CHEM 240

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Lecture 21. Reaction mechanisms

Monday 15 March 2010

Reaction mechanisms. Elementary reactions, molecularity. Relationship between reaction mechanism, elementary reactions, and rate law. Intermediates and transition states. Kinetics and chemical equilibrium. The rate-determining step of a mechanism.

Reading: (Supplemental)

21. Summary

Lecture 21 Summary

Here we turn to consideration of reaction mechanisms, which we can think of as the "movie" of what happens at the molecular level when the reaction proceeds. Some of these movies have only one "scene" (elementary step), featuring only one or two "actors", during which the reaction is completed. Other mechanistic movies are comprised of multiple scenes in which various players may appear and disappear before the final products are generated. Once we begin to consider reaction mechanisms, we will want to pay particular attention to the relationship between a mechanism - defined by its component elementary reactions - and the rate law. This relationship is determined when we identify the rate-determining step of a mechanism.

We have seen an example of a reaction with a very simple mechanism, one that consists of a single elementary reaction, namely the SN2 reaction which is a simple, concerted, single-step collision between two molecules. The bimolecular rate law for the single elementary step of an SN2 mechanism is the rate law for the reaction. Other mechanisms consist of many steps. For a multi-step mechanism, the rate laws for each step - depending on the relative values of rate constants and concentration of intermediates - will determine the rate law observed for the overall reaction. Typically, there is a step in the mechanism that is particularly slow, so that its rate controls the overall reaction rate. If the rate law for this rate-limiting step involves the concentrations of intermediates, we assume that these are determined by equilibrium relations in each elementary step prior to the rate-limiting step.

An example where the rate-limiting step occurs after a first fast step is given in Oxtoby, in Section 14-5, beginning on the bottom of p.627, continuing with discussion on p.628 and Example 14-8.

Topics: Relationship between reaction mechanism, elementary reactions, and rate law. The rate-determining step of a mechanism. Experimental kinetics data: Progress curves. Interpretation of experimental kinetics data in terms of rate laws. We contrast the SN2 mechanism to the SN1 mechanism, a two-step mechanism whose first step - formation of a carbocation - is the rate limiting step.

Definitions to know: Elementary reaction, molecularity (unimolecular, bimolecular, termolecular), rate-determining step (rate-limiting step), transition state, intermediate.

Reaction mechanisms

A reaction mechanism can be thought of as a detailed description of how a reaction occurs. The point of view of collision theory is useful here, as we imagine a mechanism in terms of reacting molecules colliding, with a certain energy and orientation, at which point bonds begin to break, and new bonds begin form, as atoms rearrange their bonding pattern in a specific sequence, like a movie. The complete formation of product molecules may require several different collisional rearrangements in sequence. A reaction mechanism can be quite simple and occur in one step, called an elementary reaction (or elementary step). The number of molecules involved in a collision corresponding to an elementary step is called the molecularity of that step. For example, the SN2 reaction is a nucleophilic substitution reaction whose rate depends on both the concentration of the nucleophile and the alkyl halide "substrate". The underlying mechanism that explains the kinetics of an SN2 reaction is a simple, concerted, single-step encounter between a nucleophile and the substrate bearing the leaving group. This is said to be a bimolecular step, and it leads to the rate law as described because the rate of reaction will be limited by the rate of collision of the two reactant molecules. This collision rate is dependent upon both their concentrations.

Reaction mechanisms can often be complex, consisting of multiple elementary steps. Worth bearing in mind here are the following points.

  • Most elementary steps will be unimolecular or bimolecular. A step with three species colliding, a termolecular step, is invoked infrequently. A step involving the simultaneous collision of four reacting species is unheard of.
  • The sum of the equations of all the elementary steps of a mechanism must be the net overall reaction. All reaction equations must be balanced.
  • The rate law from an elementary step can be written directly from the chemical equation. Note that this is not generally true for equations representing net reactions.
  • For multistep mechanisms, often there is a step that is significantly slower than all the ort her steps. In such cases this rate-determining step will determine the observed kinetics.

Reaction coordinate diagrams and the transition state

Here we develop the key conceptual relationship between activation energy of a reaction and what chemists refer to as the transition state for an elementary step. We make use of reaction coordinate diagrams, which are a representation of the "energy landscape" of a reaction mechanism. As an initial illustration of these concepts, we choose the SN2 reaction, familiar from the study of organic chemistry. The SN2 mechanism represents the simplest type of mechanism, in that it occurs in one step. The overall reaction is the same as the elementary reaction of the single, bimolecular step.

The lower half of the figure at right (click on it to see a larger version) shows a reaction coordinate diagram for an SN2 mechanism between the nucleophile ammonia and methyl iodide. The energy of the reacting molecules is plotted along the vertical axis as they progress along a horizontal "reaction coordinate" in going from reactant to product species. Above the reaction coordinate diagram are three corresponding frames from a movie of the reaction between two molecules. The first frame shows the two molecules moving toward one another in the right orientation and with sufficient energy to react upon collision. The next frame shows the transition state for the reaction (symbolized by a double dagger, ‡), which is the highest energy species along the reaction coordinate. Note that the activation energy for the forward reaction is the difference in energy between the reactants and the transition state. The movie frame shows the bonds between the carbon and nitrogen and carbon and iodine as lengthened dashed lines, indicating partially forming and breaking bonds. The final frame shows the iodide ion leaving group and the protonated methylamine product moving apart. If we run the movie backwards, we will see the reverse reaction with iodide the nucleophile attacking methylammonium cation.   Reaction coordinate diagram for an SN2 reaction  

In this case, the activation energy for the reverse reaction, Ea, rev, is the difference in energy between the products and the transition state.

Finally, the energy change for the reaction, DErxn, is the energy change for the overall reaction, Eproducts,Ereactants, which we could measure by calorimetry. It is also the difference between the activation energies, Ea, fwd,Ea, rev. This is an important idea that provides a basis for the understanding of the relationship between kinetics and thermodynamics in chemistry.

It is crucial for us to distinguish between "liitle k" (rate constant) and "big K" (equilibrium constant), but as we have noted already there is a relationship between them that stems from the fact that a reaction is at equilibrium when the rates of the forward and reverse reactions are equal. For the simple SN2 mechanism above, we can formulate this quite simply as Keq  =  kfwd / krev.

Other mechanisms - the SN1 mechanism

As a second example of a reaction mechanism, this one a little more complicated, we can look at another reaction from organic chemistry. The SN1 mechanism is a mechanism that consists of at least two elementary steps. For example, in the reaction of tert-butyl chloride and hydroxide ion to form tert-butyl alcohol and chloride ion, the first step for the mechanism is the relatively slow formation of a tert-butyl carbocation. This high-energy intermediate then rapidly reacts with hydroxide to form the alcohol in the second step.

We can represent the mechanism as follows:

(1)  (CH3)3CCl  =  (CH3)3C+  +  Cl       (slow)
(2)  (CH
3)3C+ +  OH  =  (CH3)3OH       (fast)

The corresponding reaction coordinate diagram is shown at right. Note how the slow carbocation-forming step has a much higher activation energy than the fast step in which the carbocation reacts with hydroxide ion to form the alcohol.

(Clicking on the figure will take you to a larger version that is posted to the reaction coordinate diagrams page.)

  Reaction coordinate diagram for an SN1 reaction mechanism

Another feature to note is that each elementary step of a mechanism will have a corresponding transition state. Thus, there are two transition states shown in the reaction coordinate diagram for a two-step mechanism.

Does the mechanism given explain the experimentally-observed rate law for this reaction? The observed rate law is

rate = k [(CH3)3CCl]

The rate is first-order overall, first-order in the tert-butyl chloride concentration, and does not depend at all on the hydroxide ion concentration. Since the first step in the mechanism is so much slower than the second step, the first step is rate-limiting. The rate at which the first step takes place determines the rate of the overall reaction. Since the rate law for the first step can be written as rate(1) = k1 [(CH3)3CCl], where is a first order rate constant for step 1 in the forward direction, we see that the mechanism is consistent with experiment, and the rate constant for the reaction is essentially that for the first elementary step, i.e. k = k1.

We look at another example of a mechanism, this time a two-step mechanism in which the first step is fast and the second step is rate limiting (See also Example 14-8 in Oxtoby).

We had the following mechanism:

(1) NO(g)  + O2(g)  =  NO3(g)       (fast)
(2) NO3(g) +  NO(g) →  2 NO2(g)     (slow)

Let k1 be the second-order rate constant for step 1 in the forward direction, k1 be the first-order rate constant for the reverse of step 1, and k2 the forward second-order rate constant for the rate-limiting step, step 2. Then we can write the rate laws for the steps as follows

(rate 1)fwd  =  k1[NO][O2]

(rate 1)rev  =  k1[NO3]

(rate 2)fwd  =  k2[NO3][NO]

Since step 2 is rate-limiting, its rate determines the observed rate of the overall reaction

V0, obs  =  k2[NO3][NO]

We must express the rate law as a constant times terms with reactant concentrations, whereas [NO3] is the concentration of an intermediate. To replace [NO3] with reactant concentration terms, we make the assumption that step 1 is so much faster than step 2 that the reaction in step 1 is essentially at equilibriun. Below, we work out the consequences of this assumption.

At equilibrium, (rate 1)fwd  =  (rate 1)rev

k1[NO]eq[O2]eq  =  k1[NO3]eq.

k1 / k1 =  [NO3]eq / [NO]eq[O2]eq    or    [NO3]eq  =  (k1/ k1)[NO]eq[O2]eq

If we substitute this latter expression for [NO3] as follows,

V0, obs  =  k2[NO3][NO]  =  k2(k1/ k1)[NO]2[O2]

where the observed rate constant, kobs, a third-order rate constant, is

kobs  =  k2(k1/ k1)



Learning objectives

Page update in progress, Jan 2010 (01-01-10)


  1. Chemistry: Science of Change (4th edition) Oxtoby, Freeman, Block

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