13C NMR of isoamyl alcohol: 200 180 160 140 120 100 ppm 09-0 80 60.970 50 60 40 8- 20 -0 41.750 0 24.830 22.680 13C NMR of isoamyl acetate: 171.080 200 180 160 440 140 120 100 ppm - 80 63.130 60 -60 - 40 -37.480 25.180 22.510 -20.960 -0 20 0
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can you validate the synthesis of isoamyl acetate through the esterification reaction of isoamyl alcohol and acetic acid using the attached Carbon-13 NMR spectra.
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- 4. (Chapter 15 - 472) An aromatic compound C3H;Br has the following H NMR spectrum and a peak at 820 cm in its IR spectrum. Answer the following questions. Chem. Rel. shit area 3.00 2.00 120 2.58 207 2.00 739 2.00 TMS 5 Chemical shift (8) O ppm 10 1 ing 4(a). The IR peak at 820 cm1 in its IR spectrum indicates that the compound is. disubstituted = 4(b) The name of the compound is = IntensityProvide the correct answer choice/choices for the following:Chemistry H NMR AND C13NMR ??? IN 3 CI 5 A 20= B CI 7 9
- 16-17 Propylamine (bp 48°C), ethylmethylamine (bp 37°C), and trimethylamine (bp 3°C) are constitutional isomers with the molecular formula C3HgN. Account for the fact that trimethylamine has the lowest boiling point of the three and propylamine has the highest.500 20 8.0 Interpret the 'H-NMR spectrum and assign the correct structure. Please be sure to explain your work. C₂H₁, CL 3.0 7.0 400 4.0 6.0 5.0 -B00 5.0 6.0 4.0 200 7.0 E 3.0 8.0 MA 20 100 9.0 1.0 10.0 (ppm) 가 O Hz 08 (ppm)A 13C NMR spectrum is shown for a molecule with the molecular formula of C.HgO2. Draw the structure that best fits this data. 180 160 140 120 100. 80 60 40 20 "PPM Drawing
- A 13C NMR spectrum is shown for a molecule with the molecular formula of CsH11Cl. Draw the structure that best fits this data. 60 50 40 30 PPM 20 10 Q2. (Chapter 13 - Q58b) The compound whose H NMR spectrum is shown has the molecular formula C7H7B1. Follow the following questions to predict the unknown structure. Chem. shift Rel. area 2.31 1.50 701 1.00 7.35 1.00 TMS 10 7 6 O ppm 4 3 2 Chemical shift (8) e20s Cenge leaming 2(a) Degree of the unsaturation of this compound is= 2(b) The two distinct peaks in the aromatic region of the H NMR indicate that compound is .disubstituted = 2(c) The splitting pattern of the peak at 2.31 ő is = 2(d) The group that corresponds to the splitting pattern in 1(c) is = 2(e) This compound has the plane of symmetry Yes or No = 2(f) The name of the unknown compound = Intensity1% 76 70- 60- 50 40- 30- 20 10- -of 4000 3505.81cm-1 acetophenone 3500 O 2-methylcyclohexanone benzoyl chloride benzyl acetate O butyrolactone O cyclopentanone methyl benzoate 3523.83cm-1 methyl propionate 2992 510m-1 3000 2817.24cm-1 2500 1769.54cm-1 cm-1 2000 N 930.88cm-1 1377.30pm-1 1500 1167.95cm-1 1000 1036.26cm-1 500
- An unknown compound has the following data. Deduce the structure of the unknown compound and upload your answer. IF YOU DRAW MULTIPLE STRUCTURES, ONLY THE STRUCTURE WORTH THE LEAST POINTS WILL BE CONSIDERED. Molecular Formula: C10H120 13C-NMR data: there are eight signals 1H-NMR data: Chem. Rel. shift area 1.84 3.00 3.76 3.00 6.09 1.00 6.36 1.00 6.82 2.00 7.23 2.00 TMS 10 9. 8 7 3. 2 1 0 ppm Chemical shift (8) 6.5 6.4 6.3 6.2 6.1 6.0 Intensityexplain the peaks on Sebacoyl Chloride C -13 NMR (thousandths) 50.0 0'0 0'05 20.0 10.0 180.0 170.0 173.848 160.0 150.0 140.0 130.0 X: parts per Million : Carbon13 Sebacoyl Chloride 120.0 110.0 100.0 90.0 CDC13 Solvent 80.0 77416 77.160- 806'94 70.0 60.0 2 50.0 40.0 47.105 n 30.0 20.0 AL 28.8258 28.327 100'SZ 10.034. Pentan-1-amine vs. Pentanamide When comparing pentanal and pentan-3-one: 1. How do the amount of peaks differ? 2. Compare the range of PPM 3. Are there any other differences or similarities? Which carbon would be present at 173 В. NH2 4. С. Е. ppm? Spectra for pentane Spectra for pentanamide 10 20 PPM 35 30 25 15 180 160 120 40 34 100 PPM 140 80 60 20