Question

Benzene undergoes the following sequence of reactions:

Find the percentage of nitrogen in compound \( (E) \).

Hint:

For multistep organic conversions: \begin{itemize} \item Identify directing effects before predicting substitution \item Protect functional groups when required (e.g., acetylation of amines) \item Always compute percentage composition from molecular formula \end{itemize}

Answer 1

Correct Answer: 20.29

To determine the percentage of nitrogen in compound \(E\), follow these steps:

1. Nitration of Benzene: Benzene reacts with conc. HNO₃ and conc. H₂SO₄ to form nitrobenzene, compound \(A\) (C₆H₅NO₂).

2. Reduction: Nitrobenzene (\(A\)) is reduced using Sn/HCl to form aniline, compound \(B\) (C₆H₅NH₂).

3. Acetylation: Aniline (\(B\)) reacts with acetic anhydride (CH₃CO)₂O to form acetanilide, compound \(C\) (C₆H₅NHCOCH₃).

4. Nitration of Acetanilide: Acetanilide (\(C\)) undergoes nitration to form a nitro derivative, compound \(D\) (C₆H₄(NO₂)NHCOCH₃).

5. Hydrolysis: Compound \(D\) is hydrolyzed in acidic conditions to produce compound \(E\), p-nitroaniline (C₆H₄(NO₂)NH₂).

6. Calculate Molar Mass: 
Molar mass of C₆H₄(NO₂)NH₂:
C: 6×12 = 72
H: 6×1 = 6
N: 2×14 = 28
O: 2×16 = 32
Total = 138 g/mol

7. Percentage of Nitrogen:
Percentage of nitrogen = \(\frac{\text{Mass of nitrogen}}{\text{Total mass}} \times 100 = \frac{28}{138} \times 100 \approx 20.29\%\).

Conclusion: The percentage of nitrogen in compound \(E\) is approximately 20.29%.