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Chapter 13: NMR

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Chapter 13: NMR

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  1. Chapter 13: NMR CARBON-13 (13C) NMR SPECTROSCOPY Dr. Sivappa Rasapalli Chemistry and Biochemistry University of Massachusetts Dartmouth

  2. Carbon-13 NMR Spectroscopy both give us information about the number of chemically nonequivalent nuclei (nonequivalent hydrogens or nonequivalent carbons) both give us information about the environment of the nuclei (hybridization state, attached atoms, etc.) H1 and C13 NMR Spectroscopy

  3. SPIN PROPERTIES OF ATOMIC NUCLEI • What is spin? The Simple explanation • Spin is a fundamental property of nature like electrical charge or mass. • Spin is a measure of angular momentum (rotation about an axis) hence the term • Spin comes in multiples of 1/2 (0, 1/2, 1, 3/2, 2, 5/2…) and can be + or -. • Protons, electrons, and neutrons possess spin. • Individual unpaired electrons, protons, and neutrons each possesses a spinof 1/2 • Atomic nuclei composed of neutrons and protons may also possess spin. • The spin of an atomic nucleus is determined by the number of protons and neutrons in the nucleus. • Atoms with an odd number of protons will have spin • Atoms with an odd number of neutrons will have spin • Atoms with an odd number of both protons and neutrons will have spin • Atoms with an even number of both protons and neutrons will not have spin • The value of nuclear spin is represented by the symbol I, the nuclear spin quantum number. (I = 0, 1/2, 1, 3/2, 2, 5/2….) • A nucleus with spin of I can exist in (2I+1) spin states. • The shell model for the nucleus tells us that nucleons (protons and neutrons), just like electrons, fill orbitals. When the number of protons or neutrons equals 2, 8, 20, 28, 50, 82, and 126, orbitals are filled. Because nucleons have spin, just like electrons do, their spin can pair up when the orbitals are being filled and cancel out. Odd numbers mean unfilled orbitals, that do not cancel out.

  4. A Basic Concept in Electromagnetic Theory- Direct Application to NMR A moving perpendicular external magnetic field will induce an electric current in a closed loop An electric current in a closed loop will create a perpendicular magnetic field

  5. Carbon-13 NMR Spectroscopy Where’s Waldo?

  6. 13C NMR Spectroscopy Bo = external magnetic field strength • = magnetogyric ratio 1H= 26,752 13C= 6.7 gBo h 2 p One carbon in 3 molecules of squalene is 13C DE=

  7. 13C Transition Energy

  8. Magnetic alignment Bo RANDOM ORIENTATION = g h / 4p Add a strong external field (Bo) and the nuclear magnetic moment: aligns with (low energy) against (high-energy) In the absence of external field, each nuclei is energetically degenerate

  9. Nuclear Spin (cont.) radiofrequency energy source Bo external magnetic field -spin state +1/2 (lower energy) -spin state -1/2 (higher energy) 1. At zero external magnetic field, spins are degenerate! 2. Apply an external magnetic field – spins states will differ in energy depending upon relative orientation with respect to external field. -nuclei with I = ½ will adopt two specific orientations with respect to an externally-applied magnetic field... to convert lower energy spin state into higher energy spin state, require external energy source .........irradiate with radiofrequency (rf) radiation!

  10. Spins Orientation in a Magnetic Field (Energy Levels) • radiofrequency required depends on E • E depends on strength of Bo -spin: -1/2 higher energy h  E apply magnetic field no external magnetic field -spin: +1/2 lower energy increasing Bo 1. At zero external magnetic field, spins are degenerate! 2. Apply an external magnetic field – spins states will differ in energy depending upon relative orientation with respect to external field. -nuclei with I = ½ will adopt two specific orientations with respect to an externally-applied magnetic field... to convert lower energy spin state into higher energy spin state, require external energy source .........irradiate with radiofrequency (rf) radiation!

  11. Spins Orientation in a Magnetic Field (Energy Levels) • radiofrequency required depends on E • E depends on strength of Bo -spin: -1/2 higher energy h  E apply magnetic field no external magnetic field -spin: +1/2 lower energy increasing H0 • Difference in energy between the two states is given by: • DE = g h Bo / 2p (DE = ghn) • where: Bo=external magnetic field • h = Planck’s constant; g – gyromagnetic ratio Bo =0 Bo >0

  12.  = Bo/2 The value, , is the magnetogyric ratio

  13. Spins Orientation in a Magnetic Field (Energy Levels) • Transition from the low energy to high energy spin state occurs through an absorption of a photon of radio-frequency (RF) energy RF Frequency of absorption: n = g Bo / 2p

  14. Nuclear Spin (cont.) ENERGY OF A PHOTON E = h SPIN STATE ENRGY DIFFERENCE E = hB0/2 WHEN E = E, SPIN FLIP OCCURS h hB0/2 THE NECESSARY FREQUENCY IS:  B0/2 • Difference in energy between the two nuclear spin states: depends on strength of external magnetic field • For nucleus of H atom (proton), spin energy differences: -1/2  (MHz) 100 200 300 360 500 +1/2 2.34 4.73 6.35 8.46 11.75 0 H0(Tesla) Thus, at H0 = 4.7 T (Tesla), use rf radiation of 200 MHz for 1H Nuclei

  15. Nuclear Spin (cont.) ENERGY OF A PHOTON E = h SPIN STATE ENRGY DIFFERENCE E = hB0/2 WHEN E = E, SPIN FLIP OCCURS h hB0/2 THE NECESSARY FREQUENCY IS:  B0/2 • Difference in energy between the two nuclear spin states: depends on strength of external magnetic field • For nucleus of H atom (proton), spin energy differences: -1/2  (MHz) 25 50 75 90 125 +1/2 2.34 4.73 6.35 8.46 11.75 0 H0(Tesla) Thus, at B0= 4.7 T (Tesla), use rf radiation of 50 MHz etc for 13C nuclei.

  16. Fourier Transform NMR 13C-spectra of CH3CH2CH2CH2CH2OH after one scan average of 200 scans

  17. Features of 13C NMR Spectra • Each unique C in a structure gives a single peak in the spectrum; there is rarely any overlap. • The C NMR spectrum spans over 200 ppm; chemical shifts only 0.001 ppm apart can be distinguished; this allows for over 2x105 possible chemical shifts for carbon. • The intensity (size) of each peak is NOT directly related to the number of that type of carbon. Other factors contribute to the size of a peak: • Peaks from carbon atoms that have attached hydrogen atoms are bigger than those that don’t have hydrogens attached. • Carbon chemical shifts are usually reported as downfield from the carbon signal of tetramethylsilane (TMS).

  18. H1 and C13 NMR Spectroscopy Number of peaks Chemical shifts Integration Spin-Spin Splitting Number of peaks Chemical shifts Integration Spin-Spin Splitting

  19. 13C NMR

  20. 13C NMR

  21. Predicting 13C Spectra

  22. Predicting 13C NMR

  23. Symmetry in C-13 NMR Each unique carbon in a molecule gives rise to a 13C NMR signal. Therefore, if there are fewer signals in the spectrum than carbon atoms in the compound, the molecule must possess symmetry. Examples:

  24. How many signals would you expect?

  25. EnantiotopicvsDiastereotopiccarbons’s * * * * *

  26. Predicting 13C NMR Determine the number of signals in the proton-decoupled C-13 NMR spectrum of each of the following compounds:

  27. 13C signals are spread over a much wider range than 1H signals making it easier to identify and count individual nuclei The following slides show the 1H NMR and the 13C spectrumof 1-chloropentane. It is much easier to identify the compound as 1-chloropentane by its 13C spectrum than by its 1H spectrum. 1H and 13C NMR compared:

  28. Proton Spectrum 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0 1H CH3 ClCH2 ClCH2CH2CH2CH2CH3 Chemical shift (, ppm)

  29. Carbon Spectrum 200 180 160 140 120 100 80 60 40 20 0 13C a separate, distinct peak appears for each of the 5 carbons ClCH2CH2CH2CH2CH3 CDCl3 Chemical shift (, ppm)

  30. Intensity of 13C NMR signals 128.0 128.5 132.8 128.0 137.1 128.5 132.8 17.8 13.9 40.5 TMS CDCl3 200.3 137.1

  31. 13C-NMR: Integration time for nucleus to relax from excited spin state to ground state • 1H-NMR: Integration reveals relative number of hydrogens per signal • 13C-NMR: Integration reveals relative number of carbons per signal • Rarely useful due to slow relaxation time for 13C

  32. electronegativity of groups attached to carbon hybridization state of carbon 13C Chemical shifts are most affected by:

  33. 13C NMR Chemical Shifts Several functionalities appear directly on 13C NMR which are not ‘visible’ in 1H NMR: - Quaternary carbons - ipso carbons - Carbonyl carbons sp3 carbon sp3-EWG alkyne carbons alkene carbons aromatic carbons carbonyl carbons downfield d (ppm) upfield deshielded shielded higher DE lower DE

  34. 13C NMR

  35. Chemical Shift - 13C-NMR • Trends • RCH3 < R2CH2 < R3CH • Electronegative atoms cause downfield shift • Pi bonds cause downfield shift • C=O 160-210 ppm

  36. Chemical Shift - 13C-NMR Chemical Shift Range of 13C Note the carbonyl range

  37. Carbonyl Carbon Chemical Shifts anhydrides nitriles acid chlorides amides esters carboxylic acids aldehydes conj. Ketones ketones

  38. Alkane: 2-methylpentane

  39. Alcohol: 2-hexanol

  40. Alkyl Halide: 3-bromopentane

  41. Alkene: 1-hexene

  42. Aromatic Ring: eugenol

  43. Carboxylic Acid: pentanoic acid

  44. Ester: ethyl valerate

  45. Amide: pentanamide

  46. Ketone: 3-methyl-2-pentanone

  47. Aldehyde: 2-methylpentanal

  48. Carbon-13 NMR Spectrum of Geraniol ppm Carbon #139.07 1 131.62 2 124.07 3 123.71 4 59.16 5 39.64 6 26.51 7 25.66 8 17.66 9 16.24 10 8 9

  49. Spin-Spin Coupling in 13C NMR Homonuclear coupling of 13C-13C is possible in theory. However, due to the low natural abundance of 13C, it is rare to find two 13C’s in the same molecule, let alone adjacent to one another. No need to consider 13C-13C coupling except for enrichment studies! Heteronuclear coupling between 13C and the 1H atoms attached to them is observed (1H abundance ~99%). Because the 1H atoms are directly attached, the coupling constants (1J)are large, typically 100-250 Hz. When such spectra are observed, they are referred to as proton coupled spectra (or non-decoupled spectra).

  50. Carbon-13 NMR Spectroscopy Where’s Waldo?