Question

Which of these intermediates is formed during the Reimer-Tiemann reaction of phenol with chloroform and NaOH?

• A. Nitrene
• B. Free radical
• C. Carbonium ion
• D. Dichlorocarbene

Hint:

The Reimer–Tiemann reaction puts an aldehyde group (–CHO) onto phenol to make salicylaldehyde. To understand it you must know which reactive intermediate does the attacking.

Answer 1

The Correct Option is D

Step 1: What is the Reimer-Tiemann reaction?
The Reimer-Tiemann reaction converts phenol to an ortho-hydroxybenzaldehyde (salicylaldehyde) using chloroform (CHCl₃) and sodium hydroxide (NaOH). It is a classic electrophilic aromatic substitution.

Step 2: What happens to CHCl₃ in strong base?
NaOH removes all three chloride ions from CHCl₃ stepwise. First one H is removed to give a trichloromethyl carbanion, then a chloride ion leaves, generating a highly reactive neutral species called dichlorocarbene (:CCl₂).

Step 3: Role of the intermediate.
Dichlorocarbene is an electrophilic carbene. It attacks the electron-rich ortho or para position of the phenoxide ion (phenol in NaOH). This addition, followed by hydrolysis, gives the aldehyde group.

Step 4: Why not the other options?
A nitrene is a nitrogen analogue of carbene (produced in Curtius/Hofmann reactions, not here). A free radical mechanism would give random substitution without regioselectivity. A carbonium ion (carbocation) is produced in electrophilic reactions driven by Lewis acids, not strong base conditions.

Step 5: Confirm the answer.
The key intermediate uniquely formed from CHCl₃ + NaOH in this reaction is dichlorocarbene. All textbooks confirm this.

Answer: Option (4) — Dichlorocarbene