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Solutions to Problems . 423 Seven resonance forms make this carbanion especially stable. It is also aromatic, having 14 electrons in an unbroken loop of p orbitals. 43. (a) One solution, perhaps a bit roundabout: CH₂CH₃ BrCHCH₃ NBS, hv CH=CH₂ CH₂CH₂Br HBr, ROOR (Anti-Markovnikov) CH₃ CH₃ COOH I. FeCl₃ 1 2. NH3 (b) product CI CH₃ COOH COCI KMnO₄. 1 (c) O C CH₃ II C COOH product CH₃ CH₃ Br Br H2SO4 2 1 (d) CH₃ Br Br product