Question

Aniline on direct nitration at 288 K gives 51% (A), 47% (B) and 2% (C). ‘B’ on diazotisation, followed by reaction with CuCN | KCN gives a compound X. The percentage of nitrogen in X is:

• A. 28.92
• B. 18.92
• C. 38.92
• D. 48.92

Hint:

Ensure accuracy in empirical formula calculations by double-checking molecular structures and stoichiometry, especially in complex organic reactions involving multiple functional group transformations.

Answer 1

The Correct Option is B

Step 1: Understanding the reaction sequence. 
1. Nitration of Aniline: 
Direct nitration of aniline gives a mixture of p-nitroaniline (51%), o-nitroaniline (47%), and m-nitroaniline (2%). 
Here, B is o-nitroaniline. 
2. Diazotisation and Sandmeyer Reaction: 
o-Nitroaniline (B) undergoes diazotisation with NaNO\(_2\) + HCl, forming o-nitrobenzenediazonium chloride. 
This undergoes Sandmeyer reaction with CuCN/KCN, replacing \(-N^{+} \equiv N\) with \(-CN\), giving it{o}-nitrobenzonitrile (X). 
Step 2: Calculating the percentage of nitrogen in X. 
Molecular formula of o-nitrobenzonitrile (X): 
\[ C_7H_4N_2O_2 \] Molar mass of X: \[ (7 \times 12) + (4 \times 1) + (2 \times 14) + (2 \times 16) = 84 + 4 + 28 + 32 = 148 { g/mol} \] Mass of nitrogen in X: \[ 2 \times 14 = 28 { g} \] Percentage of nitrogen in X: \[ \frac{28}{148} \times 100 = 18.92% \] Thus, the correct answer is 18.92%.