Question:medium

When phenol is treated with chloroform (\(\text{CHCl}_3\)) in the presence of an aqueous sodium hydroxide solution followed by acidification, a prominent aromatic aldehyde is generated. What is the name of this organic reaction?

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Use the reagents to easily distinguish between these similar reactions of phenol: Phenol + \(\text{CHCl}_3 / \text{NaOH} \rightarrow\) Salicylaldehyde (Reimer-Tiemann). Phenol + \(\text{CO}_2 / \text{NaOH} \rightarrow\) Salicylic Acid (Kolbe's).
Updated On: May 30, 2026
  • Kolbe's Reaction
  • Reimer-Tiemann Reaction
  • Rosenmund Reduction
  • Friedel-Crafts Acylation
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The Correct Option is B

Solution and Explanation

Step 1 : Understanding the Question:
The topic of this question is Named Organic Reactions involving Phenols. Phenols are highly reactive toward electrophilic aromatic substitution due to the strongly activating $-OH$ group. There are several famous named reactions that introduce different functional groups onto the phenol ring. The question describes a specific reaction using chloroform and sodium hydroxide to produce an aromatic aldehyde and asks us to identify the name of this process.
Step 2 : Key Formulas and approach:
The approach is to match the reagents and products to known named reactions:
1. Phenol + Chloroform ($CHCl_3$) + $NaOH \rightarrow$ Salicylaldehyde ($o$-hydroxybenzaldehyde).
2. Identify the active intermediate: Dichlorocarbene ($:CCl_2$).
3. Distinguish from similar reactions: For example, using $CO_2$ instead of $CHCl_3$ would lead to a different reaction.
Step 3 : Detailed Explanation:

The reaction described is the Reimer-Tiemann Reaction. It is a classic method for the ortho-formylation of phenols.

In the first step, Chloroform ($CHCl_3$) reacts with the base ($NaOH$) to produce a highly reactive, neutral electrophile called dichlorocarbene ($:CCl_2$).

This dichlorocarbene attacks the electron-rich phenoxide ion (formed by phenol and $NaOH$) at the ortho position.

After subsequent hydrolysis of the intermediate and final acidification, a $-CHO$ (formyl) group is successfully attached to the ring, producing Salicylaldehyde.

Let's evaluate the other options:
- Kolbe's Reaction (Option A) uses $NaOH$ and Carbon Dioxide ($CO_2$) to produce Salicyllic Acid (not an aldehyde).
- Rosenmund Reduction (Option C) involves the hydrogenation of acyl chlorides to aldehydes using a poisoned catalyst.
- Friedel-Crafts Acylation (Option D) uses acyl halides and $AlCl_3$ to add ketone groups to an aromatic ring.

Since the reagents are $CHCl_3$ and $NaOH$, the reaction is uniquely identified as the Reimer-Tiemann Reaction.

Step 4 : Final Answer:
The introduction of a formyl group using chloroform and alkali is known as the Reimer-Tiemann Reaction. The correct option is (B).
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