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Intext Questions · 7.9

Q.Write the equations involved in the following reactions:

(i) Reimer-Tiemann reaction
(ii) Kolbe's reaction
Yanam BieapTextbookSubjective· 2mImportance★★★★★
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Both reactions are electrophilic aromatic substitution (EAS) on phenol, where a strong base generates a phenoxide ion that activates the ring, and a special electrophile attacks the ortho position. In the Reimer–Tiemann reaction, the electrophile is dichlorocarbene (:CCl₂), giving salicylaldehyde. In Kolbe’s reaction, the electrophile is carbon dioxide (CO₂), giving salicylic acid.


The Core Idea: Why Phenol is Special

Phenol (C6H5OHC_6H_5OH) is unusually reactive toward electrophilic substitution because the –OH group donates electrons into the ring by resonance. But in both reactions below, we first treat phenol with a strong base (NaOH or KOH). This deprotonates the –OH to give the phenoxide ion (C6H5O−C_6H_5O^-), which is an even stronger activating group. The negative charge on oxygen pushes electrons into the ring so powerfully that the ortho and para positions become highly nucleophilic — ready to attack even weak or unusual electrophiles.


(i) Reimer–Tiemann Reaction

What happens: Phenol is treated with chloroform (CHCl₃) in the presence of aqueous NaOH (or KOH) at about 60–70 °C. After work‑up with acid, the product is salicylaldehyde (2‑hydroxybenzaldehyde).

Step‑by‑step reasoning

  1. Generation of the electrophile The base (NaOH) deprotonates chloroform to form the trichloromethyl carbanion (:CCl3−:CCl_3^-), which then loses a chloride ion to give dichlorocarbene (:CCl2:CCl_2). This is a highly reactive, electron‑deficient species — a powerful electrophile.

CHCl3+OH−⟶:CCl3−+H2O\text{CHCl}_3 + \text{OH}^- \longrightarrow :\text{CCl}_3^- + \text{H}_2\text{O}

:CCl3−⟶:CCl2+Cl−:\text{CCl}_3^- \longrightarrow :\text{CCl}_2 + \text{Cl}^-

  1. Attack on the phenoxide ion The phenoxide ion (formed from phenol + NaOH) attacks the electrophilic carbene at the ortho position (the para position is sterically hindered by the –OH group in the transition state). This gives an intermediate dichloromethyl‑substituted phenol.

C6H5O−+:CCl2⟶ortho-(Cl2CH)C6H4O−\text{C}_6\text{H}_5\text{O}^- + :\text{CCl}_2 \longrightarrow \text{ortho-}(\text{Cl}_2\text{CH})\text{C}_6\text{H}_4\text{O}^-

  1. Hydrolysis to aldehyde The dichloromethyl group (–CHCl2–CHCl_2) is hydrolysed by the aqueous base. Two successive nucleophilic substitutions (by OH⁻) replace both chlorines with –OH groups, which then tautomerise to give the aldehyde.

Ar–CHCl2+2 OH−⟶Ar–CH(OH)2+2 Cl−\text{Ar–CHCl}_2 + 2\,\text{OH}^- \longrightarrow \text{Ar–CH(OH)}_2 + 2\,\text{Cl}^-

Ar–CH(OH)2→tautomerisationAr–CHO+H2O\text{Ar–CH(OH)}_2 \xrightarrow{\text{tautomerisation}} \text{Ar–CHO} + \text{H}_2\text{O}

  1. Acidification Finally, dilute acid is added to protonate the phenoxide back to phenol, yielding salicylaldehyde.
Watch out

A common mistake is to think the carbene attacks the –OH group directly. It does not — the attack is on the ring at the ortho carbon. Also, the product is always the ortho isomer; para‑hydroxybenzaldehyde is not formed in significant amounts under these conditions.

Overall equation:

CX6HX5OH+CHClX3+3 NaOH→60−70°C2-HOCX6HX4CHO+3 NaCl+2 HX2O\ce{C6H5OH + CHCl3 + 3 NaOH ->[60-70°C] 2-HOC6H4CHO + 3 NaCl + 2 H2O}


(ii) Kolbe’s Reaction (Kolbe–Schmitt Reaction)

What happens: Sodium phenoxide is heated with carbon dioxide under pressure (about 125 °C, 4–7 atm), then acidified to give salicylic acid (2‑hydroxybenzoic acid).

Step‑by‑step reasoning

  1. Activation of the ring

    Phenol is first converted to sodium phenoxide (C6H5ONaC_6H_5ONa) using NaOH. The phenoxide ion is a much stronger nucleophile than phenol itself.

  2. Attack by CO₂

    Carbon dioxide is a weak electrophile (the carbon is partially positive due to the two electronegative oxygens). The ortho carbon of the phenoxide ion attacks the carbon of CO₂, forming a carboxylate group at the ortho position. …

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