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(B) The reaction of an amide with $Br_2$ and $KOH$ is the $Hofmann$ bromamide degradation reaction,which produces a primary amine. $C_6H_5CONH_2 + Br_

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Benzamide

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(B) The reaction of an amide with $Br_2$ and $KOH$ is the $Hofmann$ bromamide degradation reaction,which produces a primary amine. $C_6H_5CONH_2 + Br_2 + 4KOH 	o C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O$ The product $C_6H_5NH_2$ (aniline) is an oily liquid. When aniline is treated with acetic anhydride,it undergoes acetylation to form acetanilide $(CH_3CONHC_6H_5)$,which is a well-known antipyretic drug. $C_6H_5NH_2 + (CH_3CO)_2O 	o CH_3CONHC_6H_5 + CH_3COOH$ Therefore,the starting compound is benzamide.

List-$I$ (Amine)	List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline	$I$. $9.30$
$B$. Aniline	$II$. $8.92$
$C$. $N$-Ethylethanamine	$III$. $9.38$
$D$. $N$-Methylaniline	$IV$. $3.00$

$A$ nitrogen-containing organic compound gave an oily liquid on heating with bromine and potassium hydroxide solution. On shaking the product with acetic anhydride,an antipyretic drug was obtained. The reactions indicate that the starting compound is

Similar Questions

$A$ student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from $9.3 \ g$ of aniline. However,the student has managed to obtain $11 \ g$ of dry acetanilide. The $\%$ yield of this reaction is $.....$ (in $\%$)

In the given reaction,the end product $(C)$ is:

Match the following:
List-$I$ (Amine) List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline $I$. $9.30$
$B$. Aniline $II$. $8.92$
$C$. $N$-Ethylethanamine $III$. $9.38$
$D$. $N$-Methylaniline $IV$. $3.00$

The correct answer is

The product $(Y)$ of the following reaction is:

In a reaction of aniline,a coloured product $C$ was obtained as shown below:
$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{N,N-dimethylaniline, \text{cold}} C$
The structure of $C$ would be:

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