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Q7.18(ii):
Explain the following with an example. Reimer-Tiemann reaction.

Solution :

Theoretical Foundation: The Reimer-Tiemann Reaction

The Reimer-Tiemann reaction is a form of substitution reaction used for the ortho-formylation of phenols. When phenol is treated with chloroform ($CHCl_3$) in the presence of an aqueous base (typically $NaOH$ or $KOH$), an aldehyde group ($-CHO$) is introduced at the ortho-position of the benzene ring. The major product formed is salicylaldehyde ($2$-hydroxybenzaldehyde).

Step 1: Generation of the Electrophile

The base ($OH^-$) abstracts a proton from chloroform ($CHCl_3$), leading to the formation of a trichloromethyl anion. This anion subsequently undergoes $\alpha$-elimination to generate the highly reactive, neutral electrophile, dichlorocarbene ($:CCl_2$).

$CHCl_3 + OH^- \rightarrow :CCl_2 + H_2O + Cl^-$ [Per the mechanism of $\alpha$-elimination]

Step 2: Electrophilic Attack on Phenoxide Ion

In the presence of $NaOH$, phenol is converted to the more nucleophilic phenoxide ion ($C_6H_5O^-$). The phenoxide ion undergoes electrophilic attack by the dichlorocarbene at the ortho-position. This breaks the aromaticity temporarily to form an intermediate.

$C_6H_5O^- + :CCl_2 \rightarrow [Intermediate \ Complex]$

Step 3: Hydrolysis and Regeneration of Aromaticity

The intermediate is subjected to hydrolysis by the aqueous base. The geminal dichloride group ($-CHCl_2$) is converted into an aldehyde group ($-CHO$) via a gem-diol intermediate, and the aromatic system is restored by the loss of a proton.

Chemical Equation Representation

The transformation is summarized as follows:

$C_6H_5OH + 3NaOH + CHCl_3 \xrightarrow{65-70^\circ C} C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$

Phenol 1) CHCl3/NaOH 2) H3O+ Salicylaldehyde

Example: Synthesis of Salicylaldehyde

Reactant Reagent Product
Phenol ($C_6H_5OH$) $CHCl_3 / aq. NaOH$ $2$-hydroxybenzaldehyde (Salicylaldehyde)

The reaction serves as a vital method for the introduction of a formyl group into phenolic compounds, which are otherwise resistant to standard Friedel-Crafts formylation due to the formation of complexes between $AlCl_3$ and the phenolic oxygen.

Final Solution: The Reimer-Tiemann reaction is the electrophilic substitution of phenol with dichlorocarbene (generated from $CHCl_3$ and $NaOH$) to produce salicylaldehyde.


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