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(D) In the presence of strong acid $(H_2SO_4)$,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$.The $-NH_3^+$ group is meta-directing

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Reaction is possible and compound $(A)$ will be the major product.

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(D) In the presence of strong acid $(H_2SO_4)$,aniline is protonated to form the anilinium ion $(C_6H_5NH_3^+)$.The $-NH_3^+$ group is meta-directing due to its strong electron-withdrawing nature.As a result,a significant amount of the $m$-nitroaniline (compound $(B)$) is formed,along with the $p$-nitroaniline (compound $(A)$) and $o$-nitroaniline (compound $(C)$).Experimental data shows that $p$-nitroaniline $(51\%)$ is the major product,followed by $m$-nitroaniline $(47\%)$ and $o$-nitroaniline $(2\%)$.

The correct statement about the given chemical reaction is:

Similar Questions

The major products obtained from the following sequence of reactions are
$(CH_3)_2CHCH_2N(CH_2CH_3)_2$ $\xrightarrow{CH_3I}$ $\xrightarrow[H_2O]{Ag_2O}$ $\xrightarrow{\Delta} \text{products}$

Primary nitro compounds react with $HNO_2$ to form crystalline solids which,on treatment with $NaOH$,give:

Which of the following has the highest value of $pK_b$?

Aniline with solid $KOH$ and $CS_2$ gives :

The major product formed in the following reaction is:

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