Imagine you have a carboxylic acid — say, vinegar (acetic acid). You know it has a carboxyl group (−COOH) at one end. Now, what if you could snap off that carboxyl group and join the two remaining hydrocarbon pieces together? That is exactly what Kolbe electrolysis does: it takes two carboxylic acid molecules, removes their CO2 groups, and couples the leftover alkyl fragments into a longer hydrocarbon chain.
The reaction happens in an electrolytic cell — the same kind of setup you use to split water into hydrogen and oxygen. But here, the "fuel" is a concentrated solution of a carboxylate salt (the conjugate base of the acid), and the electrodes are usually platinum.
The Core Idea in One Sentence
2RCOO−electrolysisR−R+2CO2+2e−
The carboxylate ions lose electrons at the anode, lose CO2, and the two alkyl radicals (R⋅) combine to form a dimer (R−R).
Step-by-Step Mechanism (Anode Only — That's Where the Action Is)
At the anode (oxidation):
The carboxylate ion RCOO− gives up one electron to the electrode, forming a carboxyl radical:
RCOO−→RCOO⋅+e−
Decarboxylation (loss of CO2):
The carboxyl radical is unstable. It immediately loses CO2 to produce an alkyl radical:
RCOO⋅→R⋅+CO2
Dimerization:
Two alkyl radicals meet and couple:
2R⋅→R−R
The net result: two carboxylate ions become one alkane (the dimer) and two molecules of CO2.
Note
The cathode reaction is usually the reduction of water (or the solvent) to hydrogen gas and hydroxide ions. It is not special to Kolbe electrolysis — the real chemistry is at the anode.
What You Actually See in the Lab
Starting material: A concentrated aqueous or methanolic solution of the sodium or potassium salt of a carboxylic acid (e.g., sodium acetate, CH3COONa).
Electrodes: Inert platinum (carbon works too, but can get messy).
Products at anode: The alkane dimer bubbles out (if short-chain) or deposits as a solid (if long-chain), along with CO2 gas.
Products at cathode: Hydrogen gas and hydroxide ions (the solution becomes basic).
For sodium acetate (R=CH3), the product is ethane (CH3−CH3).
For sodium propionate (R=CH3CH2), the product is butane (CH3CH2−CH2CH3).
The Precise Statement (Exam-Ready)
Important
Kolbe electrolysis is the anodic decarboxylative dimerization of carboxylate ions. When an aqueous solution of a sodium or potassium salt of a carboxylic acid is electrolysed using platinum electrodes, the carboxylate ion loses an electron at the anode, undergoes decarboxylation to form an alkyl radical, and two such radicals couple to give a symmetrical alkane (the dimer). Carbon dioxide is evolved at the anode, and hydrogen gas at the cathode.
Key Conditions and Limitations
Concentration matters: The solution must be concentrated. In dilute solution, the carboxylate radical may instead react with water to form an alcohol or aldehyde (the Hofer–Moest reaction).
No other oxidisable groups: If the alkyl chain has functional groups that are easier to oxidise (like −OH, −NH2, or double bonds), those will react first — the reaction fails.
Only symmetrical dimers: You get R−R from RCOO−. If you mix two different carboxylates (RCOO− and R′COO−), you get a statistical mixture of R−R, R−R′, and R′−R′ — not useful for a single product. …
The key idea is decarboxylation with sodalime (§9.2.2, "From carboxylic acids"): heating the sodium salt of a carboxylic acid with sodalime (NaOH + CaO) removes the carboxylate carbon as carbonate, giving an alkane with one carbon fewer than the acid.
Step 1: Propane (CX3HX8) has three carbons, so the starting acid must have 3+1=4 carbons — butanoic acid, CHX3CHX2CHX2COOH. Its sodium salt is sodium butanoate, CHX3CHX2CHX2COONa.
Sodalime decarboxylation removes exactly one carbon from a carboxylic acid's sodium salt. Propane has 3 carbons, so the salt must come from the 4-carbon acid — butanoic acid: CHX3CHX2CHX2COONa+NaOHCaO,ΔCHX3CHX2CHX3+NaX2COX3.
The concept: decarboxylation
Section 9.2.2 gives a standard laboratory route from carboxylic acids to alkanes: heat the sodium salt of the acid with sodalime — a mixture of sodium hydroxide and calcium oxide, written NaOH (CaO). The carboxylate group is eliminated as carbonate, a process called decarboxylation. The essential bookkeeping is that the product alkane always contains one carbon atom fewer than the parent acid, because the carboxyl carbon is the one that leaves.
Step-by-step reasoning
Count the carbons the product needs. Propane is CHX3CHX2CHX3 — three carbons.
Work backwards to the acid. Since decarboxylation removes one carbon, the acid must have four: CHX3CHX2CHX2COOH, butanoic acid. The salt actually heated is its sodium salt, sodium butanoate, CHX3CHX2CHX2COOX−NaX+.
Write the reaction. The CaO does not appear in the equation — it keeps the mixture dry and porous and acts as a heat-transfer medium, which is why it is written over the arrow:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2024Set B-21 markMCQ
Q.Sodium ethanoate on heating with soda lime give ‘X’. Electrolysis of aqueous solution of sodium ethanoate gives ‘Y’, ‘X’ and ‘Y’ respectively are :
(A) Methane and Ethane
(B) Methane and Methane
(C) Ethane and Methane
(D) Ethane and Ethane
›Reveal solutionSolution
Sodium ethanoate undergoes two different reactions: decarboxylation with soda lime gives methane, while Kolbe's electrolysis of its aqueous solution gives ethane. So the pair is methane and ethane.
The key here is recognising that the same starting material — sodium ethanoate — is subjected to two completely different chemical environments. One is a dry, high-temperature reaction with a strong base; the other is an electrolytic process in water. Each follows a distinct mechanism, and the products are not the same.
Let’s walk through each reaction separately.
Reaction with soda lime (decarboxylation)
Soda lime is a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO). When sodium ethanoate (CHX3COONa) is heated with it, the carboxylate group is removed as carbon dioxide, and the alkyl group picks up a hydrogen from the base.
The reaction is:
CHX3COONa+NaOHCaO, ΔCHX4+NaX2COX3
The product ‘X’ is methane (CHX4). This is a classic decarboxylation — the −COONa group is replaced by −H.
Electrolysis of aqueous sodium ethanoate (Kolbe's electrolysis)
In aqueous solution, sodium ethanoate dissociates into CHX3COOX− and NaX+. At the anode, acetate ions lose an electron to form acetate radicals, which then decarboxylate to give methyl radicals. Two methyl radicals couple to form ethane.
The anode reaction is: …
Q.2-Methyl propane can be prepared by Wurtz reaction. The haloalkanes taken along with metallic sodium and dry ether are :
(A) chloromethane and 2-chloropropane
(B) chloroethane and chloromethane
(C) chloroethane and 1-chloropropane
(D) chloromethane and 1-chloropropane
›Reveal solutionSolution
Cut the target alkane at a C–C bond; the two fragments tell you which two alkyl halides to feed into the Wurtz coupling.
Step 1 — The concept: what Wurtz actually does
The Wurtz reaction joins alkyl groups, not whole molecules. Sodium in dry ether strips the halogen and the resulting alkyl fragments couple:
2R-X+2Nadry etherR-R+2NaX
With a mixture of two different halides R-X and R′-X you get the cross product R-R′ (along with the two symmetrical by-products R-R and R′-R′).
Step 2 — Draw the target and find the bond to break
2-Methylpropane (isobutane) is
CH3−∣CCH3H−CH3≡(CH3)2CH−CH3
The C–C bond that Wurtz would have made is the one joining the isopropyl group(CH3)2CH− to a methyl group−CH3.
(CH3)2CHCH3
Step 3 — Name the two halides
The isopropyl fragment (CH3)2CH− comes from (CH3)2CH−Cl = 2-chloropropane.
The methyl fragment −CH3 comes from CH3−Cl = chloromethane.