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.
KEAM 2026Set eng-2026-04174 marksMCQ
Q.The alkane that cannot be prepared in pure state by Kolbe's electrolysis is
(A) ethane
(B) 2,3-dimethyl butane
(C) propane
(D) n-butane
(E) n-hexane
›Reveal solutionSolution
Kolbe electrolysis of RCOONa gives the symmetrical dimer R−R. Propane is an odd-carbon, unsymmetrical alkane needing two different radicals, so it cannot be obtained pure.
Kolbe's electrolysis decarboxylates a carboxylate to an alkyl radical R∙, and two such radicals combine to form R−R.
Ethane: from CH3COONa (CH3−CH3) — possible.
2,3-dimethylbutane: coupling two isopropyl radicals — possible.
n-butane: from propionate, C2H5−C2H5 — possible.
n-hexane: from butyrate, C3H7−C3H7 — possible. …
Q.Which one of the following reaction is called Kolbe's method?
(A) Hydrogenation of propyne with Pt/Pd/Ni
(B) Chlorination of chloroform.
(C) Treatment of alkyl halides with sodium metal in dry ethereal solution.
(D) Electrolysis of an aqueous solution of potassium carboxylates.
(E) Isomerization of n-hexane to 2-methylpentane in presence of anhy.AlCl3/ HCl.
›Reveal solutionSolution
Kolbe's method = electrolysis of aqueous potassium/sodium salts of carboxylic acids → alkanes.
In Kolbe's electrolytic method, an aqueous solution of a sodium or potassium salt of a carboxylic acid is electrolysed. At the anode the carboxylate is oxidised and decarboxylates to give alkyl free radicals, which couple to form the alkane:
Q.In Kolbe's electrolytic method, when sodium acetate is electrolysed, the gases generated at anode are
(A) ethane and H2
(B) H2 and CO2
(C) methane and ethane
(D) ethane and CO2
(E) methane and H2
›Reveal solutionSolution
Kolbe electrolysis of sodium acetate liberates ethane and CO2 at the anode.
Concept and Intuition
In Kolbe's electrolytic method, the carboxylate ion (CH3COO^-) migrates to the anode, loses an electron to form an acetate radical, which decarboxylates to a methyl radical (releasing CO2). Two methyl radicals couple to give ethane. Both gases evolve at the anode; hydrogen appears at the cathode, not the anode.
Q.Which of the following statement is INCORRECT with Kolbe's electrolytic process?
(A) Ethane can be prepared by this method.
(B) Presence of alkyl groups in α-position decrease the yield of alkanes.
(C) The reaction proceeds via methyl free radical.
(D) At anode alkane and CO2 gas is formed.
(E) An alkane obtained at anode contains odd number of carbon atoms.
›Reveal solutionSolution
2RCOO−→2R∙+2CO2; 2R∙→R−R gives an even-carbon alkane.
At the anode the carboxylate is oxidised to a radical which loses CO2; two such radicals combine to give a symmetrical alkane. Coupling two identical R groups always yields an alkane with an even number o …
Q.Which of the following sodium salt of carboxylic acid is used for the preparation of n-hexane by Kolbe's electrolytic method?
(A) CH3CH2COONa
(B) CH3COONa
(C) HCOONa
(D) CH3CH2CH3CH2COONa
(E) CH3CH2CH2COONa
›Reveal solutionSolution
Kolbe's electrolysis of a sodium carboxylate RCOONa gives the coupled hydrocarbon R–R. For n-hexane (C6H14), R must be propyl, so the salt is sodium butanoate CH3CH2CH2COONa.
In Kolbe's electrolytic method the carboxylate is oxidised at the anode to a radical R∙ after loss of CO2, and two such radicals dimerise:
Q.Which of the following statement is incorrect with Kolbe's electrolytic method?
(A) It gives an alkane with even number of carbon atoms at the anode.
(B) At anode decarboxylation and formation of methyl radical occurs.
(C) Methane cannot be prepared by this method.
(D) At anode acetate ion accepts electrons to give acetate free radical.
(E) At cathode hydrogen gas is liberated.
›Reveal solutionSolution
In Kolbe's electrolysis, the acetate ion is oxidised at the anode — it loses an electron to give the acetate free radical. Statement (D), that it accepts electrons, is incorrect.
Kolbe's electrolytic method electrolyses an aqueous solution of a carboxylate salt. At the anode: CH3COO−→CH3COO∙+e− (oxidation, loss of an electron), then decarboxylation gives CH3∙, and two methyl radicals couple to form ethane (an alkane with an even number of carbons). At the cathode, H2 is liberated. Methane cannot be made this wa …