Q.72. Consider the following sequence of reactions. [Figure: Reaction scheme starting with a diazoni

Q: What is the correct answer for JEE Main Chemistry Q72?

The correct answer is 21. The reaction sequence describes the synthesis of an azo dye. 1. The starting material is a diazonium salt (implied from the structure -N=N-NH-). The first step (i) HCl is for forming the diazonium salt from an aromatic amine (e.g., aniline + NaNO$_2$/HCl). 2. The diazonium salt reacts with aniline (in excess) in step (ii). This is an azo coupling reaction. The diazonium salt (e.g., benzenediazon

Q: Q.72. Consider the following sequence of reactions. [Figure: Reaction scheme starting with a diazonium salt, reacting with (i) HCl, (ii) Aniline (in excess), (iii) Warm 40-45°C, (iv) Cool, acetic…

The correct answer is 21. The reaction sequence describes the synthesis of an azo dye. 1. The starting material is a diazonium salt (implied from the structure -N=N-NH-). The first step (i) HCl is for forming the diazonium salt from an aromatic amine (e.g., aniline + NaNO$_2$/HCl). 2. The diazonium salt reacts with aniline (in excess) in step (ii). This is an azo coupling reaction. The diazonium salt (e.g., benzenediazon

Question

Q.72. Consider the following sequence of reactions. [Figure: Reaction scheme starting with a diazonium salt, reacting with (i) HCl, (ii) Aniline (in excess), (iii) Warm 40-45°C, (iv) Cool, acetic acid to give Yellow Product (X)]. The percentage of nitrogen in the yellow product (X) formed is ______ %. (Nearest Integer) (Given Molar mass in g mol H : 1, C : 12, N : 14)

FirstInTest · Accepted Answer

The reaction sequence describes the synthesis of an azo dye.
1.  The starting material is a diazonium salt (implied from the structure -N=N-NH-). The first step (i) HCl is for forming the diazonium salt from an aromatic amine (e.g., aniline + NaNO$_2$/HCl).
2.  The diazonium salt reacts with aniline (in excess) in step (ii). This is an azo coupling reaction. The diazonium salt (e.g., benzenediazonium chloride) reacts with aniline to form diazoaminobenzene (Ph-N=N-NH-Ph).
3.  Step (iii) Warm 40-45°C and (iv) Cool, acetic acid: Diazoaminobenzene undergoes rearrangement in the presence of acid (acetic acid) to form p-aminoazobenzene (Ph-N=N-C$_6$H$_4$-NH$_2$, where NH$_2$ is para to the azo group). This is called the Wallach rearrangement or diazoamino rearrangement.

The yellow product (X) is p-aminoazobenzene.
Structure of p-aminoazobenzene: C$_6$H$_5$-N=N-C$_6$H$_4$-NH$_2$

Let's calculate the molar mass of p-aminoazobenzene:
Number of Carbon atoms = 6 + 6 = 12
Number of Hydrogen atoms = 5 (from C$_6$H$_5$) + 4 (from C$_6$H$_4$) + 2 (from NH$_2$) = 11
Number of Nitrogen atoms = 2 (from N=N) + 1 (from NH$_2$) = 3

Molar mass of C$_{12}$H$_{11}$N$_3$:
C: 12 $	imes$ 12 = 144
H: 11 $	imes$ 1 = 11
N: 3 $	imes$ 14 = 42
Total Molar Mass = 144 + 11 + 42 = 197 g/mol

Mass of nitrogen in the compound = 42 g

Percentage of nitrogen = $\frac{	ext{Mass of Nitrogen}}{	ext{Molar Mass of X}} 	imes 100$
= $\frac{42}{197} 	imes 100 \approx 21.319 \%$

Rounding to the nearest integer, the percentage of nitrogen is 21%.

Related Questionsin Chemistry

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions