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Chem 263 Sept. 6 , 201 6 Nuclear Magnetic Resonance (NMR) Spectroscopy Light and Energy E = hv = h c / λ Where: E = energy v = frequency c = speed of light = 3 x 10^8 meter/second h = Planck’s constant = 6.6 x 10-^34 Joule•second λ = wavelength Electromagnetic Spectrum Rotation High E High v Short! Low E Low v Long! gamma rays x-rays ultra violet visible infrared microwave radiowaves 4000 Å 8000 Å Diffraction Absorption Vibration Rotation NMR λ Short Wavelength, High Energy, High Frequency λ Long Wavelength, Low Energy, Low Frequency
Nuclear Magnetic Resonance (NMR) If an atom has an odd number of protons and/or an odd number of neutrons, there is a nuclear magnetic moment (aka nuclear spin ). Examples:
- 1 H has atomic number 1 (therefore 1 proton), no neutrons, and nuclear spin ½; can be detected by NMR
- 2 H (deuterium) has atomic number 1, number of neutrons 1, and nuclear spin 1
- 3 H (tritium) has atomic number 1, number of neutrons 2, and nuclear spin ½
- 12 C has atomic number 6 (6 protons), number of neutrons 6, and no nuclear spin
- 13 C (only 1% of all carbon) has atomic number 6, number of neutrons 7, and a nuclear spin of ½ (and can be detected by NMR)
- 14 C has atomic number 6, number of neutrons 8, no nuclear spin, radioactive and used for “carbon dating” MRI (Magnetic Resonance Imaging): Based on the principles of NMR and developed by Paul Lauterbur NMR Continued (focusing on 1 H NMR) An NMR experiment requires a magnet (e.g.14 Tesla), and a radio frequency transmitter. A tube containing the sample is placed inside the magnet – Below is a crude drawing of the setup. (http://www.chemistry.ccsu.edu/glagovich/teaching/316/nmr/instrumentation.html)
downfield deshielded upfield shielded CH 3 H 3 C Si CH 3 CH 3 R H O H R C H O R C H N CH 2 CH 3 (^109876543210) = TMS (ppm) C (^) C H Downfield Deshielded Upfield Shielded Examples: (^1) C C 2 H H CH 3 3 CH 3 2 - methyl- 1 - propene The intensity (area under the peak) depends on the number of hydrogen atoms, in this case 2:6 which is the same as 1: 1 - chloroethene chloroethylene vinyl chloride Cl 5 1.5 ppm Intensity: 1 (2H’s) 3 (2 CH 3 ’s) HA HC HB HA is deshielded by the electron-withdrawing Cl. HB and HC are different due to the distance from the Cl. HB is trans to the Cl, while HC is cis. HC is more deshielded than HB because it is closer to Cl. 10 8 6 5 0 ppm C C HB HC Cl HA
Diastereotopic Hydrogens C 2 (S) C 3 H 3 C 1 CH 3 4 H HA HB HO (S)- 2 - butanol butan- 2 - ol The methyl group shows only one signal because the C-C bond can freely rotate and NMR cannot distinguish between the Hs on that carbon. The barrier to bond rotation is only a few kilocalories per mole (1-3 kcal/mole) and about 15 to 20 kilocalories per mole of energy are available at room temperature. Hence rotation is very rapid and the methyl hydrogens are all in the same chemical environment. C-2 is a stereogenic centre, and there is no plane of symmetry in the molecule. HA and HB are in different chemical environments and have different chemical shifts, therefore they are diastereotopic. If you substitute HA or HB with deuterium (labeled as D (^2 H; an isotope of hydrogen with one proton and one neutron)) you get diastereomers. These two hydrogens are called diastereotopic hydrogens. Note: The chemical shifts of two diastereotopic protons may not always be very different. C 2 C 3 H 3 C 1 CH 3 4 H HA D HO C 2 C 3 H 3 C 1 CH 3 4 H D HB HO diastereomers SS isomer SR isomer Example: OH (R)- 2 - dodecanol Every CH 2 group in this molecule has two diastereotopic protons. What kinds of H does it have? O-H H at C 2 CH 3 at C 1 CH 3 at C 4 HA (Pro-R) HB (Pro-S)
Bo HA HA HB spin J AB = Coupling constant between atoms A and B, measured in Hertz (Hz) The HA coupling constant corresponds to the energy differences caused by the HB proton in alignment with and in opposition to the field. This energy difference is equal to the effect that the HA proton has on HB, so both of these coupling constants are equal. The coupled signal is centered around the same chemical shift value that the signal would have if there were no coupling. Limitations of Coupling
- 2 to 3 bonds separating nuclei (coupling is typically not seen between nuclei that are further apart than 3 bonds)
- Usually no coupling across O, N, S, C=O Example: In the following structure, HA and HB would not couple due to being 4 bonds apart (too far) and oxygen is in between C O C HA HB Note: There are exceptions to these rules
