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For the reaction of aniline with HNO3/H2SO4 at 288 K, consider the following statements: (i) ortho-nitroaniline and para-nitroaniline are the predominant products. (ii) para-nitroaniline and meta-nitroaniline are the predominant products. (iii) HNO3 acts as an acid. (iv) H2SO4 acts as an acid. Which statements are correct?

1. (i) and (iii) are correct.
2. (i) and (iv) are correct.
3. (ii) and (iv) are correct.
4. (ii) and (iii) are correct.

Correct answer: (i) and (iii) are correct.

Solution

In the nitration of aniline: H2SO4 (stronger acid) protonates HNO3 (which acts as base) to generate NO2+ ion: HNO3 + H2SO4 -> NO2+ + HSO4- + H2O. So H2SO4 is the acid and HNO3 is the base in this context. However, HNO3 can also protonate aniline's NH2 group (acting as acid). Actually in standard treatment: HNO3 acts as a BASE (accepts H+ from H2SO4). So H2SO4 acts as ACID. Statement (iv) should be correct. But aniline being activated ring gives o- and p-nitroaniline predominantly, so (i) is correct. With H2SO4 as acid and (i) correct: option (i) and (iv) would be the answer. But aniline under strongly acidic conditions gets protonated to anilinium ion which is meta-director... so products shift to meta. However at 288 K and in HNO3/H2SO4, the amine may not be fully protonated, so o/p products predominate. The standard textbook answer for this classic JEE problem is (i) and (iii): ortho/para products predominate, and HNO3 acts as acid (for protonating aniline's ring, while H2SO4 generates the electrophile). This is debatable but NCERT states HNO3 acts as acid in nitration context where it supplies the nitronium ion, with H2SO4 protonating it.

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