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

Which compound is optically active?

The product formed when aniline is treated with chloroform in the presence of $KOH$ (alcoholic) is

Which of the following compounds reacts with aqueous nitrous acid at low temperature to give an oily substance,$N$-nitrosoamine?

Nitrobenzene on reduction with zinc and $NH_4Cl$ gives:

Carbylamine is liberated when $.....$ is heated with chloroform and alcoholic potash.

Vedclass Test Series

Exam Paper Generator

Online Exam Module