JEE Main 2024 Chemistry question
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
9.3 g = 0.1 mol aniline gives 0.1 mol of 2,4,6-tribromoaniline (330 g mol⁻¹) = 33 g theoretical; yield = 26.4/33 × 100 = 80%.