Identify the reaction conditions

The question refers to nitration of aniline using the usual nitrating mixture: $\text{aniline} + HNO_3/H_2SO_4$

What happens to aniline in strongly acidic medium?

Aniline is basic, so in the presence of strong acid it gets protonated: $C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+$

Thus, under nitrating conditions, a significant amount of aniline exists as the anilinium ion.

Directing effect of the protonated amino group

• Free $-NH_2$ group is ortho/para-directing.
• But protonated $-NH_3^+$ group is deactivating and meta-directing.

Therefore, in the strongly acidic nitration medium, nitration occurs predominantly at the meta position.

However, some free aniline may still be present, giving a little para product. In standard JEE interpretation, the major products are taken as:

• $p$-nitroaniline
• $m$-nitroaniline

Hence statement (B) is correct, while (A) is not.

Role of $HNO_3$ and $H_2SO_4$

In the nitrating mixture: $HNO_3 + H_2SO_4 \rightarrow NO_2^+ + HSO_4^- + H_2O$

Here, $H_2SO_4$ being the stronger acid protonates $HNO_3$, so:

• $H_2SO_4$ acts as an acid
• $HNO_3$ acts as a base

Therefore:

• Statement (C) "$HNO_3$ acts as an acid" is false
• Statement (D) "$H_2SO_4$ acts as an acid" is true

Evaluate the given statements

• (A) $o$-Nitroaniline and $p$-nitroaniline are predominant products. False
• (B) $p$-Nitroaniline and $m$-nitroaniline are predominant products. True
• (C) $HNO_3$ acts as an acid. False
• (D) $H_2SO_4$ acts as an acid. True

Choose the correct option

Correct statements are (B) and (D).

Therefore, the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer = C.

Our derived answer also = C. So they agree.