Reimer-Tiemann Reaction: Mechanism, Dichlorocarbene and the Ortho Product

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

The Reimer-Tiemann reaction puts a CHO group on the ring of phenol, next to the OH. Phenol is warmed with chloroform and aqueous sodium hydroxide, and the mixture is then acidified: the main product is salicylaldehyde (2-hydroxybenzaldehyde). The electrophile is dichlorocarbene, $: CC l_{2}$ , which the alkali makes from chloroform. The reaction is in the Class 12 NCERT chapter on alcohols, phenols and ethers, and is asked in NEET and JEE Main.

At a glance

In the alkali:
$C_{6} H_{5} OH + CHC l_{3} + 4NaOH 340 K NaO C_{6} H_{4} CHO + 3NaCl + 3 H_{2} O$
Then acid:
$NaO C_{6} H_{4} CHO + HCl HO C_{6} H_{4} CHO + NaCl$
Electrophile: dichlorocarbene, $: CC l_{2}$ , a neutral carbon with six valence electrons.
Type: electrophilic substitution on the phenoxide ion (a formylation: it adds CHO).
Intermediate: 2-(dichloromethyl)phenoxide, a substituted benzal chloride, whose $CHC l_{2}$ group the alkali hydrolyses to CHO.
Product: mainly ortho (salicylaldehyde), some para (4-hydroxybenzaldehyde).
With $CC l_{4}$ instead of $CHC l_{3}$ : the ring gets COOH: salicylic acid.

1. Mechanism

1.1 Chloroform gives dichlorocarbene

Chloroform's hydrogen is slightly acidic, because three chlorines pull electrons away from the carbon. Hydroxide removes it, and the trichloromethyl anion then loses a chloride ion.

Figure 1: Chloroform and hydroxide give dichlorocarbene,

: CC l_{2}

. Its carbon has only six electrons, so it is the electrophile of the reaction.

1.2 The phenoxide ring attacks the carbene

In the same alkali, phenol is present as the phenoxide ion. Its negative oxygen makes the ortho and para carbons rich in electrons, and the ortho carbon attacks the electron-poor carbene carbon.

Figure 2: The phenoxide ring adds to

: CC l_{2}

at the ortho position; after the proton transfers, the

CHC l_{2}

group is hydrolysed by the alkali to CHO.

Carbene. $H O^{-}$ removes $H^{+}$ from $CHC l_{3}$ ; $CC l_{3}^{-}$ loses $C l^{-}$ to give $: CC l_{2}$ .
Attack. The ortho carbon of the phenoxide ion bonds to the carbene carbon; the ring loses its aromaticity for a moment.
Proton transfer. Through water and hydroxide, the $CC l_{2}^{-}$ carbon gains $H^{+}$ and the attacked ring carbon loses its $H^{+}$ , so the ring becomes aromatic again: 2-(dichloromethyl)phenoxide.
Hydrolysis. The alkali replaces both chlorines by OH; the gem-diol loses water to give CHO. Acid then frees the phenol OH: salicylaldehyde.

Exam Trick

"Reimer adds CHO, Kolbe adds COOH." Both start from sodium phenoxide and both go mainly ortho. Chloroform with aqueous alkali on a phenol gives an aldehyde; chloroform with alcoholic KOH on a primary amine gives an isocyanide (carbylamine reaction). Both run through $: CC l_{2}$ .

2. Ortho, para and other variations

Starting compound	Reagents	Main product
Phenol	$CHC l_{3}$ , aq. NaOH, then $H_{3} O^{+}$	Salicylaldehyde (ortho); 4-hydroxybenzaldehyde (minor)
Phenol	$CC l_{4}$ , aq. NaOH, then $H_{3} O^{+}$	Salicylic acid (2-hydroxybenzoic acid)
4-Methylphenol (p-cresol)	$CHC l_{3}$ , aq. NaOH, then $H_{3} O^{+}$	2-Hydroxy-5-methylbenzaldehyde (para is blocked)
Aniline (a 1° amine)	$CHC l_{3}$ , alc. KOH, warm	Phenyl isocyanide (carbylamine reaction), not an aldehyde

Why ortho. There are two ortho carbons and only one para carbon, and the sodium ion held by the phenoxide oxygen keeps the reagent near the ortho positions. When both ortho positions are taken, the CHO group goes para. Anisole and other phenol ethers do not react: they cannot form a phenoxide ion.

With carbon tetrachloride. $CC l_{4}$ has no hydrogen, so no carbene forms. The ring ends up with a $CC l_{3}$ group instead, which the alkali hydrolyses to COOH, giving salicylic acid.

Separating the two isomers. In salicylaldehyde the OH and CHO are close enough to form an intramolecular hydrogen bond, so the molecule does not hold on to its neighbours and is volatile in steam. 4-Hydroxybenzaldehyde forms intermolecular hydrogen bonds instead and stays behind, so steam distillation separates the two.

JEE Advanced

A neutral electrophile. Dichlorocarbene has no charge, yet it is an electrophile: its carbon has an empty p orbital and only six electrons, with its lone pair in an sp $^{2}$ orbital. The chlorine lone pairs feed a little electron density into the empty p orbital, so it is a weak electrophile and needs the very reactive phenoxide ion. The lone pair it keeps becomes the negative charge on the $CC l_{2}^{-}$ carbon right after the ring attacks.

Why no reaction without alkali. Without base there is neither carbene nor phenoxide ion, and neither phenol nor chloroform alone is reactive enough.

3. Worked examples

Solved Example 1

What is the electrophile in the Reimer-Tiemann reaction of phenol?

Solution:

Dichlorocarbene, $: CC l_{2}$ . It is formed when hydroxide removes the H of chloroform and the $CC l_{3}^{-}$ ion loses $C l^{-}$ . It is not a dichloromethyl cation or a formyl cation.

Solved Example 2

4-Methylphenol is warmed with chloroform and aqueous NaOH and then acidified. Name the main product.

Solution:

The para position is taken by $C H_{3}$ , so CHO goes ortho to OH: 2-hydroxy-5-methylbenzaldehyde.

Solved Example 3

Phenol gives a mixture of ortho and para hydroxybenzaldehydes with

CHC l_{3}

and aqueous KOH. Which is the major product, and how are they separated?

Solution:

The ortho isomer, salicylaldehyde, is the major product. It is steam volatile (intramolecular hydrogen bond), while 4-hydroxybenzaldehyde is not, so steam distillation separates them.

4. Exam traps

Watch out

The electrophile is neutral dichlorocarbene, $: CC l_{2}$ , not a cation and not $CHC l_{2}^{+}$ .
The main product is ortho (salicylaldehyde); para is the minor product.
$CC l_{4}$ in place of $CHC l_{3}$ gives salicylic acid, not the aldehyde.
Chloroform and base with a 1° amine is the carbylamine test (isocyanide), not Reimer-Tiemann.
The ring attacked is the phenoxide ion, so the reaction needs the alkali.
Salicylaldehyde is steam volatile because of its intramolecular hydrogen bond; the para isomer is not.
Reimer-Tiemann adds CHO; Kolbe's reaction adds COOH.

Past-paper questions on Reimer-Tiemann reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 2 questions in them that test this:

JEE Main 1 (2026)
NEET 1 (2018)

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

Properties of PhenolNotes, formulas and examples →

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