*19. Following are the NMR spectra of three isomeric esters with the formula C;H14O2, all derived from propanoic acid. Provide a structure for each.
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- The 1H-NMR spectrum of lycopene is provided below. Comment on the general features of the NMR spectrum.Propose a structure for the molecule with the formula of C5H10O from the H NMR, C NMR, IR and mass spectrometry spectrums.Please provide the IR, 1H NMR and UV Spectra values of Benzoic Acid, 3-bromo-, methyl ester. And discuss how the results were obtain. For example the result for 3-(3-methoxyphenyl) propanoic acid would be written like this: 1 H NMR spectrumd (ppm in CDCI3 at 60 MHZ: 11.75 (Singlet, 1H, #7), 7.40-7.23 (Multiplet, 1H, #2), 6.85 (quartet, 3H, #1-3-6), 3.86 (Singlet, 3H, #5), 3.15-2.53 (Multiplet, 4H, #9-8). The discussion would be like this: the unknown molecule contained 12H. Revealing the presence of a peak of medium intensity at 11.75 ppm that we can associate with a carbocyclic acid. It also showed a high intensity peak at 3.86 ppm that we can associate with an ether, an alcohol or an amino group. Since the signal also shows that 3 protons are attached to it, it is therefore considered an ether (OCH3) in the unknown molecule. We also see two average peaks around 7.23 ppm and 6.85 ppm which show the presence of an aromatic.
- The tabulated 1H NMR and 13C NMR spectral data acquired for an unknown compound with an atomic composition of C11H20O2is detailed below. Using this data, propose a molecular structure for the unknown compound and, based on this structure, hypothesise a molecular structure of at least one major fragment appearing in the corresponding EI mass spectrum of the unknown compound, which is shown below; 1H NMR (500 MHz, Chloroform-d) δ 5.55 (s, 1H), 5.11 (d, J = 5.8 Hz, 1H), 4.16 (q, J = 6.3 Hz, 2H), 2.22 (d, J = 6.1 Hz, 2H), 2.18 (d, J = 6.1 Hz, 2H), 2.13 (s, 3H), 1.65 (s, 3H), 1.61 (s, 3H), 1.27 (t, J = 6.3 Hz, 3H). 13C NMR (125 MHz, Common NMR Solvents) δ 166.72 (C), 158.99 (C), 131.99 (C), 123.58 (CH), 116.45 (CH), 59.71 (CH2), 38.94 (CH2), 26.53 (CH2), 24.98 (CH3), 19.87 (CH3), 18.90 (CH3), 14.53 (CH3).The 1H- and 13C-NMR data of an ester of molecular formula C6H10O2 are given below. Also shown are the COSY and HETCOR NMR spectra of the ester. Draw the structure of the ester, explaining how you reach your conclusion. 1H-NMR: 7.20-6.90 (1H), 5.85 (1H), 4.16 (2H), 1.88 (3H), 1.31 (3H) ppm 13C-NMR: 166.7, 144.5, 123.0 , 60.2, 18.0, 14.3 ppmShown on the next two pages are the mass, IR and NMR spectra for anunknown organic molecule. Based on the spectra provided, determine thestructure of the molecule. You do not need to assign spectral peaks.Remember to ignore the 13 C NMR triad at 77 ppm that comes from the NMRsolvent.
- Compound A is a hydrocarbon with a molar mass of 96g/mol, with the given C13 spectral data. When compound A reacts with BH3 followed by the treatment with basic H2O2 it is converted to compound B. Propose structures for A and B, explain your analysis.Compound A- Proton decoupled C NMR: 26.8, 28.7, 35.7, 106.9, 149.7 δ.DEPT-90: No peak.DEPT-135: No positive peaks; negative peaks at 26.8, 28.7, 35.7, 106.9 δ.Compound B- Proton decoupled C NMR: 26.1, 26.9, 29.9, 40.5, 68.2 δ.DEPT-90: 40.5 δ.DEPT-135: positive peak at 40.5 δ; negative peaks at 26.1, 26.9, 29.9, 68.2 δGiven the information about the unknown: Unknown: MP: 101-106 oC White solid IR: 2837, 1701, 1602, 1391, 1139 cm-1 ** Using the Reference sheet, The IR and the given H-NMR, and C-NMR. Deduce, what the unknown is. Please give the name of uknownIn the attached 1H NMR spectrum of ethyl-3-coumarincarboxylate, draw the structure on the spectrum and assign the various resonances (peaks) to the hydrogen nuclei responsible for them.
- Predict 1H NMR (with splitting patterns) and 13C NMR chemical shifts for the benzene Alkene derivative below .Is it possible to distinguish between the following compounds using the spectroscopic techniques covered in CHE331? If so, discuss how. If not, discuss why.Benzonitrile vs Benzaldehyde(a) UV spectroscopy(b) IR spectroscopy(c) Mass spectrometry(d) 13C NMR spectroscopy(e) 1H NMR spectroscopyIn the presence of a small amount of acid, a solution of acetaldehyde (CH3CHO) in methanol (CH3OH) was allowed to stand and a new compound L was formed. L has a molecular ion in its mass spectrum at 90 and IR absorptions at 2992 and 2941 cm−1. L shows three signals in its 13C NMR at 19, 52, and 101 ppm. The 1H NMR spectrum of L is given below. What is the structure of L?