1. Figure 1. Phenethylamine 1H FT-NMR Spectrum in CDCl3 at 400 MHz.
Analyzing the Structure of an Unknown Compound using FT-NMR Spectroscopy
Nikole M. Gomez Department of Chemistry, Southern Oregon University, Ashland, Oregon
Introduction
NMR Spectroscopy allows one to determine the molecular structure of a compound. 1H, 13C, and DEPT-135 FT-NMR data were acquired to determine the structure of an unknown organic compound. The 1H FT-NMR spectrum provides information on the chemical environment of hydrogens, the number of hydrogens, the number of hydrogens on nearby atoms, and the number of chemically distinct hydrogens. The 13C FT-NMR spectrum provides information on the chemical environment of carbon-13 and the number of chemically distinct carbons while the DEPT-135 FT-NMR data provides information on the number of hydrogens bonded to a certain carbon when compared to the 13C FT-NMR spectrum. Positive peaks on the DEPT-135 FT-NMR spectrum are methine or methyl carbons while negative peaks are methylene carbons. Utilizing the spectral data, the molecular structure of an unknown organic compound can be determined.
Figure 2. Phenethylamine 13C and DEPT FT-NMR Spectrum in CDCl3 at 400 MHz. c b d
2-phenylethylamine e f
Analysis of Figure 1
Peaks A and B are at 7.29 and 7.20 ppm are aromatic hydrogens and the integration indicates a total of five hydrogens indicating a monosubstituted benzene ring. Peaks C and D are two triplets, each one has two hydrogen neighbors, that display peak leaning towards each other at 2.73 and 2.95 ppm indicating that they are coupled to each other and the integration shows that the peaks each contain two hydrogens. Methylene hydrogen peaks are usually from ppm, but peaks C and D are further downfield indicating that hydrogens are being deshielded by a neighboring electron withdrawing group. If there was a methyl group to end the carbon chain, there would be a triplet peak at approximately ppm. On the spectrum there is a broad wide singlet at 1.13 ppm. The broadness of the peak is caused by hydrogen bonding interactions which only could be nitrogen, fluorine or oxygen. With information provided from the instructor, fluorine was not in the compound. Looking at the integration of peak E, only nitrogen would have two hydrogens bound to it.
Figure 1. Phenethylamine 1H FT-NMR Spectrum in CDCl3 at 400 MHz. b c b c d
B, 3H, m
A, 2H, m
C, 2H, t d e f
B, 3H, m
D, 2H, t
CDCl3 a
C, 2H, t
A, 2H, m
D, 2H, t
E, 2H, s
TMS
Analysis of Figure 2
On the 13C FT-NMR peaks a, b, c, d, are carbons from the benzene ring. Peaks e and f are the methylene carbons. There is no peak for the nitrogen because the 13C FT-NMR only provides information regarding carbon-13.
Peak a is not present on the DEPT spectrum because the carbon has no hydrogens attached to it. Peaks b, c, and d, are all positive peaks which are methines and not methyls because methyl peaks on the 13C FT-NMR are from 8-30 ppm. Peaks e and f are negative because they are methylenes which supports peaks C and D from the 1H FT-NMR.
Conclusion
After analyzing all three spectra, the structure was determined to be 2-phenylethylamine.
Acknowledgements
Thank you Hala Schepmann Ph.D. Professor of Bioorganic Chemistry
References
Pavia, D. L. et al. (2015). Introduction to Spectroscopy.