Question

Write short notes on Etard reaction and Gattermann-Koch reaction. (2 + 2)

Hint:

Etard: toluene + chromyl chloride \((CrO_2Cl_2)\) then hydrolysis. Gattermann-Koch: benzene + CO + HCl with anhydrous AlCl₃/CuCl. Both give benzaldehyde.

Answer 1

Step 1: Both reactions are methods to make the aromatic aldehyde benzaldehyde; one starts from toluene, the other from benzene.
Step 2 (Etard reaction): In this reaction the CH₃ group of toluene is partially oxidised. The oxidising agent chromyl chloride, \(CrO_2Cl_2\), attacks the methyl group. To stop the reaction at the aldehyde stage the reaction is carried out in an inert solvent (carbon disulphide or carbon tetrachloride). A brown chromium complex is first formed at the methyl carbon.
\[ C_6H_5CH_3 + CrO_2Cl_2 \rightarrow \text{brown chromium complex} \]
Step 3: Decomposition of this complex with dilute acid or water releases benzaldehyde:
\[ \text{complex} + H_2O \rightarrow C_6H_5CHO \]
Step 4 (Gattermann-Koch reaction): This is a Friedel-Crafts type formylation. A mixture of carbon monoxide and hydrogen chloride behaves like the formylating agent (equivalent to formyl chloride, HCOCl). In presence of the Lewis acid anhydrous \(AlCl_3\) (with CuCl aiding CO uptake) the formyl group is placed on benzene:
\[ C_6H_6 + CO + HCl \xrightarrow{AlCl_3,\,CuCl} C_6H_5CHO + HCl \]
Step 5: Thus toluene via Etard and benzene via Gattermann-Koch both deliver the same product.
\[\boxed{\text{Product} = \text{benzaldehyde}}\]