The Intuition: Why Would Chloroform Add an Aldehyde?
Imagine you have a phenol molecule — a benzene ring with an –OH group. That –OH is not just sitting there; it's a powerful electron-donating group. It pushes electron density into the ring, especially onto the ortho and para positions. This makes those positions unusually reactive toward electrophiles (species that love electrons).
Now, chloroform (CHCl3) in the presence of a strong base like aqueous NaOH does something dramatic. The base pulls off a proton from chloroform, generating a highly reactive species called dichlorocarbene (:CCl2). This carbene is a fierce electrophile — it has a sextet of electrons and desperately wants two more.
The phenol's ortho position, rich in electrons, is the perfect target. The carbene attacks there, and a cascade of reactions follows, ultimately converting that –CHCl₂ group into an aldehyde (–CHO). The product is salicylaldehyde (2-hydroxybenzaldehyde).
Note
The reaction is ortho-selective because the –OH group directs the incoming electrophile to the ortho position. Para substitution is possible but much less common under these conditions.
The Precise Statement
Reimer–Tiemann Reaction:
When phenol is treated with chloroform (CHCl3) and aqueous sodium hydroxide (NaOH) at about 60–70 °C, followed by acidification, the formyl group (–CHO) is introduced at the ortho position relative to the –OH group. The major product is salicylaldehyde.
Generation of dichlorocarbeneNaOH deprotonates chloroform:
CHClX3+OHX−CClX3X−+HX2O
The trichloromethyl anion loses a chloride ion to form the electrophilic carbene:
CClX3X−:CClX2+ClX−
Attack on phenoxide ion
Phenol first reacts with NaOH to form the more nucleophilic phenoxide ion (CX6HX5OX−). The carbene attacks the ortho carbon of the phenoxide ring.
Rearrangement and hydrolysis
The intermediate undergoes ring-opening to a dichloromethyl phenol derivative, which then hydrolyzes under basic conditions to give the aldehyde.
Acidification
After the reaction, adding dilute acid converts the sodium salt of salicylaldehyde back to the free aldehyde.
Watch out
A common mistake: thinking the –CHO group comes directly from chloroform. It does not — the carbon of the aldehyde is the carbon from chloroform, but it arrives via the carbene intermediate, not as a pre-formed formyl group.
Key Points for Exams
Reagents: Phenol + CHCl3 + aqueous NaOH (not alcoholic NaOH — that would give a different reaction).
Temperature: ~60–70 °C (reflux). Too low, the carbene doesn't form; too high, side reactions dominate.
Product: Salicylaldehyde (ortho-hydroxybenzaldehyde). A small amount of para-hydroxybenzaldehyde may also form, but ortho is the major product. …
This is the Reimer-Tiemann reaction, in which dichlorocarbene attacks the electron-rich phenoxide ring at the ortho position to introduce a formyl group. …
Phenol + chloroform + aqueous NaOH is the classic Reimer–Tiemann reaction, which introduces a –CHO group ortho to the –OH, giving (after workup) the sodium salt of salicylaldehyde.
Mechanism outline: NaOH generates the phenoxide ion, and CHCl3 + NaOH generates dichlorocarbene (:CCl2). The electron-rich phenoxide ring attacks this electrophilic carbene preferentially at the ortho position, giving a dichloromethyl intermediate that is then hydrolysed by the alkaline medium to an aldehyde group. Because the phenoxide oxygen strongly activates the ortho/para positions and the ortho product is favoured (chelation/ring strain factors), the ma …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set ANNUAL1 mark
Q.Assertion (A): Reimer-Tiemann reaction of phenol with chloroform in presence of NaOH at 340K gives salicylaldehyde as the major product.
Reason (R): The reaction occurs through intermediate formation of dichlorocarbene.
›Reveal solutionSolution
Both statements are true, and dichlorocarbene formation is exactly the mechanistic reason salicylaldehyde forms.
Assertion: In the Reimer–Tiemann reaction, phenol is treated with CHCl3 and NaOH at about 340 K; the product (after hydrolysis) is predominantly the ortho-hydroxybenzaldehyde, i.e. salicylaldehyde. True.
This is the classic Reimer-Tiemann reaction, in which phenol reacts with chloroform and KOH to introduce a -CHO group ortho to the -OH, giving salicylaldehyde.
Q.When phenol is treated with CHCl3 and NaOH, the product formed is
(a) benzaldehyde
(b) salicylaldehyde
(c) salicylic acid
(d) benzoic acid
›Reveal solutionSolution
This is the Reimer–Tiemann reaction: phenol reacts with chloroform under strongly basic conditions to install a formyl (−CHO) group at the ortho position, giving salicylaldehyde after hydrolysis.
Mechanism outline
NaOH deprotonates CHCl3 to form the trichloromethyl carbanion :CCl3−, which rapidly loses Cl− to generate the highly reactive electrophile dichlorocarbene (:CCl2).
The phenoxide ion (from phenol + NaOH) attacks :CCl2 preferentially at the ortho position (directed by the strongly activating −O− group), giving a dichloromethyl-substituted intermediate.
Hydrolysis of the −CHCl2 group under the alkaline reaction conditions converts it to −CHO.
Dichlorocarbene, generated from CHCl3+NaOH, formylates phenol's activated ortho ring position, and hydrolysis of the resulting dichloromethyl intermediate reveals the aldehyde.
In the Reimer–Tiemann reaction, phenol is treated with chloroform and concentrated NaOH. Base generates dichlorocarbene (:CCl₂) from CHCl₃ (by α-elimination), which is attacked by the electron-rich phenoxide ring (ortho position preferred), forming a ortho-substituted dichloromethyl-phenol intermediate; hydrolysis of this gem-dihalide under the basic conditions (then acidification) re …
Phenol reacts with CHCl3/NaOH via a dichlorocarbene intermediate to install a -CHO group ortho to the -OH.
In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl3) and concentrated aqueous sodium hydroxide, heated at around 340 K. NaOH first generates the electrophilic species dichlorocarbene (:CCl2) from CHCl3, which attacks the electron-rich phenoxide ring (mainly at the position ortho to -OH). After hydrolysis of the resulting intermediate, the final product is salicylaldehyde (2-hydroxyb …
Q.By which of the following reactions Phenol is converted into salicyl aldehyde?
(a) Etard reaction
(b) Kolbe's reaction
(c) Reimer-Tiemann reaction
(d) Cannizzaro's reaction
›Reveal solutionSolution
Phenol + CHCl3 + NaOH (Reimer-Tiemann) introduces a -CHO group ortho to -OH, giving salicylaldehyde.
In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl3) in the presence of aqueous NaOH, followed by acidification. Dichlorocarbene (:CCl2), generated in situ, attacks the ring mainly at the ortho position, and after hydrolysis a -CHO group is introduced:
Phenol --CHCl3 / NaOH, then H+--> salicylaldehyde (2-hydroxybenzaldehyde)
Q.Salicylaldehyde is a product obtained by the action of CHCl3 on C6H5OH in presence of aq. KOH. What is the name of the reaction ?
›Reveal solutionSolution
Treating phenol with chloroform and aqueous KOH, then hydrolysing, introduces an −CHO group ortho to −OH — this is the Reimer–Tiemann reaction.
In the Reimer–Tiemann reaction, phenol is treated with chloroform in the presence of aqueous sodium/potassium hydroxide; the reagent generates dichlorocarbene (:CCl2) in situ, which attacks the electron-rich phenoxide ring (predominantly at the ortho position), and subsequent hydrolysis converts the resulting dichloromethyl intermediate into an aldehyde group …