Question

Correct statement about the given chemical reaction is : 

• A. -NH$_2$ group is ortho and para directive, so product (B) is not possible.
• B. Reaction is possible and compound (B) will be the major product.
• C. The reaction will form sulphonated product instead of nitration.
• D. Reaction is possible and compound (A) will be the major product.

Hint:

The nitration of aniline is a classic exception to the usual directing rules. Always remember that in a strong acid medium, aniline exists largely as the meta-directing anilinium ion, leading to a high percentage of the meta product along with the expected para/ortho products.

Answer 1

The Correct Option is D

Step 1: Identify the reaction The reaction shown is the nitration of aniline using a nitrating mixture ($\mathrm{HNO_3 + H_2SO_4}$) at 288 K. Step 2: Nature of the \(-NH_2\) group In neutral conditions, the \(-NH_2\) group is:
strongly activating,
ortho- and para-directing.
However, the nitration reaction is carried out in a strongly acidic medium. Step 3: Protonation of aniline In the presence of $\mathrm{H_2SO_4}$, aniline gets protonated: \[ \mathrm{C_6H_5NH_2 + H_2SO_4 \rightleftharpoons C_6H_5NH_3^+ + HSO_4^-} \] The protonated form (\(-NH_3^+\), anilinium ion) is:
strongly deactivating,
meta-directing.
Step 4: Reason for formation of all three isomers In solution, an equilibrium exists between:
free aniline (\(-NH_2\)) → gives ortho and para products,
anilinium ion (\(-NH_3^+\)) → gives meta product.
Therefore, nitration of aniline produces a mixture of ortho, meta, and para nitroanilines. Step 5: Relative proportions of products At 288 K, the approximate distribution is:
p-nitroaniline (A) ≈ 51% (major product),
m-nitroaniline (B) ≈ 47%,
o-nitroaniline (C) ≈ 2%.
The para product is major due to:
less steric hindrance compared to ortho position,
resonance stabilization from the \(-NH_2\) group.
Step 6: Evaluation of options
Option (A): Incorrect — meta product (B) is formed due to protonation of aniline.
Option (B): Incorrect — meta product is not the major product.
Option (C): Incorrect — sulphonation occurs at higher temperatures, not at 288 K.
Option (D): Correct — reaction occurs and para product (A) is formed in maximum amount.
Conclusion: The reaction is feasible, produces all three isomers, and the para-nitroaniline (A) is the major product. \[ \boxed{\text{Correct answer: (D)}} \]