Kolbe Reaction: Phenol to Salicylic Acid (Kolbe-Schmitt) and Kolbe Electrolysis

ChemistryClass 12, Alcohols, Phenols and EthersFor JEE and NEET aspirants

Kolbe's reaction turns phenol into salicylic acid: sodium phenoxide is heated with carbon dioxide under pressure, and the product is acidified. It is also called the Kolbe-Schmitt reaction and is in the Class 12 NCERT chapter on alcohols, phenols and ethers. The same chemist's name is on a second reaction that exams also ask about, Kolbe electrolysis (Class 11, hydrocarbons): electrolysing the sodium salt of a carboxylic acid joins two alkyl groups into an alkane. This page covers both, for JEE Main, JEE Advanced and NEET.

At a glance

Kolbe's reaction:
$C_{6} H_{5} ONa + C O_{2} 400 K 4-7 atm HO C_{6} H_{4} COONa$
Then acid:
$HO C_{6} H_{4} COONa + H^{+} HO C_{6} H_{4} COOH + N a^{+}$
Product: salicylic acid (2-hydroxybenzoic acid): the COOH goes mainly ortho to the OH.
Why phenoxide: $C O_{2}$ is a weak electrophile; only the very electron-rich phenoxide ion reacts.
Kolbe electrolysis:
$2C H_{3} COONa + 2 H_{2} O electrolysis C H_{3} C H_{3} + 2C O_{2} + H_{2} + 2NaOH$
Rule: two R groups join: RCOONa gives R–R at the anode, so methane cannot be made this way.

1. Kolbe's reaction: phenol to salicylic acid

Phenol is first dissolved in NaOH. The phenoxide ion is far more reactive than phenol towards electrophiles, because the negative oxygen pushes electron density into the ring at the ortho and para positions. Even carbon dioxide, a weak electrophile, can then attack the ring.

Figure 1: Kolbe's reaction.

C O_{2}

is a weak electrophile, so it needs the very electron-rich phenoxide ion; the ortho carbon attacks, and acidification of sodium salicylate gives salicylic acid.

Attack. The ortho carbon of the phenoxide ion attacks the carbon of $C O_{2}$ ; one C=O $π$ bond of $C O_{2}$ moves onto oxygen. The ring loses its aromaticity for a moment.
Rearomatisation. A proton is lost from the attacked carbon and the oxygen is protonated (tautomerism, keto to enol), giving the aromatic ring back and the salicylate ion.
Acidification. Dilute acid turns sodium salicylate into salicylic acid.

Why ortho. With the sodium salt the ortho product is the main one: the sodium ion holds the phenoxide oxygen and the incoming $C O_{2}$ close together. Potassium phenoxide heated with $C O_{2}$ at a higher temperature (about 510 K) gives mainly 4-hydroxybenzoic acid, the more stable para product; the larger potassium ion does not hold $C O_{2}$ next to the oxygen.

Tests on salicylic acid. Its phenolic OH gives a violet colour with neutral $FeC l_{3}$ , and its COOH gives effervescence ( $C O_{2}$ ) with $NaHC O_{3}$ . Salicylic acid is then acetylated with ethanoic anhydride to make aspirin, 2-acetoxybenzoic acid (the phenolic OH becomes an ester), a pain-killer that blocks prostaglandin synthesis.

HO C_{6} H_{4} COOH + (C H_{3} CO)_{2} O H^{+} C H_{3} COO C_{6} H_{4} COOH + C H_{3} COOH

Exam Trick

"Kolbe adds COOH, Reimer adds CHO." Both start from sodium phenoxide and put the new group mainly ortho to OH: $C O_{2}$ gives salicylic acid (Kolbe), chloroform gives salicylaldehyde (Reimer-Tiemann).

2. Kolbe electrolysis: carboxylate salts to alkanes

When an aqueous solution of the sodium or potassium salt of a carboxylic acid is electrolysed, the carboxylate ions are oxidised at the anode. Each loses one electron, then $C O_{2}$ , and the alkyl radicals left behind pair up. The cathode gives hydrogen and the solution becomes alkaline.

Figure 2: Kolbe electrolysis. Each fish-hook arrow moves one electron; two methyl radicals join, so the alkane has twice the carbons of the acid's alkyl group. Anode:

2RCO O^{-} \to R - R + 2C O_{2} + 2 e^{-}

; cathode:

2 H_{2} O + 2 e^{-} \to H_{2} + 2O H^{-}

Salt electrolysed	Alkane at the anode	Carbons
Sodium ethanoate, $C H_{3} COONa$	Ethane, $C H_{3} - C H_{3}$	2
Sodium propanoate, $C_{2} H_{5} COONa$	Butane, $C_{2} H_{5} - C_{2} H_{5}$	4
Sodium butanoate, $C_{3} H_{7} COONa$	Hexane, $C_{3} H_{7} - C_{3} H_{7}$	6
Sodium 3-methylbutanoate, $(C H_{3})_{2} CHC H_{2} COONa$	2,5-Dimethylhexane	8
Mixture of $C H_{3} COONa$ and $C_{2} H_{5} COONa$	Ethane, propane and butane	2, 3, 4

One salt always gives an alkane with an even number of carbons, twice the carbons of its alkyl group. A mixture of two salts gives three alkanes, as in the Wurtz reaction, so the method suits symmetrical alkanes.

JEE Advanced

Gas volumes. For sodium ethanoate the anode gives one $C_{2} H_{6}$ and two $C O_{2}$ for every $H_{2}$ at the cathode, so the volumes of gas at the anode and the cathode are in the ratio 3 : 1.

Salicylic acid against benzoic acid. Salicylic acid ( $p K_{a} \approx 3.0$ ) is stronger than benzoic acid ( $p K_{a} \approx 4.2$ ): in the salicylate ion the ortho OH hydrogen-bonds to the carboxylate and stabilises it.

Salts of dicarboxylic acids. Both ends lose $C O_{2}$ and the two radicals on one molecule join as a $π$ bond: sodium succinate, $NaOOCC H_{2} C H_{2} COONa$ , gives ethene, and sodium maleate or fumarate gives ethyne.

From electrolysis to benzene. Sodium butanoate gives hexane, which is aromatised to benzene over $C r_{2} O_{3}$ (or $V_{2} O_{5}$ , $M o_{2} O_{3}$ ) at 773 K and 10-20 atm.

3. Worked examples

Solved Example 1

Phenol is heated with NaOH and

C O_{2}

under pressure and then acidified, giving X. Is X more or less acidic than phenol?

Solution:

X is salicylic acid, a carboxylic acid ( $p K_{a} \approx 3$ ), so it is much more acidic than phenol ( $p K_{a} \approx 10$ ). Unlike phenol, it gives $C O_{2}$ with $NaHC O_{3}$ .

Solved Example 2

Which alkane is formed at the anode when aqueous sodium propanoate is electrolysed? What forms at the cathode?

Solution:

Two ethyl radicals join: butane, $C H_{3} C H_{2} C H_{2} C H_{3}$ , with $C O_{2}$ at the anode. Hydrogen gas forms at the cathode and NaOH stays in solution.

Solved Example 3

Can methane be prepared by Kolbe electrolysis?

Solution:

No. The alkane always contains two alkyl groups from two carboxylate ions. The smallest, from sodium ethanoate, is ethane. Methane is made instead by heating sodium ethanoate with soda lime (decarboxylation).

4. Exam traps

Watch out

Kolbe's reaction needs the phenoxide ion (phenol in NaOH); phenol itself does not react with $C O_{2}$ .
The main product is ortho (salicylic acid), not para, with the sodium salt.
Kolbe's reaction adds $COOH$ ; Reimer-Tiemann adds $CHO$ . Both go mainly ortho to the OH.
Aspirin is 2-acetoxybenzoic acid, an ester of the phenolic OH, not 2-acetylbenzoic acid.
Kolbe electrolysis gives R–R, twice the carbons of the alkyl group: sodium ethanoate gives ethane, not methane.
Soda-lime decarboxylation removes one carbon (RCOONa gives RH); Kolbe electrolysis joins two R groups.
A mixture of two different salts gives three alkanes.
The alkane and $C O_{2}$ form at the anode; hydrogen forms at the cathode.
Salicylic acid is a stronger acid than phenol and gives effervescence with $NaHC O_{3}$ .

Past-paper questions on Kolbe reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 7 questions in them that test this:

JEE Main 2 (2026)
JEE Advanced 5 (2022-2023, 2026)

Each question was read and checked by hand (list checked 2026-10-14). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

Learn it with the chapter

Related named reactions

Practise questions like these

Concept-wise questions with instant solutions, and full mock tests, on Fundamenthol.

Practice questions →Take a mock test