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232 Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY (h) C₁₀H₁₄ = 2(10) + 2 = 22; degree of unsaturation = (22 - 14)/2 = 4 62. (a) = 2(7) + 2 = 16; degree of unsaturation = (16 12)/2 = 2 (b) = 2(8) + 2 + 1 (for the N) = 19; degree of unsaturation = (19 - 7)/2 = 6 (c) = 2(6) + 2 - 6 (for the Cl's) = 8; degree of unsaturation = (8 0)/2 = 4 (d) = 2(10) + 2 = 22; degree of unsaturation = (22 22)/2 = 0 (e) = 2(6) + 2 = 14; degree of unsaturation = (14 - 10)/2 = 2 (f) = 2(18) + 2 = 38; degree of unsaturation = (38 - 28)/2 = 5 63. Begin by using the molecular mass to determine the molecular formula. Given that the unknown is a hydrocarbon, first determine which combinations of carbon and hydrogen can give an approximate mass of 96. Although 96 is the mass of 8 carbon atoms, that option is not possible (no mass left for hydrogen atoms). Reduce the number of carbon atoms and add the necessary number of hydrogens to arrive at a feasible formula. Thus, 7 carbons + 12 hydrogens also gives a mass of 96, and by using the data in Table 11-5, we quickly find a good match for the exact mass measurement: 7(12.000) + 12(1.0078) = 96.0936. Thus we are working with a formula of C₇H₁₂: two degrees of unsaturation. The unknown compound contains a total of two rings or bonds. The IR is useful: The peaks marked indicate at least one of the degrees of unsatura- tion is from a C=C bond. Also, the sharp band at 888 cm⁻¹ is diagnostic for a R₂C=CH₂ group. Can you decide whether the unknown has two double bonds, or one double bond and one ring? Hydrogenation gives you C₇H₁₄: One degree of unsaturation still remains, suggesting one ring in the original compound. The ¹H NMR supports this conclusion: The integration of the alkene H signal (at δ = 4.8) is 2 H. That limits the structure of one R₂C=CH₂ group. What else can you learn from the ¹H NMR? The alkene signal is a quintet (five lines). On the basis of the N+ 1 rule, you can try to make a structure where four equivalent neighbor H's split the alkene H's: -CH₂ C=CH₂ This structure fits the signal pattern, and the J of 3 Hz is right for allylic coupling, too (Table 11-2). The two CH₂ groups fit the ¹H NMR signal for 4 H at δ = 2.2 as well. This takes care of four carbons and six hydrogens, leaving C₃H₆ unaccounted for. Try the simplest way to make a ring: add three CH₂ groups to get CH₂ Does this structure fit the rest of the spectrum? The structure contains two more equivalent CH₂ groups, and the unique CH₂ at the opposite side of the ring from the double bond, fitting the NMR, and the ¹³C NMR reflects the symmetry with five peaks. This is indeed the answer. 64. A saturated 60-carbon alkane has the formula C₆₀H₁₂₂. "bucky-ball" possesses 122/2 = 61 degrees of unsaturation. In its hydrogenation C₆₀H₃₆, there are (122 36)/2 = 43 degrees of unsaturation. Therefore, there are at least 61 43 = 18 bonds in C₆₀. (As you will see later, there are actually 30 π but not all undergo hydrogenation.) 65. First list the information you can derive from the data: (a) Molecular formula: There is one degree of unsaturation. (b) 'H NMR: There are three methyl groups; two (at δ = 1.63 and 1.71 ppm) probably are attached to alkenyl the other split by one neighbor, as A group