The correct statement about the given chemica | vedclass

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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

What is formed when nitrobenzene is reduced using zinc and alkali?

$(A)$ $\xrightarrow[(1) \text{ excess } CH_3I/K_2CO_3, (2) \text{ } Ag_2O/H_2O, (3) \text{ } \Delta]{} (B)$ $\xrightarrow[(i) \text{ excess } CH_3I/K_2CO_3, (ii) \text{ } Ag_2O, (iii) \text{ } \Delta]{} H_2C=CH-CH(CH_3)-CH=CH_2$. Identify $(A)$.

Identify the final product $D$ in the following reaction sequence: $CH_3CH_2Br$ $\xrightarrow[\Delta]{Aq. KOH} A$ $\xrightarrow[\Delta]{KMnO_4/H^+} B$ $\xrightarrow[\Delta]{NH_3} C$ $\xrightarrow[alkali]{Br_2} D$

In the reaction $C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow A + 3KCl + 3H_2O$,what is '$A$'?

The correct increasing order of basic strength for the following compounds is:

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