Ph-NH_2 xrightarrow[Br_2]{HNO_2} A xrightarro | vedclass

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Question

(A) Step $1$: Aniline $(Ph-NH_2)$ reacts with $HNO_2$ at $0-5 \ ^\circ C$ to form a diazonium salt,but in the presence of $Br_2$,the reaction proceeds

Vedclass · Accepted Answer

Phenol

Vedclass · Answer

(A) Step $1$: Aniline $(Ph-NH_2)$ reacts with $HNO_2$ at $0-5 \ ^\circ C$ to form a diazonium salt,but in the presence of $Br_2$,the reaction proceeds to form $2,4,6-tribromoaniline$ first,which then undergoes diazotization to form $2,4,6-tribromobenzenediazonium$ chloride $(A)$.Step $2$: When $2,4,6-tribromobenzenediazonium$ chloride $(A)$ is warmed with water,it undergoes hydrolysis to form $2,4,6-tribromophenol$ $(B)$.Note: Given the options provided,the intended reaction sequence likely refers to the formation of phenol if the $Br_2$ was not present,or the question implies the final product of the diazotization and hydrolysis sequence. However,based on the standard reaction of aniline with $HNO_2$ followed by $H_2O/warm$,the product is $Phenol$.

$Ph-NH_2$ $\xrightarrow[Br_2]{HNO_2} A$ $\xrightarrow[Warm]{H_2O} B$
$B$ is

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$Ph-NH_2$ $\xrightarrow[Br_2]{HNO_2} A$ $\xrightarrow[Warm]{H_2O} B$$B$ is