Question

Consider the following 
Statement-I: In the nitration of aniline, more amount of m-nitroaniline is formed than expected.
Statement-II: In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta directing.
 

• A. Both statement-I and statement-II are correct
• B. Both statement-I and statement-II are not correct
• C. Statement-I is correct, but statement-II is not correct
• D. Statement-I is not correct, but statement-II is correct

Hint:

When considering the electrophilic substitution reactions of aromatic compounds with basic functional groups (like amines), always consider the effect of the reaction medium. If the medium is acidic, the basic group might get protonated, changing its directing and activating/deactivating nature. \(\operatorname{NH}_2\) group: strongly activating, ortho-para directing. \(\operatorname{NH}_3^+\) group: strongly deactivating, meta-directing.

Answer 1

The Correct Option is A

Let's analyze each statement regarding the nitration of aniline. 
Statement-I: In the nitration of aniline, more amount of m-nitroaniline is formed than expected.
Aniline has a strongly activating and ortho-para directing amino (-\(\operatorname{NH}_2\)) group. Based on this, one would expect the nitration to yield predominantly ortho- and para-nitroaniline. However, experimental observations show that a significant amount of m-nitroaniline (around 47%) is also formed during the nitration of aniline. This yield of meta product is unusually high for an ortho-para directing group.
Therefore, Statement-I is correct. 
Statement-II: In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta directing.
Nitration is carried out using a nitrating mixture, typically concentrated nitric acid and concentrated sulfuric acid, which provides a strongly acidic medium. Aniline is a basic compound. In a strongly acidic environment, the amino group (-\(\operatorname{NH}_2\)) of aniline gets protonated to form the anilinium ion (\(\operatorname{C}_6\operatorname{H}_5\operatorname{NH}_3^+\)).
The \(\operatorname{NH}_3^+\) group is a positively charged group. Such groups are strong electron-withdrawing groups and are meta-directing and deactivating. This means that electrophilic substitution on the anilinium ion will predominantly occur at the meta position. 
Therefore, Statement-II is correct. Conclusion:
Both Statement-I and Statement-II are correct. Moreover, Statement-II provides the correct explanation for the observation described in Statement-I. The formation of the meta-directing anilinium ion in acidic medium leads to the unexpected high yield of m-nitroaniline. 
The final answer is Both statement-I and statement-II are correct