acetone CH 3 I + Cl _ methyl iodide (reacts rapidly) 3 C

Transcription

1 82 CHAPTER 9 THE CHEMSTRY OF ALKYL HALDES Some equilibria that are not too unfavorable can be driven to completion by applying Le Châtelier s principle (Sec. 4.9, p. 171). For example, alkyl chlorides normally do not react to completion with iodide ion because iodide is a weaker base than chloride; the equilibrium favors the formation of the weaker base, iodide. However, in the solvent acetone, it happens that potassium iodide is relatively soluble and potassium chloride is relatively insoluble. Thus, when an alkyl chloride reacts with K in acetone, KCl precipitates, and the equilibrium compensates for the loss of KCl by forming more of it, along with more alkyl iodide. R L Cl + K R L + KCl (precipitates) acetone (9.1) PROLEM 9.4 Tell whether each of the following reactions favors reactants or products at equilibrium. (Assume that all reactants and products are soluble.) (a) Cl + F (b) Cl + (c) Cl + N F + Cl + Cl N + Cl (Hint: the pk a of HN is 4.72.) (d) Cl + O O + Cl 9. REACTON RATES The previous section showed how to determine whether the equilibrium for a nucleophilic substitution reaction is favorable. Knowledge of the equilibrium constant for a reaction provides no information about the rate at which the reaction takes place (Sec. 4.8A). Although some substitution reactions with favorable equilibria proceed rapidly, others proceed slowly. For example, the reaction of methyl iodide with cyanide is a relatively fast reaction, whereas the reaction of cyanide with neopentyl iodide is so slow that it is virtually useless: C' N + H CL H C LC ' N + methyl iodide (reacts rapidly) (9.14) C' N + H H LC LCH L C ' N + C LC LCH 2 L C 2 neopentyl iodide (reacts very slowly) Why do reactions that are so similar conceptually differ so drastically in their rates? n other words, what determines the reactivity of a given alkyl halide in a nucleophilic substitution reaction? ecause this question deals with reaction rates and the concept of the transition state, you should review the introduction to reaction rates and transition-state theory in Sec A. Definition of Reaction Rate 1 1 (9.15) The term rate implies that something is changing with time. For example, in the rate of travel, or the velocity, of a car, the car s position is the something that is changing. change in position velocity = y = (9.16) 1 1

2 9. REACTON RATES 8 The quantities that change with time in a chemical reaction are the concentrations of the reactants and products. reaction rate = (9.17a) change in reactant concentration =- (9.17b) (The signs in Eqs. 9.17a and 9.17b differ because the concentrations of the reactants decrease with time, and the concentrations of the products increase.) n physics, a rate has the dimensions of length per unit time, such as meters per second. A reaction rate, by analogy, has the dimensions of concentration per unit time. When the concentration unit is the mole per liter (M), and if time is measured in seconds, then the unit of reaction rate is. The Rate Law change in product concentration concentration = mol 1 L = M s 1 (9.18) time s For molecules to react with one another, they must get together, or collide. ecause molecules at high concentration are more likely to collide than molecules at low concentration, the rate of a reaction is a function of the concentrations of the reactants. The mathematical statement of how a reaction rate depends on concentration is called the rate law. A rate law is determined experimentally by varying the concentration of each reactant (including any catalysts) independently and measuring the resulting effect on the rate. Each reaction has its own characteristic rate law. For example, suppose that for the reaction A + LT C the reaction rate doubles if either [A] or [] is doubled and increases by a factor of four if both [A] and [] are doubled. The rate law for this reaction is then rate = k[a][] (9.19) f in another reaction D + E LT F, the rate doubles only if the concentration of D is doubled, and changing the concentration of E has no effect, the rate law is then rate = k[d] (9.20) The concentrations in the rate law are the concentrations of reactants at any time during the reaction, and the rate is the velocity of the reaction at that same time. The constant of proportionality, k, is called the rate constant. n general, the rate constant is different for every reaction, and it is a fundamental physical constant for a given reaction under particular conditions of temperature, pressure, solvent, and so on. As Eqs and 9.20 show, the rate constant is numerically equal to the rate of the reaction when all reactants are present under the standard conditions of 1 M concentration. The rates of two reactions are compared by comparing their rate constants. An important aspect of a reaction is its kinetic order. The overall kinetic order for a reaction is the sum of the powers of all the concentrations in the rate law. For a reaction described by the rate law in Eq. 9.19, the overall kinetic order is two; the reaction described by this rate law is said to be a second-order reaction. The overall kinetic order of a reaction having the rate law in Eq is one; such a reaction is thus a first-order reaction. The kinetic order in each reactant is the power to which its concentration is raised in the rate law. Thus, the reaction described by the rate law in Eq is said to be first order in each reactant. A reaction with the rate law in Eq is first order in D and zero order in E.

3 84 CHAPTER 9 THE CHEMSTRY OF ALKYL HALDES The units of the rate constant depend on the kinetic order of the reaction. With concentrations in moles per liter, and time in seconds, the rate of any reaction has the units of M s 1 (see Eq. 9.18). For a second-order reaction, then, dimensional consistency requires that the rate constant have the units of M 1 s 1. rate = k A Further Exploration 9.1 Reaction Rates M s 1 = M 1 s 1 M M Similarly, the rate constant for a first-order reaction has units of s 1. (9.21) Further Exploration 9.2 Absolute Rate Theory C. Relationship of the Rate Constant to the Standard Free Energy of Activation According to transition-state theory, which was discussed in Sec. 4.8, the standard free energy of activation, or energy barrier, determines the rate of a reaction under standard conditions. n Sec. 9. we showed that the rate constant is numerically equal to the reaction rate under standard conditions that is, when the concentrations of all the reactants are 1 M. t follows, then, that the rate constant is related to the standard free energy of activation DG. f DG is large for a reaction, the reaction is relatively slow, and the rate constant is small. f DG is small, the reaction is relatively fast, and the rate constant is large. This relationship is shown in Fig Table 9.2 illustrates the quantitative relationship between the rate constant and the standard free energy of activation. (For the derivation these numbers, see Further Exploration 9.2.) Table 9.2 also translates these numbers into practical terms by giving the time required for the completion of a reaction with a given rate constant. This time is approximately 7Ük. (This is also justified in Further Exploration 9.2.) We ll most often be interested in the relative rates of two reactions. That is, we ll be comparing the rate of a reaction to that of a standard reaction. A relative rate is defined as the ratio of two rates. The relationship between the relative rate of two reactions A and under standard DG8 DG8 larger DG8 slower reaction smaller k reaction coordinate (a) smaller DG8 faster reaction larger k reaction coordinate (b) Figure 9.1 Relationship among the standard free energy of activation (DG ), reaction rate, and rate constant,k.(a) A reaction with a larger DG has a smaller rate and a smaller rate constant. (b) A reaction with a smaller DG has a larger rate and a larger rate constant.

4 9. REACTON RATES 85 TALE 9.2 Relationship between Rate Constants, Standard Free Energies of Activation, and Reaction Times for First-Order Reactions DG Rate constant (s 1 ) (T 298 K) Time to completion* kj mol 1 kcal mol years days hours minutes seconds milliseconds microseconds microseconds nanosecond 0 0 *Time required for 99% completion of reaction ' 7/k conditions (that is, 1 M in all reactants) and their standard free energies of activation, first presented as Eq. 4.a (p. 160), is given in Eq. 9.22a: relative rate = r atea = 10 DG DG A Ü2.RT rate (9.22a) ecause the rate constant is numerically equal to the rate under standard conditions, the relative rate is the ratio of rate constants: or relative rate = k A = 10 DG DG A Ü2.RT k log (relative rate) = log k A k = DG - DG A 2.RT (9.22b) (9.22c) This equation says that each increment of 2.RT (5.7 kj mol 1 or 1.4 kcal mol 1 at 298 K) in the DG difference for two reactions corresponds to a one log unit (that is, 10-fold) factor in their relative rate constants. PROLEMS 9.5 For each of the following reactions, (1) what is the overall kinetic order of the reaction, (2) what is the order in each reactant, and () what are the dimensions of the rate constant? (a) an addition reaction of bromine to an alkene with the rate law rate = k[alkene][r 2 ] 2 (b) a substitution reaction of an alkyl halide with the rate law rate = k[alkyl halide] 9.6 (a) What is the ratio of rate constants k A Ük at 25 C for two reactions A and if the standard free energy of activation of reaction A is 14 kj mol 1 (.4 kcal mol 1 ) less than that of reaction? (b) What is the difference in the standard free energies of activation at 25 C of two reactions A and if reaction is 450 times faster than reaction A? Which reaction has the greater DG?

5 86 CHAPTER 9 THE CHEMSTRY OF ALKYL HALDES 9.7 What prediction does the rate law in Eq make about how the rate of the reaction changes as the reactants D and E are converted into F over time? Does the rate increase, decrease, or stay the same? Explain. Use your answer to sketch a plot of the concentrations of starting materials and products against time. 9.4 THE S N 2 REACTON A. Rate Law and Mechanism of the S N 2 Reaction Consider now the nucleophilic substitution reaction of ethoxide ion with methyl iodide in ethanol at 25 C. C 2 H 5 O + H CL C 2 H 5 O L + C 2 H 5 OH The following rate law for this reaction was experimentally determined for this reaction: (9.2) rate = k[ ][C 2 H 5 O ] (9.24) with k = 6.0 X 10 4 M 1 s 1. That is, this is a second-order reaction that is first order in each reactant. The rate law of a reaction is important because it provides fundamental information about the reaction mechanism. Specifically, the concentration terms of the rate law indicate which atoms are present in the transition state of the rate-limiting step. Hence, the rate-limiting transition state of reaction 9.2 consists of the elements of one methyl iodide molecule and one ethoxide ion. The rate law excludes some mechanisms from consideration. For example, any mechanism in which the rate-limiting step involves two molecules of ethoxide is ruled out by the rate law, because the rate law for such a mechanism would have to be second order in ethoxide. The simplest possible mechanism consistent with the rate law is one in which the ethoxide ion directly displaces the iodide ion from the methyl carbon: C 2 H 5 O21 H C L 21 H d d C 2 H 5 O21 C ) 21 C 2 H 5 O21 L 21 H H transition state + ) (9.25) Mechanisms like this account for many nucleophilic substitution reactions. A mechanism in which electron-pair donation by a nucleophile to an atom (usually carbon) displaces a leaving group from the same atom in a concerted manner (that is, in one step, without reactive intermediates) is called an S N 2 mechanism. Reactions that occur by S N 2 mechanisms are called S N 2 reactions. The meaning of the nickname S N 2 is as follows: substitution S N 2 nucleophilic bimolecular (The word bimolecular means that the rate-limiting step of the reaction involves two species in this case, one methyl iodide molecule and one ethoxide ion.) Notice that an S N 2 reaction, because it is concerted, involves no reactive intermediates.