9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product 'P'. The mass of product 'P' obtained is 26.4 g. The percentage yield is ......... %.
Correct Answer: 80
Approach Solution - 1
beginfigure[h]
93 g of aniline produces 330 g of 2,4,6-tribromoaniline.
Theoretical yield for 9.3 g of aniline:
\[ \frac{330}{93} \times 9.3 = 33 \, \text{g} \]
Percentage yield:
\[ \text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \]
\[ \text{Percentage yield} = \frac{26.4}{33} \times 100 = 80\% \]
Approach Solution -2
Aniline reacts with bromine water at room temperature to give the white precipitate 2,4,6-tribromoaniline via electrophilic aromatic substitution (three brominations on the activated ring). The overall stoichiometry is 1:1 in moles between aniline and product P:
C6H5NH2 + 3 Br2 → C6H2Br3NH2 + 6 HBr.
Step 2: Moles of aniline taken
Given mass of pure aniline = 9.3 g. Molar mass of aniline (C6H7N):
M(aniline) = 6×12.011 + 7×1.008 + 14.007 ≈ 93.13 g·mol⁻¹.
n(aniline) = 9.3 / 93.13 ≈ 0.0999 mol ≈ 0.100 mol.
Step 3: Molar mass of product P (2,4,6-tribromoaniline)
Use formula C6H4Br3N or adjust aniline by adding 3×(Br − H):
M(P) ≈ 93.13 + 3×(79.904 − 1.008) = 93.13 + 3×78.896 ≈ 93.13 + 236.688 ≈ 329.82 g·mol⁻¹.
Step 4: Theoretical mass of product
Stoichiometry 1:1 ⇒ n(P)_theor = n(aniline) ≈ 0.100 mol.
m(P)_theor = n×M ≈ 0.100 × 329.82 ≈ 32.98 g.
Step 5: Percentage yield
Actual mass obtained = 26.4 g.
% yield = (26.4 / 32.98) × 100 ≈ 80.0 %.
Final answer
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