Question

Which of the following is the Reimer-Tiemann reaction?

• A. \( \text{Formylation of phenols using chloroform and alkali to form ortho-hydroxybenzaldehyde.} \)
• B. \( \text{Bromination of phenols in the presence of bromine water.} \)
• C. \( \text{Oxidation of phenols to quinones using an oxidizing agent.} \)
• D. \( \text{Nitration of phenols in the presence of concentrated nitric acid.} \)

Hint:

The Reimer-Tiemann reaction is a selective method for formylating phenols, producing ortho-hydroxybenzaldehyde. It requires chloroform (\( \text{CHCl}_3 \)) and a strong base like NaOH to generate the reactive dichlorocarbene intermediate.

Answer 1

The Correct Option is A

The Reimer-Tiemann reaction, a key electrophilic aromatic substitution, introduces a formyl group (\( -CHO \)) at the ortho position of phenols. This process utilizes chloroform (\( \text{CHCl}_3 \)) and a strong base (e.g., NaOH).
Mechanism:- Chloroform reacts with a strong base (such as NaOH or KOH) to generate dichlorocarbene (\( :\text{CCl}_2 \)), an electrophilic intermediate.- Dichlorocarbene attacks the ortho position of the phenol ring, forming an intermediate compound.- Hydrolysis of this intermediate yields ortho-hydroxybenzaldehyde as the final product.
Example:The Reimer-Tiemann reaction involving phenol (\( \text{C}_6\text{H}_5\text{OH} \)) and chloroform (\( \text{CHCl}_3 \)) in the presence of NaOH proceeds as follows:\[\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} \rightarrow \text{o-HO-C}_6\text{H}_4\text{CHO} + 3\text{NaCl} + 2\text{H}_2\text{O}.\]Ortho-hydroxybenzaldehyde (\( \text{o-HO-C}_6\text{H}_4\text{CHO} \)) is the predominant product.
Conclusion:The Reimer-Tiemann reaction specifically formylates the ortho position of phenols, making \( \mathbf{(A)} \) the correct answer.