Question

When HCHO is treated with ${{C}_{6}}{{H}_{5}}CHO$ in presence of KOH, the product are:

• A. $HCOONa$ and ${{C}_{6}}{{H}_{5}}C{{H}_{2}}OH$
• B. $C{{H}_{3}}OH$ and ${{C}_{6}}{{H}_{5}}COONa$
• C. $C{{H}_{3}}OH$ and ${{C}_{6}}{{H}_{5}}COONa$
• D. ${{C}_{6}}{{H}_{5}}C{{H}_{2}}OH$ and ${{C}_{6}}{{H}_{5}}COONa$

Answer 1

Answer: (A)

$HCOONa$ and ${{C}_{6}}{{H}_{5}}C{{H}_{2}}OH$

Answer 2

This reaction is also called crossed-Cannizzaro reaction. In this reaction, formaldehyde is oxidised to acid salt while benzaldehyde is reduced to benzyl alcohol.) $HCHO+{{C}_{6}}{{H}_{5}}CHO\xrightarrow{dil.\,KOH}$ $\underset{\begin{smallmatrix} \text{Potassium} \\\ \,\,\,\text{formate} \end{smallmatrix}}{\mathop{HCOOK}}\,+\underset{\text{Benzyl}\,\text{alcohol}}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH}}\,$