Malonic and acetoacetic ester — Carbonyl & Enolate Breadth

Two named sequences, one idea. Both take a compound whose α hydrogen sits between two carbonyls, alkylate that carbon, and then remove one of the carbonyls to leave a product that could not have been made directly.

The carbonyl that gets removed was never wanted. It was there to make the hydrogen acidic, and once the alkylation is done it is thrown away. That is the whole trick, and it is worth naming: a group installed to enable a step and then deleted is an activating group, and this is the clearest example of one in the course.

Why two carbonyls

A plain ester has an α hydrogen with a pKa near 25 — too high to deprotonate with anything convenient, and high enough that a strong base would attack the ester instead. Put a second carbonyl on the other side and the resulting carbanion is delocalized onto two oxygens:

Compound	α pKa	Deprotonated by
Ester	~25	LDA, and even then awkwardly
Ketone	~20	LDA
Diethyl malonate	~13	Sodium ethoxide
Ethyl acetoacetate	~11	Sodium ethoxide

Thirteen against twenty-five is twelve orders of magnitude, and it is the difference between needing an exotic base and needing the alkoxide that matches your solvent.

Malonic ester synthesis → a carboxylic acid

What the second carbonyl is for. An ester’s α hydrogen sits at pKa 25, where the only bases strong enough will also attack the ester; flanking that carbon with a second carbonyl delocalizes the carbanion onto a second oxygen and drops it to 11 to 13, where the alkoxide matching your solvent does the job.The carbonyl that made this possible is gone from the product. It was installed to acidify one hydrogen and is removed by hydrolysis and decarboxylation once the alkylation is done — which is what an activating group is, and the clearest example of one in the course. Note also that the decarboxylation works only because the intermediate is a β-keto acid, able to reach a six-membered cyclic transition state.

Start from diethyl malonate, CH₂(CO₂Et)₂.

NaOEt removes the α hydrogen.
R–X alkylates that carbon by S_N2. Repeating both steps installs a second group if you want one.
H₃O⁺, heat hydrolyzes both esters to acids and then decarboxylates.

The net result is a carboxylic acid carrying whatever you alkylated with: R–CH₂–COOH. Reading it backwards, any carboxylic acid with at least one α hydrogen can be disconnected at the bond β to the carboxyl.

Acetoacetic ester synthesis → a methyl ketone

The same three steps from ethyl acetoacetate, CH₃COCH₂CO₂Et, give a methyl ketone: R–CH₂–CO–CH₃. Only the ester is hydrolyzed and lost; the ketone survives, because a ketone has no carboxyl to lose.

One starting material decides the product class. Malonate has two esters, loses one carboxyl and keeps the other as the acid. Acetoacetate has one ester and one ketone, loses the ester's carboxyl and keeps the ketone. Learn which compound goes with which product and the rest of both sequences is identical.

Why the decarboxylation works

Carboxylic acids are not generally unstable to heat. One carrying a second carbonyl β to the carboxyl is, and the reason is a six-membered cyclic transition state: the carboxyl O–H hydrogen reaches the other carbonyl's oxygen, CO₂ leaves, and an enol forms which tautomerizes to the product.

That geometry requires the second carbonyl to be exactly β, and it does not care whether that carbonyl is a ketone or another carboxyl. The acetoacetic route passes through a genuine β-keto acid; the malonic route passes through a substituted malonic acid, which is a 1,3-diacid. Same six-membered transition state, same outcome, two different names. An α- or γ-keto acid reaches neither, and neither does a plain carboxylic acid.

The S_N2 limit, which is the real constraint

The alkylation step is an S_N2 by a stabilized enolate, so the halide must be methyl, primary, or allylic/benzylic. A secondary halide gives mostly elimination and a tertiary one gives nothing else, because the enolate is a base as well as a nucleophile.

So a question asking for a carboxylic acid with a tertiary group on the α carbon cannot be answered with a malonic ester synthesis, however neatly the disconnection looks. That is the first thing to check, and it is checked on the halide rather than on the target.

What carries forward

Deprotonate, alkylate, hydrolyze and decarboxylate. Malonic ester gives a carboxylic acid, acetoacetic ester gives a methyl ketone, and both are limited by the S_N2. Retrosynthetically: a carboxylic acid with an α substituent, or a methyl ketone with one, should send you looking for the halide that installed it.