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Alcohols Phenols and Ethers question

2025 · 23 Jan · Shift 1 · Q3
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Alcohols Phenols and Ethers question

2025 · 23 Jan · Shift 1 · Q3

JEE MainChemistryAlcohols Phenols and EthersMCQ+4 / −1
What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol? (Given molar mass in gmol−1\mathrm{g} \mathrm{mol}^{-1}gmol−1 of C,H,O,Br\mathrm{C}, \mathrm{H}, \mathrm{O}, \mathrm{Br}C,H,O,Br are 12,1,16,8012,1,16,8012,1,16,80 respectively )
  1. A
    6.0 g
  2. B
    10.22 g
  3. C
    20.44 g
  4. D
    4.0 g
View written solutionFree

Correct answer: B

  1. Write the reaction

Phenol undergoes bromination to form 2,4,62,4,62,4,6-tribromophenol:

C6H5OH+3Br2→C6H2Br3OH+3HBr\mathrm{C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr}C6​H5​OH+3Br2​→C6​H2​Br3​OH+3HBr

So, 1 mole of phenol requires 3 moles of Br2Br_2Br2​.

  1. Calculate molar mass of phenol

Phenol = C6H5OH=C6H6O\mathrm{C_6H_5OH} = \mathrm{C_6H_6O}C6​H5​OH=C6​H6​O

M=6(12)+6(1)+16=72+6+16=94 g mol−1M = 6(12) + 6(1) + 16 = 72 + 6 + 16 = 94\ \mathrm{g\,mol^{-1}}M=6(12)+6(1)+16=72+6+16=94 gmol−1
  1. Calculate moles of phenol in 2 g
n(phenol)=294 moln(\text{phenol}) = \frac{2}{94}\ \mathrm{mol}n(phenol)=942​ mol
  1. Calculate moles of bromine required

Using stoichiometry:

n(Br2)=3×294=694=347 moln(Br_2) = 3 \times \frac{2}{94} = \frac{6}{94} = \frac{3}{47}\ \mathrm{mol}n(Br2​)=3×942​=946​=473​ mol
  1. Calculate molar mass of Br2Br_2Br2​
M(Br2)=2×80=160 g mol−1M(Br_2) = 2 \times 80 = 160\ \mathrm{g\,mol^{-1}}M(Br2​)=2×80=160 gmol−1
  1. Calculate mass of bromine required
m(Br2)=347×160=48047≈10.21 gm(Br_2) = \frac{3}{47} \times 160 = \frac{480}{47} \approx 10.21\ \mathrm{g}m(Br2​)=473​×160=47480​≈10.21 g

Rounding appropriately,

10.22 g\boxed{10.22\ \mathrm{g}}10.22 g​
  1. Option check
  • A: 6.0 g6.0\,\mathrm{g}6.0g ❌
  • B: 10.22 g10.22\,\mathrm{g}10.22g ✅
  • C: 20.44 g20.44\,\mathrm{g}20.44g ❌
  • D: 4.0 g4.0\,\mathrm{g}4.0g ❌

Therefore, the correct option is B.

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