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JEE Chemistry 2025 Question with Solution

What amount of bromine will be required to convert 2 g2 \, \text{g} of phenol into 2,4,62, 4, 6-tribromophenol? (Given molar mass in g mol−1\text{g mol}^{-1} of C, H, O, Br are 1212, 11, 1616, 8080 respectively)

  • A

    10.22 g10.22 \, \text{g}

  • B

    6.0 g6.0 \, \text{g}

  • C

    4.0 g4.0 \, \text{g}

  • D

    20.44 g20.44 \, \text{g}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: mass of phenol = 2 g2 \, \text{g}

Find: amount of bromine required to convert phenol into 2,4,62,4,6-tribromophenol.

The balanced reaction is

C6H5OH+3Br2→C6H2Br3OH+3HBr\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}

So, 11 mole of phenol reacts with 33 moles of Br2\text{Br}_2.

Molar mass of phenol:

6(12)+6(1)+16=94 g mol−16(12) + 6(1) + 16 = 94 \, \text{g mol}^{-1}

Molar mass of bromine:

2(80)=160 g mol−12(80) = 160 \, \text{g mol}^{-1}

Moles of phenol:

nphenol=294≈0.0213 moln_{\text{phenol}} = \frac{2}{94} \approx 0.0213 \, \text{mol}

From stoichiometry, moles of bromine required:

nBr2=3×0.0213=0.0639 moln_{\text{Br}_2} = 3 \times 0.0213 = 0.0639 \, \text{mol}

Mass of bromine needed:

mBr2=0.0639×160=10.224 gm_{\text{Br}_2} = 0.0639 \times 160 = 10.224 \, \text{g}

Therefore, the amount of bromine required is 10.22 g10.22 \, \text{g}. The correct option is A.

Stepwise Stoichiometric Calculation

Given: phenol = C6H5OH\text{C}_6\text{H}_5\text{OH}, mass = 2 g2 \, \text{g}

Find: required mass of Br2\text{Br}_2.

First calculate the molar mass of phenol:

M(C6H5OH)=6×12+6×1+16=72+6+16=94 g mol−1\begin{aligned} M(\text{C}_6\text{H}_5\text{OH}) &= 6 \times 12 + 6 \times 1 + 16 \\ &= 72 + 6 + 16 \\ &= 94 \, \text{g mol}^{-1} \end{aligned}

Now calculate moles of phenol:

2 g94 g mol−1≈0.0213 mol\frac{2 \, \text{g}}{94 \, \text{g mol}^{-1}} \approx 0.0213 \, \text{mol}

Use the reaction stoichiometry:

C6H5OH+3Br2→C6H2Br3OH+3HBr\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}

Thus, bromine needed = 33 times the moles of phenol.

So,

nBr2=3×0.0213=0.0639 mol\begin{aligned} n_{\text{Br}_2} &= 3 \times 0.0213 \\ &= 0.0639 \, \text{mol} \end{aligned}

Molar mass of bromine is

M(Br2)=2×80=160 g mol−1M(\text{Br}_2) = 2 \times 80 = 160 \, \text{g mol}^{-1}

Hence,

mBr2=0.0639×160=10.224 g\begin{aligned} m_{\text{Br}_2} &= 0.0639 \times 160 \\ &= 10.224 \, \text{g} \end{aligned}

Therefore, the required bromine mass is 10.22 g10.22 \, \text{g}.

Common mistakes

  • Using the molar ratio as 1:11:1 instead of 1:31:3 is incorrect because the balanced equation shows that one mole of phenol reacts with three moles of Br2\text{Br}_2. Always read the stoichiometric coefficients from the balanced reaction first.

  • Calculating the molar mass of phenol incorrectly as 9393 or 95 g mol−195 \, \text{g mol}^{-1} leads to a wrong answer. Phenol is C6H5OH\text{C}_6\text{H}_5\text{OH}, so total hydrogen count is 66, giving molar mass 94 g mol−194 \, \text{g mol}^{-1}.

  • Using atomic bromine mass 8080 directly instead of molecular bromine mass 160 g mol−1160 \, \text{g mol}^{-1} is wrong because the reacting species is Br2\text{Br}_2. Convert moles to mass using the molar mass of the actual molecule present in the equation.

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