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Solutions to Problems . 405 Perhaps a CH₂ and an NH₂? (Signals at δ = 1.3 and 1.4). MS: m/z 114 is [M - CH₃]⁺, 72 is [M and 58 is most likely an iminium ion. Before guessing, notice that there are no NMR H signals in the δ = 2.7 region, where you might expect to find -C-N signals. So, most likely the N is attached to a tertiary carbon. Possible pieces: 2 All atoms in the formula are present, so put it together CH₃ CH₃ The m/z 58 fragment therefore is with 32. As you do each of these, keep the C₆H₁₅N formula in mind. and (a) NMR: The δ = 23.7 peak may correspond to one or more than one equivalent groups. and the peak at δ = 45.3 is one or more equivalent CH- units (attached to N due to chemical shift). IR: A secondary amine, -NH-. No other signals are present, so attach as many of each as are necessary. The answer is is of (b) NMR: Now you have only and groups (the latter attached to N); IR: A tertiary amine. So, the answer is (CH₃CH₂)₃N. (c) NMR: groups. groups not attached to and groups that are attached to N. Amine is secondary. So the answer is (d) NMR: One and five Primary amine This is (e) NMR: Two different types. one (δ = 38.7) attached to N: also a C attached to N (δ = 53.2). IR: A tertiary amine. Remembering the formula. you can construct the molecule: 25.6 CH₃ CH₃ 38.7 53.2 33. Figure 21-5 is for comparison purposes. Look in each case for important fragments from cleavage to make iminium ions. (a) 72 is important. which is - 29]* or loss of The only amine that should easily lose an ethyl group from those in Problem 28 is [see (c)]. This is the answer.