An Introduction to the Aldol Reaction (addition & condensation)
Introduction
The aldol reaction involves the addition of the enolate (or equivalent functional group) of one carbonyl-containing compound (aldehyde or ketone) to a second carbonyl-containing compound (aldehyde or ketone) to give a β-hydroxy carbonyl compound. This is commonly called the aldol addition. There can be a subsequent elimination reaction that leads to a α,β-unsaturated carbonyl compound (enal or enone). This is known as the aldol condensation.
The aldol addition of a ketone with an aldehyde gives a β-hydroxyketone. Under the correct conditions this will undergo a second reaction, an elimination reaction or the aldol condensation.
The aldol addition is one of the cornerstones of organic synthesis. It is a powerful reaction that creates a C–C bond while installing up to two new stereocenters. The addition leaves the product with two functional groups, a carbonyl group and an alcohol. The molecule is still functionalized and can participate in more reactions; being able to do more chemistry is a good thing! The value of the aldol reaction is demonstrated by its extensive use in the total synthesis of Natural Products, and fact that there are multiple enzymes that permit organisms to use this reaction in the synthesis of essential biomolecules.
This is a simple introduction to both the aldol addition and aldol condensation. It will cover the formation of the C–C bond (addition) and then the C=C bond (condensation). Normally, this is all that is required for 100 and 200 level undergraduate chemists. In subsequent summaries, I will expand on this to show some of the variants of the aldol reaction that use other functional groups rather than aldehydes or ketones, the use of enolate equivalents in the aldol addition, and the stereochemical implications of the aldol addition (how you can control the diastereomers and even the enantioselectivity). But all of that comes later, what about the simple addition reaction?
The Aldol Reaction (Addition and Elimination (or condensation))
If you react an aldehyde with a base such as the hydroxide anion you will establish an equilibrium that favors the aldehyde but contains a small amount of enolate (a much stronger base is required if you want to force the equilibrium to the side of the enolate, and I’ll cover this in a subsequent summary). This means you have formed a mixture of nucleophilic enolate, the anionic species, and electrophilic aldehyde. Nucleophiles react with electrophiles.
Deprotonation of an aldehyde leads to an equilibrium in which there is a very small amount of enolate. The equilibrium favors the left-hand side, meaning that there is still a high concentration of aldehyde and only a small quantity of enolate. But this small quantity is enough to give you a mixture of a nucleophilic enolate and an electrophilic aldehyde. A more accurate representation of this diagram should not have the second molecule of aldehyde, and instead should show the aldehyde on the left-hand side and the enolate on the right connected by an equilibrium arrow, but too many students seem to think that this means all the aldehyde has reacted and not that the mixture contains every molecule in the reaction diagram.
Don’t get confused by the scheme above, in an equilibrium all the reactants are present at the same time (and I have now redrawn the scheme to show this even though the diagram is now kind of wrong/ugly). This means we have both the nucleophile and electrophile, in the form of an enolate and an aldehyde respectively, in the same flask at the same time. The nucleophilic enolate will attack the highly polarized aldehyde and a new bond is formed. The resulting alkoxide is a stronger base than hydroxide and will either deprotonate the water present or deprotonate more of the aldehyde. Either way the product β-hydroxy aldehyde is formed. Every step of this reaction is reversible, and you must be careful as the β-hydroxy aldehyde can return to two equivalents of aldehyde by the retro-aldol reaction.
A mixture containing a nucleophile and an electrophile leads to a reaction. The nucleophilic enolate attacks the electrophilic aldehyde to give an addition reaction; all atoms of the starting materials are found in the product. The reaction leads to a β-hydroxy aldehyde. The resulting alkoxide is more basic than hydroxide so will either deprotonate more aldehyde, converting it to enolate or will deprotonate water.
The product or β-hydroxy aldehyde is an aldehyde and an alcohol or ald-ol. This is the aldol addition as all atoms found in the starting materials are included in the product. The example above is the reaction of an aldehyde, the same reaction occurs equally well for ketones (and no, it isn't called a ketol reaction. It's still the aldol reaction).
The aldol reaction of two ketones in the presence of sodium hydroxide. Reaction given in the box at the top while the mechanism is below it. The reaction starts with deprotonation to give an enolate. This participates in nucleophilic attack on the remaining ketone to form a new C–C bond and an alkoxide. A second proton transfer gives the final product.
If you are not careful, and add either an excess of base or start to heat the mixture, the alcohol will be eliminated in a second reaction. This leads to the formation of an α,β-unsaturated aldehyde or ketone called an enal or enone. This reaction is the aldol condensation as water is liberated (it is also sometimes called the aldol elimination). It is a favorable process as the product is conjugated and delocalization confers greater stability on the molecule. Below shows the acid-catalyzed aldol condensation.
The aldol condensation reaction commences with the aldol addition reaction. This involves formation of an enolate that acts as a nucleophile and attacks the ketone. A proton transfer then gives the product of the aldol addition. A second proton transfer results in the formation of a new enolate. This reforms the ketone with the creation of a double C=C bond and elimination of hydroxide (overall this is loss of water). The reaction is not E2 elimination, it must go through the second enolate.
The aldol condensation reaction proceeds through the aldol addition reaction with the first step being deprotonation of the ketone to give a nucleophilic enolate. Addition of the enolate to an electrophilic ketone leads to (a badly drawn) alkoxide. Proton transfer gives the β-hydroxy ketone. This time this compound is not the product but is only an intermediate on the way to the final product. There is still base in the reaction, either hydroxide or the alkoxide and this deprotonates the β-hydroxy ketone to give a new enolate. The electrons flow back down, forming the carbonyl group then eliminating hydroxide to give the enone.
The reaction is not E2 elimination (see HERE). The hydroxide anion is a poor leaving group. In these reactions it is forced to be eliminated by the formation of a β-anion or the enolate, which effectively means the negative charge hangs around for long enough to allow the disfavored process to occur. This is an example of E1cB or Elimination Unimolecular Conjugate Base elimination mechanism. This is the anionic equivalent of an E1 elimination. It requires the formation of a relatively stable anion, in this case stablized by delocalization, in the presence of a poor leaving group in the β-position.
If an aldehyde (or non-symmetrical ketone) is used in an aldol condensation then there is a choice of stereoisomers. The reaction invariably favors formation of the trans-product as this minimizes steric hindrance by placing the bulky groups far apart. This is only the first of many issues of selectivity when you dig into either aldol addition or aldol condensation reactions.
The aldol condensation of an aldehyde can lead to two stereoisomers, the cis and trans (Z & E) forms of the enal. The major product will be the sterically least hindered or more stable compound. The mechanism is exactly the same as described for the ketone.
During the elimination step, the C–C bond between the enolate and the β-carbon with the hydroxyl group is free to rotate. The elimination will occur when the π orbital of the enolate is parallel with the σ* antibonding orbital of the C–O bond. This can occur from either face of the enolate and so the bond rotates to minimize steric interactions within the molecule.
Both the aldol addition and the aldol condensation can be promoted by acid instead of the base used above. While the product is the same, the order of the steps within the mechanism changes. The aldol addition involves protonation of the carbonyl group to form an oxonium ion. A second proton transfer, this time deprotonation, forms the enol. The enol is less reactive than an enolate but it will add to an activated carbonyl or the oxonium ion. This leads to β-hydroxy ketone product. The mechanism for the acid-catalyzed aldol condensation is shown below:
Acids can catalyze both the aldol addition and the aldol condensation. The reaction proceeds through acid-catalyzed enol formation and then nucleophilic addition to an activated electrophile, the protonated carbonyl group. After another proton transfer, this leads to the β-hydroxy ketone, the product of the addition reaction. Yet another proton transfer creates a good leaving group in the form of a hydronium ion. This undergoes elimination to give a carbocation. The final proton transfer gives the enone.
Acid-catalyzed E1 elimination gives the enone. First, proton transfer creates a good leaving group in the form of a hydronium ion. This is eliminated to form a carbocation. Deprotonation of the acidic α-proton leads to an alkene in conjugation with carbonyl group. Again, the stepwise mechanism means the most stable enone will be formed, so if there is a question of stereochemistry, the least sterically congested alkene will be formed.
The acid-catalyzed condensation reaction is normally said to go through the E1 elimination but it is possible that two other mechanisms operate depending on the structure of the leaving group; substrates that stabilize a cation will be E1 but other substitution patterns may favor a different mechanism. E2 elimination is possible. Alternatively, you could imagine a mechanism similar to E1cB, where the alcohol is protonated and the enol is formed. Deprotonation occurs during enol formation, and then the electrons flow from the oxygen of the enol causing the double bond to kick out leaving group. Personally, I prefer the E1 mechanism but understand that it is a simplification.
The aldol addition involves the addition of one carbonyl containing compound to either an aldehyde or a ketone resulting in the formation of a β-hydroxy aldehyde or ketone. The addition reaction can be promoted by either an acid or a base. The aldol condensation occurs when the β-hydroxy aldehyde or ketone undergoes an elimination reaction to give either an enal or an enone. The reaction can either be base promoted, in which case it proceeds by an E1cB mechanism with formation of an anion (the enolate) prior to elimination. Alternatively, the acid catalyzed reaction proceeds via the carbocation and an E1 mechanism.
A quick summary of the aldol addition and the aldol condensation.
The Crossed or Mixed Aldol Reaction
All the examples are above have been boring as they have involved the combination of two identical aldehydes or ketones, the carbonyl compounds have reacted with themselves (self-condensation). The aldol addition and condensation reactions are more powerful when they combine two different carbonyl compounds. Reacting two different aldehydes or ketones is known as a crossed aldol or mixed aldol reaction. Such a reaction has a potential problem, which molecule will act as the nucleophile and which as the electrophile? This can lead to a mess of products.
The crossed or mixed aldol addition involves the reaction of two different aldehydes (or ketones). If you do not take any care to control the reaction you can get a complex mixture of products.
The problem is that you have two different potential nucleophiles and two different potential electrophiles. If you only consider the reaction of aldehydes this leads to at least four possible products. Two come from the combination of two identical aldehydes (the top line) while the other two result from a crossed aldol or mixed aldol, where the different aldehydes add together (bottom line). One mixed aldol is formed if the black aldehyde is the nucleophile and the red aldehyde is the electrophile while the other mixed aldol is formed when you reverse the reactivity.
And in this case, I stressed that you were reacting aldehydes, if you had used a ketone then regiochemistry can be an issue. In fact, the situation is even worse than you think! I've totally ignored stereochemistry. Just look at the molecules above, each has two stereocentres and can exist as four stereoisomers, so I should have drawn sixteen compounds.
How can you control the product of the crossed or mixed aldol reaction?
There are many different ways to control the selectivity. Most involve making a single enolate at low temperature. As long as enolization (deprotonation) is faster than nucleophilic addition the reaction will be selective. Only after all of the carbonyl compound is converted to enolate is the electrophile added and this leads to a single combination of nucleophile and electrophile. Other solutions involve the use of enol or enolate equivalents, such as silyl enol ethers or enamines. These are compounds that can behave like enols or enolates but permit greater control by allowing formation to occur in a separate step before the electrophile has been added. I will cover these methods in a different summary.
In this summary, I only want to cover the simplest solution; picking the correct substrates so that only a single reaction is viable. This is achieved if only one aldehyde or ketone has α-protons, meaning only one substrate can become the nucleophile. The other substrate should be more electrophilic to encourage crossed aldol. Examples of the strategy are found in almost every university organic chemistry course. The reaction below is the one I’ve being demonstrating in lab for 17 years!
An example of the crossed aldol reaction taken from an undergraduate laboratory manual.
The reaction involves the addition of a ketone to a benzaldehyde derivative. The ketone is the only compound that has α-protons so it is the only molecule that can be converted to an enolate. Only one side of the ketone can be deprotonated as the other side has no α-protons (I know I'm laboring a point, but this is the only reason that this reaction can give a good yield). Deprotonation of the aldehyde is not possible under these reaction conditions.
The mechanism of the crossed aldol condensation is identical to that of a normal aldol condensation. The difference is that there are different aldehydes and ketones. In this example, the ketone will form the nucleophile as it is the only compound with protons in the α position. Without these protons it is impossible to form the enolate. The aldehyde is more electrophilic than the ketone so addition predominantly occurs to this molecule.
Under the reaction conditions, only a small quantity of enolate is formed and there are still two electrophiles present, the ketone and the aldehyde. The reaction predominantly occurs at the aldehyde because it is the better electrophile. Aldehydes possess only a single electron donating carbon substituent so the polarity of carbonyl group is not diminished much. This means the carbon is strongly partially positive and a good electrophile. Ketones have two electron donating carbon substituents and this pushes electrons onto the partially positive carbonyl carbon. This reduces the electrophilicity of the ketone. Aldehydes are less sterically demanding, which means it is easier for the nucleophile to approach. See HERE for more explanation. Both effects mean reaction occurs at the aldehyde.
Once the initial crossed aldol reaction has occurred, the condensation process is rapid. Elimination and formation of an enone leads to the molecule being fully conjugated with delocalization across the entire molecule. This is good and a favorable process.
The success of this strategy for the crossed aldol reaction relies on only one coupling partner having α-protons and the other substrate being more electrophilic.
Conclusion
The aldol addition is one of the most useful reactions in synthesis, and is employed by both chemists and nature. The β-hydroxy ketone or 1,3-diol (formed from the reduction of the carbonyl group) is common in natural products. The aldol condensation is a reliable method for the formation of activated alkenes.
The aldol addition starts with the formation of a nucleophile, an enolate or enol, that attacks an electrophilic carbonyl group. After proton transfer this leads to a β-hydroxy aldehyde or ketone. Under either acid or basic conditions, it is possible to force the reaction further and get the aldol condensation. This involves dehydration of molecule to give an enal or enone.
Both reactions can occur between identical molecules or between different carbonyl-containing compounds in what is known as the crossed or mixed aldol reaction. The crossed aldol reaction will lead to a mixture of compounds unless you find a method to control the addition. The simplest strategy involves having a single reactant that has α-protons so only one of the coupling partners can form a nucleophilic enolate or enol. The other coupling partner must be more electrophilic than the first to prevent competitive self-condensation.
In subsequent summaries I will address other, more useful, strategies for controlling the crossed aldol reaction as well as looking at variants of the aldol reaction that employ other functional groups rather than just aldehydes and ketones.