A is a neutral organic compound (M.F.: C_8H_9 | vedclass

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(C) $1$. The molecular formula $C_8H_9ON$ and the reaction with $Br_2 / HO^{(-)}$ (Hofmann bromamide degradation) suggest that $A$ is an amide. $2$. T

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(C) $1$. The molecular formula $C_8H_9ON$ and the reaction with $Br_2 / HO^{(-)}$ (Hofmann bromamide degradation) suggest that $A$ is an amide. $2$. The product $B$ is an amine, which is soluble in dilute acid. $3$. The reaction sequence $B$ $\xrightarrow{NaNO_2/HCl} C$ $\xrightarrow{CuCN} D$ $\xrightarrow{H_3O^+} E$ is the standard conversion of an amine to a carboxylic acid via a diazonium salt and nitrile. $4$. $E$ is also obtained from the hydrolysis of $A$, confirming $A$ is an amide. $5$. $E$ on oxidation with $KMnO_4$ gives $F$. If $E$ is $p$-toluic acid, oxidation gives terephthalic acid $(F)$, which has two types of hydrogen atoms (aromatic and carboxylic). $6$. Based on the reaction path, $A$ must be $p$-methylbenzamide.

List-$I$ (Mixture of Compounds)	List-$II$ (Reagent used to distinguish)
$A$. Diethyl amine + Ethyl amine	$I$. Bromine water
$B$. Acetaldehyde + Acetone	$II$. $CHCl_3 + KOH, \Delta$
$C$. Ethanol + Phenol	$III$. Neutral $FeCl_3$
$D$. Benzoic acid + Cinnamic acid	$IV$. Ammonical silver nitrate

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