Question

Write the reaction involved in the following:

Reimer-Tiemann reaction

Hint:

The Reimer-Tiemann reaction introduces a \(-CHO\) group at the ortho position of phenol using chloroform and a base.

Answer 1

Step 1: Aim of the reaction.
The Reimer-Tiemann reaction puts a \(-CHO\) (aldehyde) group onto the benzene ring of phenol, mainly at the ortho position, giving salicylaldehyde.

Step 2: Reagents used.
Phenol is treated with chloroform \((CHCl_3)\) and aqueous sodium hydroxide \((NaOH)\) at about 340 K.

Step 3: Formation of the attacking species.
The \(NaOH\) reacts with \(CHCl_3\) to generate dichlorocarbene \((:CCl_2)\), which is the electron-seeking species that attacks the ring.

Step 4: Attack on the ring.
Phenol first forms phenoxide ion in the basic medium. The dichlorocarbene attacks the highly reactive ortho position of the phenoxide ring.

Step 5: Final hydrolysis.
The intermediate formed is then hydrolysed by the alkali to give the ortho-hydroxybenzaldehyde (salicylaldehyde) after acidification.

Step 6: Overall reaction.
\[ C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\,K} o\text{-}HOC_6H_4CHO + 3NaCl + 2H_2O \]

Answer: Phenol with \(CHCl_3\) and aqueous \(NaOH\) at 340 K, followed by acidification, gives salicylaldehyde (2-hydroxybenzaldehyde) as the main product. This is the Reimer-Tiemann reaction.