Stir ethyl acetate with sodium ethoxide, NaOEt, then add dilute acid. Two ethyl acetate molecules join into one: ethyl acetoacetate (ethyl 3-oxobutanoate), CH₃COCH₂CO₂Et. One ethoxy group is lost along the way.
This is the Claisen condensation. A condensation joins two molecules and releases a small one, here ethanol. In a Claisen, the enolate of one ester attacks a second ester.
The product is a beta-ketoester: an ester with a ketone C=O on its beta carbon. Count from the ester's C=O carbon. The next carbon along is the alpha carbon, and the one after it is the beta carbon. In ethyl acetoacetate the beta carbon is the ketone carbon.
Enolate meets ester
The first move is the one from Aldol reactions: a base removes an alpha hydrogen and makes an enolate. The difference is what the enolate attacks. In an aldol it attacks an aldehyde or ketone. In a Claisen it attacks the C=O carbon of an ester.
That ester C=O carbon carries a leaving group, the OEt. So the attack runs as nucleophilic acyl substitution, the addition–elimination mechanism of Carboxylic Acids & Derivatives. The enolate adds, then the alkoxide leaves and the C=O comes back. An aldehyde or ketone has no leaving group, so an aldol stops at the addition, and its oxygen ends up as an alcohol.
That is the whole structural difference. An aldol gives a beta-hydroxy carbonyl: an OH on the beta carbon. A Claisen gives a beta-keto ester: a second C=O in that spot. In both, the new C–C bond joins the alpha carbon of one partner to the C=O carbon of the other.
Step by step
Follow ethyl acetate through the four steps in the figure below.
Step 1. Ethoxide removes an alpha hydrogen from the CH₃ of ethyl acetate. The C–H electrons form a C=C, and the C=O electrons move onto oxygen. The result is the ester enolate. Its negative charge is shared between the alpha carbon and the oxygen.
The base is nearly always the alkoxide that matches the ester's own alkoxy group: NaOEt for an ethyl ester. Any other alkoxide would swap itself into the ester by transesterification, which is acyl substitution with an alkoxide as the nucleophile. That would scramble the ethyl ester into a mixture of esters. When ethoxide does the swap, it just gives back the same ethyl ester.
Step 2. The enolate's alpha carbon attacks the C=O carbon of a second ethyl acetate. This makes the new C–C bond. The attacked carbon becomes a tetrahedral intermediate: it now holds an O⁻, a CH₃, the new CH₂CO₂Et group and its original OEt.
Step 3. The intermediate collapses. A lone pair on the O⁻ re-forms the C=O, and the C–OEt bond breaks, so ethoxide leaves. What remains is ethyl acetoacetate, the beta-ketoester.
Step 4. The ethoxide just released removes a proton from the CH₂ between the two C=O groups. This step is what makes the reaction work at all. The next section shows why.
Workup. Dilute aqueous acid at the end puts that proton back and gives neutral ethyl acetoacetate.
Why the final deprotonation drives everything
Steps 1 to 3 are all reversible. Their overall equilibrium is not favorable, because a beta-ketoester is not much more stable than two esters. If it stopped there, the reaction would not go.
Step 4 rescues it. Look at the CH₂ of ethyl acetoacetate. It sits between two C=O groups. When a base removes one of its hydrogens, the negative charge spreads over three atoms: the central carbon, the ketone oxygen and the ester oxygen. Two of those three places are oxygens, which hold negative charge well. This is the doubly stabilized 1,3-dicarbonyl anion from Alpha hydrogens & enolates.
So that CH₂ has a pKa of about 11. Ethoxide's conjugate acid, ethanol, has a pKa of about 16. The product is five pKa units more acidic, so the equilibrium lies about 10⁵ to one toward the anion. Ethoxide removes the proton essentially completely.
Taking the product out of the equilibrium this way pulls every earlier step forward. That is Le Châtelier's principle: remove a product and the equilibrium shifts to make more of it. The anion then waits in the flask until the acid workup gives back its proton.
Which esters work: two alpha hydrogens
Compare two esters. Ethyl propanoate, CH₃CH₂CO₂Et, has a CH₂ as its alpha carbon, so it has two alpha hydrogens. Ethyl 2-methylpropanoate, (CH₃)₂CHCO₂Et, has a CH there, so it has only one.
With NaOEt, ethyl propanoate condenses normally. Making the enolate uses one of its two alpha hydrogens. The other one ends up on the carbon between the two C=O groups of the product, ethyl 2-methyl-3-oxopentanoate, and step 4 removes it.
Ethyl 2-methylpropanoate forms its enolate perfectly well, and it can even condense. But its only alpha hydrogen was spent making the enolate. The product's central carbon carries two methyl groups and no hydrogen, so step 4 cannot happen. Nothing pulls the equilibrium forward, and NaOEt gives essentially no product.
Crossed Claisens: two different partners
A crossed Claisen joins two different esters. Take ethyl benzoate, C₆H₅CO₂Et, with ethyl acetate. In ethyl benzoate the C=O carbon is bonded directly to the benzene ring, so there is no alpha hydrogen to remove. It cannot form an enolate, so it can only be the electrophile. Ethyl acetate can still condense with itself, though. So ethyl benzoate is used in excess and ethyl acetate is added slowly. That way, ethyl acetate's enolate nearly always meets ethyl benzoate, and the product is ethyl benzoylacetate (ethyl 3-oxo-3-phenylpropanoate), C₆H₅COCH₂CO₂Et.
The general problem is the one from crossed aldols. If both esters have alpha hydrogens, either one can form the enolate and either one can be attacked, so you get a mixture. There are two ways out. Use one ester with no alpha hydrogens, such as ethyl formate, diethyl oxalate or ethyl benzoate, in excess, so it is nearly always the partner that gets attacked. Or deprotonate one partner completely with a strong base before adding the other.
A ketone's enolate can also attack an ester. This is a crossed Claisen of a different kind, and it works cleanly. A ketone's alpha hydrogen (pKa 20) is more acidic than an ester's (pKa 25), so the base deprotonates the ketone first. The acidities pick the roles for you.
Reagents: cyclohexanone, (EtO)₂C=O, NaOEt, then dilute acid.
Step 1 — assign the roles. Diethyl carbonate has no alpha hydrogens, so it can only be the electrophile. Cyclohexanone's alpha C–H (pKa 20) is the most acidic C–H in the flask, so it is the nucleophile. The roles are forced, so no crossed mixture forms.
Step 2 — attack. The enolate carbon is ring carbon 2, next to the C=O. It attacks the carbonyl carbon of (EtO)₂C=O.
Step 3 — collapse. The tetrahedral intermediate expels EtO⁻. A CO₂Et group is left on that ring carbon.
Step 4 — the drive. Ring carbon 2 now sits between a ketone and an ester, at pKa 11, and ethoxide takes its proton irreversibly.
After acidic workup: ethyl 2-oxocyclohexane-1-carboxylate, a beta-ketoester built from a ketone rather than from two esters. Swap the diethyl carbonate for ethyl formate, HCO₂Et, and the same sequence puts a CHO in that position instead.
The Dieckmann condensation: closing a ring
When both ester groups belong to the same molecule, the enolate at one end can reach the ester at the other end. The Claisen then closes a ring. This intramolecular version is the Dieckmann condensation.
Substrate: EtO₂C–CH₂CH₂CH₂CH₂–CO₂Et. Number the chain C1 to C6. The two ester carbonyls are C1 and C6.
Step 1: NaOEt removes an alpha hydrogen from C2 (or from C5, which gives the same result by symmetry). This makes the ester enolate.
Step 2: C2 attacks C6. The ring that closes is C2, C3, C4, C5, C6: five atoms. C1 stays outside the ring as the ester.
Step 3: the tetrahedral intermediate expels EtO⁻, and the C=O at C6 comes back, now as a ketone inside the ring.
Step 4: EtO⁻ removes the hydrogen on C2, which now sits between the ring ketone and the ester it kept. This is the irreversible step, and it drives the ring closure exactly as it drives the two-molecule Claisen. If the attacking carbon had no hydrogen left at this point, the diester would have the same problem as ethyl 2-methylpropanoate.
Workup: dilute acid gives ethyl 2-oxocyclopentane-1-carboxylate, a cyclopentanone with a CO₂Et group on the carbon next to the ketone.
Count the ring before you draw anything. As in Alkanes & Conformations, five- and six-membered rings form readily. Three- and four-membered rings are too strained to form. Rings of seven or more carry little strain, but the two ends rarely meet, so the two-molecule Claisen competes and yields drop.
So diethyl heptanedioate, one CH₂ longer than adipate, closes a six-membered ring. Its product is ethyl 2-oxocyclohexane-1-carboxylate, the same compound the diethyl carbonate example made. Diethyl pentanedioate, one CH₂ shorter, would have to close a four-membered ring, and it does not react this way.
The same reaction in your cells
Fatty acid biosynthesis is a Claisen condensation, run once for every two carbons added to the chain. The electrophile is a thioester, R–CO–SR′: an ester with sulfur in place of the single-bonded oxygen. In the first round it is an acetyl thioester that comes from acetyl-CoA; in later rounds it is the growing chain. A malonyl thioester provides the nucleophile.
Two details differ from the flask. Thioesters are used rather than ordinary esters because their leaving group is better. And the nucleophile forms when the malonyl thioester loses CO₂, not by deprotonation. Cells arrived at the same reactivity order of acid derivatives that Carboxylic Acids & Derivatives set out.
What carries forward
With the Claisen in hand you have most of this course's carbon–carbon bond-forming reactions: acetylide alkylation, Grignard and organolithium addition, cuprate conjugate addition, cross-coupling, the Wittig, the Diels–Alder, the aldol and now the Claisen, with the Michael reaction still to come at the end of this chapter. Between them they can build most carbon skeletons. Recognizing which one made a given bond is the central skill of retrosynthesis, working backward from a target molecule to the reactions that could build it.
Hold on to ethyl acetoacetate. It is the starting material for the acetoacetic ester synthesis, one of the classical routes to substituted ketones, covered in Malonic and acetoacetic ester later in this chapter. There you alkylate its stabilized anion, then hydrolyze the ester to a beta-keto acid and warm it. The acid loses CO₂ through the six-membered cyclic transition state of Carboxylic Acids & Derivatives. The malonic ester synthesis starts from a beta-dicarbonyl too.