Phenol + CHCl₃ + NaOH → salicylaldehyde (o-hydroxybenzaldehyde). Mechanism: NaOH+CHCl₃ generates dichlorocarbene (:CCl₂), which attacks ortho position of phenoxide.
-- NCERT Class 12 Chemistry, Ch. 7, p. 213Reimer Tiemann
8 MCQs9-step worked example
Source: NCERT Alcohols, Phenols, Ethers and Carbonyl CompoundsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Reimer Tiemann, explained for NEET
Reimer-Tiemann Reaction
The Reimer-Tiemann reaction introduces an aldehyde group (–CHO) at the ortho position of phenol using chloroform (CHCl₃) and aqueous NaOH, followed by acid hydrolysis. The product from phenol is salicylaldehyde (2-hydroxybenzaldehyde).
Mechanism in brief:
- CHCl₃ reacts with strong base (NaOH) to generate dichlorocarbene (:CCl₂), an electrophilic carbene intermediate. This is the rate-determining step.
- Dichlorocarbene attacks the electron-rich ortho position of the phenoxide ion (phenol deprotonated by NaOH).
- The intermediate dichloromethyl phenol undergoes hydrolysis in aqueous alkali to yield the –CHO group.
Key points to lock in:
- The reactive species is dichlorocarbene (:CCl₂), not trichloromethyl anion or free CHCl₃.
- The reaction is specific to phenols — ordinary alcohols or aromatic hydrocarbons without –OH do not undergo it.
- Substitution occurs preferentially at the ortho position relative to –OH. If both ortho positions are blocked, the para product (4-hydroxybenzaldehyde) forms.
- If CCl₄ is used instead of CHCl₃, the product is a salicylic acid derivative (Kolbe-type carboxylation path) — this is a common NEET swap distractor.
- The reaction proceeds under alkaline conditions (aqueous NaOH), not acidic.
NEET trap to watch: Questions may offer "trichloromethyl carbanion" or "CHCl₃ directly" as the electrophile. The correct intermediate is always dichlorocarbene (:CCl₂). A second common confusion: swapping the CHCl₃ product (aldehyde) with the CCl₄ product (carboxylic acid).
Reference: NCERT Class 12 Chemistry, Chapter 7 (Alcohols, Phenols and Ethers), Part 2, page 213.
Can you answer these Reimer Tiemann MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the Reimer-Tiemann reaction, phenol is heated with chloroform and aqueous NaOH. What is the electrophilic intermediate generated in this reaction?
Show answer and why every option is right or wrong
Answer: A. A is correct. CHCl₃ reacts with strong base to lose HCl and form dichlorocarbene (:CCl₂), the electrophilic species that attacks phenoxide. NCERT Class 12 Chemistry, Chapter 7 Part 2, page 213 explicitly identifies dichlorocarbene as the reactive intermediate.
Why B is wrong: B is wrong. The trichloromethyl carbanion (⁻CCl₃) is a transient species in the deprotonation step, not the electrophilic intermediate that attacks phenol. The electrophile is the neutral carbene :CCl₂ formed after α-elimination of Cl⁻.
Why C is wrong: C is wrong. CHCl₃ itself is not electrophilic enough to attack the aromatic ring. It must first undergo base-promoted α-elimination to generate dichlorocarbene.
Why D is wrong: D is wrong. CCl₄ is a different reagent entirely. When CCl₄ is used with NaOH and phenol, the product is a salicylic acid derivative (Kolbe-related path), not an aldehyde.
MCQ 2Easy RecallPractice
What is the major organic product when phenol undergoes the Reimer-Tiemann reaction with CHCl₃/NaOH followed by acidification?
Show answer and why every option is right or wrong
Answer: B. B is correct. The Reimer-Tiemann reaction with CHCl₃ introduces an aldehyde group at the ortho position of phenol, yielding salicylaldehyde after hydrolysis. NCERT Class 12 Chemistry, Chapter 7 Part 2, page 213.
Why A is wrong: A is wrong. Salicylic acid is the product of the Kolbe reaction (phenol + CO₂/NaOH under pressure), not the Reimer-Tiemann reaction with CHCl₃. Confusing these two named reactions is a common distractor swap.
Why C is wrong: C is wrong. Anisole is formed by Williamson ether synthesis (phenoxide + methyl halide). The Reimer-Tiemann reaction does not produce ethers.
Why D is wrong: D is wrong. Benzaldehyde lacks the –OH group on the ring. The Reimer-Tiemann reaction preserves the phenolic –OH and introduces –CHO at the ortho position, giving 2-hydroxybenzaldehyde, not unsubstituted benzaldehyde.
MCQ 3Easy RecallPractice
The Reimer-Tiemann reaction is specific to which class of organic compounds?
Show answer and why every option is right or wrong
Answer: C. C is correct. The Reimer-Tiemann reaction requires the phenoxide ion's electron-rich aromatic ring for electrophilic attack by dichlorocarbene. Ordinary alcohols and hydrocarbons without a phenolic –OH do not undergo this reaction. NCERT Class 12 Chemistry, Chapter 7 Part 2.
Why A is wrong: A is wrong. Primary aliphatic alcohols (like ethanol) do not have an activated aromatic ring. Dichlorocarbene requires the electron-rich phenoxide ring for electrophilic substitution.
Why B is wrong: B is wrong. Aromatic hydrocarbons like benzene and toluene lack the –OH group that generates the phenoxide ion in base. Without the strong activating effect of –O⁻, the ring is not sufficiently nucleophilic for carbene attack under these conditions.
Why D is wrong: D is wrong. Secondary aliphatic alcohols have no aromatic ring and therefore cannot participate in electrophilic aromatic substitution by dichlorocarbene.
MCQ 4Direct ApplicationPractice
When 2,6-dimethylphenol is treated with CHCl₃ and aqueous NaOH, the formyl group (–CHO) is introduced at which position?
Show answer and why every option is right or wrong
Answer: A. A is correct. The Reimer-Tiemann reaction normally gives ortho substitution. When both ortho positions are blocked (by methyl groups at positions 2 and 6), the –CHO group is directed to the para position. The reaction still proceeds because the para carbon is activated by the phenoxide oxygen.
Why B is wrong: B is wrong. The meta position is not activated by the –OH group. Electrophilic substitution on phenol follows ortho/para directing, not meta.
Why C is wrong: C is wrong. Both ortho positions (2 and 6) are already occupied by methyl groups. Steric hindrance prevents dichlorocarbene from attacking these positions.
Why D is wrong: D is wrong. The reaction does proceed. When ortho positions are blocked, the para position remains available and activated. The product is 4-hydroxy-3,5-dimethylbenzaldehyde.
MCQ 5Direct ApplicationPractice
A student performs the Reimer-Tiemann reaction but accidentally uses CCl₄ instead of CHCl₃ with phenol and NaOH. Which product is most likely obtained after acidification?
Show answer and why every option is right or wrong
Answer: C. C is correct. With CCl₄ in place of CHCl₃, the ring is substituted at the ortho position by a trichloromethyl (–CCl₃) group instead of a dichloromethyl one, and hydrolysis of –CCl₃ gives –COOH rather than –CHO. The product is salicylic acid.
Why A is wrong: A is wrong. Salicylaldehyde is the product when CHCl₃ is used (dichloromethyl intermediate → –CHO on hydrolysis). With CCl₄, the extra chlorine means the trichloromethyl intermediate hydrolyses to –COOH, not –CHO. This CHCl₃ vs CCl₄ swap is a common NEET distractor.
Why B is wrong: B is wrong. Phenyl formate (an ester) is not a product of this pathway. The Reimer-Tiemann mechanism proceeds via electrophilic aromatic substitution, not esterification.
Why D is wrong: D is wrong. Chlorobenzene would require a Sandmeyer-type or direct halogenation reaction. The Reimer-Tiemann mechanism introduces a carbon-containing group (–CHO or –COOH), not a halogen, onto the ring.
MCQ 6Direct ApplicationPractice
In the Reimer-Tiemann reaction, which of the following conditions is essential for the generation of dichlorocarbene from CHCl₃?
Show answer and why every option is right or wrong
Answer: D. D is correct. Dichlorocarbene is generated by α-elimination: NaOH deprotonates CHCl₃ to give ⁻CCl₃, which rapidly loses Cl⁻ to form :CCl₂. A strong base is required. NCERT Class 12 Chemistry, Chapter 7 Part 2, page 213.
Why A is wrong: A is wrong. Acidic conditions would protonate the phenol rather than generating phenoxide, and would not promote α-elimination of CHCl₃. The reaction requires a strong base.
Why B is wrong: B is wrong. A neutral medium does not provide the hydroxide ion concentration needed for α-elimination of CHCl₃. Without base, no dichlorocarbene is generated.
Why C is wrong: C is wrong. AlCl₃ is a Lewis acid catalyst used in Friedel-Crafts reactions, not in the Reimer-Tiemann reaction. The Reimer-Tiemann mechanism requires base-promoted carbene formation, not Lewis acid catalysis.
MCQ 7Concept TrapPractice
A student claims that the Reimer-Tiemann reaction can be used to convert ethanol to ethanal by treatment with CHCl₃/NaOH. Which statement correctly evaluates this claim?
Show answer and why every option is right or wrong
Answer: B. B is correct. The Reimer-Tiemann reaction is an electrophilic aromatic substitution. Dichlorocarbene attacks the electron-rich aromatic ring of phenoxide. Ethanol has no aromatic ring and cannot undergo this reaction. The reaction is specific to phenols.
Why A is wrong: A is wrong. The Reimer-Tiemann reaction is not an oxidation reaction. It is an electrophilic aromatic substitution that introduces a formyl group onto the phenol ring via dichlorocarbene. It has no applicability to aliphatic alcohols.
Why C is wrong: C is wrong. Dichlorocarbene does not insert into O–H bonds under Reimer-Tiemann conditions. Its reactivity in this context is as an electrophile attacking the π-electron-rich aromatic ring of phenoxide.
Why D is wrong: D is wrong. The reaction fails for all aliphatic alcohols regardless of classification (primary, secondary, or tertiary). The requirement is an activated aromatic ring from phenoxide, which no aliphatic alcohol possesses.
MCQ 8Concept TrapPractice
Consider the following two reactions:
(I) Phenol + CHCl₃ + NaOH → Product A
(II) Phenol + CO₂ + NaOH (high pressure, 125°C) → Product B
Which statement about Products A and B is correct?
Show answer and why every option is right or wrong
Answer: B. B is correct. Reaction (I) is the Reimer-Tiemann reaction, which introduces –CHO at the ortho position → salicylaldehyde. Reaction (II) is the Kolbe reaction (Kolbe-Schmitt), which introduces –COONa at ortho → sodium salicylate (salicylic acid after acidification). Distinguishing these two named reactions is a common NEET discriminator.
Why A is wrong: A is wrong. While Product A is indeed salicylaldehyde (Reimer-Tiemann), Product B from the Kolbe reaction is salicylic acid, not salicylaldehyde. The Kolbe reaction uses CO₂ under pressure, not CHCl₃.
Why C is wrong: C is wrong. This reverses the products. The Reimer-Tiemann reaction (CHCl₃/NaOH) gives aldehyde (–CHO), not acid. The Kolbe reaction (CO₂/NaOH/pressure) gives carboxylic acid (–COOH). Swapping Reimer-Tiemann and Kolbe products is a high-frequency error.
Why D is wrong: D is wrong. Product A from the Reimer-Tiemann reaction is salicylaldehyde (–CHO), not salicylic acid. Only the Kolbe reaction yields salicylic acid.
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How do you solve a Reimer Tiemann question? A worked example
- 1
Given
Phenol is treated with CHCl₃ in the presence of aqueous NaOH, followed by acid hydrolysis.
- 2
Required
Identify the product and name the reaction.
- 3
Concept
The Reimer-Tiemann reaction converts phenol to salicylaldehyde via electrophilic aromatic substitution by dichlorocarbene (:CCl₂), generated in situ from CHCl₃ and NaOH.
- 4
Formula / reaction
C₆H₅OH + CHCl₃ + 3NaOH → 2-HOC₆H₄CHO + 3NaCl + 2H₂O
- 5
Substitution
Phenol → phenoxide ion (in NaOH). CHCl₃ → :CCl₂ (by α-elimination). :CCl₂ attacks ortho carbon of phenoxide → dichloromethyl intermediate → hydrolysis → –CHO.
- 6
Calculation
No numerical calculation required. The product is identified by the reaction mechanism: ortho-formylation of phenol.
- 7
Final answer
The product is salicylaldehyde (2-hydroxybenzaldehyde). The reaction is the Reimer-Tiemann reaction.
- 8
Common trap
Confusing the Reimer-Tiemann product (aldehyde, from CHCl₃) with the Kolbe reaction product (carboxylic acid, from CO₂). Also: misidentifying the electrophile as CHCl₃ itself rather than dichlorocarbene (:CCl₂).
- 9
Similar NEET-style question
"When p-cresol (4-methylphenol) is heated with CHCl₃ and aqueous NaOH, the product is (a) 2-hydroxy-5-methylbenzaldehyde, (b) 4-methylbenzoic acid, (c) 4-methylbenzaldehyde, (d) anisole."
Answer: (a). The –CHO is introduced ortho to –OH (which is position 2; position 4 is occupied by –CH₃), yielding 2-hydroxy-5-methylbenzaldehyde.
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What to remember before solving Reimer Tiemann questions
More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
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