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Solutions to Problems . 271 (c) The molecular formula is C₅H₈O. How? Briefly: Carbon. 71.4% of 84 = 60; 60/12 (atomic mass of C) = 5 Hydrogen. 9.6% of 84 = 8; 8/1 (atomic mass of H) = 8 Oxygen. 19% (the remainder) of 84 = 16; 16/16 (atomic mass of O) = 1 Double check using exact masses from Table 11-5: 5(12.00000) + 8(1.00783) + 15.9949 = 84.05754 = 10 + 2 = 12; degree of unsaturation = (12 8)/2 = 2 bonds or rings. IR: stretch shows at 2100 cm⁻¹, broad band from 3200-3500 cm⁻¹ suggest -0-H. NMR, focus on the signals with the simplest splitting patterns first: δ = 1.8 (broad 1 H) OH, broad singlet gives it away δ = 3.7 (t, 2 H) = CH₂, next OH (chemical shift tells you this), and also next to another CH₂ (triplet splitting tells you that) δ = 1.9 (t, 1 H) C=CH (narrowness of splitting is typical), "long-range" coupled to a CH₂ on the other side of the triple bond. Let's see what we know so far. We have figured out that the molecule contains the two pieces and Add them up and you get C₅H₈O: that's all there is in the so just put them together: The two middle CH₂ groups are responsible for the two signal sets that we didn't bother to try to interpret because they were more complicated. Try to figure out on your own why they look the way they do. 35. =C-H of terminal alkyne has ≈ 3300 cm⁻¹. (a) (b) C=C-D (c) Before reaction, m₁ is H (mass = 1) and m₂ is C₉H₁₁ (mass = 119). Rewrite the Hooke's law equation as v² = + So = k²f(120/119), or = 1.1 X 10⁷. Because k and f are assumed to be constant, use this value for to predict v² for the product. Now m₁ is D (mass = 2), so v² = (1.1 X = 5.6 X 10⁶ and predicted = 2366 cm⁻¹. The discrepancy of about 10% is typical and due to changes in k and f. 36. (a) (b) (after aqueous work-up) R Cl H CH₃ Rotate C-C (c) (R = (anti) H H H R R H R R E R Meso R H (d) The reverse of the meso compound, this gives R Z 37. (a) trans-3-octene, via two sequential one-electron reductions as described in Section 13-6: