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← JEE Advanced 2018 · Paper 2
ChemistryNumerical+3Question 29 of 54

JEE Advanced 2018 · Paper 2, Chemistry question 11

In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is ____.

(Atomic weights in g mol−1^{-1}: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis)

Acetophenone → [NaOBr; H3_3O+^+] → A (60%) → [NH3_3, Δ] → B (50%) → [Br2_2/KOH] → C (50%) → [Br2_2 (3 equiv), AcOH] → D (100%)

Acetophenone → (NaOBr, then H₃O⁺) A (60%) → (NH₃, Δ) B (50%) → (Br₂/KOH) C (50%) → (Br₂, 3 equiv, AcOH) D (100%).
Answer and worked solution

Official answer: 490

  • A: the haloform reaction (NaOBr, then HX3OX+\ce{H3O+}) gives benzoic acid (60%).
  • B: NHX3\ce{NH3}, Δ gives benzamide (50%).
  • C: BrX2/KOH\ce{Br2/KOH} (Hofmann degradation) gives aniline (50%).
  • D: BrX2\ce{Br2} (3 equiv) gives 2,4,6-tribromoaniline (100%).

Moles of D.
10×0.6×0.5×0.5×1=1.5 mol.10\times0.6\times0.5\times0.5\times1=1.5\text{ mol}.

Mass. M(CX6HX4BrX3N)=72+4+240+14=330M(\ce{C6H4Br3N})=72+4+240+14=330, so
1.5×330=495 g.1.5\times330=495\text{ g}.

Answer: 495 (key: 490 to 500)

Topics:AminesNamed reactionsAcid derivativesStoichiometry
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