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.

Column-$I$	Column-$II$
$A$. Benzenesulphonyl chloride	$1$. Zwitter ion
$B$. Sulphanilic acid	$2$. Hinsberg reagent
$C$. Alkyl diazonium salts	$3$. Dyes
$D$. Aryl diazonium salts	$4$. Conversion to alcohols

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 order of base strength in benzene as a solvent for the given amines is:
$I: CH_3CH_2NH_2$
$II: (CH_3CH_2)_2NH$
$III: (CH_3CH_2)_3N$

Which of the following is not correct regarding basic strength?

Match the compounds given in Column-$I$ with the items given in Column-$II$.

Column-$I$ Column-$II$

$A$. Benzenesulphonyl chloride $1$. Zwitter ion

$B$. Sulphanilic acid $2$. Hinsberg reagent

$C$. Alkyl diazonium salts $3$. Dyes

$D$. Aryl diazonium salts $4$. Conversion to alcohols

The reaction of a primary amine with an aldehyde gives $......$.

The reagent with which the following reaction is best accomplished is:

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