Add a little sodium hydroxide to cold acetaldehyde, CH₃CHO. Two molecules join into one: 3-hydroxybutanal, CH₃CH(OH)CH₂CHO. Two molecules of two carbons each have become one molecule of four.
That is the aldol reaction. It joins the alpha carbon of one carbonyl compound to the carbonyl carbon of another. The alpha carbon is the carbon bonded directly to the C=O carbon, as in Alpha hydrogens & enolates.
The aldol is one of the main ways chemists build carbon–carbon bonds. Living cells use it too: photosynthesis runs aldols forwards to build sugars, and cells run them backwards to break sugars down.
The enolate as a carbon nucleophile
Hydroxide removes an alpha hydrogen from acetaldehyde and leaves an enolate: an anion whose negative charge is shared between the alpha carbon and the oxygen. Alpha hydrogens & enolates draws its two resonance contributors.
Most of the charge sits on oxygen, but the enolate forms its new bond at carbon. So the alpha carbon behaves like the carbon nucleophiles you already know, such as an acetylide or a Grignard reagent from Organometallic Bonding & Grignard Reagents. Given an electrophile, it attacks.
In the aldol, the electrophile is another carbonyl molecule. One compound plays both roles at once: some of its molecules become enolates and act as the nucleophile, while the rest stay as neutral carbonyl molecules and act as the electrophile.
The aldol reaction, step by step
Follow acetaldehyde through the three steps. The figure draws each one with its curved arrows.
Step 1. Hydroxide removes an alpha hydrogen from one acetaldehyde. This gives the enolate and a molecule of water.
Step 2. The enolate's alpha carbon attacks the carbonyl carbon of a second acetaldehyde. This is ordinary nucleophilic addition: the C=O pi electrons move onto oxygen. The attacked carbon is now tetrahedral (sp³), and its oxygen carries the negative charge. An O⁻ bonded to carbon like this is an alkoxide.
Step 3. The alkoxide takes a proton from water (or from an alcohol solvent). Hydroxide comes back out, so the base acts as a catalyst and is not used up.
Now number the product from the carbonyl carbon that survived. That carbon is C1. The alpha carbon that did the attacking is C2. The new OH sits on C3, the carbon after the alpha carbon, which is called the beta carbon. So the product is a beta-hydroxy carbonyl (β-hydroxy carbonyl). Every aldol addition gives this pattern, whatever the starting compounds.
The base is weak on purpose. Hydroxide or an alkoxide turns only a small fraction of the carbonyl compound into enolate, so plenty of neutral carbonyl is left to be the electrophile. Alpha hydrogens & enolates shows why these bases give only a little enolate.
Every step can run backwards. The reverse reaction, the retro-aldol, splits a beta-hydroxy carbonyl into two carbonyl compounds. It is not just a textbook idea: the enzyme aldolase runs it in glycolysis, cutting a six-carbon sugar into two three-carbon pieces.
Where the equilibrium sits depends on the carbonyl. For an aldehyde, the addition is downhill, and the beta-hydroxy aldehyde can be isolated in good yield. For a ketone, it is uphill. Acetone with catalytic hydroxide gives only a few percent of its aldol product, diacetone alcohol, at equilibrium. That product is crowded, and a ketone C=O is more stable than an aldehyde C=O from the start.
So most ketone aldols are run hot, all the way to the condensation in the next section. The dehydration is strongly downhill, so it pulls the unfavorable addition along with it.
Aldol condensation: heat pushes further
Warm the 3-hydroxybutanal in base, and it loses a molecule of water. The product is (E)-but-2-enal (crotonaldehyde), CH₃CH=CHCHO. Its new C=C sits right beside the C=O.
The whole sequence, aldol addition and then loss of water, is the aldol condensation. A condensation is any reaction that joins two molecules and expels a small one, here water.
The loss of water follows a pattern called E1cb, short for elimination, unimolecular, conjugate base. "Conjugate base" means the molecule loses a proton first, and the anion that forms then loses the leaving group. That order sets it apart from E1 and E2, which you met in Substitution & Elimination.
In the aldol, E1cb takes two steps. First, hydroxide removes the alpha hydrogen, the one on the carbon between the C=O and the carbon that carries the OH. That gives an enolate. Second, the enolate's electrons move back toward the alpha carbon and push hydroxide off the beta carbon.
Hydroxide is normally a very poor leaving group, too poor for E1 or E2. It leaves here only because the enolate pushes it out, and that enolate can form in the first place because the carbonyl stabilizes it.
The H always comes from the alpha carbon and the OH from the beta carbon. So the new C=C always forms between Cα and Cβ, directly beside the C=O.
The product is an enone, a C=C conjugated with a C=O. A more exact name is an α,β-unsaturated carbonyl: the double bond joins the alpha and beta carbons. This chapter uses "enone" loosely for aldehydes as well as ketones. (E)-But-2-enal, for example, is an α,β-unsaturated aldehyde, not a ketone.
The driving force is conjugation. The C=C and C=O pi systems overlap, so the enone is more stable than the beta-hydroxy carbonyl it came from. That extra stability pulls the dehydration forward.
Conditions decide where you stop. A low temperature and a short reaction time give the beta-hydroxy carbonyl. Heat gives the enone.
The condensation gives almost only the E alkene, with the two larger groups on opposite sides of the C=C. The transition state that leads to it is the less crowded one.
Target: 4-phenylbut-3-en-2-one, PhCH=CH–CO–CH₃. Ph stands for phenyl, a benzene ring used as a substituent.
Step 1 — find the C=C conjugated to the C=O, and put the water back. That C=C came from the dehydration. Add OH to the carbon further from the C=O (the beta carbon) and H to the nearer one (the alpha carbon). That gives PhCH(OH)–CH₂–CO–CH₃.
Step 2 — cut the alpha–beta bond. It is the only bond the aldol made. The alpha side keeps its carbonyl and takes back a hydrogen. On the beta side, the C–OH becomes a C=O again.
Step 3 — read off the roles. The alpha side is acetone, CH₃COCH₃. It supplied the alpha carbon, so it was the nucleophile. The beta side is benzaldehyde, PhCHO. Its carbonyl carbon was attacked, so it was the electrophile.
Check that the reaction makes sense. Benzaldehyde has no alpha hydrogens, so it can only ever be the electrophile. That makes this a clean crossed aldol, the kind the section after next explains. Answer: acetone and benzaldehyde, NaOH, heat.
The acid-catalyzed aldol
Acid reaches the same products by a mirror-image route. Take acetone in dilute aqueous acid. An acid solution contains no strong base, so no enolate forms. The nucleophile is the neutral enol instead: the tautomer with a C=C and an OH, from Alpha hydrogens & enolates.
The enol is only a trace. For acetone, about one molecule in 10⁸ is enol at any moment. Acid keeps making more as the enol is used up, so the reaction still runs.
Acid also improves the electrophile. H₃O⁺ puts a proton on the oxygen of a second acetone, and a protonated carbonyl is far more electrophilic than a neutral one.
The enol's C=C attacks the carbonyl carbon of the protonated acetone. The enol's own oxygen is left as C=OH⁺, and water takes that proton. The result is the neutral beta-hydroxy ketone, 4-hydroxy-4-methylpentan-2-one, and H₃O⁺ comes back.
Acid also removes water from that product, by a route of its own. The ketone forms its enol again, and H₃O⁺ protonates the beta OH so it can leave as water. The enol's C=C pushes in as the water leaves. Water then takes the last O–H proton, giving the enone 4-methylpent-3-en-2-one (mesityl oxide). (A tertiary β-hydroxy ketone like this one can also lose water through a carbocation; both paths give the same enone.) In practice, acid almost always runs straight through to the enone.
Set the two routes side by side. Base improves the nucleophile by taking a proton off. Acid improves the electrophile by putting one on.
Crossed aldols and why they need control
Mix acetone and propanal, CH₃CH₂CHO, with sodium hydroxide. Both have alpha hydrogens, so both form enolates. Each enolate can attack either carbonyl, so four different aldol products form.
An aldol between two different carbonyl compounds is a crossed aldol. Run this way, it gives a mixture that is hard to separate, so it is not a useful reaction. Two strategies give control.
Start with one fact: an aldehyde is the better electrophile. It has one alkyl group on the carbonyl carbon where a ketone has two, so its carbonyl carbon is less hindered and less electron-rich. That preference alone does not give one product, though: an aldehyde that has alpha hydrogens can still form its own enolate and attack another aldehyde.
Strategy 1: use one partner with no alpha hydrogens. Benzaldehyde, formaldehyde and 2,2-dimethylpropanal, (CH₃)₃CCHO, cannot form enolates, so they can only be the electrophile. Mix benzaldehyde with a ketone and base, and one product forms cleanly. The ketone's enolate could attack another ketone, but that addition is uphill, and the aldehyde is the better electrophile anyway. The enone that forms is conjugated with the ring as well, so it dehydrates readily.
This case has its own name, the Claisen–Schmidt condensation. It is named after the same Claisen as the next section, but it is an aldol, not a Claisen.
Strategy 2: make one enolate first, with LDA. Treat one partner with LDA at −78 °C. LDA turns it completely and irreversibly into its enolate; Alpha hydrogens & enolates explains why. Only then add the second partner. The first compound is now all enolate, so none of it is left to be an electrophile. The second compound arrives after the base is used up, so it never forms an enolate. This is the general, modern solution.
Intramolecular aldol: the same reaction, closing a ring
Hexanedial, OHC–CH₂CH₂CH₂CH₂–CHO, has an aldehyde at each end. In base, an enolate at one end can reach the carbonyl at the other end. The aldol then closes a ring instead of joining two molecules. A reaction between two groups in the same molecule is called intramolecular.
Nothing about the mechanism changes. The only new question is which enolate attacks which carbonyl.
The answer is ring size, and the preference is the one from Alkanes & Conformations. Five- and six-membered rings close readily. Three- and four-membered rings are too strained. Rings of seven or more need two chain ends to find each other, which is unlikely, so they do not compete. So work out every enolate the substrate can form, count the ring each one would close, and keep the five or the six.
Substrate: OHC–CH₂CH₂CH₂CH₂–CHO. Number it C1 to C6, with the two CHO carbons as C1 and C6.
Step 1: the alpha carbons are C2 and C5. Take the C2 enolate.
Step 2: C2 attacks C6. Count the ring that closes: C2, C3, C4, C5, C6. That is five atoms, so it works.
The alternatives: C2 attacking C1 makes no ring at all, since the two are already bonded. The C5 enolate is the same by symmetry: attacking C1, it closes the same five-membered ring.
Step 3: protonation gives 2-hydroxycyclopentane-1-carbaldehyde, a five-membered ring carrying an OH, with a CHO on the carbon next to it.
With heat: E1cb dehydration puts the new C=C inside the ring, conjugated with the CHO. The product is cyclopent-1-ene-1-carbaldehyde.
Learn to recognize a ring with an enone in it as the mark of an intramolecular aldol condensation. It is the single most useful pattern in synthesis questions on this chapter.
Conjugate addition: the other place a nucleophile can attack
The enone from an aldol condensation is itself an electrophile, and it has two electrophilic carbons. Take but-3-en-2-one, CH₂=CH–CO–CH₃.
The carbonyl carbon is the first site. A nucleophile that attacks it does 1,2-addition, the ordinary carbonyl addition. The numbers count from oxygen: O is atom 1, the carbonyl carbon 2, the alpha carbon 3 and the beta carbon 4.
The beta carbon is the second site. A nucleophile that attacks it does 1,4-addition, also called conjugate addition. It is the same numbering idea as 1,2- vs 1,4-addition to dienes, now counted from oxygen.
Why is the beta carbon electrophilic? Push the C=C electrons toward the carbonyl, and the C=O pi electrons onto oxygen. The resonance contributor you get has its positive charge on the beta carbon.
Both contributors describe the same molecule, so both sites are electrophilic at once. The question is which one a given nucleophile takes.
In that contributor, the negative charge sits on oxygen. So when a nucleophile adds at the beta carbon, the anion it makes is an enolate. That enolate can then be protonated, or it can attack an electrophile of its own.
The choice depends on whether the nucleophile is hard or soft. A hard nucleophile is small and highly reactive, with its charge packed onto one atom. Organolithiums and hydride reagents are hard, and they tend to attack the carbonyl carbon directly.
A soft nucleophile is less reactive, with its electrons spread over several atoms or held on a large atom. Enolates, thiolates (RS⁻) and the organocopper reagents called cuprates (R₂CuLi) are soft, and they tend to add at the beta carbon. Amines also add 1,4, largely because their 1,2-addition is reversible. Organolithiums, Cuprates & Cross-Coupling worked through why cuprates choose the beta carbon, and the same argument applies to these enones.
Two named reactions come from this, and the rest of this chapter uses both. A conjugate addition whose nucleophile is an enolate is the Michael reaction. A Michael reaction followed by an intramolecular aldol condensation is the Robinson annulation, the standard way to build a new six-membered ring onto a molecule (an annulation is a reaction that builds a ring). Michael and Robinson covers both.
What carries forward
The aldol follows a simple pattern: make an enolate, let it attack a carbonyl, and join two fragments. The Claisen condensation in the next section follows the same pattern with an ester as the electrophile, and so does a great deal of biological chemistry.
The choice between 1,2- and 1,4-addition is one more competition between two sites on one molecule. Another comes in Aromatic Chemistry, where a group already on a benzene ring decides which ring position the next group takes.