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.

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In a reaction of aniline,a coloured product $C$ was obtained as shown below:
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