When phenol is treated with CHCl₃ and NaOH, the product formed is

Question

The major product formed in the following reaction is-

Answer 1

The reaction of phenol with chloroform (\( \text{CHCl}_3 \)) and sodium hydroxide (\( \text{NaOH} \)) is known as the Reimer-Tiemann reaction. This reaction is used to introduce a formyl group (\(-\text{CHO}\)) into the ortho-position of an aromatic compound.

Let's analyze the steps involved in the Reimer-Tiemann reaction leading to the formation of salicylaldehyde from phenol:

In the presence of \(\text{NaOH}\), phenol (\( \text{C}_6\text{H}_5\text{OH} \)) is converted into phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^- \)), which is a stronger nucleophile.
\(\text{CHCl}_3\) reacts with the base (\(\text{NaOH}\)) to form a dichlorocarbene (\(:\text{CCl}_2\)) intermediate.
The nucleophilic ortho-position of the phenoxide ion attacks the dichlorocarbene, resulting in an intermediate compound.
Hydrolysis of this intermediate during work-up leads to the formation of salicylaldehyde (\( \text{C}_6\text{H}_4(\text{OH})\text{CHO} \)).

The correct answer is Salicylaldehyde because it matches the expected product of this chemical reaction based on the Reimer-Tiemann mechanism. Other products like Benzaldehyde, Salicylic acid, and Benzoic acid are not formed in this particular reaction.

To conclude, the treatment of phenol with \(\text{CHCl}_3\) and \(\text{NaOH}\) yields salicylaldehyde via the Reimer-Tiemann reaction.

Answer 2

The reaction of phenol with chloroform (\( \text{CHCl}_3 \)) and sodium hydroxide (\( \text{NaOH} \)) is known as the Reimer-Tiemann reaction. This reaction is used to introduce a formyl group (\(-\text{CHO}\)) into the ortho-position of an aromatic compound.

Let's analyze the steps involved in the Reimer-Tiemann reaction leading to the formation of salicylaldehyde from phenol:

In the presence of \(\text{NaOH}\), phenol (\( \text{C}_6\text{H}_5\text{OH} \)) is converted into phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^- \)), which is a stronger nucleophile.
\(\text{CHCl}_3\) reacts with the base (\(\text{NaOH}\)) to form a dichlorocarbene (\(:\text{CCl}_2\)) intermediate.
The nucleophilic ortho-position of the phenoxide ion attacks the dichlorocarbene, resulting in an intermediate compound.
Hydrolysis of this intermediate during work-up leads to the formation of salicylaldehyde (\( \text{C}_6\text{H}_4(\text{OH})\text{CHO} \)).

The correct answer is Salicylaldehyde because it matches the expected product of this chemical reaction based on the Reimer-Tiemann mechanism. Other products like Benzaldehyde, Salicylic acid, and Benzoic acid are not formed in this particular reaction.

To conclude, the treatment of phenol with \(\text{CHCl}_3\) and \(\text{NaOH}\) yields salicylaldehyde via the Reimer-Tiemann reaction.