Q. Compound $A$ undergoes Cannizaro reaction and $B$ undergoes positive iodoform test. Therefore,
Delhi UMET/DPMTDelhi UMET/DPMT 2009Aldehydes Ketones and Carboxylic Acids
Solution:
According to Cannizaro, aldehyde or ketones having no any a-hydrogen atoms undergo intermolecular oxidation-reduction to form an alcohol and acid.
Again, in iodoform test, any alcohol, aldehyde or ketone, when treated with iodine $ (I_{2}) $ and sodium hydroxide $(NaOH)$ give iodoform.
Therefore, the correct combination of $A$ and $B$ must be formaldehyde and $2$-pentanone.
Their reactions are as follows :
(i) $ \underset{\text{formaldehyde}}{2HCHO} + NaOH \to CH_3OH $
(ii) $NaOH + I_2 \to NaOI +NaI + H_2O$
$\xrightarrow{NaOH} C_3H_7COONa + \underset{\text{iodoform}}{CHI_3}$
