Question

3. Propose a structure corresponding to the spectral data which are provided. CsH8N203. Put your final answer in the box. Hint: The proton at 10.3 ppm is NOT what you would suspect at that region. There are two coupling constants given for the triplets around 7-8 ppm because they are really distorted doublets of doublets. Explain your answer to receive full credit. Based on your answer you can assign a structure to problem 9 in Part I which is isomeric. 00- 180 3373 1343 1704 1 | 1513 otrophiftyp 25 50 75 100 e 150 o + 125 175 8.76 (1H, dd, J= 8.6; 1.5 Hz) 8.20 (1H, dd, J-8.4; 1.4 Hz) 7.65 (1H, td, J = 8.6; 6.0 and 1.4 Hz) 7.16 (1H, td, J = 8.4; 6.0 and 1.5 Hz) 136.5 (OH) 135.8 (17) 134.8 (OH) 125.6 (1H) 123.3 (1H) 122.3 (1H) 169.2 (OH) 24.5 (3H) 2.30 (3H, S) 10.3 (1H, s) TTTTTTTTTTT 180 160 200 140 120 100 80 60 40 20 0 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Answer 1

For C8H8N203 equivivalent unsaturated hydrocarbon formula is C13H16 equivivalent saturated hydrocarbon formula is C13H28 unsaturation index (28- 16)/2 6 From the IR peak at 1703 is corresponding to carbonyl carbon, and peak at 1343 and 1513 corresponds to the presence of nitro as functional groups From the mass, the peak at m/z 43 indicate presence of CH3CO group. The proton NMR spectrum indicates the presence of disubstituted aromatic ring. All the four protons on aromatic ring have a ortho coupling, and two of them shows the triplet indicate two adjescent protton to each of those two protons showing triplet. Therefore, the ortho-substituted benzene ring. The distorted coupling of 8.4 and 6 Hz indicate the donating and withdrawing substituent on ring, respectively. н R" From the IR spectra one of the functional group was -NO2 Н R Н NO2 The peak at 2.3 (s) indicate the -CH3 group attached to either carbonyl or aromatic ring. The peak at 10 ppm is not due to acid therefore could be because of NH attached with carbonyl i.e., presence of -NH- CO-CH3 group Therefore final structure is Н CH3 NO2 Н Н