Claisen reactions — Enolate Chemistry

The Claisen condensation is the aldol reaction's ester counterpart, and the difference between them is exactly the difference between addition and substitution. Same nucleophile, same first step, different fate for the tetrahedral intermediate — which is a good final illustration that carbonyl chemistry has fewer independent ideas in it than it appears to.

Enolate meets ester

The Claisen reacts an ester's enolate with a second ester molecule as the electrophile. Because the electrophile is an ester and carries a leaving group, the attack proceeds by nucleophilic acyl substitution — the addition–elimination mechanism of Module 10 — rather than by simple addition. The alkoxide leaves, the carbonyl reforms, and the product retains a ketone where the aldol would have left an alcohol.

That is the whole structural difference. An aldol gives a beta-hydroxy carbonyl; a Claisen gives a beta-keto ester. Both join two molecules with a new C–C bond two carbons apart.

Step by step

Step 1. Base removes an alpha hydrogen, forming the ester enolate. The base is nearly always the alkoxide matching the ester's own alkoxy group — NaOEt for an ethyl ester — and the reason is practical: using a different alkoxide would scramble the ester by transesterification.

Step 2. The enolate carbon attacks a second ester's carbonyl carbon, forming a tetrahedral intermediate that now carries both the new carbon substituent and the original OR.

Step 3. The intermediate collapses, expelling the alkoxide and reforming the carbonyl. The product is a beta-ketoester.

Step 4. The alkoxide just released deprotonates the product — and this step is what makes the reaction work at all.

The alkoxide released in step 3 is the same base consumed in step 1, so the sequence is nominally catalytic in base — right up until step 4, which consumes a full equivalent permanently. This is why a Claisen is run with a stoichiometric amount of alkoxide, not a catalytic one, and why textbooks that call it base-catalyzed are being imprecise.

Why the final deprotonation drives everything

Claisen is the aldol with an ester on the receiving end — so instead of stopping at a tetrahedral intermediate, the alkoxide gets expelled and a second carbonyl survives. The product is a β-keto ester, and the hydrogens between its two carbonyls are the most acidic thing in the flask at pKa 11. The base takes one, and because that step does not go back, it drags the whole reversible sequence forward with it.Every step before this one is reversible and none of them is downhill by much. What makes Claisen go is the LAST step: the product has a hydrogen flanked by two carbonyls, acidic enough that the base removes it irreversibly — and that pulls the whole equilibrium forward. It is also why you need a full equivalent of base.

Every step up to this point is reversible, and the equilibrium is not particularly favourable — a beta-ketoester is not much more stable than two esters. Left alone, the reaction would not go.

What rescues it is the product's acidity. A beta-ketoester's alpha hydrogen sits between two carbonyls, so its enolate is doubly delocalized, and its pKa is about 11. The alkoxide base present, whose conjugate acid is an alcohol at pKa 16, deprotonates it essentially completely — five pKa units, an equilibrium of about 10⁵ to one.

That deprotonation removes the product from the equilibrium entirely, pulling the whole reversible sequence forward by Le Châtelier. An aqueous acid workup at the end reprotonates the stabilized anion and reveals the neutral beta-ketoester.

A Claisen requires two alpha hydrogens on the nucleophilic ester. The first is removed to form the enolate; the second must survive into the product so that step 4 can remove it. An ester with only one alpha hydrogen — such as one with a fully substituted alpha carbon — can form an enolate and can even condense, but the product cannot be deprotonated, the equilibrium is never pulled forward, and the yield is negligible. This is a favourite exam question precisely because the reason is mechanistic rather than steric.

Worked example — ethyl acetate with itself

Step 1: NaOEt removes an alpha H from CH₃CO₂Et, giving ⁻CH₂CO₂Et.

Step 2: that carbon attacks a second ethyl acetate's carbonyl.

Step 3: the tetrahedral intermediate expels EtO⁻, giving ethyl acetoacetate, CH₃COCH₂CO₂Et.

Step 4: EtO⁻ deprotonates the product between its two carbonyls. Irreversible, and the reaction is driven.

Workup: dilute acid gives neutral ethyl acetoacetate.

This compound is the starting material for the acetoacetic ester synthesis, one of the classical routes to substituted ketones — alkylate the stabilized anion, then hydrolyze and decarboxylate.

Crossed and intramolecular variants

A crossed Claisen between two different esters needs the same kind of control as a crossed aldol, and the same solutions apply: use one ester with no alpha hydrogens — ethyl formate, diethyl oxalate, ethyl benzoate — so it can only be the electrophile, or deprotonate one partner completely with a strong base before adding the other.

When both ester groups belong to the same molecule, the intramolecular version is a Dieckmann condensation and it forms a ring. As in Module 4, five- and six-membered rings form readily and larger or smaller ones do not, because the transition state has to bring the two ends together without strain.

A ketone's enolate can also attack an ester, which is a crossed Claisen of a different kind — and it works cleanly because a ketone (pKa 20) is deprotonated preferentially over an ester (pKa 25), so the roles are assigned by acidity rather than by design.

The same reaction in your cells

Fatty acid biosynthesis is a Claisen condensation, run once per two carbons added. The thioester of acetyl-CoA is the electrophile, a malonyl thioester provides the nucleophile, and the chain grows two carbons at a time. Thioesters are used rather than ordinary esters because their alpha hydrogens are more acidic and their leaving group is better — evolution reached the same conclusion about the reactivity ladder that Module 10 did.

What carries forward

The Claisen completes this course's set of carbon–carbon bond-forming reactions: acetylide alkylation, Grignard addition, aldol, Michael, Claisen. Between them they can build most carbon skeletons, and recognizing which one made a given bond is the central skill of retrosynthesis. The beta-ketoester products are also the entry point to the acetoacetic and malonic ester syntheses, if your course covers them.