Malonic ester synthesis: practice and mechanism
The malonate enolate is alkylated, then hydrolysed and decarboxylated. Overall the alkyl group of the haloalkane is attached to CH₂COOH: R–X becomes R–CH₂–COOH.
Example
The malonate enolate is alkylated by the haloalkane via SN2. Acidic hydrolysis gives the malonic acid, which loses CO₂ on heating (decarboxylation). Result: a carboxylic acid with two carbons more than the haloalkane (R–CH₂–COOH).
Mechanism step by step
- 1
Ethoxide deprotonates diethyl malonate to the stabilised enolate.
- 2
The enolate attacks the haloalkane via SN2: an alkylated malonate.
- 3
Heating in acid hydrolyses the esters to the β-keto acid or malonic acid, which loses CO₂ through a six-membered transition state (decarboxylation).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Heating in acid releases one CO₂ (decarboxylation of the β-diacid). In the end only one COOH group remains.
Often confused with: Acetoacetic ester synthesis · Carboxylic acid from Grignard and CO₂ · SN2 with cyanide (chain extension)
Frequently asked questions
Why is the CH₂ between two C=O groups so acidic?
The negative charge of the enolate spreads over both carbonyl oxygens. The pKa is about 11 to 13, so ethoxide is enough for complete deprotonation.
How do I find the product quickly?
R–X becomes R–CH₂–COOH. With two alkylations (base and R–X twice) R(R′)CH–COOH forms.