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Solutions_Manual_for_Organic_Chemistry_6th_Ed 469

Solutions_Manual_for_Organic_Chemistry_6th_Ed 469 - 1 9 58...

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19-58 Mass spectrum: -molecular ion at 87 = odd mass = odd number of nitrogens present -if one nitrogen and no oxygens molecular formula C S H13N R t CH2NH2 -base peak at mJz 30 structure must include this fragment 30 IR spectrum: -two peaks in the 3300-3400 cm- 1 region 1 ° amine NMR spectrum: -singlet at 8 0.9 for 9H must be a t-butyl group -2H signal at 8 1.0 exchanges with D20 must be protons on N or ° 8 1.0 80.9 { CIl3_ H3 CH2- l , H3 t 82.4 H H \ I -. . C = N+ / \ H H mJ z 3 0 Note that the base peak in the MS arises from cleavage to give these two, relatively stable fragments: 19-59 (a tough problem) CH3 H H I \ / CH3 -C· + C = N + 1 / \ CH3 H H mJz 30 molecular formula C II Hl6N2 has 5 elements of unsaturation, enough for a benzene ring; no oxygt:ns precludes
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