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 \underline\hspace2cm %.

Compounds Containing Nitrogen PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q28 with Solution

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Correct answer: 80

Aniline reacts with bromine water at room temperature to form white precipitate of 2,4,6-tribromoaniline:

\ce{C6H5NH2 + 3Br2 -> C6H2Br3NH2 + 3HBr}

So, the mole ratio is:

\text{aniline} : \text{product} = 1:1

Molar mass of aniline, $\ce{C6H5NH2}$ :

6\times 12 + 7\times 1 + 14 = 72 + 7 + 14 = 93\,\text{g mol}^{-1}

Given mass of aniline = $9.3\,$ g

\text{moles of aniline} = \frac{9.3}{93} = 0.1\,\text{mol}

Hence, theoretical moles of product formed = $0.1$ mol.

Product is 2,4,6-tribromoaniline: $\ce{C6H4Br3N}$

Its molar mass:

6\times 12 + 4\times 1 + 3\times 80 + 14

= 72 + 4 + 240 + 14 = 330\,\text{g mol}^{-1}

\text{theoretical mass} = 0.1 \times 330 = 33\,\text{g}

Actual mass obtained = $26.4\,$ g

\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}}\times 100

= \frac{26.4}{33}\times 100 = 80\%

\boxed{80}

The derived answer matches the stored correct answer.