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JEE Main 2017 · Chemistry · Hydrocarbons · hard

A hydrocarbon C8H14 decolourises bromine water and undergoes reductive ozonolysis (O3; Zn/H2O) to a single compound C8H14O2 which does not reduce…

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A hydrocarbon C8H14\mathrm{C_8H_{14}} decolourises bromine water and undergoes reductive ozonolysis (O3\mathrm{O_3}; Zn/H2O\mathrm{Zn/H_2O}) to a single compound C8H14O2\mathrm{C_8H_{14}O_2} which does not reduce Tollens' reagent but gives a positive iodoform test. The number of moles of iodoform obtained per mole of C8H14O2\mathrm{C_8H_{14}O_2} on treatment with excess I2/NaOH\mathrm{I_2/NaOH} is
  1. A. 3
  2. B. 4
  3. C. 2
  4. D. 1
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Answer: (C) 2

C8H14\mathrm{C_8H_{14}} has 2 degrees of unsaturation. One is a C=C\mathrm{C=C}; for a single ozonolysis product with the same carbon count and no new fragments, the other must be a ring (the ring opens to give a diketone). The product is a diketone, not an aldehyde, containing two CH3C=O\mathrm{CH_3C=O} groups: CH3CO(CH2)4COCH3\mathrm{CH_3CO(CH_2)_4COCH_3} (octane-2,7-dione), derived from 1,2-dimethylcyclohexene. Each methyl ketone group gives CHI3\mathrm{CHI_3}, so 2 mol.
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