In base, bromination is different and more complicated because it usually won’t stop with the introduction of one halogen atom. We’ll use the bromination of acetone as our example: the first step will now be a base-catalysed enolization to give the enolate ion instead of the enol. The enolate ion can attack a bromine molecule in a very similar way to the attack of the enol on bromine. The enolate will, of course, be even more reactive than the enol .

The problem is that the reaction does not stop at this point. The first step was the removal of a proton and the protons between the carbonyl group and the bromine atom in the product are more acidic than those in the original acetone because of the electron-withdrawing bromine atom. Bromoacetone forms an enolate faster than acetone does.

Dibromoacetone is formed. Now we have one remaining proton in between the carbonyl group and two bromine atoms. It is even more acidic and so forms a new enolate ion even more quickly. The first product observable in any amount is tribromoacetone.

But even this is not the end of the story. To see why, we need to backtrack a bit. You may already have asked yourself, ‘Why doesn’t the hydroxide ion, being a nucleophile, attack the carbonyl group?’ This is a general question you might ask about all enolizations in base. The answer is that it does. The reaction is shown in the margin. A tetrahedral intermediate forms. What can happen now? This tetrahedral intermediate will revert to a carbonyl compound by expelling the best leaving group. Me− can never act as a leaving group: the only possible leaving group is the hydroxide ion (pKa of water = 15.7), so it just drops out again. This state of affairs continues until we reach the tribromoketone. The CBr₃ − group now has a chance to be a leaving group since the carbanion is stabilized by three bromine atoms. A real reaction occurs:

These initial products exchange a proton to reveal the true products of the reaction—the anion of a carboxylic acid and tribromomethane (CHBr₃).

The same thing happens with iodine, and we can summarize the whole process with iodine using a general structure for a carbonyl compound bearing a methyl group. It must be a methyl group because three halogens are necessary to make the carbanion into a leaving group. This reaction is often called the ‘iodoform’ reaction. Iodoform was an old name for tri iodomethane, just as chloroform is still used for trichloromethane. It is one of the rare cases where nucleophilic substitution at a carbonyl group results in the cleavage of a C–C single bond.

●Acid conditions are best for halogenation of carbonyl compounds should be carried out in acid solution. Attempts in basic solution lead to multiple substitutions and C–C bond cleavage.

Why does acid-catalysed halogenation work better?

The reason why halogenation in base continues until all the hydrogens have been replaced is clear: each successive halide makes the remaining proton(s) more acidic and the next enolization easier. But why does acid-catalysed halogena tion stop after the introduction of one halogen? It would be more accurate to say that it can be made to stop after one halogen is introduced if only one equivalent of halogen is used. Acid-catalysed halogenation will continue if there is more halogen available.

However, the second halogen goes on the other side of the carbonyl group, if it can. It is evidently the case that the second halogenation is slower than the first. Most of the intermediates are positively charged and hence destabilized by the presence of a halogen. The bromoketone is less basic than acetone so less of the reactive protonated form is present. This slows down any further electrophilic attack.

addition of a second bromine to the same position in acid solution (this does not usually happen)

The second step is the rate-determining step, and the presence of a bromine atom at the α position slows this step down still further: if a proton can be lost from a different α position—one without a Br atom—it will be. The transition state for proton removal illustrates why bromine slows this step down. The part of the structure close to the bromine atom is positively charged.

We can add a useful piece of evidence to this weak-sounding explanation. The halogenation of an unsymmetrical dialkyl ketone gives different results in acid and in base. In base halogenation occurs preferentially on a methyl group, that is, on the less highly substituted side. In acid solution by contrast, the fi rst (and only) halogenation occurs on the more substituted side of the carbonyl group. Alkyl groups have the opposite effect to bromine atoms—they stabilize positive charges. So the reactions of an enol, with a positively charged transition state, are faster at more highly substituted positions. Enolates react through negatively charged transition states and are faster at less highly substituted carbon atoms.

مواضيع ذات صلة

The Robinson annelation

Intramolecular aldol reactions

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

Halogenation

Racemization

Thermodynamically stable enols: 1,3-dicarbonyl compounds