In learning about organic synthesis, the key is to familiarize oneself with as many transformations
as possible, so as to be able to look at some arbitrary functional group and know of a way
to transform it into some other functional group that is more desirable for a particular synthetic
pathway. When it comes to carbonyl-containing compounds, there are plenty of transformations
we can discuss. We could oxidize or reduce, and with aldehydes and carboxylic acids in particular,
as they sit at a terminus of a molecule, there are many transformations we can employ
to produce acid chlorides and other such functional groups.
But much more difficult is achieving such transformations on ketones,
which are internal carbonyls, because apart from simple reduction to the secondary alcohol,
many desirable transformations would involve breaking one of the carbon-carbon bonds stemming
from the carbonyl carbon, which can be difficult to do. But there is one such reaction that is
quite useful in this regard, and it is called the haloform reaction. Let’s take a look at this now.
The haloform reaction is a method of converting methyl ketones into carboxylic acids. So by methyl
ketones we are referring to a ketone in which the alkyl group to one side of the carbonyl is
a methyl group. The reason it must be a methyl group is that we will need all three of these
alpha protons to be present in order for this to work. So the way this goes is as follows. Let’s
react this methyl phenyl ketone with excess strong base. As we know from learning all about enolate
chemistry earlier in the series, this will deprotonate at an alpha carbon to produce the
enolate anion, and since there are no alpha protons in the direction of the phenyl group,
we must deprotonate at the methyl. For the ease of displaying this particular mechanism, we will
display the resonance structure for the enolate that has the negative charge on the carbon, though
we know that another exists with the negative charge on oxygen. Now we introduce excess diatomic
halogen, which we will represent generally as X2, which can represent chlorine, bromine, or iodine.
We know that this can act as an electrophile due to the polarizability of the halogen-halogen bond,
so the enolate will attack this halogen and kick off the other as a halide ion.
Now because both the strong base and halogen are in excess, and because there are two more alpha
protons on this carbon, we can do precisely the same thing two more times. In fact,
the acidity of the substrate is only increased by the presence of the electron-withdrawing halogen.
So more base extracts the proton, the enolate attacks and gets halogenated a second time,
and then this happens yet a third time to give us this.
Now the interesting thing about this structure is that the trihalomethyl anion is actually a
good leaving group, because the resulting negative charge on the carbon is stabilized via induction
by the three neighboring halogens. Since we have still more base lying around, this means that
hydroxide can attack the carbonyl, kick that pi bond up, which will then come back down to reform
the carbonyl and kick off the trihalomethyl anion. This anion, though stabilized, is still
rather basic, so it will take the carboxyl proton from the newly formed carboxylic acid,
and we are left with a carboxylate anion and something called a haloform, which is the name
given to a molecule of methane where three hydrogens are replaced by halogens. Aqueous
acidic workup will protonate the carboxylate to get the carboxylic acid we wanted, and the
transformation is complete. We have successfully converted a methyl ketone into a carboxylic acid.
As an additional application, when performing this with iodine, it can act as a qualitative
test to see if an unknown substrate is in fact a methyl ketone, because the iodoform that is
produced is a yellow solid that is extremely easy to identify. This is called the iodoform test.
But in general, the things to keep in mind are that this requires excess strong base and
halogen so as to ensure reaction to completion, and that this reaction works best when there are
no alpha protons on the other carbon, so as to ensure deprotonation at the methyl carbon,
which is why we utilized a phenyl group in this example.
And with that, we have another transformation in our organic synthesis bag of tricks.