



The bromination of aniline is an important reaction where the amine group on aniline activates the benzene ring towards electrophilic substitution, specifically making it more susceptible to halogenation. During this reaction, aniline undergoes bromination more readily compared to benzene due to the electron-donating effect of the -NH2 group.
In the process of bromination, the arenium ion intermediates are formed. These intermediates determine the reaction pathway and ultimately the position at which bromination occurs.
In the provided question, we need to identify which arenium ion intermediate is not involved in the bromination of aniline. Let's evaluate the possible arenium ions that arise during this process:
When aniline is brominated, the major positions affected are the ortho and para due to the activating effect of the amine group. Hence, the arenium ions corresponding to ortho and para positions are most commonly formed.
Analyzing each option:
Therefore, the arenium ion in Option C is the one not involved in the bromination of aniline.
Thus, the correct answer is Option C.
Since the −NH2 group is ortho/para directing, the arenium ion will not be formed by attack at the meta position. This is due to the electron-donating nature of the −NH2 group, which stabilizes the carbocation at the ortho and para positions relative to itself.
Thus, the arenium ion corresponding to the meta substitution, as shown in answer (3), does not participate in the bromination of aniline.
So, the correct answer is: Option (3)


Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
