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Solutions to Problems 231 Looking at the rest of the spectrum, the base peak is at m/z 59, or (M loss of CH₃CH₂. CH₃ + OH (CH₃)₂COH + CH₃CH₂· m/z 59 CH₃ This is, all together, good evidence for isomer C being 2-methyl-2-butanol, as shown. Isomer B also has a peak at m/z 73 for loss of CH₃. Its base peak 45) corresponds to loss of 43, or CH₃CH₂CH₂. These signals are what you might expect for 2-pentanol. OH CH₃CH₂CH₂CH + + a m/z 73 b OH CH₃CH + CH₃CH₂CH₂· + m/z 45 Both fragmentations give cations stabilized by resonance from an oxygen lone pair. This is, in fact, the correct answer. Isomer A does not lose CH₃ or CH₃CH₂ (no peaks at m/z 73 or 59). That pretty much rules out any tertiary or secondary alcohol structure as a possibility. (Any example that you can write should show those fragmentations.) How about possible primary alcohol structures? Look again to intense fragment peaks for clues. The m/z 70, loss of water, doesn't really help much, except to rule out which has no ß-hydrogens and therefore cannot dehydrate. That leaves three possibilities for A: CH₃ CH₃ The data that you have are in fact quite consistent with either of the first two (the third is difficult to maneuver into a fragment with m/z 42). If you got this far, you are doing well! (The actual spectrum of isomer A is that of 1-pentanol, by the way.) 61. (a) C₇H₁₄ = 2(7) + 2 = 16; degree of unsaturation = (16 14)/2 = 1 (b) C₃H₅Cl = 2(3) + 2 - 1 (for the = 7; degree of unsaturation = (7 5)/2 = (c) C₇H₁₂ = 2(7) + 2 = 16; degree of unsaturation = (16 12)/2 = 2 (d) C₅H₆ = 2(5) + 2 = 12; degree of unsaturation = (12 - 6)/2 = 3 (e) C₆H₁₁N = 2(6) + (for the N) = 15; degree of unsaturation = (15 11)/2 = 2 (f) C₅H₈O = 2(5) + 2 = 12; degree of unsaturation = (12 8)/2 = 2 (g) C₅H₁₀O = 2(5) + 2 = 12; degree of unsaturation = (12 10)/2 = 1