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Solutions to Problems 229 52. = 10 + 2 = 12; degree of unsaturation = (12 10)/2 = (a) The only way for five carbons to be equivalent is to make a ring: is the answer. (b) Three CH₃'s, and a -CH=C CH₃ CH₃ (c) Two CH₃'s, one CH₂, and is the answer (stereochemistry is ambiguous). 53. Lower, because the vibrational frequency varies inversely with the square of the "reduced" mass involving the atoms about the bond. So, bonds involving heavier atoms have lower energies associated with vibrational excitation. Typically, ≈ 700 cm⁻¹, ≈ 600 cm⁻¹, and 500 cm⁻¹. 54. 10,000/v = µm. (a) 5.81 µm; (b) 6.06 µm; (c) 3.03 µm; (d) 11.24 µm; (e) 9.09 µm; (f) 4.42 µm. 55. A-(b) (saturated alkane) (alcohol band at 3300 cm⁻¹) C-(a) (alkene band at 1640 cm⁻¹ in addition to alcohol band) (alkene band at 1665 cm⁻¹ but no alcohol band) 56. (i) Both alkene (1660) and alcohol (3350) products have formed. (ii) Only alkene (1670) forms. (iii) Only alcohol (3350) forms. (a) Conclusions: Isomer C is probably a primary bromoalkane, which gives a primary alcohol product Isomer B is probably a tertiary bromoalkane, which gives only alkene as product (E2). Isomer A is probably a secondary bromoalkane, which gives a mixture of SN2 and E2 products. (b) A possibilities: CH₃CHBrCH(CH₃)₂, or CH₃CH₂CHBrCH₂CH₃ B possibilities: (CH₃)₂CBrCH₂CH₃ (only tertiary isomer) C possibilities: or but probably not (CH₃)₃CCH₂Br (too hindered to give SN2 reaction) 57. Process of elimination: we begin by noticing the absence of absorptions in certain regions of the spectrum, and conclude that the corresponding functional groups cannot be in the actual molecule. So, no (no strong broad band around 3300 cm⁻¹), no (nothing around 1700 cm⁻¹ or so), no C=C-H (bands near 2100 and 3300 cm⁻¹), and no C=C (1680 cm⁻¹). All that are left as possibilities are the alkane and the ether. The spectrum shows strong bands between 1000 and 1200 cm⁻¹, strongly suggesting the pres- ence of bonds. The fact that such absorptions do not appear in the sample spectra for alkanes in the O chapter (see Figures 11-18 and 19) confirms that the correct answer is the ether, 58. Begin by noting the masses of the most prominent ions in each mass spectrum. Then try to predict how each of the alkanes might be most likely to fragment, using as a guiding principle a preference for forming more rather than less stable carbocations upon bond cleavage. The three compounds are constitutional iso- mers, all with the molecular formula C₆H₁₄ and a molecular weight of 86.