Question
Phenol reacts with chloroform in the presence of aqueous sodium hydroxide, followed by acidification. Name the reaction and identify the product. Write the mechanism and explain why the formyl group (-CHO) goes to the ortho position.
Solution — Step by Step
This is the Reimer-Tiemann reaction — a classic electrophilic aromatic substitution that converts phenol to an aldehyde. The reagents are CHCl₃ (chloroform) + KOH (or NaOH, aqueous) + heat, followed by acid workup. The product is salicylaldehyde (2-hydroxybenzaldehyde).
The real electrophile here is not CHCl₃ itself — it’s dichlorocarbene (:CCl₂). Here’s how it forms:
NaOH abstracts the acidic proton from CHCl₃ to give a trichloromethyl carbanion (CCl₃⁻), which immediately loses Cl⁻ to become the neutral, highly electrophilic dichlorocarbene. This is the step students always miss — the carbene is the true attacking species, not chloroform.
In strongly alkaline conditions, phenol loses its –OH proton to form the phenoxide ion (C₆H₅O⁻). This makes the ring electron-rich, especially at the ortho and para positions — exactly the positions where electrophilic attack is favored.
:CCl₂ attacks the ortho position of phenoxide. The resulting intermediate has a –CHCl₂ group attached. Under the hot alkaline conditions, this dichloromethyl group rapidly hydrolyses:
The two C–Cl bonds are sequentially hydrolysed, going through –CH(OH)Cl, then –CHO (the aldehyde).
The product after hydrolysis exists as the sodium salt of salicylaldehyde. Acid workup (dilute HCl or H₂SO₄) protonates the phenoxide, giving the free product:
The final product is salicylaldehyde (2-hydroxybenzaldehyde). The –CHO group is exclusively at the ortho position (minor para product also forms, but ortho dominates due to intramolecular hydrogen bonding stabilization in the transition state).
Why This Works
The key to this reaction is understanding that NaOH plays a dual role — it activates the phenol ring by forming phenoxide, AND it generates the electrophile (dichlorocarbene) from chloroform. Without strongly alkaline conditions, neither step happens.
Dichlorocarbene is a neutral but extremely electron-deficient species. It attacks the ortho carbon of phenoxide (which carries high electron density) preferentially. The alkaline medium also ensures the subsequent hydrolysis of –CHCl₂ proceeds quickly, so you don’t isolate any intermediate.
The ortho selectivity is explained by a hydrogen-bonded cyclic transition state — the phenoxide oxygen coordinates with the carbene during attack, directing it to the adjacent ortho carbon. This is a well-accepted explanation in NCERT and JEE literature.
Alternative Method
You can remember the Reimer-Tiemann in two parts using this breakdown:
Reimer-Tiemann = Formylation of phenol
| Step | What happens |
|---|---|
| CHCl₃ + NaOH | Generates :CCl₂ (dichlorocarbene) |
| Phenol + NaOH | Generates phenoxide C₆H₅O⁻ |
| :CCl₂ + phenoxide | Electrophilic substitution at ortho |
| –CHCl₂ + NaOH/H₂O | Hydrolysis → –CHO |
| + H⁺ | Acid workup → salicylaldehyde |
If CCl₄ (carbon tetrachloride) is used instead of CHCl₃, a carboxyl group (–COOH) is introduced instead of –CHO — this gives salicylic acid. That variant is called the Kolbe reaction, not Reimer-Tiemann. Both are NCERT-level, and examiners love testing which reagent gives which product.
Scoring shortcut: In JEE Main and CBSE boards, the Reimer-Tiemann question almost always asks: (a) name the reaction, (b) write the product, (c) state the role of NaOH. Cover all three in your answer — that’s full marks. The mechanism detail (carbene formation) is more for JEE Advanced.
Common Mistake
Writing CHCl₃ as the direct electrophile. Many students draw chloroform attacking the ring directly — this is wrong and will lose marks. CHCl₃ itself is not electrophilic enough to attack an aromatic ring. The actual electrophile is :CCl₂ (dichlorocarbene), generated by NaOH abstracting the proton from CHCl₃. Always show carbene formation as a separate step in the mechanism. Another slip: writing the product as para-hydroxybenzaldehyde — the major product is the ortho isomer (salicylaldehyde).