Question

Write short notes on the following: (A) Reimer–Tiemann Reaction
(B) Kolbe’s Reaction
(C) Williamson Ether Synthesis

Hint:

Reimer–Tiemann = aldehyde formation, Kolbe = acid formation, Williamson = ether formation.

Answer 1

(A) Reimer–Tiemann Reaction:
The Reimer–Tiemann reaction is used to introduce a formyl group ($-CHO$) at the ortho position of phenol. Phenol reacts with chloroform (CHCl$_3$) and sodium hydroxide (NaOH) to give salicylaldehyde.
\[ C_6H_5OH + CHCl_3 + 3NaOH \;\; \longrightarrow \;\; o\!-\!HOC_6H_4CHO + 3NaCl + 2H_2O \] (B) Kolbe’s Reaction:
The Kolbe reaction is the carboxylation of phenol to produce salicylic acid. Sodium phenoxide reacts with carbon dioxide under high temperature and pressure, followed by acidification.
\[ C_6H_5ONa + CO_2 \;\; \xrightarrow{373K, \; 4-7 atm} \;\; o\!-\!HOC_6H_4COONa \] \[ o\!-\!HOC_6H_4COONa + HCl \;\; \longrightarrow \;\; o\!-\!HOC_6H_4COOH \] This method is important for synthesizing salicylic acid (a precursor of aspirin). (C) Williamson Ether Synthesis:
This is a laboratory method for preparing ethers. An alkoxide ion reacts with a primary alkyl halide in a nucleophilic substitution reaction to produce ether.
\[ R\!-\!ONa + R'X \;\; \longrightarrow \;\; R\!-\!O\!-\!R' + NaX \] Example: \[ C_2H_5ONa + CH_3I \;\; \longrightarrow \;\; C_2H_5OCH_3 + NaI \] This reaction is widely used for the synthesis of both symmetrical and unsymmetrical ethers. Conclusion:
- Reimer–Tiemann introduces $-CHO$ group in phenol.
- Kolbe introduces $-COOH$ group in phenol.
- Williamson ether synthesis is the standard method to prepare ethers.