Question

Consider the following reaction sequence.

Hint:

Protecting the \(NH_2\) group as acetanilide prevents poly-nitration and oxidation during the nitration step, ensuring the para-isomer is the major product.

Answer 1

Correct Answer: 21

   Step 1: Identify Product T
% Option (A) Nitration of Benzene \(\rightarrow\) P (Nitrobenzene, \(Ph-NO_2\)). % Option (B) Reduction (\(Sn/HCl\)) \(\rightarrow\) Q (Aniline, \(Ph-NH_2\)). % Option (C) Acetylation (\(Ac_2O\)) \(\rightarrow\) R (Acetanilide, \(Ph-NH-COCH_3\)). % Option (D) Nitration of R: The \(-NHAc\) group is ortho/para directing. Para is major. Product: p-nitroacetanilide. % Option (E) Hydrolysis (\(OH^-, \Delta\)): The amide bond cleaves to regenerate the amine. Product T: p-nitroaniline (\(NH_2-C_6H_4-NO_2\)). Step 2: Calculate Percentage of Nitrogen
Formula of p-nitroaniline: \(C_6H_6N_2O_2\). \begin{itemize} \item Molar Mass = \(6(12) + 6(1) + 2(14) + 2(16)\) \item \(= 72 + 6 + 28 + 32 = 138 \text{ g/mol}\). \end{itemize} Total Mass of Nitrogen = \(2 \times 14 = 28 \text{ g}\). \[ \% N = \frac{28}{138} \times 100 \approx 20.28 \% \] Final Answer: Nearest integer is 20.