Entropy dominates equilibrium constants in the difference between inter- and intramolecular reactions. In Chapter 6 we explained that hemiacetal formation is often an equilibrium, with neither starting materials nor products strongly favoured. The addition of ethanol to acetaldehyde shown below on the left, for example, has an equilibrium constant not far from 1. Overall, ΔG must therefore be approximately 0 (in fact it’s very slightly positive). The enthalpy change associated with the reaction is the result of the change in bonding: in this case, a C=O double bond becomes two C–O single bonds, and these two single bonds are marginally more stable than the C=O double bond, therefore ΔH is slightly negative. But working against this is the fact that every molecule of hemiacetal that forms consume two molecules of starting material. Decreasing the number of molecules (and moving from a mix ture of aldehyde and alcohol towards pure hemiacetal) leads to an increase in the order of the mixture — in other words a decrease in entropy. ΔS is negative, so the –TΔS is positive, just about counterbalancing the small negative ΔH, and giving a slightly positive ΔG.

The reaction on the right is different because it is an intramolecular reaction: the hydroxyl group and aldehyde lie in the same molecule. ΔH will have essentially the same value as in the intermolecular reaction on the left, but as the intramolecular reaction progresses, one molecule stays one molecule—there is consequently a much less significant decrease in entropy. Our TΔS term no longer weighs against the negative ΔH term, making ΔG negative overall and allowing the equilibrium to lie to the right. In Chapter 11 we showed you how acetals can be used as base-stable protective groups to prevent nucleophiles attacking carbonyl groups. The acetals we chose to use were cyclic com pounds known as dioxolanes, for a very good reason: cyclic acetals are more resistant to hydrolysis than their acyclic counterparts. They are also easier to make—they form quite readily, even from ketones. Again, we have entropic factors to thank for their stability. For the formation of an acyclic acetal (below on the left), three molecules go in and two come out, but for a cyclic one, a cyclic acetal, two molecules go in (ketone plus diol) and two molecules come out (acetal plus water), so the usually unfavourable ΔS factor is no longer against us.

Overcoming entropy:

orthoesters There is a neat way of sidestepping the entropic problem associated with making acyclic acetals: we can use an ortho ester as a source of alcohol. Orthoesters can be viewed as the ‘acetals of esters’, which are hydrolysed by water, when catalysed by acid, to an ordinary ester and two molecules of alcohol.

Here is the mechanism for the hydrolysis—you should be feeling quite familiar with this sort of thing by now.

Ketones or aldehydes undergo acetal exchange with orthoesters. The mechanism starts off as if the orthoester is going to hydrolyse but the alcohol released adds to the ketone and acetal formation begins. The water produced is taken out of the equilibrium by hydrolysis of the orthoester, and we get two molecules from two: entropy is no longer our enemy.

مواضيع ذات صلة

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

Some reactions are reversible on heating : cracking

Equilibrium constants vary with temperature

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