CH_3CHO + HCHO text{ (excess)} xrightarrow{te | vedclass

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(D) The reaction of acetaldehyde $(CH_3CHO)$ with excess formaldehyde $(HCHO)$ in the presence of concentrated $NaOH$ is a classic example of a sequen

Vedclass · Accepted Answer

$C(CH_2OH)_4$

Vedclass · Answer

(D) The reaction of acetaldehyde $(CH_3CHO)$ with excess formaldehyde $(HCHO)$ in the presence of concentrated $NaOH$ is a classic example of a sequence involving aldol condensation followed by a crossed Cannizzaro reaction.$1$. Acetaldehyde has three $\alpha$-hydrogens. It undergoes three successive aldol additions with three molecules of formaldehyde to form $CH_3-C(CH_2OH)_3-CHO$.$2$. This intermediate product,$CH_3-C(CH_2OH)_3-CHO$,has no $\alpha$-hydrogens remaining.$3$. In the presence of excess formaldehyde and concentrated $NaOH$,this intermediate undergoes a crossed Cannizzaro reaction where the aldehyde group $(-CHO)$ is reduced to a primary alcohol $(-CH_2OH)$ and the formaldehyde is oxidized to sodium formate.$4$. The final product is pentaerythritol,$C(CH_2OH)_4$.

$CH_3CHO + HCHO \text{ (excess)} \xrightarrow{\text{Conc. } NaOH} \text{Product}$. The product is:

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