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 %.
Answer
80
Solution
[figure]
∵ –NH2 strongly activates o-, p-positions ⇒ 2,4,6-tribromoaniline
Molar mass: aniline = 93 g mol-1
P (C6H4NBr3) = 330 g mol-1
Moles of aniline = 9.393 = 0.1 mol ⇒ theoretical P = 0.1 × 330 = 33 g
% yield = 26.433 100 = 80%
Answer: 80