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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.

$A$ 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

The product $(A)$ of the reaction is:

What are $X$ and $Y$ respectively in the following set of reactions?

The relative rate of reaction of the following amines with methyl iodide is:

The correct option$(s)$ for the following sequence of reactions is(are):
$(A)$ $Q = KNO_2, W = LiAlH_4$
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Which among the following represents reagent $A$ in the given reaction?
$C_6H_5N_2^+Cl^- + A \xrightarrow{NaOH} \text{Product (Orange-red dye)}$

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