and free-radical inhibitors are known to slow it or stop it. Your job is to develop a mechanism, using the following reaction as the example. trace Br2, h. v + + The following bond-dissociation enthalpies may be helpful: o H 3 97 kJ/moi o B r -Br Br 2 85 kJ/moi H -B r B U3Sn -H BU3Sn - Br 1 92 kJ/mol 368 kJ/mo! 3 10 kJ/mo! 552 kJ/mo! (a) Propose initiation and propagation steps to account for this reaction. ( b) C alculate values of D.. H for your proposed steps to show that they are energetically feasible. (Hint: A trace of Br2 and light suggests it's there only as an initiator, to create Br' radicals. Then decide which atom can be abstracted most favorably from the starting materials by the Br' radical. That should complete the initiation. Now decide what energetically favored propagation steps will accomplish the reaction.) 1 66 * 4-55 C hapter 4: The Study of Chemical Reactions When healthy, Earth's stratosphere contains a low concentration of ozone ( 03 ) t hat absorbs poten tially harmful ultraviolet (UV) radiation by the cycle shown at right. C hlorofluorocarbon refrigerants, such as Freon 1 2 ( CF2 CI 2 ) , are stable in the lower atmos phere, but in the stratosphere, they absorb high-energy UV radiation to generate chlorine radicals. The presence of a small number of chlorine radicals appears to lower ozone concentrations dramati- heat c ally. The following reactions are all known to be exothermic (except the one requiring light) and to have high rate constants. Propose two mechanisms to explain how a small number of chlorine radicals can destroy large numbers of ozone molecules. Which of the two mechanisms is more likely when the concentration of chlorine atoms is very smaIl? CI - O - O - CI * 4-56 hv -----,> + o hv CI - O ' 2 CI - 0' +0 O 2 + CI' C I - O - O - Cl Deuterium (D) is the hydrogen isotope of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C - D bond is slightly stronger than the C - H bond by 5 . 0 kllmol ( 1 2 kcal/mol). Reaction rates tend to be slower if a C - D bond (as opposed to a C - H bond) is broken in a rate- limiting step. This effect, called a kinetic isotope effect, is clearly seen in the chlorination of methane. Methane undergoes free radical chlorination 1 2 times as fast as tetradeuteriomethane ( CD4 ) . Faster: Slower: CH4 + CI ' CD4 + CI' C H3Cl + HCI C H3CI + DCI relative rate relative rate = = 12 (a) Draw the transition state for the rate-limiting step of each of these reactions, showing how a bond to hydrogen or deuterium is being broken in this step. ( b) Monochlorination of deuterioethane ( C2HsD ) leads to a mixture containing 93% C2H4DCl and 7% C 2 HsCI. Calculate the relative rates of abstraction per hydrogen and deuterium in the chlorination of ethane. (c) Consider the thermodynamics of the chlorination of methane and the chlorination of ethane, and use the Hammond postulate to explain why one of these reactions has a much larger isotope effect than the other. ... View Full Document

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