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Solutions to Problems 175 :+ + HO: H⁺ HOH H C + H + + (e) or Rearrangement H from secondary to tertiary HO: +: + + (f) (CH₃)₃COH (CH₃)₃C H 12+ (CH₃)₃CO: (CH₃)₃CO: 35. The water-free conditions allow quantitative conversion of the alcohol to its alkyloxonium ion form, in the presence of high Br⁻ concentration: RCH₂OH Conc. RCH₂OH₂ + Br⁻ (SN2) RCH₂Br Ca. 100% In concentrated aqueous HBr, the main acid present is which is a weaker acid than the alkyloxonium ion. The first equilibrium lies well to the left, so the overall reaction rate is much lower (there is much less protonated alcohol to react with Br⁻): RCH₂OH RCH₂OH₂ + Br⁻ (SN2) RCH₂Br Ca. 99% Ca. 1% CH₃ 36. (a) A likely choice, after 2° (secondary) 3° (tertiary) carbocation rearrangement. (b) Both (CH₃)₃CCH₂I and (CH₃)₂CICH₂CH₃. CH₂ CH₃ (c) OH (d) 37. (a) through (e) are the same, via the SN2 displacement of a phosphite leaving group by bromide ion. Br (f) CH₃