Reimer–Tiemann Reaction

Introduction

Phenol can gain an aldehyde group on its aromatic ring when treated with chloroform and aqueous base, followed by hydrolysis and acid work-up. The main introductory product has –CHO ortho to –OH and is called salicylaldehyde. This Reimer–Tiemann reaction first generates dichlorocarbene from chloroform; the eventual aldehyde carbon is that chloroform carbon, not a carbon from the phenol ring.

Core explanation

Base converts some phenol to phenoxide, C₆H₅O⁻. As in other reactions of phenoxide, oxygen donation increases electron density at ortho and para aromatic carbons. Strong base also removes the relatively acidic H from CHCl₃. The resulting trichloromethyl anion loses Cl⁻ to form dichlorocarbene, :CCl₂. A carbene is a neutral carbon species with only six valence electrons; in this context :CCl₂ is an electrophilic reactive intermediate generated in situ rather than stored as a bottle of reagent.

An activated ring carbon of phenoxide forms a bond to the carbene carbon. The reaction is often described as an electrophilic aromatic substitution sequence: attachment disturbs aromaticity, then loss of a ring proton restores it. The directly introduced carbon still bears two chlorines, so the first ring-substitution intermediate is not yet an aldehyde. Under the aqueous basic and subsequent work-up conditions, its dichloromethyl group is hydrolysed to –CHO. The final organic structure is hydroxybenzaldehyde, generally with the formyl group at the ortho position relative to OH.

For unsubstituted phenol, salicylaldehyde is 2-hydroxybenzaldehyde, HO–C₆H₄–CHO with OH and CHO on adjacent ring carbons. Para-hydroxybenzaldehyde may also arise under some conditions, so "ortho major" is more careful than "only ortho." If the ortho positions are blocked by substituents, product selection changes and para placement may become more important. In a basic exam prediction, however, phenol plus CHCl₃/NaOH followed by work-up is mapped to salicylaldehyde.

Atom tracking separates Reimer–Tiemann from Kolbe–Schmitt. Both start from phenoxide and both commonly functionalise ortho to OH. Reimer–Tiemann adds the carbon of CHCl₃ and ends with –CHO, an aldehyde. Kolbe–Schmitt adds the carbon of CO₂ and ends with –COOH after acidification, a carboxylic acid. The reagent's carbon oxidation state and subsequent hydrolysis govern that difference. A student who memorises only "phenol gets a carbon at ortho" may write the wrong functional group.

In a full mechanism, avoid showing OH⁻ simply substituting chlorine on chloroform to make the aromatic aldehyde in one step. The crucial first transformations are deprotonation of chloroform, alpha elimination of chloride to give :CCl₂, attachment to activated phenoxide, and hydrolysis of the dichloromethyl group. These stages explain why the product requires both the aromatic activation and the chloroform-derived electrophile.

Salicylaldehyde has both phenolic and aldehyde functions. Its aldehyde can take part in nucleophilic addition and condensation reactions, while its phenolic oxygen can undergo acid–base or etherification chemistry. This dual reactivity makes the product useful as a synthetic intermediate; the reaction installs a new functional handle without removing the original OH.

Step-by-step reasoning

First identify phenol, CHCl₃ and strong aqueous base. Convert phenol to phenoxide and chloroform to :CCl₂ by deprotonation plus Cl⁻ loss. Mark an activated ortho ring carbon and attach the carbene carbon there, restoring aromaticity by proton loss. Convert the attached –CHCl₂ equivalent to –CHO by hydrolysis and work-up. Name the product 2-hydroxybenzaldehyde.

Visual explanation

Draw two parallel preparation arrows: C₆H₅OH → C₆H₅O⁻ and CHCl₃ → CCl₃⁻ → :CCl₂. Bring the phenoxide ortho carbon and carbene carbon together. Show the first ring substituent with two chlorine labels, then a separate hydrolysis arrow turning that same carbon into –CHO. Keep the phenolic oxygen on the ring throughout.

Real-world analogy

Imagine preparing a temporary tagged building block before attaching it to a marked seat on a ring. The tag is changed after attachment to reveal the final aldehyde handle. Chloroform supplies the tagged carbon as dichlorocarbene; hydrolysis later changes its two chlorine attachments. The analogy helps keep carbon installation separate from final functional-group conversion.

Real-world example

An undergraduate synthesis problem may ask for conversion of phenol to salicylaldehyde. CHCl₃/aqueous NaOH followed by acid work-up gives the Reimer–Tiemann route. The isolated product's adjacent OH and CHO groups make it useful for further Schiff-base formation with amines, demonstrating how one named reaction creates a multifunctional aromatic intermediate.

Why?

Why is phenoxide a useful partner for the weakly selective electrophile :CCl₂? Its oxygen donor places substantial electron density at ortho and para ring positions, making ring-carbon attack more favourable than with benzene. The same activation pattern helps explain ortho and para products, while conditions and steric effects influence their proportions.

Common misconception

"CHCl₃ and base add a carboxyl group just like CO₂." Chloroform-derived dichlorocarbene leads after hydrolysis to a formyl, –CHO, group. CO₂ in the Kolbe–Schmitt reaction supplies a carboxyl, –COOH, group. The two products have different oxygen counts and oxidation states.

Worked example

Question: Predict the principal product from phenol treated with CHCl₃ and aqueous NaOH, then acidified.

Reasoning: Base makes phenoxide and generates :CCl₂ from chloroform. The phenoxide ring reacts mainly at ortho, and hydrolysis changes the installed dichloromethyl-derived carbon into an aldehyde group.

Answer: Salicylaldehyde, 2-hydroxybenzaldehyde, HO–C₆H₄–CHO with adjacent substituents.

Quick check

1. What reactive electrophile is generated from chloroform in Reimer–Tiemann chemistry? Answer: Dichlorocarbene, :CCl₂, is generated by base-promoted deprotonation and chloride loss.

Exam focus

Recognise CHCl₃/base as the formylation reagent pair, write :CCl₂ as intermediate, and keep phenol's OH on the ring. Show ortho –CHO after hydrolysis and acid work-up. Compare with CO₂/phenoxide carboxylation to avoid confusing salicylaldehyde with salicylic acid.

Advanced insight

The introduced carbon passes through different functional states: CHCl₃, electrophilic :CCl₂, ring-bound dichloromethyl derivative and finally CHO. Tracking that single carbon is a useful antidote to name-reaction memorisation. Substrate substitution can block ortho approach and change the regioisomer distribution, so the familiar salicylaldehyde product belongs to the simple phenol case.

Summary

Reimer–Tiemann reaction uses chloroform and base to form dichlorocarbene, which functionalises phenoxide's activated aromatic ring. Hydrolysis changes the installed carbon into an aldehyde; acid work-up of phenol commonly gives ortho-hydroxybenzaldehyde, salicylaldehyde. The reaction adds –CHO, whereas Kolbe–Schmitt CO₂ carboxylation adds –COOH.

Practice questions

1. Which starting reagent supplies the carbon of the new aldehyde group? Answer: Chloroform, CHCl₃, supplies that carbon through dichlorocarbene. 2. What is the principal positional relationship between OH and CHO from phenol? Answer: They are ortho, on adjacent aromatic carbons, in salicylaldehyde. 3. Why is the first ring-bound carbon intermediate not yet –CHO? Answer: The attached carbene-derived carbon initially bears chlorine atoms that must be hydrolysed. 4. What product group would CO₂/phenoxide give after acid work-up instead? Answer: A carboxylic acid group, usually salicylic acid under standard Kolbe–Schmitt conditions.