Question

Write the reaction involved in Reimer-Tiemann reaction.

Hint:

Reimer-Tiemann = Phenol + Chloroform \(\rightarrow\) Salicylaldehyde.
If $CCl_4$ is used instead of $CHCl_3$, Salicylic acid is formed.

Answer 1

The Reimer-Tiemann reaction is a chemical reaction that involves the formylation of phenols. It is used to introduce an aldehyde group (-CHO) at the ortho or para position to the hydroxyl group (-OH) on an aromatic ring. The reaction uses chloroform (CHCl₃) and a strong base like sodium hydroxide (NaOH) to produce salicylaldehyde.

The general reaction is as follows:

\[ \text{Phenol} + \text{CHCl}_3 + \text{NaOH} \rightarrow \text{Salicylaldehyde} \]

This reaction proceeds by the formation of a dichlorocarbene intermediate, which then reacts with the phenol to give the ortho and para products, typically with the para product being the major one.