Malonic and acetoacetic ester

Say you want butanoic acid, CH₃CH₂CH₂COOH, and you plan to make it by putting an ethyl group on the α carbon of acetic acid. The α carbon is the carbon bonded to the C=O. The obvious starting ester is ethyl acetate, but its α hydrogens are hard to remove: their pKa is near 25. Only a very strong, bulky base such as LDA (lithium diisopropylamide) removes them fully, at −78 °C. A smaller base such as ethoxide removes only a trace, and that enolate attacks another ester molecule instead (a Claisen condensation).

Diethyl malonate gets round this. Its CH₂ sits between two ester carbonyls, so sodium ethoxide removes one of its hydrogens easily. That carbon takes the ethyl group. Later, one of the two esters is removed completely, as CO₂, and butanoic acid is left.

The carbonyl that leaves was never wanted in the product. It was there only to make one hydrogen acidic. A group put into a molecule to make one step possible, and removed once that step is done, is called an activating group. This is the clearest example of one in the course.

This section covers two sequences built on that idea. The malonic ester synthesis starts from diethyl malonate and makes carboxylic acids. The acetoacetic ester synthesis starts from ethyl acetoacetate and makes methyl ketones.

Why two carbonyls

Compare the three esters in the figure. In ethyl acetate the α carbon touches one carbonyl. In diethyl malonate and ethyl acetoacetate it sits between two. When a base removes a hydrogen from that carbon, the negative charge spreads onto both carbonyl oxygens instead of one. A more spread-out charge means a more stable anion, and so a more acidic hydrogen. The anion is an enolate, like those in Alpha hydrogens & enolates, but with two carbonyls to share the charge. That chapter draws its resonance forms.

The highlighted carbon in each is the α carbon.

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

Each pKa unit is a factor of ten. So 13 against 25 is twelve units, which makes the malonate hydrogen about 10¹² times more acidic. That is the difference between needing LDA at −78 °C and getting by with the alkoxide that matches your solvent.

Malonic ester synthesis → a carboxylic acid

Here is the whole route to butanoic acid, one step per panel of the figure below.

NaOEt removes one α hydrogen (panel 1). Use the alkoxide that matches the ester: ethoxide with an ethyl ester, methoxide with a methyl one. A different alkoxide swaps itself into the ester (transesterification) and turns the starting material into a mixture.
CH₃CH₂Br then reacts with this stabilized enolate in an S_N2 (panel 2). The lone pair on the α carbon forms a bond to the CH₂ of the halide, and bromide leaves. The product, diethyl ethylmalonate, still has one α hydrogen (panel 3).
H₃O⁺, heat hydrolyzes both esters to COOH groups (panel 4). The product, ethylmalonic acid, has two COOH groups on one carbon. It is a 1,3-diacid: number its three-carbon backbone and the two acid carbons are C1 and C3.
The same heating then removes one COOH as CO₂ (panels 5 and 6). Losing CO₂ this way is called decarboxylation. The next section explains how it works.

Diethyl malonate plus ethyl bromide gives butanoic acid. Follow the highlighted α carbon: it takes the ethyl in panel 2, keeps it through the hydrolysis, and loses one of its two carboxyls in panel 5.

The figure draws the anion with its negative charge on carbon, because carbon is the atom that forms the new bond. In the real anion that charge is shared with both carbonyl oxygens.

Some courses write the hydrolysis as two operations: NaOH in water to saponify the esters, then H₃O⁺ and heat. That comes to the same thing, because the acid has to be free before it can decarboxylate.

The general product is R–CH₂–COOH, where R is the group from the halide. Malonate's CH₂ has two acidic hydrogens, so you can repeat NaOEt and a second halide, R′–X, before the hydrolysis. That gives R(R′)CH–COOH. Two groups is the limit.

Why the decarboxylation works

Ordinary carboxylic acids survive heating. Ethylmalonic acid does not. Look at panel 5 above. The molecule curls so that six atoms form a ring: the H of one COOH, that COOH's oxygen and carbon, the α carbon, and the other carboxyl's carbon and C=O oxygen.

This is not the molecule falling apart in the heat. It is a cyclic proton transfer: three pairs of electrons move at once, all the same way round that ring.

The C=O π electrons of the carboxyl that stays reach the H and make a new O–H bond.
The old O–H electrons become the second C=O bond of CO₂.
The C–C bond from the α carbon to the leaving carboxyl breaks. Its electrons move over to make a C=C between the α carbon and the carbon of the carboxyl that stays.

CO₂ leaves as a gas. What remains is an enol: a C=C with OH on one of its carbons (here two OH groups, because that carbon came from a carboxyl). The enol quickly shifts to the more stable C=O form, and the α carbon picks up a hydrogen as it does (panel 6). That shift is the keto–enol tautomerization from Alpha hydrogens & enolates.

You met this ring for β-keto acids in Carboxylic acids. It needs a second carbonyl β to the carboxyl, which means bonded to the α carbon. That carbonyl can be a ketone or another carboxyl. The acetoacetic route passes through a true β-keto acid. The malonic route passes through a substituted malonic acid, which is a 1,3-diacid, not a keto acid. The ring and the outcome are the same; only the names differ.

A carbonyl in any other position does not work. The electrons of the breaking C–C bond must flow into a C=O right next to them. In an α-keto acid the ring would have only five atoms, and in a γ-keto acid seven. Neither has a second C=O beside the α carbon to take the electrons, and a plain carboxylic acid has none at all.

The highlighted bond is the one that would break. The Greek letters count carbons outward from the COOH.

This is also why the esters must be hydrolyzed first. The ring closes through the carboxyl O–H, and an ester has no O–H.

Two alkylations, one dihalide: a ring

Treat diethyl malonate with two equivalents of NaOEt and one of 1,4-dibromobutane, Br–CH₂CH₂CH₂CH₂–Br.

The first alkylation uses one end of the chain (panel 1 below). It is intermolecular: it happens between two molecules. The α carbon now carries a four-carbon chain with a bromine on its far end.

The second NaOEt removes the last α hydrogen. The new anion attacks the far end of its own chain, and bromide leaves (panel 2). Both ends of this S_N2 are in one molecule, so it is intramolecular, and it closes a ring.

Hydrolysis and heat then give cyclopentanecarboxylic acid (panel 3).

The numbers 1 to 5 follow the same atoms through all three panels. Atom 1 is the malonate α carbon.

Count the ring before you answer. It holds the malonate α carbon plus every carbon of the dihalide chain. So 1,4-dibromobutane gives a five-membered ring, and 1,5-dibromopentane gives a six.

Acetoacetic ester synthesis → a methyl ketone

Ethyl acetoacetate, CH₃COCH₂CO₂Et, has a ketone on one side of its CH₂ and an ester on the other. It is the product of the Claisen condensation of ethyl acetate (Claisen reactions).

Run the same steps on it with ethyl bromide. NaOEt, then CH₃CH₂Br, puts the ethyl on the CH₂ (panel 1 below). H₃O⁺ hydrolyzes the ester to a COOH, which makes a β-keto acid. Heat removes that COOH as CO₂ through the same six-atom ring, but this time the ketone oxygen takes the H (panel 2). The enol then becomes pentan-2-one (panel 3).

Ethyl acetoacetate plus ethyl bromide gives pentan-2-one. The ketone is never touched: only the ester side is hydrolyzed and lost as CO₂.

The ketone survives because it cannot be hydrolyzed. Only an ester has an OR group that water can replace with OH, so only the ester side becomes a carboxyl and leaves.

The general product is a methyl ketone, R–CH₂–CO–CH₃: a ketone with a CH₃ on one side. The CH₃–CO and the CH₂ come from acetoacetate, and R comes from the halide. As with malonate, the CH₂ can take two groups.

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.

The S_N2 limit, which is the real constraint

Try diethyl malonate with tert-butyl bromide, (CH₃)₃C–Br. The carbon that holds the bromine carries three methyl groups, which block the back-side approach an S_N2 needs (SN2). So the anion takes a hydrogen from one of the methyls instead. The electrons of that C–H bond form the new C=C, and bromide leaves. This is an E2 elimination (E2). You get 2-methylpropene, and the malonate back unchanged.

Three curved arrows, one E2. The anion ends up with a hydrogen, not a new carbon group.

The general rule follows. The alkylation is an S_N2 by a stabilized enolate, so the halide must be methyl, primary, allylic or benzylic. Allylic means the carbon next to a C=C; benzylic means the carbon next to a benzene ring. A secondary halide reacts slowly, and elimination competes. A tertiary halide gives only elimination, because the anion 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 neat the disconnection looks. Check the halide, not the target.

Reading a target backwards

Worked example — make 4-methylpentanoic acid, (CH₃)₂CH–CH₂–CH₂–COOH

Find the α carbon. It is the CH₂ next to the COOH. It carries two hydrogens and one substituent, so this needs one alkylation.

Cut the bond from it to its substituent (top of the figure). The piece that keeps the COOH, CH₂–COOH, is what malonate supplies. The other piece, –CH₂CH(CH₃)₂ (isobutyl), names the halide: 1-bromo-2-methylpropane.

Check the halide. It is primary. It branches at the next carbon along, so its S_N2 is slower than an unbranched primary halide's. It is not secondary, though, so the reaction still works.

Write the sequence. (1) NaOEt, EtOH; (2) (CH₃)₂CHCH₂Br; (3) H₃O⁺, heat.

Check the trap. 3,3-Dimethylbutanoic acid, (CH₃)₃C–CH₂–COOH, differs only in where the branching sits. Its cut hands you a tert-butyl group (bottom of the figure), so it would need tert-butyl bromide. That halide is tertiary, so this target has no route.

The dashed mark cuts the bond from the α carbon to its substituent. Top: a route. Bottom: the near miss with no route.

The general method: find the α carbon of the target acid. It must still carry at least one hydrogen, because the decarboxylation puts one there. Cut the bond from it to each substituent. Each cut names one alkyl halide. Whatever is left, the α carbon with its hydrogen and its COOH, is always malonate's. A methyl ketone reads the same way: cut the substituents off the carbon next to the C=O on the side away from the CH₃, and CH₂–CO–CH₃ (CH–CO–CH₃ after two alkylations) is left for acetoacetate.

What carries forward

Every route runs the same four steps: deprotonate, alkylate, hydrolyze, decarboxylate. The S_N2 step limits all of them. When a target is a carboxylic acid or a methyl ketone with a substituent on the α carbon, look for the halide that installed it.