The first-order kinetics of this unusual substitution reaction is here to illustrate a point, but it should not distract you from the fact that most nucleophilic substitutions of carboxylic acid derivatives (the reactions you met in Chapter 10) are bimolecular reactions with rate- determining formation of the tetrahedral intermediate.

However, something different again happens when we come to reactions of amides. Because of the delocalization of the nitrogen lone pair into the carbonyl group, nucleophilic attack on the carbonyl group is very difficult. In addition, the leaving group (NH2 −, with pKa of NH3 about 35) is very bad indeed.

What happens as a consequence is that in the hydrolysis of amides the second step—the breakdown of the tetrahedral intermediate—becomes rate determining. But this offers the opportunity for base catalysis of this step. If a second hydroxide ion removes the proton from the tetrahedral intermediate, the loss of NH2 from what is now a dianion is made easier, and a stable carboxylate ion is formed directly.

Notice that in the fi rst mechanism just one hydroxide ion is involved, whereas now two are involved: one is consumed to form product, but the second is in fact regenerated when the product NH₂ − anion reacts with water—in other words the second hydroxide ion is a catalyst. The rate equation for the amide hydrolysis reflects this involvement of two hydroxide ions: the rate depends on the square of the hydroxide ion concentration and it is third order. We’ll label the rate constant k₃ to emphasize this:

rate = k3[MeCONH₂] × [HO−]²

But you may be asking yourself where this third-order kinetics comes from, since the hydroxide ions are not actually involved in the rate-determining step. In fact, third-order kinetics hardly ever mean the real simultaneous termolecular collision of three molecules at once—such events are just too rare. The rate-determining step here is actually unimolecular—the collapse of the dianion. So we expect

rate = k[dianion]

We don’t know the concentration of the dianion but we do know that it’s in equilibrium with the monoanion—we’ll call this equilibrium constant K₂:

and so [dianion] = K2[monanion][HO−]. This sort of helps, but we still don’t know what [monoanion] is, other than that it’s again in equilibrium, this time with the amide—we’ll call this equilibrium constant K1:

and so [monoanion] = K1[amide][HO−]. Substituting these values in the simple rate equation we discover that

rate = k[dianion] becomes

 rate = kK₁K₂[amide][HO−]²

The third-order kinetics result from two equilibria starting with the amide and involving two hydroxide ions, followed by a unimolecular rate-determining step, and the ‘third-order rate constant’ k3 is actually a product of the two equilibrium constants and a fi rst-order rate constant:

k₃ = k × K₁ × K₂

This often happens with reactions with late rate-determining steps: the rate constant can depend on the concentrations of any species involved before the rate-determining step (although not necessarily in that step itself) but never depends on species involved after the rate-determining step.

Just because a proposed mechanism gives a rate equation that fi ts the experimental data, it does not necessarily mean that it is the right mechanism; all it means is that it is consistent with the experimental facts so far, but there may be other mechanisms that also fi t. It is then up to the experimenter to design cunning experiments to try to rule out other possibilities. Mechanisms are given throughout this book—eventually you will learn to predict what mechanism to expect for a given type of reaction, but this is because earlier experimentalists have worked out the mechanisms by a study of kinetics and other methods.

مواضيع ذات صلة

Protons on saturated carbon atoms Chemical shifts are related to the electronegativity of substituents

Regions of the proton NMR spectrum

Integration tells us the number of hydrogen atoms in each peak

The differences between carbon and proton NMR

Kinetic versus thermodynamic products

Catalysis in carbonyl substitution reactions

Rate equations

Catalysis

Reaction intermediates

Why some reactions are done at low temperature

The route from reactants to products: the transition state

How fast do reactions go? Activation energies