Question

I'm stuck on this Spect. problem, could you help with this?

112119 IN 120.14 (9.52%) a. 52/1-1= 8.65 78 or ac m+2:121-12 (40%) 7 Nos,cl, or Br Example Problem #3 Mass Spectrometry Data 1 m/z 119.074 (M; 100%), m/z = 120.14 (9.52%), m/z = 121.12 (0.40%) Molecule does NOT contain any fluorine or iodine atoms. 3420260 N-H Formula ON IH 1 'H NMR Data 7.05 ppm (d, 1H), 6.72 ppm (t. 1H), 6.70 ppm (t. 1H), 7.02 ppm (d, 1H), 4.01 ppm (s. 1.0H), 3.05 ppm (t. 2.0H), 3.55 ppm (t, 2 H) oporen If c8:8x12 - 96 7119.074 - 96 =23 for Hon I Formula Comments Co He rensic asistent wit me 24: 9x12 =108 -7 111.074-10-11.07 for H . un ONIH Ferrucos courronsonable in mt Cl inceniccinnt with it Assignment / Chem. Shika Split 7.65 d 6.72 t 6-70 t 7.02. d 4.0 S 3.05 t 3.55 + Int|##'s 1 Structure 1 1# i I4 H-N-Cthe 1 14 1 14 1 24 2 24 123 2H

Answer 1

Problem - B 1/2 - 119.074 at olo odd no. of Molecular weight 119.079 so one nitrogen is present. Ah 13) ligt (with 117 cutando la Molecular formula One nitrogen is predent toN - CH₂ x D x 8 h eqH, -ch, tN - C HqN Molecular formule is a Cg HqN HNMR data i en no of unsaturated bund. One aromatich ring e - 8- 2 9-0+1+1 2 2 One cyclic rin = 5

C rib Ha = 2Hgt at 3.55ppm 4. 2 24t, 3.05 ppm He ) 4.01 ppre 33 (s, 1014) He= 6.70ppm (t, HD Hf = 6.72 ppm 1 t, 1H) Hd = (a, 1H) 7.02 ppm Hg 7 19, 1H) 7.05 ppm IR => show N-14 - es secondary amine which a how only one peck 2950, 300r for spe hybrid CH stretching - 1650 ont fer cac streeking c-c- loro umt streching.