Nuclear magnetic resonance partial lecture notes

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  • Resonate – absorb at the transition frequency.

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  • 1. MC-511, Part 3Nuclear Magnetic Resonance (NMR) Partial lecture notes 1
  • 2. Suggested reading material…• Spectrometric identification of organic compounds by Silverstein, R.M.; et al (the Bible of spectroscopy, with some good problems at the end).• Modern NMR spectroscopy, a Guide for Chemists, by Snyder (a highly advanced text book, but very helpful).• http://orgchem.colorado.edu/hndbksupport/spect.html - a website run by University of Colorado, which gives detailed theory and some good problems in NMR and IR.• http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/h1/h1_fs_soln.h tml - more problems.• http://www.chem.ucla.edu/cgi-bin/webspectra.cgi - some more problems! Note: This list will be updated! 2
  • 3. Spin I…• Electrons – spin about their own axes – Spin quantum number of + ½ or – ½. – Effect of electron spin – magnetic moment, also called a magnetic dipole (direction?) – Remember – a charged body spinning about its own axis generates a magnetic dipole (moment) along its axis.• Similarly, some nuclei (not all!) have spin! – Examples: 1H, 13C, 19F, 31P, etc. – Nucleus – positive charge; hence spinning charge generates a magnetic dipole. – Hence, each nucleus acts as a tiny magnet. 3
  • 4. Spin II…• Spin (nuclear or electronic) determined by the spin quantum number, S. – The rules for determining net spin: • If # of neutrons and protons are even – no spin. • If # of neutrons + protons is odd, the nucleus has a half integer spin (1/2, 3/2, 5/2…) • If # of neutrons + protons is even, the nucleus has an integer spin (1, 2, 3…) – Number of spin states (or overall spin) given by formula (2S + 1). – Examples: consider a hydrogen atom, 1H. • Only one proton in the nucleus • Hence, sum of protons and neutrons = 1 • Nuclear spin = 1/2. • Number of spin states = 2(1/2) + 1 • +1/2 and -1/2 4
  • 5. Spin III – For 2H, S = 1; # of spin states = 3 – Nuclei with S = 0 are not NMR active (examples, 12C, 18F, 18O, etc.)• Spin states: – A nucleus of spin ½, can have two possible orientations. – In absence of a magnetic field, these orientations will be of equal energy. – In a magnetic field, the magnetic moment created by the spinning charge can line up with or against the field. – Alignment with the magnetic field is a lower energy state ( state) than against ( state). – Difference in the energy between these two states depends on the strength of the applied magnetic field. – Population of energy levels governed by Boltzmann distribution: there is always a finite excess of nuclei in the lower energy state than in the higher energy state. 5
  • 6. NMR phenomenon I…• Imagine a nucleus of spin ½ in a magnetic field, in the lower energy level.• In a magnetic field, the axis of rotation of the nucleus will precess around the magnetic field. – Precess – change in orientation of the rotation axis of a rotating body. 6
  • 7. NMR phenomenon II…• If energy is now absorbed by the nucleus, the magnetic moment is now „flipped‟ so that it now opposes the applied field (higher energy state) – resonance!• This absorbed energy depends on the applied magnetic field – quantized! 7
  • 8. Transition energy I…• Magnetic moment of the nucleus is proportional to its spin, S. – Where, = magnetic moment, = „magnetogyric‟ or „gyromagnetic‟ ratio, a fundamental nuclear constant, dependent on nucleus. h = Planck‟s constant• Energy of a particular energy level is; B – Where, B is strength of magnetic field at the nucleus (not equal to the applied field). 8
  • 9. Transition energy II…• The difference in energy levels (transition energy) can be obtained from :• Features of the equation: – If magnetic field, B increases, E increases. – Greater the , greater E. – is the ratio of magnetic moment to angular momentum. – If increases, precession increases. – Hence, greater energy required for the „flip‟. 9
  • 10. Relaxation I…• Only a small proportion of nuclei in the state can get excited and absorb radiation.• At some point, the population in the and states become equal.• No further absorption of radiation – saturated spin system.• Relaxation – return of nuclei to the lower energy state. – Spin-lattice relaxation: • The NMR sample – called as lattice. • Nuclei in the lattice are in vibrational and rotational motions, giving rise to magnetic field, called lattice field. • If nuclear precession frequency is equal in phase and frequency to lattice field, the nucleus in the state can transfer its energy to lattice and return to state. • Results in a slight warming of the sample. 10
  • 11. Relaxation II… – Spin-spin relaxation: – Interact with neighboring nuclei with identical precession frequencies. – However, nuclei in both states can interact! – No net change in populations, but lifetime of a nucleus in the state will decrease – line broadening in the spectrum – not good!• Relaxation time T1: – Average lifetime of nuclei in the higher energy state. – Depends on of the nucleus and mobility of the lattice. – As mobility increases, vibrational and rotational frequencies increase, increasing the probability of interaction with excited nuclei. 11
  • 12. Chemical shifts I…• Magnetic field at nucleus is not equal to the applied field.• Electrons around the nucleus shield it from the applied field.• Nuclear shielding – difference between applied magnetic field and field at the nucleus.• Consider s-electrons in a molecule: – Symmetry? – Circulate in the applied field – produce a magnetic field, which opposes the applied field. – Applied field strength must increase, for the nucleus to resonate. – Upfield shift, also called as diamagnetic shift. 12
  • 13. Chemical shifts II… – If the electron density around the nucleus is reduced considerably (how?), applied field strength must decrease for resonance. – Nuclear deshielding, also called downfield shift. – Electrons in p-orbitals have no spherical symmetry. – They produce comparatively large magnetic fields at the nucleus. – Deshielding or paramagnetic shift.• In proton NMR, p-orbitals play no part (why?). – Small range of chemical shift (10 ppm) observed. – Effect of s-electrons on chemical shift – look at substituted methanes. – CH3X – as X becomes more electronegative, what happens to: • Shielding? • Chemical shift? 13
  • 14. Chemical shift IV…• If two scientists want to compare data using two different field strengths, that correction has to be applied.• Hence, chemical shifts!• Definition: nuclear shielding in an applied magnetic field.• A function of the nucleus and its environment.• Measured relative to a reference compound.• For 1H and 13C NMR, usually use TMS (Me4Si) as internal standard.• We will see later about use of NMR solvents as internal standards. 14
  • 15. Chemical shift VI…• Information on what kinds of protons are present in the molecule. – Aromatic, aliphatic, primary, secondary, tertiary, vinylic, allylic, benzylic, acetylenic, adjacent to halogens or hetero atoms, etc.• Shifts observed due to shielding and deshielding of nuclei.• Circulation of electrons nearby can generate induced magnetic field (what kind of compounds?) that can either reinforce or oppose the applied magnetic field: – When induced magnetic field opposes – shielding. – Shielded protons require higher applied field for resonance – upfield shift. – When induced magnetic field reinforces – deshielding. – Deshielded protons require lower applied field for resonance – downfield shift. 15
  • 16. Measurement of chemical shift…• Denoted by , units are ppm.• Reference material: TMS (tetramethylsilane)• Why TMS? – Inert, wouldn‟t react with the sample! – All equivalent protons, so only one signal. – Volatile (b.p. = 27 0C). – Soluble in common organic solvents. – Can be used as an external standard when using D2O as solvent. – Low electronegativity of silicon, hence highly shielded protons. – Most compounds absorb downfield to TMS. – Large number protons means only a drop of standard is required. 16
  • 17. Calculation of chemical shift…• Denoted by units are ppm.• If a proton absorbs at 60 Hz in a 60 MHz instrument, = 1 ppm• is independent of operating frequency of instrument. The same signal above will absorb at 100 Hz in a 100 MHz instrument. Thus, = 1 ppm• Remember, depends on the environment of the nucleus!• All these calculations are done by the computer. 17
  • 18. Magnetic anisotropy I…• Anisotropy – non-uniform.• Non-uniform magnetic field.• Recall: – Circulation of electrons nearby can generate induced magnetic fields that can either reinforce or oppose the applied magnetic field• Nearby protons can experience 3 fields: – Applied field – Shielding field of the valence electrons – Field due to the system• Depending on the position in the third field, the proton can be: – Shielded (smaller ) – Deshielded (larger ) 18
  • 19. Magnetic anisotropy II…• Acetylene: – Shape of molecule? – Triple bond symmetrical about the axis. – If axis is aligned with the magnetic field, electrons of the triple bond circulate perpendicular to applied field. – Induce their own magnetic field, opposing the applied field. – Protons lie along the magnetic axis – the induced field shields them. – Hence, for acetylenic protons are more upfield than expected. 19
  • 20. Magnetic anisotropy III…• Now, let‟s look at benzene: – “Ring current effect” in play here. Field lines aligned with applied field – So, aromatic protons are deshielded, more downfield then expected and hence larger . – A proton held directly above or below the ring would be heavily shielded. 20
  • 21. Magnetic anisotropy IV…• Finally, let‟s look at ethylene: – Double bond oriented perpendicular to the applied field. – electrons circulating at right angles. – Induced magnetic field lines are parallel to the external field at the location of the alkene protons. – Hence, downfield shift. 21
  • 22. Number of signals…• Indicates the kinds of protons in the molecule.• Equivalent protons: protons in the same environment.• Non-equivalent protons: protons in different environments.• Example and some problems: Cl H * * Cl H O F O 22
  • 23. Kinds of protons I…• Homotopic protons: – Proton, when substituted by a deuterium, leads to the same structure. – Always equivalent, and will give one signal in the NMR spectrum. Ha D Ha F F F Hb F Hb F D F Replace Ha Replace Hb Same compounds, hence Ha and Hb are homotopic. 23
  • 24. Types of protons II…• Enantiotopic protons: – Proton, when substituted by a deuterium, leads to a pair of enantiomeric structures. – Appear to be equivalent and usually, give one signal. – In a chiral environment, can be made non-equivalent and give two signals. Ha D Ha F F F Hb Cl Hb Cl D Cl Replace Ha Replace Hb Enantiomers, hence Ha and Hb are enantiotopic. 24
  • 25. Kinds of protons III…• Diastereotopic protons: – Proton, when substituted by deuterium, leads to a pair of diastereomeric structures. – Not equivalent, and usually, give two signals in the spectrum. Ha D Ha F F F Hb R* Hb R* D R* R* is a chiral center, which Replace Ha Replace Hb undergoes no change. Diastereomers, hence Ha and Hb are diastereotopic. 25
  • 26. Splitting I…• Just chemical shift information alone wouldn‟t be useful.• Splitting of peaks is what adds extra value to NMR.• Splitting due to 1H-1H coupling, also called spin-spin coupling or J coupling.• How does it work? – Imagine a molecule with two different protons, HA and HB. – How many signals would you expect? – HA feels the presence of HB and vice-versa. – Recollect, these protons are tiny magnets, oriented with or against the applied magnetic field. – When HB reinforces the magnetic field, HA feels a slightly stronger field; when it opposes the applied field, HA feels a slightly weaker field. – So, we see two signals for HA. – Same can be applied for HB. 26
  • 27. Splitting II…• How does it work? – Overall, we see two „doublets‟ for the two kinds of protons. – So, when there is only one proton adjacent, we see 2 peaks due to that proton.For this line, H B is lined up For this line, H B is lined up w ith the magnetic f ield against the magnetic f ield (adds to the overall (subtracts from the overall magnetic field, so the line magnetic field, so the linecomes at higher frequency) comes at low er f requency) HA HB C C HA HB HA is split into tw o lines because HB is split into tw o lines because it f eels the magnetic field of H B. it f eels the magnetic field of H A. 27
  • 28. Splitting III…• When there is more than one proton in the neighboring carbon? – More lines! – Consider Cl2CHCH2Cl. – Look at CH – it „feels‟ the two protons from CH2. – The 2 protons: • Both are aligned with the field • Both oppose the field • One proton is aligned and the other is against the field • The reverse of the above case – Because the two protons in CH2 are the same, the last two cases add up. – Hence, CH has three lines in the ratio of 1:2:1 (triplet). 28
  • 29. Splitting IV… Applied field Signal from uncoupled proton Ho Spin combinations for adjacent -- CH2Spin-spin coupling: coupling with two protons give a 1 : 2 : 1 triplet. 29
  • 30. Splitting V…• Generally, the “n+1” rule is followed.• If there are “n” (equivalent) protons in the neighboring carbon(s), the proton of interest will be split into n+1 peaks.• Intensity of the lines is dictated by “Pascal‟s triangle”.• Example: – Doublet – 1:1 – Triplet – 1:2:1 – Quartet – 1:3:3:1 – Pentet – 1:4:6:4:1• Applies only to simple systems!• Most „real world‟ systems are much more complex! 30
  • 31. Pascal‟s triangle… 31
  • 32. NMR spectrum of ethanol… 32
  • 33. NMR spectrum of ethyl acetate… H O H H H H O H H H 33
  • 34. Coupling constants I…• The line separation within a given multiplet is the coupling constant.• Measure of interaction between a pair of protons – structural information!• Indicated by J.• Units of J is Hz, and is magnetic field independent.• Example: 34
  • 35. Coupling…• Interaction between „related‟ protons.• Three types:• Vicinal coupling• Geminal coupling• Long range coupling 35
  • 36. Vicinal coupling…• Denoted as 3J.• Coupling transmitted through three bonds.• Magnitude depends on the dihedral angle: – Maximum (about 16 Hz) at 180 0 – About 10 Hz at 0 0 – Minimum (close to 0 Hz) at 90 0. – To get the dihedral angle, draw the Newman structures and compute the angle. – Staggered conformation – 60 or 180 0; eclipsed – 0 0; Gauche - ~ 60 0. 36
  • 37. Vicinal coupling II… CH3 H CH3CH3H H H H H H H H H H H H H CH3 HH H H HH Br H H H H Br Br Br 37
  • 38. Karplus curve…• Variation of coupling constant with change in dihedral angle.• Remember: these values are approximate! 38
  • 39. Vicinal coupling – problem…• Based on the Karplus curve, predict the approximate coupling constants of the indicated protons in the following molecules: – Trans-1,2-dimethylcyclohexane. Assume that the two methyl groups are axial. Predict 3J between hydrogens on C1 and C2. – Trans-1,2-dimethylcyclohexane. Assume that the two methyl groups are equatorial. Predict 3J between hydrogens on C1 and C2 – Cis-2-butene – 3J between hydrogens on C2 and C3. – Trans-2-butene – 3J between hydrogens on C2 and C3 39
  • 40. Geminal coupling…• Denoted as 2J.• Coupling between protons on the same carbon.• Note: these two protons must be non-equivalent!• Again, value of J depends on H-C-H coupling.• Normal values – about 10 – 18 Hz; at about 125 0, 2J = 0; maximum at about 100 0, 2J = 35 Hz.• Particularly important in terminal vinyl systems. a H b H 40
  • 41. Long range coupling…• Coupling beyond three bonds (> 3J).• Normally, observed up to 4 – 5 bonds.• With polyalkynes, this can be observed as far as 9 bonds!• Typical coupling constants are in the range 0 – 4 Hz.• Two types: – Allylic coupling – W coupling 41
  • 42. Allylic coupling… c H a b H H a H a H• Why are Hb and Hc non-equivalent? – 4Jab = 3 Hz – 4Jac = 3.5 Hz 42
  • 43. W coupling… a H H a H a H b b O H H H b H4J ab (meta) = 1 – 3 Hz 4J ab = 0 – 2 Hz Bicyclo[2.2.1]hexane 4J = 7 Hz ab 43
  • 44. More couplings…• Consider 1,1,2-trichloropropane: H Cl Cl H Cl H H H – Look at the proton on C2. – Expected splitting pattern is pentet (why?). – Can get a pentet, only if J1-2 and J2-3 are identical. – Actually, get a quartet of doublet (why?). – The larger coupling is mentioned first. – Denoted as: 4.30, 1H, J = 6.6, 3.8 Hz. 44
  • 45. 1,1,2-trichloropropane… 45
  • 46. Coupling tree… 46
  • 47. Another example…• Consider trimethylsilyl ethylene: SiMe 3 H H H – Ignore the methyls in the silyl group. – How many kinds of protons in the double bond? – How are they split? – Approximate coupling constants? 47
  • 48. Integration…• The area under each peak is obtained by integration of the signal.• Proportional to the number of hydrogen nuclei giving rise to the signal.• Sometimes, integral shown as a step function at the top of each peak, with the height proportional to the area.• Error in integration can be high – up to 10 %; depends upon instrument optimization.• Usually, all integration done by the instrument / computer.• Normalized values are shown.• Integration gives a measure of the proton count, adjusted for molecular symmetry. 48
  • 49. Molecular symmetry I…• Consider the spectrum of 2-butanone: – Symmetry? – Can get the actual proton count: 3 + 2 + 3. 49
  • 50. Molecular symmetry II…• Consider diethyl ether, CH3CH2OCH2CH3, a total of 10 protons. – Symmetry? – Two peaks in the ratios 3 : 2. 50
  • 51. Leaning of peaks I…• Consider ethanol (again!) 51
  • 52. Leaning of peaks II…• Triplet not a „perfect‟ triplet; quartet not a „perfect‟ quartet.• Coupled peaks lean towards each other.• Sometimes, helpful in complex systems. 52
  • 53. Advantage of higher field I…• Separation of different sets of protons is proportional to field strength.• However, coupling constants do not change!• Consider spectrum of benzyl alcohol, recorded at 90 and 400 MHz.• 90 MHz: – A broad strong signal at d = 7.24 ppm; characteristic of aromatic protons. – Chemical shifts of the 5 protons are identical; no spin coupling is observed.• 400 MHz: – The aromatic peaks are more dispersed. – Spin coupling of adjacent protons are now seen. 53
  • 54. Advantage of higher field II…http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr2.htm 54
  • 55. Advantage of higher field III…• In an instrument of field strength X MHz, the distance between two units on the scale equals X Hz.• Example: – In a 90 MHz instrument, this difference is 90 Hz. – In a 400 MHz instrument, this difference is 400 Hz.• However, J and remain the same!• Hence, greater field strength translates to greater dispersion. – Dispersion – resonances with different chemical shifts are further apart. 55
  • 56. Structural elucidation…• To determine structure – a suggested approach: – Usually, molecular formula, IR, NMR and MS information will be given. – For now, only molecular formula and NMR! – Calculate the degree of unsaturation from the molecular formula. – Look at the NMR spectrum to determine the connectivity. – Draw some possible structures and see if they “work” with the data given. – Approach this as a jigsaw puzzle, where you have all the pieces of information – just need to put them together in the correct order! – It is a lot of fun! 56
  • 57. Degree of unsaturation I…• Also known as index of hydrogen deficiency.• Can determine the number of rings, double or triple bonds…• Doesn‟t give the exact number rings / double / triple bonds.• Sum of number of rings and double bonds + twice the number of triple bonds.• Formula: – where, • H = # hydrogens • X = # halogens • N = # nitrogens 57