So far, we have been rather vague about the term ‘stability’, just saying things like ‘this compound is more stable than that com pound’. What we really mean is that this compound has less energy than that one. For example, as you know from Chapters 4 and 7 alkenes can come in two forms, we can call cis and trans. In general, trans-alkenes are more stable than cis-alkenes. How do we know? Well, we can convert both cis- and trans-butene to the same alkane, butane, by adding a molecule of hydrogen. Energy is given out during the reaction, and if we measure how much energy we get from hydrogenation of trans-butene and compare it with the amount we get from cis-butene, we find that the cis-alkene give us about 2 kJ mol−1 more. Cis-butene is higher in energy, and must therefore be less stable. We can rep resent this in the energy profile diagram on the right. The two red lines show the energies of the molecules, and the black arrows the amount of energy released when hydrogen is added. This comparison of energy is most interesting when two compounds can interconvert. For example, as you saw in Chapter 7, rotation about the C–N bond of an amide is slow because delocalization of the N lone pair gives it some double-bond character. The C–N bond can rotate, but the rotation is slow and can be measured by NMR spectroscopy. We might expect to find two forms of an amide of the type RNH–COR: one with the two R groups trans to one another, and one with them cis. Depending on the size of R we should expect one form to be more stable than the other and we can represent this on an energy profile diagram showing the relationship between the two molecules in energy terms.

This time there is an axis along the bottom indicating the extent of rotation about the C–N bond. The two red lines show the energies of the molecules and the curved black line shows what must happen in energy terms as the two forms interconvert. Energy goes up as the C–N bond starts to rotate and reaches a maximum at point X when rotation by 90° has removed the conjugation (the nitrogen lone pair can’t delocalize into the C=O bond because it is perpendicular to the C=O π* orbital) before falling again as the conjugation is regained. The relative energies of the two states will depend on the nature of R. The situation we have shown, with the cis arrangement being much less stable than the trans, would apply to large R groups. We can defi ne an equilibrium constant K for this process. For large R groups, K will be very large:

At the other extreme is the case when both substituents on nitrogen are H. Then the two arrangements would have equal energies. The process which interconverts the structures is the same but there is now no difference between them. If you could measure an equilibrium constant, it would now be exactly K = 1.

In more general terms, amide rotation is a simple example of an equilibrium reaction. If we replace ‘amount of C–N bond rotation’ with ‘reaction coordinate’ we have a picture of a typical reaction in which reagents and products are in equilibrium.

مواضيع ذات صلة

Energy, enthalpy, and entropy: ΔG, ΔH, and ΔS

Typical method for making an ester

How to make the equilibrium favour the product you want The direct formation of esters

A small change in ΔG makes a big difference in K

The sign of ΔG tells us whether products or reactants are favoured at equilibrium

How the equilibrium constant varies with the difference in energy between reactants and products

Substitution of C=O for C=C: a brief look at the Wittig reaction

Cyanide will attack iminium ions: the Strecker synthesis of amino acids

Lithium aluminium hydride reduces amides to amines

Iminium ions can react as electrophilic intermediates

Secondary amines react with carbonyl compounds to form enamines

Iminium ions and oxonium ions