A C-H bond is broken on the carbon adjacent to the carbonyl carbon (i.e. the “alpha-carbon”). The carbons further away from the carbonyl are never deprotonated.
The reaction is reversible and equilibrium favors the starting materials by about 100:1 [ Note 1 ] this is still a pretty surprising result, given all that we know about C-H bonds. After all, alkanes are incredibly resistant to base (pKa = 50) so something here is making the aldehyde or ketone about 30 more pKa units (= 10 30 !) more acidic than it usually would be.
Why would this be?
To understand acidity, the first place to look is to examine the stability of the conjugate base. Way back in Org 1 we looked at various factors that affect acidity and noted that the negative charges are stabilized by
(See how all these concepts we learn back in the first quarter of organic chemistry continue to be relevant many chapters later! )
If we compare the conjugate base of an alkane versus the conjugate base of an aldehyde or ketone, two things should stand out.
Note that if we form the negative charge on the beta– or gamma– carbon, resonance delocalization isn’t possible.
In order for an enolate to form, there must be a hydrogen on the alpha carbon. If an aldehyde or ketone lacks a proton on the alpha carbon we call it “non-enolizable”.
Also avoid this common mistake: the aldehyde C-H bond is not acidic! [ Note 2 ]
As it turns out, enolate ions are very important intermediates for many reactions and we’ll spend a considerable amount of time getting to know them.
We’ll start out by covering some of their simple reactions, and then finish up by exploring various factors that affect their stability (and reactivity).
Enolates can be thought of as the conjugate bases of aldehydes and ketones (among others) but they are also the conjugate bases of enols.
Keto-enol tautomerism can occur through addition of either acid or base to an aldehyde or ketone. If you need a review, go back and see this post : Keto-Enol Tautomerism]
Generally speaking the keto tautomer is favored at equilibrium [mostly because the C-O pi bond is about 20 kcal/mol stronger than the C-C pi bond].
When keto-enol tautomerism occurs through the addition of base, it proceeds through the formation of an intermediate enolate before being protonated and converted to the enol form.
Since the aldehyde / ketone (pKa = 17-19) is a weaker acid than H2O (pKa 15) the equilibrium lies mostly to the left, meaning that the enolate is a minor component of the reaction mixture.
You’ll be forgiven if this doesn’t seem very exciting.
All that’s happening is that base is being used to form a compound (the enol) which is going to be outnumbered at least 100:1 at equilbrium in most cases. Who cares?
The exciting thing is that the enolate itself is a great nucleophile and is very reactive with other electrophiles if they happen to be present, such as:
We’ll get to these! But first, some consequences of enolate formation.
Enolates have two prominent resonance forms and can be drawn with a negative charge on carbon or a negative charge on oxygen.
They are flat, so their most “correct” form is to draw them with their negative charge on oxygen.
However, they tend to react as if they were carbanions, attacking most electrophiles via the carbon atom. There are ways to get the oxygen to be more reactive instead of carbon, but for now we will defer this issue to more advanced courses [ Note 3 ]. For our purposes here we will draw them with the negative charge on carbon.
The flatness of enolates has an important consequence.
If you start with a chiral center on the alpha carbon and treat it with base, the alpha carbon, being flat, will lose its chirality. If the enolate then reacts with another electrophile (e.g. H+) it can do so from either face of the flat enolate, resulting in racemization at that carbon. [To be more specific, we usually call this process epimerization [ Note 4 ] since it won’t result in a racemic mixture of enantiomers if another chiral center is present.
(Chiral centers on the beta- or gamma- carbon are not affected since these C-H bonds are not acidic).
Just for fun, if we use a deuterated solvent like D2O instead of H2O (recall that D is the heavy isotope of hydrogen) then we will replace our C-H bonds with C-D bonds. This can be a useful way of incorporating deuterium.
Halogenation
If we make an enolate in the presence of a halogen such as Br2, Cl2, or I2, a new C-halogen bond will form. This is called halogenation of enolates.
One key thing to watch out for with halogenation under basic conditions is that the reaction has Cookie Monster characteristics (the Cookie Monster never stops at eating just one cookie).
Electron withdrawing groups make adjacent C-H bonds more acidic since they stabilize negative charge. Therefore the resulting product will have a more acidic C-H bond, and be even more likely to be removed by base. It leads to a runaway reaction where multiple halogens can be installed on the same carbon. [see post: The Haloform Reaction]
The better (and more controlled) way to do this is to irreversibly form the enolate with a super strong base like LDA and then add (for example) Br2. No runaway reaction.
Aldol Reaction
Enolates can react with other aldehydes or ketones! This is called the Aldol Reaction (see post: Aldol Reaction)
The aldol reaction has many cousins that go by various names like the Claisen condensation, Henry reaction, Knovenagel and whatever. They all boil down to the same essential thing: an enolate nucleophile attacking a carbonyl carbon electrophile.
So if aldehydes and ketones are reasonably acidic (pKa 17-19) what happens to the acidity of the alpha-carbon when we add an electron withdrawing group to it?
Click to FlipRemember that any factor that stabilizes the conjugate base will increase the acidity. So it should become a much stronger acid since the negative charge of the resulting conjugate base can be spread out over multiple oxygen atoms and also is stabilized by an additional inductive effect.
The bottom line is that enolates that have two adjacent electron withdrawing groups are much easier to form!
For example, beta-keto esters and acetoacetic esters (pKa about 11) can easily be deprotonated with a strong base such as CH3CH2O(-) [ Note 5 ] .
One useful reaction of these enolates is that they will perform SN2 reactions on alkyl halides, forming a new C-C bond. This is called, “alkylation”.
The products can then be made to lose CO2 (decarboxylation) in a process known as the malonic ester synthesis (for di-esters) or acetoacetic ester synthesis (for beta-keto esters). (See post: The Malonic Ester and Acetoacetic Ester Syntheses)
[ Note 6 – regarding this reaction with ordinary ketones]
Going in the opposite direction, what happens to the acidity of the alpha-carbon if we switch out a ketone for an ester?
In the case of an ester the alpha-carbon actually becomes less acidic. Therefore ester enolates are less stable than those of aldehydes and ketones.
This might seem surprising. Why are ester enolates less stable? After all, wouldn’t we expect that it should be *better*, since it has that electron-withdrawing OR group?
The answer is related to the reason why OH and OR are activating groups on aromatic rings instead of deactivating groups.
Yes, oxygen is electronegative, but the lone pairs mean that it is also a pi-donor, capable of forming a C-O pi bond with the carbonyl carbon. The pi-donation outweighs the electronegativity factor.
In a sense the C=O of an ester is less able to stabilize the negative charge of the enolate because the enolate anion is “competing” with the lone pairs from the -OR group.
In addition to esters, other groups capable of acting as pi-acceptors such as NO2, CN, and amides will also form enolates.
Knowing what you know now, do you think the enolates of amides (NR2) would be more or less stable than those of an ester (hint: think back to activating groups on aromatic rings)
Click to FlipOne simple way to think about it is to think of the C=O as a bucket for accepting an electron pair from the lone pair on carbon. When there is already a pi-electron donor like an OR or an NR2 adjacent to the C=O, there’s going to be less room in the bucket for the pair of electrons on the carbon, and the enolate will be less stable.
It might be helpful to put these trends together on a figure like this:
This article would not be complete without briefly mentioning some of the complications that come with ketone enolates.
In many cases treating a ketone with base could result in two different enolates.
So how do we know which one “wins?”.
We treat it the same way we do with double bonds, i.e. Zaitsev’s rule. Just as with alkenes, the more substituted enolate will be the most stable.
For this reason the more substituted enolate is called the “thermodynamic enolate” – it is thermodynamically more stable.
However, this tendency to form the more substituted enolate can interfere with our carefully-laid plans! There are many times when we would like to form an enolate on the less substituted side of a ketone, regardless of what it “thermodynamically” might want.
One way to overcome this tendency is to use a strong, but very bulky base. The base lithium di-isopropyl amide (LDA) fits the bill nicely. It is like a big hammerhead shark that has a hard time sticking its nose into tight corners. So when presented with a ketone bearing two different alpha carbons, it will overwhelmingly remove the one attached to the less substituted carbon atom.
LDA is also useful for making the enolates of esters, amides, and nitriles, but we’ll have more to say about that in a future article.
What have we learned?
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Note 1. We can calculate how favored this reaction is by looking at the four components of our acid base reaction (acid, base, conjugate acid, conjugate base) comparing the pKas of the acid and the conjugate acid. The acid is the aldehyde (pKa = 17) and since our acid is HO(-), the conjugate acid is H2O (pKa =15.7). The equilibrium constant K would therefore be roughly (16-17 = -1 ) or 10 -1 in the direction of product. So very little enolate would be present at any given time.
Note 2. The aldehyde C-H is not acidic, even though it seems like it should be!
Click to FlipNote 3. Enolates tend to react at carbon rather than oxygen. One reason is that the oxygen tends to be tightly bound to the counter-ion of whichever base is used (e.g. Li+ or Na+). One way to get the oxygen to be more reactive is to use alkali metal salts that are bigger and form a weaker ionic bond with oxygen (e.g. potassium, K+).
Note 4. When only one chiral center is present, and it’s on the alpha-carbon, the result will be racemization. When there are multiple chiral centers, and one of the is on the alpha carbon, only the chiral center on the alpha-carbon will be affected.
This can lead to the formation of epimers. Epimer is a special name we give to diastereomers that differ in their configurations at only one chiral center.
For example, treating this optically pure (2S, 3S)-dimethylcyclohexanone with base will result in a mixture of (2S, 3S)-dimethylcyclohexanone and (2R,3S)-dimethylcyclohexanone. They aren’t enantiomers since they have the same configuration at C-3.
Note 5. We have to be careful with the base we use here. If we use NaOH, we will end up performing saponification of the ester (basic hydrolysis). If we use a strong base with a different R group we could end up performing transesterification.
Note 6. Alkylation of ordinary ketones using NaOR as base and alkyl halides such as CH3I is a mixed bag. It can often lead to multiple alkylations happening. The best way to perform these reactions is to irreversibly form the enolate using the strong base LDA and then use an alkyl halide.
The classic study on this was by House and Kramar. Highly recommend.
Note 7. A typical ratio of thermodynamic:kinetic enolates under “thermodynamic conditions” (NaOR/ROH) is about 4:1 but it can vary widely. Page 2 of this article provides a great overview on forming thermodynamic enolates.
Note 8. One question I get asked a lot is why Grignard reagents are able to add to ketones without deprotonating them first. After all the pKa of an aldehyde or ketone is about 16-18 and the pKa of an alkane is around 50, so why wouldn’t you expect an acid-base reaction?
The answer is that alpha protons are acidic but they are only acidic if the C-H bonds are aligned with the p-orbitals of the C-O pi bond. That’s because resonance delocalization is only possible when the p-orbitals can overlap.
This makes deprotonation at carbon quite slow, relative to addition to the carbonyl and also relative to deprotonation of O-H which doesn’t require orbital overlap.
The classic example here is the bridgehead C-H bond of bicyclic ketones.
Note 9. Two other ways of making enolates that we encounter in introductory organic chemistry bear mention.
The first one is that alpha-halo esters can be treated with Zn to give enolates in a process called reductive formation of enolates . This is much like Grignard reagent formation.
The resulting enolates can add to aldehydes and ketones in a reaction known as the Reformatsky reaction.
Another way of making enolates is through conjugate addition. Addition of a nucleophile to an enone gives a new enolate.