Treatment of 2-butanone (1 mol) with Br2 (2 mol) in aqueous HBr gave C4H6Br2O. The 1H NMR spectrum of the product was characterized by signals
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Treatment of 2-butanone (1 mol) with Br2 (2 mol) in aqueous HBr gave C4H6Br2O. The 1H NMR spectrum of the product was characterized by signals at 1.9 ppm (doublet, 3 protons), 4.6 ppm (singlet, 2 protons), and 5.2 ppm (quartet, 1 proton). Identify this compound.
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Three dibromination products are possible from a halogenation of 2butanone O Br T C... View full answer

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