Question

Write short notes on the following: (i) Reimer–Tiemann reaction
(ii) Kolbe reaction

Hint:

Remember: Reimer–Tiemann → \(-CHO\) group (aldehyde), Kolbe → \(-COOH\) group (acid).

Answer 1

(i) Reimer–Tiemann Reaction:
The Reimer–Tiemann reaction is used to introduce a formyl group (\(-CHO\)) at the ortho position of phenol.
When phenol is treated with chloroform (CHCl$_3$) and aqueous sodium hydroxide, it gives salicylaldehyde (o-hydroxybenzaldehyde).
\[ C_6H_5OH + CHCl_3 + 3NaOH \; \longrightarrow \; o\!-\!HOC_6H_4CHO + 3NaCl + 2H_2O \] This reaction is important for preparing aromatic aldehydes. (ii) Kolbe Reaction:
The Kolbe reaction is the carboxylation of phenol to produce salicylic acid.
Phenol is first converted to sodium phenoxide, which reacts with carbon dioxide under high pressure and temperature, 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 widely used for the synthesis of salicylic acid, a precursor of aspirin. Conclusion:
Reimer–Tiemann introduces an aldehyde group at the ortho position of phenol, while Kolbe introduces a carboxyl group, both being important named reactions in aromatic chemistry.