Question

will you please help me figure out how to put the spectral pieces together to fine the molecule?

M,205(o o) DD least 1 4N m t120 (i.73 le) -73/le110.710 or11 C MtY:207(1.5s"le) er OINIH Fernuin 7H 78 Example Problem # 4 B=42 1H NMR Spectrum 3 8 PPM It Mass Spectrometry Data m/z 205.045 (M; 100 %), m/z= 206.11 (11.73 % ), m/z = 207.17 (1.55 %) ASsioNChen shieHISt Int HI Strycture 2 If 7-7 2 2 2 33 6-1 3.1 0

Answer 1

- or odd one nitrogen. 205. - It me one Mass spectrum data m/2 = 205.045 (M: 100%); m2 = 206.11 (11-73%); M =207-19 (-15% m+1 Mt2 → According to nitrogen sule, it m/z of Mis odd then it contains at least one nitrogen. 205 is odd so it contains at least one or odd number of nitrogens. - It m/2 of m+1 = 11.73 then divide this number by I'll to get number of a combons. i no. ct carbons = 1173 = 10.66 .. It may contain either 10 or 11 combous. H-NMR data: This data provides there are seven hydrogensi This up till now we have C=10, H=7, N=1 12x10+ 781+14x1 = 141 m/ of m=205-141 = 64 this could be due to four oxygen atoms (1684=64) Thus molecular formula from the given data is obtained = C10 Hz No4 IR dates :- small broad peak at - 3400 cm corresponds to on stretch. ~ 3000cm c -h stretch 1450-1600 cm e c=c stretch 1080-1360cml= (-N stretch. too canto = c= HNMR :- 0 shotch S (ppm) H 7.5 24 S s aromatic 61 24 3.0 s H attached to o from these information, the structure of the compound could HNMR assignment: H75 34 be. POH 6151 omi