In the following reaction,X is:X xrightarrow{ | vedclass

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(D) The reaction sequence is as follows:$1$. $X$ is Aniline $(C_6H_5NH_2)$.$2$. Bromination of Aniline with $Br_2$ water gives $2,4,6$-Tribromoaniline

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Aniline

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(D) The reaction sequence is as follows:$1$. $X$ is Aniline $(C_6H_5NH_2)$.$2$. Bromination of Aniline with $Br_2$ water gives $2,4,6$-Tribromoaniline $(Y)$.$3$. Treatment of $Y$ with $NaNO_2 + HCl$ at $0-5 \ ^\circ C$ yields the diazonium salt,$2,4,6$-Tribromobenzenediazonium chloride $(Z)$.$4$. Boiling $Z$ with ethanol $(C_2H_5OH)$ results in the reduction of the diazonium group to a hydrogen atom,producing $1,3,5$-Tribromobenzene.Therefore,$X$ is Aniline.

In the following reaction,$X$ is:
$X$ $\xrightarrow{\text{Bromination}} Y$ $\xrightarrow{NaNO_2 + HCl} Z$ $\xrightarrow[C_2H_5OH]{\text{Boiling}} \text{1,3,5-Tribromobenzene}$

Similar Questions

The correct stability order of the following diazonium salts is:

The major product $(B)$ in this reaction is:

Why is $m$-nitroaniline obtained from the nitration of aniline in a strongly acidic medium?

The correct statements of the following are:
$A$. Aniline forms a stable benzene diazonium chloride at $285 \ K$.
$B$. $N$-Phenylethanamide is less reactive towards nitration than aniline.
$C$. $p-CH_3C_6H_4COCl$ is Hinsberg reagent.

Which of the following is the strongest base?

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In the following reaction,$X$ is:$X$ $\xrightarrow{\text{Bromination}} Y$ $\xrightarrow{NaNO_2 + HCl} Z$ $\xrightarrow[C_2H_5OH]{\text{Boiling}} \text{1,3,5-Tribromobenzene}$