Organic Spectroscopy - NDSUcook.chem.ndsu.nodak.edu/wordpress/wp-content/uploads/2012/11/01... ·...
Transcript of Organic Spectroscopy - NDSUcook.chem.ndsu.nodak.edu/wordpress/wp-content/uploads/2012/11/01... ·...
©2013 Gregory R. Cook, NDSU
Organic Spectroscopy
Chem 744 / 754Spring 2013
Gregory R. Cook
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Resources
‣ Books on reserve in the Library
‣ Introduction to Spectroscopy 3rd Ed., Pavia, Lampman, Kriz; Saunders Publishing, 2001.
‣ Spectrometric Identification of Organic Compounds 5th Ed., Silverstein, Bassler, Morrill; Wiley, 1991.
‣ Basic one- and two-dimensional NMR Spectroscopy, Horst, Weinheim, 2005.
‣ NMR - from spectra to structures: an experimental approach, Mitchell and Costisella, Springer, 2004.
‣ Structure elucidation by modern NMR: A workbook, Duddeck, Dietrich, Toth, Springer, 1998.
‣ Other References and Texts
‣ Organic Structure Analysis, Crews, Rodríguez, Jaspars; Oxford Press, 1998
‣ Spectrometric Identification of Organic Compounds 6th Ed., Silverstein, Bassler, Morrill; Wiley, 1998.
‣ ABCs of FT-NMR, Roberts, University Science Books, 2000.
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744 Course Information
‣ Introduction
‣ This course is designed to provide a theoretical and practical working knowledge of modern spectroscopic techniques as applied to the elucidation of the structure of organic compounds.
‣ Mass spectroscopy, infrared spectroscopy, and NMR spectroscopy will be covered. If time permits, we will discuss Raman and UV spectroscopy.
‣ You are expected to have a solid understanding of physical organic chemistry and organic structure.
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744 Course Information
‣ Introduction
‣ Grading
‣ Midterm exam (Jan 31) (25%)Final exam (Feb 28) (50%)Homework (25%)
‣ Grades will be assigned as follows (subject to change): A - 85-100%B - 70-84%C - 57-69%D - 45-56%F - <45%
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754 Course Information
‣ Introduction
‣ The lab course is designed to train users on the various instruments for spectroscopic determination of organic compounds and to interpret data.
‣ Users will run known and unknown samples on the GC-Mass Spec, Infrared spectrometer, and NMR instruments.
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754 Course Information
‣ Grading
‣ Grades will be assigned as follows:
‣ A - 90-100%B - 80-89%C - 70-79%D - 60-69%F - <60%.
‣ Grades are based upon two laboratory reports;report on a known sample (40%)report on an an unknown sample (60%)
‣ In analyzing the data obtained, the following approximate percentages will be applied to the various spectroscopic techniques.Mass Spec - 20%, IR - 15%, NMR (1H, 13C, etc) - 65%.
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754 Course Information
‣ Plan
‣ Training on the instruments by March 8.
‣ You will be provided with a known sample. You must obtain all the data and analyze it. The first report is due on April 1.
‣ When you turn in your report, you will be provided with an unkown sample.
‣ The final report will be due on April 29.
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754 Course Information
‣ Reports
‣ Reports should be typewritten in manuscript style.
‣ A good model to follow would be a paper on the structure determination of a natural product found in J. Nat. Prod. or J. Am. Chem. Soc.
‣ MS - Determine the m/e peak and m+1 and m+2 if present. Explain the fragmentation patterns observed for as many peaks as possible. You should include the structure of the fragments as well as the fragmentation pathway from which it arose.
‣ IR - Identify all functional groups in your molecule.
‣ 1H NMR - Assign all resonances to the protons on the structure. All coupling constants should be calculated and splitting patterns explained.
‣ 13C NMR - Assign all resonances to the carbons on the structure. Advanced NMR experiments may be necessary to correlate the peaks.
‣ 2D NMR - Explain your analysis of the data as it pertains to your structure.
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Tentative Course Schedule
Date TopicJan 8/10 Introduction and Basics of NMR SpectroscopyJan 15/17 NMR Spin Coupling and Multiplet AnalysisJan 22/24 Multiplet Analysis and Multipulse NMRJan 29 NMR StereochemistryJan 31 MIDTERM EXAMFeb 5/7 NMR Practical Considerations and 2D NMRFeb 12/14 Mass SpectrometryFeb 19/21 Infrared SpectroscopyFeb 26 UV SpectroscopyFeb 28 FINAL EXAMWeek of Mar 4 Chem 754 - Make appointments with Dan Wanner and
John Bagu to be trained on the instruments. You will be given a known sample to work with and analyze
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How to determine the structure of molecules?
‣ Probe physical properties
‣ Elemental Analysis
‣ atomic composition (relative ratios)
‣ empirical formula
‣ Mass Spectrometry
‣ molecular formula
‣ element identification (isotopes)
‣ connectivity
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Ultraviolet - Visible Spectroscopy
‣ Electronic (UV-VIS) Spectroscopy
‣ energy to excite an electron to a higher excited state
‣ More conjugation, lower energy
220 nm 258 nm
455 nm
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Infrared Spectroscopy
‣ Vibrational (Infrared) Spectroscopy
‣ functional groups
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Nuclear Magnetic Resonance Spectroscopy
‣ NMR probes the spin transitions of nuclei
‣ energy in the radio frequency range
‣ NMR provides a wealth of information about structure
‣ Functional Groups
‣ Atom Connectivity
‣ Stereochemistry
‣ Higher Order Structure
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Magnetic Resonance Imaging
‣ NMR is the basis for MRI
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X-Ray Crystallography
‣ X-Ray Crystallography
‣ 3D positions of atoms
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Empirical Formula from Elemental Analysis
C - 63.31%H - 6.28%Cl - 16.99%N - 13.42%
Assume 100 g of analyte
% composition proportional to grams
Moles of C = 6.331 g / 12.011 g/mol = 5.27 / 0.48 = 11
Moles of H = 6.28 g / 1.008 g/mol = 6.23 / 0.48 = 13
Moles of Cl = 16.999 g / 35.453 g/mol = 0.48 / 0.48 = 1
Moles of N = 13.42 g / 14.007 g/mol = 0.96 / 0.48 = 2
Empirical Formula: C11H13ClN2
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Units of Unsaturation
Empirical Formula: C11H13ClN2
‣ Degrees of unsaturation is the number of pi-bonds and/or rings
‣ For saturated hydrocarbons: CnH2n+2
every halogen replaces one Hevery nitrogen adds one H
(2n+2) - #H - #X + #N
2UN =
(2*11+2) - 13 - 1 + 2
2= 6UN = N
N
H Cl
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Molecular Formula from Mass
‣ Rule of 13
‣ M / 13 = n+r/13 where CnHn+r
for example - take a molecule with a M=164
164 / 13 = 12 + 8/13 therefore a MF would be C12H12+8 or C12H20
(2*12+2) - 20 - 0 + 0
2= 3UN =
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Molecular Formula from Mass
‣ What if there are other atoms besides C and H?
‣ Subtract their mass equivalents of C and H from the formula
for example - take a molecule with a M=164
164 / 13 = 12 + 8/13 C12H20
O = 16 equivalent to CH4 C12H20 C11H16O
Note: Be careful of invalid formulas
e.g. M=32 32/13 = 2 + 6/13 C2H8 impossible UN = -1
must have another atom: CH4O CH6N
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Identification of a Natural Product
‣ High Res. Mass Spectrometry
‣ UV Spectroscopy
‣ IR Spectroscopy
EpibatidineJ. Am. Chem. Soc. 1992, 112, 3475
Isolated from the Ecuadorian tree frog - Epibatis Tricolor
Analgesic activity 500 times greater than morphine.
210.0764 4.4% C11H13N237Cl
208.0769 15.5% C11H13N235Cl
217 nm and 250-280 nm indicates pyridine ring
1428 and 1112 cm-1 suggests a pyridine ring
NN
H Cl
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Proton NMR
EpibatidineJ. Am. Chem. Soc. 1992, 112, 3475
NN
H Cl
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Carbon NMR
EpibatidineJ. Am. Chem. Soc. 1992, 112, 3475
NN
H Cl
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Steps in Structure Elucidation
27
Pure Compound
Molecular Formula
Functional Groups
Sub-structures
3D structure
EA, MS, NMR
NMR, IR, UV
NMR
X-RAY
Working 2D
structures
UN
examineall isomers
determine bestpossibilities
NMR, MS, IR, UV
Best 2D structures
Reasonable3D
structure
NMRmodel
TotalSynthesis
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