Question

\(\text{Reaction of aniline with conc. HNO}_3 \text{ and conc. H}_2\text{SO}_4 \text{ at 298 K will produce 47\% of:}\)

• A. m-Nitroaniline
• B. o-Nitroaniline
• C. m-Nitroaniline
• D. 2,4-Dinitroaniline

Answer 1

The Correct Option is A

The reaction of aniline with concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) at 298 K involves the nitration of the aromatic compound. In the presence of these strong acids, an aniline group initially converts to an anilinium ion due to the acidic environment:

\[\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+ + \text{HSO}_4^-\]

This conversion deactivates the aromatic ring towards electrophilic attack, affecting its reactivity. The strong electron-withdrawing effect of the anilinium ion lowers the reactivity of the ortho and para positions, making the meta position more favorable for substitution. Consequently, nitration predominantly occurs at the meta position, resulting in the formation of m-nitroaniline.

As the final outcome of this reaction in the specified conditions is a predominant formation of m-nitroaniline, 47% of the product will be: m-Nitroaniline.

Answer 2

When aniline reacts with concentrated HNO₃ and concentrated H₂SO₄ at 298 K, it undergoes nitration. However, due to the presence of the amino group (-NH₂), which is a strongly activating and ortho/para-directing group, unexpected products are formed.

Here's a breakdown of the reaction:

• Aniline reacts with the acidic medium (H₂SO₄) to form the anilinium ion (C₆H₅NH₃⁺).
• The anilinium ion is a meta-directing group, making the meta-product the major product.
• However, at lower temperatures, significant amounts of ortho and para products are also formed.
• The reaction is complex due to the protonation of the amino group, which deactivates and meta-directs the ring.

The product distribution at 298 K is approximately:

• 47% m-nitroaniline
• 51% p-nitroaniline
• 2% o-nitroaniline

Therefore, the major product is m-nitroaniline, which accounts for 47% of the product.

The correct answer is:

Option 3: m-Nitroaniline