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!['W' coupling…
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ab (meta) = 1 – 3 Hz 4J
ab = 0 – 2 Hz Bicyclo[2.2.1]hexane
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Nuclear magnetic resonance partial lecture notes
- 1. MC-511, Part 3 NuclearMagnetic 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… • Onlya 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 chemicalshift… • 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 chemicalshift… • 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 protonsI… • 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 protonsII… • 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 protonsIII… • 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… • Justchemical 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… • Howdoes 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 line comes 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… • Whenthere 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 -- CH2 Spin-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.
- 32. NMR spectrum ofethanol… 32
- 33. NMR spectrum ofethyl 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… • Denotedas 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 CH3CH3 H 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… • Variationof 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… • Denotedas 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 H 4J ab (meta) = 1 – 3 Hz 4J ab = 0 – 2 Hz Bicyclo[2.2.1]hexane 4J = 7 Hz ab 43
- 44. More couplings… • Consider1,1,2-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.
- 46.
- 47. Another example… • Considertrimethylsilyl 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 areaunder 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 peaksI… • Consider ethanol (again!) 51
- 52. Leaning of peaksII… • 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 higherfield 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 higherfield II… http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr2.htm 54
- 55. Advantage of higherfield 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… • Todetermine 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 unsaturationI… • 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
Editor's Notes
- #13 Resonate – absorb at the transition frequency.