Protecting the \(NH_2\) group as acetanilide prevents poly-nitration and oxidation during the nitration step, ensuring the para-isomer is the major product.
(A) Nitration of benzene gives P = Nitrobenzene (\(C_6H_5NO_2\)).
(B) Reduction of nitrobenzene using \(Sn/HCl\) gives Q = Aniline (\(C_6H_5NH_2\)).
(C) Acetylation of aniline using \(Ac_2O\) gives R = Acetanilide
(\(C_6H_5NHCOCH_3\)).
(D) Nitration of acetanilide: The \(-NHCOCH_3\) group is ortho/para directing.
Para substitution is the major product, giving p-nitroacetanilide.
(E) Hydrolysis of p-nitroacetanilide using \(OH^-, \Delta\) cleaves the amide bond and regenerates the amine.
Therefore, the final product T is p-nitroaniline
(\(NH_2–C_6H_4–NO_2\)).
Step 2: Calculation of Percentage of Nitrogen
Molecular formula of p-nitroaniline:
\(C_6H_6N_2O_2\)
Molar mass calculation:
\[
= 6(12) + 6(1) + 2(14) + 2(16)
\]
\[
= 72 + 6 + 28 + 32 = 138 \text{ g/mol}
\]
Mass of nitrogen = \(2 \times 14 = 28\) g
Percentage of nitrogen:
\[
\%N = \frac{28}{138} \times 100 = 20.28\%
\]
Final Answer:
Percentage of nitrogen ≈ 20% (nearest integer).
