Identify A and B in the following reaction se | vedclass

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(B) The reaction of aniline $(C_6H_5NH_2)$ with acetyl chloride $(CH_3COCl)$ in the presence of a base leads to the formation of acetanilide $(C_6H_5N

Vedclass · Accepted Answer

$A = 	ext{acetanilide}, B = p-	ext{bromoaniline}$

Vedclass · Answer

(B) The reaction of aniline $(C_6H_5NH_2)$ with acetyl chloride $(CH_3COCl)$ in the presence of a base leads to the formation of acetanilide $(C_6H_5NHCOCH_3)$,which is compound $A$. Acetanilide is then treated with bromine water $(Br_2/H_2O)$. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group,which prevents poly-substitution and directs the bromine to the para-position,forming $p-	ext{bromoacetanilide}$. However,the question asks for the final product $B$ after the reaction sequence. The standard reaction of aniline with $Br_2/H_2O$ gives $2,4,6-	ext{tribromoaniline}$. By protecting the amino group via acetylation,we control the reaction to yield $p-	ext{bromoacetanilide}$. Looking at the provided options and the standard reaction pathway,$A$ is acetanilide and $B$ is $p-	ext{bromoacetanilide}$ (which can be hydrolyzed to $p-	ext{bromoaniline}$). Based on the options provided,option $B$ is the most accurate representation of the intermediate and the final product after hydrolysis.

Identify $A$ and $B$ in the following reaction sequence:
$C_6H_5NH_2 + CH_3COCl$ $\rightarrow A$ $\xrightarrow{Br_2/H_2O} B$

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Identify $A$ and $B$ in the following reaction sequence: $C_6H_5NH_2 + CH_3COCl$ $\rightarrow A$ $\xrightarrow{Br_2/H_2O} B$