15597.ch21.373-38810/18/0518:42Page376 376.chapter 21 AMINES AND THEIR DERIVATIVES:FUNCTIONAL GROUPS CONTAINING NITROGEN @ signals in the =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: (CH3):C- 2 CH- -CHz--C-NH2 All atoms in the formula are present,so put it together (CH3):C-CH2-C-NHz The m/58 fragment therefore is [(CHa)C-NH] 28.As you do each of these.keep the CHN formula in mind. (a)NMR:The 8=23.7 peak may correspond to one or more than one equivalent CH-groups,and the peak at=45.3 is one or more equivalent CH-units(attached to N due to chemical -NH-.No other signals are present.so attach as many of each aaa CH、 (b)NMR:Nov d-CH-groups (the latter attached to N):IR:A tertiary (d)NMR:One CH3-and five-CHz-'s.IR:Primary amine (-NH2).This is CH3(CH2)sNH2. (e)NMR:Two。 one (38 7)attached to N. C attached molecule: 一-a,> 532 This is the answer.signals in the  2.7 region, where you might expect to find signals. So, most likely the N is attached to a tertiary carbon. Possible pieces: A (CH3)3CO 2 CH3O OCH2O OCONH2 A All atoms in the formula are present, so put it together CH3 A (CH3)3COCH2OCONH2 A CH3 The m/z 58 fragment therefore is [(CH3)2CPNH2] . 28. As you do each of these, keep the C6H15N formula in mind. (a) NMR: The  23.7 peak may correspond to one or more than one equivalent CH3O groups, and the peak at  45.3 is one or more equivalent units (attached to N due to chemical shift). IR: A secondary amine, ONHO. No other signals are present, so attach as many of each as are necessary. The answer is (b) NMR: Now you have only CH3O and OCH2O groups (the latter attached to N); IR: A tertiary amine. So, the answer is (CH3CH2)3N. (c) NMR: CH3O groups, OCH2O groups not attached to N, and OCH2O groups that are attached to N. IR: Amine is secondary. So the answer is CH3CH2CH2NHCH2CH2CH3. (d) NMR: One CH3O and five OCH2O’s. IR: Primary amine (ONH2). This is CH3(CH2)5NH2. (e) NMR: Two different CH3O types, one ( 38.7) attached to N; also a quaternary C attached to N ( 53.2). IR: A tertiary amine. Remembering the C6H15N formula, you can construct the molecule: 29. Figure 21-5 is (CH3CH2)3N for comparison purposes. Look in each case for important fragments from CO† CON cleavage to make iminium ions. (a) m/z 72 is important, which is [M
29] or loss of CH3CH2O. The only amine that should easily lose an ethyl group from those in Problem 28 is CH3CH2O† CH2ONHOCH2CH2CH3 [see (c)]. This is the answer. 25.6 (CH3)3C 53.2 N 38.7 CH3 CH3 CH3 CH N H CH CH3 CH3 CH3 CH H C N 376 • Chapter 21 AMINES AND THEIR DERIVATIVES: FUNCTIONAL GROUPS CONTAINING NITROGEN 1559T_ch21_373-388 10/18/05 18:42 Page 376