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%.

Answer 2

To solve the problem, we need to find the percentage of nitrogen in compound X, formed from aniline by direct nitration and reaction with CuCN and KCN.

1. Understanding the Problem:
Aniline undergoes direct nitration, resulting in the formation of:

• 51% of compound A
• 47% of compound B
• 2% of compound C

Compound B undergoes diazotization and then reacts with CuCN and KCN to form compound X. We are required to calculate the percentage of nitrogen in X.

2. Calculating the Total Mass of Nitrogen:
The molecular formula for nitrogen is \( N = 14 \, \text{g/mol} \), and the molecular formulas for compounds A, B, and C contain nitrogen. Since compound B undergoes further reactions to form compound X, we need to calculate the mass of nitrogen in compound X based on the percentages of nitrogen in the products of nitration. To calculate the percentage of nitrogen in X: \[ \text{Percentage of N in X} = \frac{\text{Mass of Nitrogen in X}}{\text{Total mass of X}} \times 100 \] From the data given and after calculations, we find that the percentage of nitrogen in X is 18.92%.

Final Answer:
The percentage of nitrogen in compound X is 18.92%.

Final Answer:
The correct option is (B) 18.92.