Now for something easy. When an aldol reaction can form a five- or six-membered ring, you need no longer worry about specifi c enols or anything like that. Equilibrium methods with weak acids or bases are quite enough to give the cyclic product by an intramolecular aldol reaction because intramolecular reactions are faster than intermolecular ones. We shall illustrate intramolecular reactions by looking at the cyclization of a series of diketones of increasing complexity, starting with one that can form four equivalent enols: cyclodeca 1,6-dione. It doesn’t matter where enolization occurs because the same enol is formed. And once the enol is formed, there is only one thing it can reasonably do: attack the other ketone to form a stable five-membered ring. It also gives a reasonably stable seven-membered ring, but that is by the way. In weak acid or base, only a small proportion of carbonyl groups will be enolized, so the chance of two being in the same molecule is very low. No intermolecular condensation is found and the yield of the bicyclic enone from the intramolecular reaction is almost 100% (96% with Na₂CO₃).

This may look like a long stretch for the enol to reach across the ten-membered ring to reach the other ketone, but the conformational drawing in the margin shows just how close they can be. You should compare this conformation with that of a decalin The key point to remember with intramolecular aldols is this:

●Intramolecular reactions giving five- or six-membered rings are preferred to those giving strained three- or four-membered rings on the one hand or medium rings (eight- to thirteen membered) on the other.

Acid-catalysed cyclization of the symmetrical diketone nona-2,8-dione could give two enols.

One enol can cyclize through an eight-membered cyclic transition state and the other through a six-membered one. In each case the product would first be formed as an aldol but would dehydrate to the cyclic enone having the same ring size as the transition state. In practice, only the less strained six-membered ring is formed and the enone can be isolated in 85% yield.

Most diketones lack symmetry, and will potentially have four different sites for enolization. Consider what might happen when this diketone is treated with KOH. There are four different places where an enolate anion might be formed as there are four different carbon atoms. There are also two different electrophilic carbonyl groups so that there are many possibilities for inter- and intramolecular condensation. Yet only one product is formed, in 90% yield.

We can deduce the mechanism of the reaction simply from the structure of the product by working backwards. The double bond is formed from an aldol whose structure we can predict and hence we can see which enolate anion was formed and which ketone acted as the electro philic partner.

Must we argue that this one enolate is more easily formed than the other three? No, of course not. There is little difference between all four enolates and almost no difference between the three enolates from CH₂ groups. We can argue that this is the only aldol reaction that leads to a stable conjugated enone in a stable six-membered ring. This must be the mechanism; the others are just too slow to compete. Protonation and dehydration follow as usual.

Now try some of the alternatives in which the same ketone forms an enolate on the other side. These reactions give unstable four-membered rings or bridged bicyclic systems that would revert to the enolate. Providing the reaction is done under equilibrating conditions, the whole process would go into reverse back to the original diketone and the observed (six-membered ring) cyclization would eventually predominate. The key point about the bridged compound in the margin is that dehydration is impossible. No enolate can form at the bridgehead because bridgehead carbons cannot be planar and the enone product cannot exist for the same reason: the carbons marked (•) in the brown structure would all have to lie in the same plane. The aldol has a perfectly acceptable conformation but that elimination is impossible. The aldol product remains in equilibrium with the alternative aldol products, but only one elimination is possible—and that is irreversible, so eventually all the material ends up as the one enone.

مواضيع ذات صلة

The Robinson annelation

How to control aldol reactions of aldehydes

How to control aldol reactions of esters

Specifi c enol equivalents from 1,3-dicarbonyl compounds

Conjugated Wittig reagents as specific enol equivalents

Silyl enol ethers in aldol reactions

Lithium enolates in aldol reactions

The Mannich reaction

Base-promoted halogenation

Halogenation

Racemization

Thermodynamically stable enols: 1,3-dicarbonyl compounds