Question

Write short notes on the following: (i) Claisen Reaction
(ii) Cannizzaro’s Reaction
(iii) Rosenmund’s Reaction

Hint:

Claisen → condensation (C–C bond), Cannizzaro → disproportionation (no $\alpha$-H), Rosenmund → acyl chloride to aldehyde.

Answer 1

(i) Claisen Reaction:
The Claisen condensation is the reaction of two esters or one ester and another carbonyl compound in the presence of a strong base (like sodium ethoxide) to form a $\beta$-keto ester or a $\beta$-diketone.
\[ 2CH_3COOC_2H_5 \; \xrightarrow{NaOC_2H_5} \; CH_3COCH_2COOC_2H_5 + C_2H_5OH \] This reaction is important in C–C bond formation in organic synthesis. (ii) Cannizzaro’s Reaction:
Aldehydes without $\alpha$-hydrogen undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali to form a primary alcohol and a carboxylate salt.
\[ 2HCHO \; \xrightarrow{Conc. NaOH} \; CH_3OH + HCOONa \] Example: Formaldehyde gives methanol and sodium formate. (iii) Rosenmund’s Reaction:
This reaction involves the hydrogenation of acid chlorides in the presence of palladium (Pd) catalyst on barium sulfate (BaSO$_4$) to produce aldehydes.
\[ RCOCl + H_2 \; \xrightarrow{Pd/BaSO_4} \; RCHO + HCl \] This method is used for the controlled synthesis of aldehydes from acyl chlorides. Conclusion:
Claisen reaction forms $\beta$-keto esters, Cannizzaro’s reaction converts aldehydes without $\alpha$-H into alcohols and acids, and Rosenmund’s reaction selectively reduces acyl chlorides to aldehydes.