Question

Identify the structure using H NMR, C13 NMR, IR spectrum, and Mass Spectroscopy. The data is given below:

PLEASE SHOW WORK!
Problem #4: 'H NMR: ppm 8.318 Hz 4159 4151 4059 4051 3041 Height 11.86 12.01 11.46 11.38 5.28 8.118 8.104 6.081 CH CH 0.97 1.00 1.06 å7 65 Ppm C5H₃ Oz 13C NMR: 5 carbons ppm 168 151 140 124 130 180 160 140 120 100 .80 ppm f

Answer 1

M+ = 166
Rule of nitrogen: As the mass is even number, the molecule contains at even number of nitrogens.
Use rule 13: Divide 166 with 13, to give n = 12 and remainder is r = 10
So the formula is C12H22
IR shows a sharp and strong peak at 3479, HNMR shows the ratio of 1:1:1, this is possible only in case if the ration is 2:2:2 so to give 4 aromatic protons with para substitution and so the NH2 protons are 6.08ppm. Also, there is a nitro group, so add 2 nitrogen by replacing with C2H4: C10H18N2
Now add 3 oxygens by replacing C3H12: C7H6N2O3:
Degree of unsaturation = ((2C+2)+N-H-X)/2 = (16+2-6)/2 = 6
4 sites of unsaturation account for benzene ring, one for amide carbonyl and one for the nitro group.

IR: 3479 and 3420: doublet: amide NH stretching
3077: sp2 C-H stretching
1678: amide and conjugated C=O stretching; lowered due to conjugation
1601: C=C stretching
1515 and 1347: -NO2 group
791: OOP C-H bending indicating a para substitution
Please note that writings on spectrums (-OH and m/z=167) are misleading!!

HNMR a: 6.08ppm: 2H, broad singlet: -NH2 b: 8.111ppm: 2H. d (J=8Hz) ortho coupling must be ortho to C=O and meta to -NO2; J = 4059-4051 = 8Hz C: 8.311ppm: 2H, d (J=8Hz) ortho coupling, must be ortho to -NO2 as it is most down field ONH2 CNMR: A: 168: C=0 cl B: 151: -NO2 bearing ring carbon Eb C: 140: CEO bearing ring carbon D: 130: ortho carbons to -NO2 E: 124: ortho carbons to -C=0 А а Mass fragments: O NH2 of NO, NH2 O NH2 o NO2 NO2 m/z = 104 m/z = 120 M+ = 166 m/z = 150 Thus molecule is confirmed: 4-nitrobenzamide