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

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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:

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