Complete the following reactions: (i) Phenetole (benzene ring bearing an -OC2H5 substituent) + HBr arrow ? (ii

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Q. Complete the following reactions:

(i) Phenetole (benzene ring bearing an -OC2H5 substituent) + HBr

\rightarrow

(ii) Phenol (benzene ring bearing an -OH substituent) +

CHCl_3

+ NaOH, at 340 K

\rightarrow

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Reimer-Tiemann Reaction

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 ( $\text{CHCl}_3$ ) in the presence of a strong base like aqueous $\text{NaOH}$ does something dramatic. The base pulls off a proton from chloroform, generating a highly reactive species called dichlorocarbene ( $:\text{CCl}_2$ ). 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 ( $\text{CHCl}_3$ ) and aqueous sodium hydroxide ( $\text{NaOH}$ ) 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.

\ce{C6H5OH + CHCl3 + 3 NaOH ->[\Delta] \underset{\text{salicylaldehyde}}{o-HOC6H4CHO} + 3 NaCl + 2 H2O}

Step-by-Step Mechanism (Why It Works)

Generation of dichlorocarbene $\text{NaOH}$ deprotonates chloroform:

\ce{CHCl3 + OH- -> CCl3- + H2O}

The trichloromethyl anion loses a chloride ion to form the electrophilic carbene:

\ce{CCl3- -> :CCl2 + Cl-}

Attack on phenoxide ion

Phenol first reacts with $\text{NaOH}$ to form the more nucleophilic phenoxide ion ( $\ce{C6H5O-}$ ). 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 + $\text{CHCl}_3$ + aqueous $\text{NaOH}$ (not alcoholic $\text{NaOH}$ — 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. …

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(i) HBr cleaves the alkyl–O bond of an aryl alkyl ether (never the aryl–O bond) via $S_N2$ , giving phenol and the alkyl halide. (ii) Phenol + $CHCl_3$ /NaOH at 340 K is the Reimer–Tiemann reaction, installing a $-CHO$ group ortho to $-OH$ .

(i) Phenetole ( $C_6H_5\text{–}O\text{–}C_2H_5$ ) + HBr:

In an aryl alkyl ether, the aryl–oxygen bond has partial double-bond character (from conjugation of the O lone pair with the ring) and does not break easily; also, an aryl cation is too unstable to form. So cleavage instead occurs at the alkyl–oxygen bond, via nucleophilic ( $S_N2$ -type) attack of $Br^-$ on the protonated ether's alkyl carbon:

$C_6H_5\text{–}O\text{–}C_2H_5 + HBr \longrightarrow C_6H_5OH + C_2H_5Br$

giving phenol and ethyl bromide.

(ii) Phenol + $CHCl_3$ + NaOH, 340 K:

…

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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 $CHCl_3$ and $NaOH$ at about 340 K; the product (after hydrolysis) is predominantly the ortho-hydroxybenzaldehyde, i.e. salicylaldehyde. True.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The reaction: Phenol + CHCl3 + KOH --(heat, delta)--> Salicylaldehyde, is$
(a) Gattermann-Koch reaction
(b) Sandmeyer reaction
(c) Reimer-Tiemann reaction
(d) Fittig reaction
›Reveal solutionSolution

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.

$\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{KOH} \xrightarrow{\Delta} \text{o-HOC}_6\text{H}_4\text{CHO (salicylaldehyde)} + \text{KCl} + \text{H}_2\text{O}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$CHCl_3$
(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 $CHCl_3$ to form the trichloromethyl carbanion $:CCl_3^-$ , which rapidly loses $Cl^-$ to generate the highly reactive electrophile dichlorocarbene ( $:CCl_2$ ).

The phenoxide ion (from phenol + NaOH) attacks $:CCl_2$ preferentially at the ortho position (directed by the strongly activating $-O^-$ group), giving a dichloromethyl-substituted intermediate.

Hydrolysis of the $-CHCl_2$ group under the alkaline reaction conditions converts it to $-CHO$ .

Overall:

$C_6H_5OH \xrightarrow[\text{NaOH}]{CHCl_3} \text{o-hydroxybenzaldehyde (salicylaldehyde)}$

…
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Salicyaldehyde can be prepared from phenol by$
(a) Schotten-Baumann reaction
(b) Kolbe's reaction
(c) Reimer-Tiemann reaction
(d) Cannizzaro reaction
›Reveal solutionSolution

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 …

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CBSE 2024Set ANNUAL1 mark
$Q. What is Reimer-Tiemann reaction?$
›Reveal solutionSolution

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 …

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CBSE 2021Set A1 markMCQ
$Q. By which of the following reactions Phenol is converted into salicyl aldehyde?$
(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)

…

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CBSE 2018Set ANNUAL1 mark
$CHCl_3$
›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 ( $:CCl_2$ ) 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 …

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Q.Complete the following reactions:

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