In the given reaction sequence,what is Z?Anil | vedclass

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(A) $1$. The reaction of aniline with acetyl chloride $(CH_3COCl)$ in the presence of a base (like pyridine) leads to the formation of acetanilide $(X

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p-Bromoaniline

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(A) $1$. The reaction of aniline with acetyl chloride $(CH_3COCl)$ in the presence of a base (like pyridine) leads to the formation of acetanilide $(X)$,which is $C_6H_5NHCOCH_3$.$2$. The acetamido group $(-NHCOCH_3)$ is ortho/para directing,but due to steric hindrance,the para-isomer is the major product. Thus,$Y$ is p-bromoacetanilide $(p-Br-C_6H_4-NHCOCH_3)$.$3$. Acidic hydrolysis $(H_3O^+)$ of the acetamido group converts it back into the amino group $(-NH_2)$.$4$. Therefore,the final product $Z$ is p-bromoaniline $(p-Br-C_6H_4-NH_2)$.

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 given reaction sequence,what is $Z$?
Aniline $\xrightarrow{CH_3COCl} X$ $\xrightarrow{Br_2/AlBr_3} Y$ $\xrightarrow{H_3O^+} Z$

Similar Questions

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

$A$ mixture of ethyl amine,chloroform,and alcoholic $KOH$ on heating gives:

If aniline is treated with a $1:1$ mixture of conc. $HNO_{3}$ and conc. $H_{2}SO_{4}$,$p$-nitroaniline and $m$-nitroaniline are formed in nearly equal amounts. This is due to:

The product of the following reaction is:

On heating benzyl amine with chloroform and ethanolic $KOH$,the product obtained is:

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In the given reaction sequence,what is $Z$?Aniline $\xrightarrow{CH_3COCl} X$ $\xrightarrow{Br_2/AlBr_3} Y$ $\xrightarrow{H_3O^+} Z$