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Solutions to Problems . 257 with one and one to be attached. As expected. the IR spectrum indicates that D has no C=C or O-H bonds; 720 is for the C-Cl; 1260 is new for you. a special oxacyclopropane band for its stretches. stretches are more normally seen around 1000-1100 cm⁻¹. (b) Electrophilic addition of Cl₂ to the double bond in gives Attack of H₂O would ordinarily be favored at the middle (2°) carbon, which should be the site of the more stable cation. However, the other (electron-withdrawing) inductively reduces the preference of 2° cation over 1°, so some chloronium ions react with H₂O at the primary carbon instead. :OH H₂O: C B + Base Base :0: D 60. (a) = 8 + 2 = 10; degree of unsaturation = (10 - 8)/2 = 1 π bond or ring. δ = 1.2 3 H): CH₃, split by a CH δ = 1.5 (broad S, 1 H): OH (IR at 3360 cm⁻¹) δ = 4.3 (quin, 1 H): CH, split by four hydrogens, attached to O δ = 5.0-6.0: Terminal alkene, (IR at 945, 1015, 1665, and 3095 cm⁻¹) The molecule is HO This is the only way to put the pieces together. (b) Upfield signals are assigned above. The alkene assignments are 5.0 H H 5.9 5.2 H