An aqueous solution of phenol is shaken with an excess of bromine water, and the white precipitate of 2,4,6-tribromophenol is collected. What mass of this precipitate is obtained from 4.7 g of phenol, assuming complete conversion? (Atomic masses: H = 1, C = 12, O = 16, Br = 80)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 16.55 g
With bromine water, three bromine atoms enter the ring of phenol at the 2-, 4- and 6-positions. One molecule of phenol gives one molecule of 2,4,6-tribromophenol, using three molecules of bromine:
\( \displaystyle \ce{C6H5OH + 3Br2 -> C6H2Br3OH + 3HBr} \)
\( \displaystyle n(\ce{C6H5OH}) = \frac{4.7\ \mathrm{g}}{94\ \mathrm{g\ mol^{-1}}} = 0.0500\ \mathrm{mol} \)
\( \displaystyle M(\ce{C6H2Br3OH}) = 6 \times 12 + 3 \times 1 + 3 \times 80 + 16 = 331\ \mathrm{g\ mol^{-1}} \)
\( \displaystyle m(\ce{C6H2Br3OH}) = 0.0500\ \mathrm{mol} \times 331\ \mathrm{g\ mol^{-1}} = 16.55\ \mathrm{g} \)