Question

9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/ conversion) is ________g. (nearest integer)

Answer 1

Correct Answer: 20

The reaction is as follows:
\[\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \xrightarrow{T<5^\circ \text{C}} \text{C}_6\text{H}_5\text{-N}_2^+\text{Cl}^-\xrightarrow{\text{Phenol}} \text{Orange Dye}\]
1 mole of aniline (C$_6$H$_5$NH$_2$) produces 1 mole of orange dye.
Molar mass of aniline = 93 g mol$^{-1}$.
Molar mass of orange dye = 199 g mol$^{-1}$.
Moles of aniline used:
\[\text{moles of aniline} = \frac{\text{mass of aniline}}{\text{molar mass of aniline}} = \frac{9.3}{93} = 0.1 \, \text{mol}.\]

Mass of orange dye produced:
\[\text{mass of orange dye} = \text{moles of aniline} \times \text{molar mass of orange dye} = 0.1 \times 199 = 19.9 \, \text{g} \approx 20 \,\text{g}.\]

Answer 2

This problem involves a two-step organic synthesis starting from aniline. The process includes diazotisation followed by a coupling reaction with phenol. We are asked to calculate the mass of the final product, an orange dye, assuming a 100% yield.

Concept Used:

The solution is based on the principles of stoichiometry and mole concept. The key reactions involved are:

1. Diazotisation of Aniline: Aniline reacts with nitrous acid (generated in situ from NaNO₂ and a mineral acid like HCl) at low temperatures (0-5 °C) to form a benzenediazonium salt. \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278 \, \text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O} \]
2. Azo Coupling Reaction: The benzenediazonium salt then acts as an electrophile and couples with an electron-rich aromatic compound like phenol. The coupling occurs at the para position of phenol to form p-hydroxyazobenzene, which is an orange dye. \[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{p-HOC}_6\text{H}_4\text{N=NC}_6\text{H}_5 + \text{HCl} \]

The calculation involves converting the mass of the starting material (aniline) to moles, using the stoichiometric ratios from the balanced equations to find the moles of the product, and then converting the moles of the product back to mass.

Step-by-Step Solution:

Step 1: Calculate the molar mass of aniline (\(\text{C}_6\text{H}_5\text{NH}_2\)).

The molecular formula of aniline is \(\text{C}_6\text{H}_7\text{N}\).

\[ \text{Molar Mass of Aniline} = (6 \times \text{Atomic Mass of C}) + (7 \times \text{Atomic Mass of H}) + (1 \times \text{Atomic Mass of N}) \] \[ = (6 \times 12) + (7 \times 1) + (1 \times 14) = 72 + 7 + 14 = 93 \, \text{g/mol} \]

Step 2: Calculate the number of moles of aniline used.

Given mass of aniline = 9.3 g.

\[ \text{Moles of Aniline} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{9.3 \, \text{g}}{93 \, \text{g/mol}} = 0.1 \, \text{mol} \]

Step 3: Determine the stoichiometry of the overall reaction.

From the balanced equations:

• 1 mole of Aniline produces 1 mole of Benzenediazonium chloride.
• 1 mole of Benzenediazonium chloride produces 1 mole of p-hydroxyazobenzene (the orange dye).

Therefore, the overall stoichiometric ratio between aniline and the final dye is 1:1. \[ 1 \, \text{mole of Aniline} \rightarrow 1 \, \text{mole of Orange Dye} \]

Since the reaction yield is 100%, the moles of orange dye produced will be equal to the moles of aniline reacted.

\[ \text{Moles of Orange Dye} = 0.1 \, \text{mol} \]

Step 4: Calculate the molar mass of the orange dye (p-hydroxyazobenzene).

The molecular formula of p-hydroxyazobenzene (\(\text{p-HOC}_6\text{H}_4\text{N=NC}_6\text{H}_5\)) is \(\text{C}_{12}\text{H}_{10}\text{N}_2\text{O}\).

\[ \text{Molar Mass of Dye} = (12 \times 12) + (10 \times 1) + (2 \times 14) + (1 \times 16) \] \[ = 144 + 10 + 28 + 16 = 198 \, \text{g/mol} \]

Step 5: Calculate the mass of the orange dye produced.

\[ \text{Mass of Dye} = \text{Moles of Dye} \times \text{Molar Mass of Dye} \] \[ = 0.1 \, \text{mol} \times 198 \, \text{g/mol} = 19.8 \, \text{g} \]

Final Computation & Result:

The calculated mass of the orange dye is 19.8 g. The question asks for the answer to the nearest integer.

Rounding 19.8 to the nearest integer gives 20.

The mass of orange dye produced is 20 g.