Question

Write short notes on the following: 
(i) Etard Reaction 
(ii) Gattermann–Koch Reaction 
(iii) Cannizzaro’s Reaction 
(iv) Aldol Condensation 
 

Hint:

Etard and Gattermann–Koch introduce $-CHO$, Cannizzaro requires no $\alpha$-H, while Aldol requires $\alpha$-H for condensation.

Answer 1

(i) Etard Reaction: 
- In this reaction, an aromatic hydrocarbon containing a $-CH_3$ group is oxidized to an aldehyde. 
- Chromyl chloride (CrO$_2$Cl$_2$) is used as the oxidizing agent. 
\[ C_6H_5CH_3 \; \xrightarrow{CrO_2Cl_2} \; C_6H_5CHO \] Example: Toluene is converted to benzaldehyde. 

(ii) Gattermann–Koch Reaction: 
- This reaction is used for formylation of aromatic rings (introduction of $-CHO$ group). 
- Benzene reacts with CO and HCl in the presence of AlCl$_3$ and CuCl as catalysts to give benzaldehyde. 
\[ C_6H_6 + CO + HCl \; \xrightarrow{AlCl_3/CuCl} \; C_6H_5CHO \] 

(iii) Cannizzaro’s Reaction: 
- Aldehydes without $\alpha$-hydrogen undergo self-oxidation and reduction in the presence of concentrated alkali. 
- One molecule is reduced to alcohol and the other is oxidized to carboxylate salt. 
\[ 2HCHO \; \xrightarrow{Conc. NaOH} \; CH_3OH + HCOONa \] 

(iv) Aldol Condensation: 
- Aldehydes or ketones containing $\alpha$-hydrogen undergo condensation in the presence of dilute alkali to form $\beta$-hydroxy aldehydes or ketones. 
- On heating, these give $\alpha,\beta$-unsaturated carbonyl compounds. 
\[ 2CH_3CHO \; \xrightarrow{NaOH} \; CH_3CH(OH)CH_2CHO \; \xrightarrow{\Delta} \; CH_3CH=CHCHO \] 

Conclusion: 
- Etard → toluene $\to$ benzaldehyde. 
- Gattermann–Koch → benzene $\to$ benzaldehyde. 
- Cannizzaro → aldehyde without $\alpha$-H $\to$ alcohol + salt. 
- Aldol → aldehyde/ketone with $\alpha$-H $\to$ $\alpha,\beta$-unsaturated compound.