Question

Question1) identify the name of the compound represented on the spectrum graph?

Question2) For CNMR, indicate how many types of carbon atoms present and the identity of characteristics peaks?

Question3) For HNMR, indicate ho many types of carbon atoms present, the different identity of characteristics peaks, and explanations of it various splitting patterns?

Question4) Also for IR, indicate the identity of it characteristics peaks

Question5) Finally for MS, Identify the parent peak, the molecular weight and it implication, the most abundant peak, and identify the peak that indicate the presence of bromine or chlorine?

200 150 260 140 120 100 80 4020 ppm 20 IO 20 30 40 50 60 "”90100110 m/z

Answer 1

M+ is 108 As per the nitrogen rule, either there are no nitrogen or there are evein nimber of nitrogen atoms in the molecule Rule of 13: Devide 108 by 13: 8 and remainder is 4. Basic CH formula is C8H8+4C8H12 As per the IR there is one hydroxyl group as per the broad peak around 3200-3500. However, there is no carbonyl peak present. Hence, the compound is not a caboxylic acid Thus we need to add one oxygen in our formula by replacing with CH4 So MF becomes: C8H12-CH4H0 C7H80 Degree of unsaturation = ((2C+2H4-H-X)/2 = (16+0-8-0)/2-4 So all four sites of unsaturation can be accounted for benzene ring. IR: 3400: broad peak:-OH stretching 3010: sp2 C-H stretching 2990 and 2850: sp3 CH stretching 1600: aromatic C-C stretching 1100: C-O stretching 750 and 850: meta di substituted benzene ring HNMR: a: 2.4p CH3 b: 5.1ppm: -oH C: 6.6ppm: ortho protons to the -OH d: 6.8ppm: para proton to the -OH group e: 7.1ppm: meta protons to the -OH group pm:3H, s: so it can be Ar.

13CNMR: A: 21ppm: methyl B: 112ppm: ortho to -OH and para to methyl C: 116ppm: ortho to -OH and methyl D: 121ppm: ortho to methyl and para to -OH E: 130ppm: meta to -OH and Me F: 140ppm: ring carbon bearing methyl group G: 155ppm: ring carbon bearing -OH group АСН за CH CH G b он он m/z 90 loss of water M+108 3-methylphenol