With respect to the following reaction, consider the given statements:

(A) o-Nitroaniline and p-nitroaniline are the predominant products.
(B) p-Nitroaniline and m-nitroaniline are the predominant products.
(C) HNO$$_3$$ acts as an acid.
(D) H$$_2$$SO$$_4$$ acts as an acid.
Choose the correct option.

Given reaction:

$$C_6H_5NH_2\xrightarrow[\;288\,K\;]{HNO_3/H_2SO_4}\text{Nitroanilines}$$

Step 1: Generation of the electrophile

$$HNO_3 + H_2SO_4 \rightarrow H_2NO_3^+ + HSO_4^-$$

$$H_2NO_3^+ \rightarrow NO_2^+ + H_2O$$

Hence,

$$H_2SO_4$$acts as an acid, while $$HNO_3$$ acts as a base.

Step 2: Protonation of aniline

$$C_6H_5NH_2 + H^+$$ $$\rightleftharpoons C_6H_5NH_3^+$$

The anilinium ion $$-NH_3^+$$ is a strong meta-directing group.

Therefore, direct nitration of aniline gives predominantly:

$$m\text{-nitroaniline}$$ along with a significant amount of $$p\text{-nitroaniline}$$

Evaluation of Statements:

For Statement (A):

$$\text{o-Nitroaniline and p-Nitroaniline are the predominant products}$$

$${\text{False}}$$

For Statement (B):

$$\text{p-Nitroaniline and m-Nitroaniline are the predominant products}$$

$${\text{True}}$$

For Statement (C):

$$HNO_3 \text{ acts as an acid}$$

$${\text{False}}$$

For Statement (D):

$$H_2SO_4 \text{ acts as an acid}$$

$${\text{True}}$$

Therefore,

$${\text{Statements (B) and (D) are correct}}$$