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Solutions to Problems 365 It's not a big deal, but use of NaOH, H₂O in (b) and (c) would hydrolyze the ester groups to carboxylic acids (-CO₂H). Using ROH where R corresponds to the alkyl group of the ester, prevents this (Chapter 20). CH₃ CH₃ (d) + 0 0 H₃C 0 60. No! The carbonyl group changes the mechanism. + :OH :0: Tautomerism CH₃CCH₂CH₂Cl Protonation occurs on oxygen, not carbon, and the result looks like it's anti-Markovnikov. It is really 1,4-addition. 61. Calculate your degrees of unsaturation first! (a) = = 12; degrees of unsaturation = 12 2 10 = 1 TT bond or ring. UV: n absorption of carbonyl. NMR: 2.1 2.3 0.9(t) (3H) (2H) (3H) These assignments are clear, giving as the answer 2-pentanone, (A). The sextet at δ = 1.5 corresponds to the C4 CH₂ group. (b) = 10 + 2 = 12; degrees of unsaturation = 12 2 8 = 2 TT bonds or rings. UV: TT π* absorption of α, ß-unsat'd carbonyl at 220 nm. O NMR: is clear. 2.1 (3H) Also two alkene hydrogens (δ = 5.8-6.9) and another three H's at δ = 1.9 Because UV indicates conjugation, there are three possibilities: CH₃ H H H H CH₃ C = C and C C H C CH₃ CH₃ C CH₃ H C CH₃ O