Imagine a carboxylic acid as a molecule with a long carbon chain that ends in a COOH group — a carbon double-bonded to one oxygen, single-bonded to an OH. That COOH is the "business end." Now picture heating that molecule so hard that the entire COOH group rips off, taking one extra carbon with it, and flies away as carbon dioxide (CO2). What remains is a hydrocarbon with one fewer carbon than you started with.
That's decarboxylation: the loss of carbon dioxide from a carboxylic acid (or its salt).
The name itself tells the story: de- (remove) + carboxyl (the COOH group) + -ation (process). You are literally removing the carboxyl group as CO2.
The Precise Statement
R-COOHΔR-H+CO2
In words: When a carboxylic acid is heated strongly (often with a catalyst or base), the carboxyl group breaks off as carbon dioxide, and the hydrogen that was on the OH ends up attached to the carbon that was next to the COOH. The product is an alkane (or sometimes an alkene, depending on conditions).
But there's a catch: plain carboxylic acids don't decarboxylate easily. You need special conditions.
The Classic Exam Reaction: Soda-Lime Decarboxylation
The most common method you'll see in Indian exams is heating a sodium salt of a carboxylic acid with soda lime (a mixture of NaOH and CaO). The calcium oxide doesn't react — it just provides a porous, high-surface-area solid that helps the reaction go smoothly.
Here's the step-by-step:
Start with a carboxylic acid, say sodium acetate: CH3COONa
Mix it with soda lime (NaOH+CaO) and heat strongly
The sodium salt reacts with NaOH to give methane and sodium carbonate:
CH3COONa+NaOHCaO,ΔCH4+Na2CO3
The CO2 that leaves is trapped as part of the sodium carbonate. The alkane (methane, in this case) escapes as a gas.
Tip
The reaction always shortens the carbon chain by one. If you start with a 3-carbon acid (propanoic acid → sodium propanoate), you get a 2-carbon alkane (ethane). Count the carbons in the product — it's a quick check for exam problems.
Why Does This Happen?
The carboxyl group is already partially "pre-organized" to lose CO2. The carbon in CO2 is in its highest oxidation state (+4), so it's very stable once formed. The driving force is the formation of that extremely stable CO2 molecule plus a new C–H bond in the alkane.
In the soda-lime method, the NaOH provides the hydrogen that ends up on the alkane. The CaO helps by keeping the mixture dry and preventing side reactions.
Heating a sodium carboxylate salt with soda lime removes the carboxyl group entirely as carbonate, replacing it with a hydrogen — a reaction known as decarboxylation. …
Heating a sodium carboxylate with soda lime (NaOH + CaO) removes CO2 and replaces -COONa with -H, giving an alkane.
This is the standard decarboxylation reaction: sodium salts of carboxylic acids, when heated with soda lime (a mixture of NaOH and CaO — CaO is added because pure NaOH would melt/corrode the tube), lose carbon dioxide (trapped as Na2CO3) and the -COONa group is replaced by -H.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2024Set A11 markMCQ
Q.Decarboxylating reagent is a mixture of :
(a) Alc. KOH+H2O2
(b) NaOH+CO2
(c) NaOH+CaO
(d) Conc. HCl+ZnCl2
›Reveal solutionSolution
The decarboxylating reagent is soda lime, NaOH+CaO — option (c).
Soda lime — a mixture of NaOH and CaO (CaO keeps NaOH dry and non-corrosive to glass) — removes CO2 from the sodium salt of a carboxylic acid, giving a hydrocarbon with one fewer carbon: …
Q.What will be the product of the following reaction? R-COONa --(NaOH + CaO, Heat)--> ? + Na2CO3
(a) R-CH2OH
(b) R-CHO
(c) R-CH3
(d) R-H
›Reveal solutionSolution
Heating a sodium carboxylate with soda lime (NaOH + CaO) removes CO2 and replaces -COONa with -H, giving an alkane.
This is the standard decarboxylation reaction: sodium salts of carboxylic acids, when heated with soda lime (a mixture of NaOH and CaO — CaO is added because pure NaOH would melt/corrode the tube), lose carbon dioxide (trapped as Na2CO3) and the -COONa group is replaced by -H.
Q.Heating a mixture of sodium benzoate with soda-lime gives:
(a) Methane
(b) Benzene
(c) Ethane
(d) Propane
›Reveal solutionSolution
Decarboxylation of sodium benzoate with soda lime replaces the −COONa group with −H, giving benzene — a standard laboratory preparation of an aromatic hydrocarbon.
Step 1 — Write the reaction:
C6H5COONa+NaOHCaO (soda lime)ΔC6H6+Na2CO3
Step 2 — What's happening chemically: This is a decarboxylation reaction — soda lime (a mixture of NaOH and CaO) removes the carboxylate group as CO2 (trapped as Na2CO3) and replaces it with a hydrogen atom on the ring. CaO acts mainly as a dehydrating/inert diluent that prevents the NaOH from fusing/charring, while NaOH does the actual decarboxylation.