What amount of bromine will be required to convert 6 g of phenol into 2,4,6-tribromophenol?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30.63 g
The –OH group activates the benzene ring so strongly that phenol reacts with bromine water at room temperature and all three ortho and para positions are brominated at once, giving a white precipitate of 2,4,6-tribromophenol. One mole of phenol needs three moles of Br2:
\( \ce{C6H5OH + 3Br2 -> C6H2Br3OH v + 3HBr} \)
Molar masses: \( \ce{C6H5OH} \) = 94 g mol–1, \( \ce{Br2} \) = 160 g mol–1.
\( \displaystyle \text{moles of phenol} = \frac{6}{94} = 0.0638\ \text{mol} \)
\( \text{moles of } \ce{Br2} = 3 \times 0.0638 = 0.1915\ \text{mol} \)
\( \text{mass of } \ce{Br2} = 0.1915 \times 160 \approx 30.6\ \text{g} \)
The bromine required is about 30.63 g.
Shortcut: 94 g of phenol needs 3 × 160 = 480 g of \( \ce{Br2} \), so 6 g needs \( \displaystyle \frac{480 \times 6}{94} \approx 30.6 \) g in one line.
NCERT (2026-27 edition):
Halogenation
