We started this chapter with a discussion of thermodynamics: the factors that govern equilibria. We then moved onto rates: the factors that determine the rate at which reactions proceed. Depending on the reaction, either may be more important, and in general:

• Reactions under thermodynamic control have outcomes that depend on the position of an equilibrium and therefore the relative stability of the possible products.

• Reactions under kinetic control have outcomes that depend on the rate at which the reaction proceeds, and therefore on the relative energies of the transition states leading to the alternative products.

Before we leave this chapter, we will introduce an example of a reaction where thermo dynamic control and kinetic control lead to different outcomes—in other words, where the fastest reaction does not give the most stable possible product. The reaction is one you have not yet met, but it is quite a simple one, and it follows an unsurprising mechanism. It is the reaction of an alkyne with hydrogen chloride in the presence of alumina (Al2O3). The reaction produces two geometrical isomers of a chloroalkene. Alkynes, like alkenes, are nucleophiles, and so the mechanism involves fi rst of all attack by the alkyne on HCl, followed by recombination of the vinyl cation, which is formed with the chloride anion.

The two alkenes are labelled E and Z. After about 2 hours the main product is the Z-alkene. However, this is not the case in the early stages of the reaction. The graph below shows how the proportions of the starting material and the two products change with time.

Points to note:

• When the alkyne concentration drops almost to zero (10 minutes), the only alkene that has been formed is the E-alkene.

• As time increases, the amount of E-alkene decreases as the amount of Z-alkene increases.

 • Eventually, the proportions of E- and Z-alkene do not change. Since it is the Z-alkene that dominates at equilibrium, this must be lower in energy than the E-alkene.

Since we know the ratio of the products at equilibrium, we can work out the differ ence in energy between the two isomers:

ratio of E:Z-alkenes at equilibrium = 1:35

ΔG = –RTlnK = –8.314 × 298 × ln(35) = –8.8 kJ mol⁻¹

that is, the Z-alkene is 8.8 kJ mol−1 lower in energy than the E-alkene. However, although the Z-alkene is more stable, the E-alkene is formed faster under these conditions: the route to the E-alkene must have a smaller activation energy barrier than trans addition. This is quite easy to understand: the intermediate cation has no double-bond geometry because the cationic C is sp hybridized (linear). When chloride attacks, it prefers to attack from the side of the H atom rather than the (bigger) methyl group.

There must then be some mechanism by which the quickly formed E-alkene is converted into the more stable Z-alkene. The conditions are acidic, so the most likely mechanism is the acid-catalysed alkene isomerization you saw earlier in the chapter:

This information can be summarized on an energy profile diagram:

Initially, the alkyne is converted into the E-alkene via the intermediate linear cation. The activation energy for this step is labelled ΔG1‡. The E-alkene can convert to the Z isomer via an intermediate, with activation energy ΔG2‡. Since ΔG1‡ is smaller than ΔG2‡, the E-alkene forms faster than it isomerizes, and all the alkyne is rapidly converted to the E-alkene. But over the course of the reaction, the E-alkene slowly isomerizes to the Z-alkene. An equilibrium is eventually reached that favours the Z-alkene because it is more stable (by 8.8 kJ mol−1, as we calculated earlier). Why doesn’t the Z-alkene form faster than the E? Well, as we suggested above, the transition state for its formation from the linear cation must be higher in energy than the transition state for formation of the E-alkene, because of steric hindrance.

●Kinetic and thermodynamic products

 • The E-alkene is formed faster and is known as the kinetic product or the product of kinetic control.

• The Z-alkene is more stable and is known as the thermodynamic product or the product of thermodynamic control.

If we wanted to isolate the kinetic product, the E-alkene, we would carry out the reaction at low temperature and not leave it long enough for equilibration. If, on the other hand, we want the thermodynamic product, the Z-alkene, we would leave the reaction for longer at higher temperatures to make sure that the larger energy barrier yielding the most stable product can be overcome.

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

Catalysis in carbonyl substitution reactions

What does third-order kinetics mean

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