Compound D shows the following IR spectrum. 100 80 60 %T 40 3037 1650 20 1739 1678 4000 3000 2000 1500 1000 500 wavenumber, cm-1 (i) Match the spectrum with any of the suitable compounds listed below. ОН OH -CEN NEC- II II IV VI (ii) Explain your answer in b(i).
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- 2) Figures 2 to 4 show the IR, 1H NMR and 13C NMR spectra of a compound with formula C3H5N 2a) Identify bands A and B in the IR spectrum 2b) Using the information from the 1H NMR spectrum, fill in the table below Peak cluster (ppm) Multiplicity Number of H in neighbouring C Proposed identification 0 1.30 2.35 2c) Propose a structure for the compound C3H5N. Justify your answer by referring to key spectral features of the 3 spectra.Part 3B Set 1. Can 1H NMR spectroscopy be used to differentiate between the two compounds? Briefly explain why or why not. Predict the 1H NMR spectrum for each compound (include integration, multiplicity, and approximate chemical shift). Put it in data table format.The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 The 13C peaks are also listedbelow. Name compound 1 and Provide a full analysis of the NMR spectra for compound 1. with a table showing peak shift ( carbon) and assigment 13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment. you should be able to assign these peaks to one of two carbon atoms in 1.
- Compound 1, was found to consist of C,H,N, and Cl with an elemental analysis of 61.1% C, 2.9% H, and 10.2% N. IR Spectra displayed absorbances at 2215, 1605. Deduce the structure of this compound. (there is only one Cl)The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 The 1H and 13C peaks are also listedbelow. Provide a analysis of the NMR spectra for compound 1 include rationalisation of COSY1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Shown 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.
- An compound that has a molecular formula of C5H11NO, has the below 1H NMR spectrum which is showen in the picture. There is a strong single peaked absorption at 3400 and 1710 cm-1and there are couple of mdium peaks between 2800 and 3000 cm-1. Which of the options below is more consistent with this spectrum?The mass spectrum of tert-butylamine follows shows an intense base peak at m>z 58, and very little else. Use a diagram toshow the cleavage that accounts for the base peak. Suggest why no molecular ion is visible in this spectrum.1A sample of a liquid from an unlabeled bottle was subjected to analysis: MS showed that M+ has m/z of 89 and IR displayed bands at 2995-2840 cm-1, 1562 cm^-1, and 1425 cm^-1. Identify the functional group in the molecule and provide two possible structures.
- Oxidation of citronellol, a constituent of rose and geranium oils, with PCC in the presence of added NaOCOCH3 forms compound A. A has amolecular ion in its mass spectrum at 154 and a strong peak in its IRspectrum at 1730 cm−1, in addition to C—H stretching absorptions.Without added NaOCOCH3, oxidation of citronellol with PCC yieldsisopulegone, which is then converted to B with aqueous base. B has amolecular ion at 152 and a peak in its IR spectrum at 1680 cm−1, inaddition to C—H stretching absorptions. a.) Identify the structures of A and B.b.) Draw a mechanism for the conversion of citronellol to isopulegone.c.) Draw a mechanism for the conversion of isopulegone to B.The 1H-NMR spectrum of lycopene is provided below. Comment on the general features of the NMR spectrum.Treatment of compound D with LiAlH4 followed by H2O forms compound E. D shows a molecular ion in its mass spectrum at m/z = 71 and IR absorptions at 3600–3200 and 2263 cm–1. E shows a molecular ion in its mass spectrum at m/z = 75 and IR absorptions at 3636 and 3600–3200 cm–1. Propose structures for D and E from these data and the given 1H NMR spectra.