If aniline is treated with a 1:1 mixture of c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In the presence of strong acids like conc. $H_{2}SO_{4}$,the $-NH_{2}$ group of aniline gets protonated to form the anilinium ion $(-NH_{3}^{+})$.

Vedclass · Accepted Answer

protonation of $-NH_{2}$ which causes the formation of the $m$-directing anilinium ion.

Vedclass · Answer

(D) In the presence of strong acids like conc. $H_{2}SO_{4}$,the $-NH_{2}$ group of aniline gets protonated to form the anilinium ion $(-NH_{3}^{+})$.The $-NH_{3}^{+}$ group is electron-withdrawing and meta-directing.However,because the reaction is an equilibrium and some unprotonated aniline (which is ortho/para-directing) remains,a mixture of products is obtained,including a significant amount of $m$-nitroaniline due to the presence of the anilinium ion.

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:

Similar Questions

What is formed when benzoyl chloride reacts with aniline in the presence of sodium hydroxide?

Identify the major product obtained when ethyl amine $(CH_3CH_2NH_2)$ is reacted with an excess of methyl iodide $(CH_3I)$?

The correct order of decreasing basic strength of the given amines is:

Which of the following reactions does $NOT$ involve the replacement of the diazonium group?

Identify $X$ and $Y$ in the following reactions:
$Nitrobenzene \xrightarrow{Zn/NH_4Cl} X$
$Nitrobenzene \xrightarrow{Zn + KOH/C_2H_5OH} Y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module