New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Pulsed NMRSpectroscopy Theory and Practice

by: Francisca Sipes DVM

Pulsed NMRSpectroscopy Theory and Practice CHEM 781

Marketplace > University of Wisconsin - Milwaukee > Chemistry > CHEM 781 > Pulsed NMRSpectroscopy Theory and Practice
Francisca Sipes DVM
GPA 3.94


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Chemistry

This 45 page Class Notes was uploaded by Francisca Sipes DVM on Tuesday October 27, 2015. The Class Notes belongs to CHEM 781 at University of Wisconsin - Milwaukee taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/230298/chem-781-university-of-wisconsin-milwaukee in Chemistry at University of Wisconsin - Milwaukee.


Reviews for Pulsed NMRSpectroscopy Theory and Practice


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/27/15
Lecture notes Part 4 SpinSpin Coupling F Holger Fb39rsterling September 29 2008 So far spins were regarded spins isolated from each other However spins do have effect on neighboring spins Direct through space Energy of spin A will depend on orientation of B relative to A The interaction will depend on distance A B and the angle of the vector A B relative to the field gure 41 H H L to B0 parallel arrangement higher energy H H B0 parallel arrangement lower in energy A A 3 h 3cos26 1 3 V V0 izjAjB4 2 r23 V0iEDAB 41 In solution fast tumbling will average over all orientations gt Dave 2 0 Dipolar coupling not directly observable in solution but important for T1 T2 relaxation NOE Broad lines in 1H NMR of larger proteins arise from short T2 due to insufficient averaging F Holger F rsterlingS eptember 2 9 2 008 1 Indirect scalar coupling spin spin interaction mediated by electron electron density at nucleus F ermi contact interaction consider a P H bond 1H 31P One electron will be closer to H the other closer to P but always with anti parallel orientation of spin Pauli principle The nucleus will prefer to be anti parallel to nearby electron spin note that e has negative charge and the magnetic moment will be aligned parallel thus anti parallel orientation of 1H and 31P will be lower in energy than the parallel arrangement 31 r P I 31P12 31P12 I l 5 v 10 Jgt0 gure 42 Scalar coupling through one bond all YN gt 0 Antiparallel orientation of nuclei is lower in energy gt J gt 0 00 H 7 5 v gure 43 Scalar coupling through two bonds electrons in neighboring orbitals tend to approach each other with equal spin Hunds rule Parallel orientation of nuclear spins lower in energy gt J lt 0 F Holger F rsterlingSeptember 29 2008 2 This interaction does NOT depend on the orientation of the A B vector to the field hence the name scalar coupling Dependence of spinspin coupling 0n structure rst Order splittings Only if the chemical shift difference is large compared to the coupling constant can the splitting be analyzed by first order rules Equivalent spins In the other extreme coupling between equivalent spins does not give rise to a splitting Higher order spectra If the difference in chemical shift is similar to magnitude of coupling higher order effects are observed One can intuitively understand that there has to be an intermediate regime between two equivalent and two very different spins As Avo is measured in Hz and thus depends on the magnetic field BO for the same ppm separation a higher order spectrum Avo JAB can transform into a first order spectrum AvO gtJAB when using a stronger magnet AV0 gtgt JAB Av0 JAB A gure 44 Spectra of two coupled protons with very different top similar middle and identical bottom shifts F Holger F brsterlingSeptember 29 2008 Multiple coupling partners If a nucleus is coupled to several different nuclei coupling patterns will superimpose CHE CHA CHC H A coupled to HB and HC lets assume JAB gt JAC VOA J I doublet of doublets in general up to 2 lines are possible if all coupling different special case JAB JAC same as CH2 group v0A Coupling to a group of equivalent spins Suppose a spin with a group of n neighbored spins I l 2 i Splitting into 2 n 1 1 l 3 3 l for I 12 intensities are given by Pascals Triangle statistics Amt 05 gt gt Note that for l 4 12 intensities will be different see multiplets of deuterated solvents in 1H and 13C spectra Note that this is just a special case of multiple coupling partners All first order coupling couplings patterns can be regarded as a a superposition of individual doublets F Holger F brsterlingSeptember 29 2008 4 Example CHCH2 group proton coupled to two equivalent nuclei The two nuclei allow four combinations of their spin states two of which are equivalent spin states 12 12 122 12 1212 1212gt 1 2221 triplet Example CHCH3 group spin states12 12E gt gt gt gt gt gt 1232321 quartet total spin of CH2 group Intensity of line N number of equivalent combinations of spin states 1 1 0 2 1 1 proton coupled to three equivalent nuclei total spin group Intensity of line N number of equivalent combinations of spin states 3 2 12 3 12 3 1 32 Exercise Sketch the 31F spectrum of PF6 F Holger F brsterlingSeptember 29 2008 Coupling t0 nuclei with I gt 12 important to understand spectra of deuterated solvents 13C spectrum of CDCl3 one coupling partner of 11 mI intensity I 1 1 0 1 1 1 v0 13C spectrum of DMSO df7 CD3 SO CD3 three coupling partners with 11 mI It0t Intensity 1 1 1 3 11010 10 1 1 2 3 111111 1110 0 10 1010 0 1 6 0 0 011010 10 11 110 10 10 11 0 7 111 1111110 0 10 10 10 0 1 6 110 10 10 11 2 3 111 3 1 Note that proton signals of deuterated solvents arise from partially protonated isotopomers The proton spectrum of deuterated DMSO will be that of CD3 SO CHD27 so there will be two coupling partners with I 1 Exercise Sketch the 19F spectrum of BF4 and Sng F Holger F brsterlingSeptember 29 2008 6 Magnitude of coupling constants does NOT depend on field B0 and is thus measured in units of Hz proportional to y values of involved nuclei 7 JCD D JCH 42 7H Therefore C D couplings in deuterated solvents are scaled compared to the ones in the equivalent protonated compound 1JHH dihydrogen H H 276 Hz gt How would one measure this 7 for dihydrogen complexes M n 2 H2 somewhat lower values are observed 220 140 Hz 1JCH depends on s orbital contribution to the bond In hydrocarbons one finds J 500 Hz s where s 025 for sp3 033 for sp2 and 05 for sp orbitals RCH3 CRH H ECH 125 Hz 140160 Hz 250 Hz Substituents have an inductive effect EN substituents cause an increase in JCH CH3 Li CH3H CH3 C1 CH2C12 CHC13 CHCF3 98 125 150 178 205 239 H2CCHX X H CH0 CN C1 F 1564 1623 178 195 200 Effect of nitrogen lone pair in imines and similar compounds 1JCH trans lt 1JCH cis with respect to the lone pair H3C H C2 0 177 Hz Ysz 163Hz OH H H3C 0H F Holger F brsterlingSeptember 29 2008 7 1J3 depends on hybrrdrsatron of connectrng Cratom H R H png V H H 710 7 16Hz 7472Hz 7372Hz 3J3 The most important one in structure detemunation H R Depends on drhedral angled 4800 600 D ZCI 0 60 BOIOO IZO 140 160180 theorellcal curve an experlmenlal range ofdata Figure 45 3I AB 0054 C 00524 A B C calculated for HCVCH fragment 7028 705 98 emprrrcal to t experrmental data 2 71 10 n r r r r r for other mfluences on couphng constant bond order substrtuent effects F Halger FmterlmgSeptember 29 2008 3JHmHN in proteins 19 l4 64 3JHapr in proteins 18 16 95 Always 3J 0 lt3J 180 H v gt in olefins3J lt3 H cis trans H H JHH7712H2 3JHH12718 Hz In cyclohexane chair conformation coupling between two axial H1 protons larger than between equatorial or axial with equatorial protons H3H2 31M gt 313e 3167e 21172 145 Hz 31173 22 Hz 31174 106 Hz H4 31273 56 Hz Hb Hb Cyclopropanes bums 0quot has 135 gt H6 3 3 a He Jcis gt Jtrans HG In exible aliphatic chains fast conformational interconversion occurs rotation about C C single bond gt average coupling of 7 8 Hz observed Similar considerations also apply to other couplings over 3 bonds 3JCH 3JPH 3100 etc F Holger F brsterlingSeptember 29 2008 9 Long range couplings presence of double bonds andor favored conformations in rings required39 allylic naphthalene or condensed aromatics selected ring systems H H H CH3 gt lti R H HAG H O H 4 4 4 JHH1Hz JHH OMHZ JHH0l5 1R1Hz H 7R X H H N R 5 R JHH 05 1Hz 5JHH 2 25 Hz X OSNH F Holger F rsterlingS eptember 2 9 2008 10 Extended Vector picture to describe couplings The two multiplet lines can be described by two vectors rotating in opposite directions assuming 90 1H If cnc D 13C1H A JCH 39 C Vo gure 46 Description of a doublet using an extended vector picture Using the relation IXCIBH IxclaH Lied H 13H 2 IXCIZH rationalizes the product operator for non observable anti phase magnetization Separation of these vectors into its X and y components yields the corresponding rotation equation for coupling assuming M M0ch at t 0 I M M0 IyC cos7zlt 1515 sin7 lt 43 Summary three basic rotations describe most N MR experiments Pulses rotation about x or y axis with frequency 01 chemical shift rotation about z axis with frequency Q scalar coupling rotation about IZAIZB interconversion of IyXAand IXmAlZB with frequency 1T AB F Holger F rsterlingS eptember 2 9 2008 1 1 LECTURE NOTES CHEM 781 PART 8 NOE and through space correlation November 17 2008 81 Introduction All 2D methods discussed so for utilized scalar couplings to establish through bond connectivities gt determination of 2D structure magnitude of scalar coupling could be used to determine dihedral angles and establish configuration and conformational information For many stereo chemical problems groups more than three bonds apart have to be related For example to as sign the two CH3 groups of camphor coupling constants are not useful Through space interaction direct distance information is often needed gt Dipolar coupling does not depend on bonds but only on distance N r 3 However in solution dipolar coupling can usually not be measured directly as it is averaged out by the fast tumbling of the molecule It can be measured by its effect on relaxation and creation of NOE as discussed for decoupled 13C spectra NOE in 13C spectra arising from decoupling of 1H are indicative of spatial proximity of C and H In the same way NOE s measured between protons can be used to assess H H To be useful one has to saturate one proton at a time without affecting the rest This can be achieved by irradiating N02 HA Ho A HB weak pulse at position H yB127 10 30 Hz Ho A 11 2 10 s I I I I reference experiment no B saturation pulse or pulse centered outside spectral region I I I I difference spectrum A B l with very low power WEI2n 10 30 Hz at the frequency of one specific signal for several seconds and then acquiring a spectrum with a high power pulse This results in a spectrum with protons in close proximity to the irradiated proton exhibit increased intensities due to NOE As the effects often are small typically the difference spectrum to a non irradiated is typically determined 82 NOE and relaxation As already mentioned when discussing 1H decoupling of 13C NMR spectra the NOE arises from a change in population caused by relaxation Consider a pair of protons close in space but not necessarily coupled BB 39 1HA HB W0 VAVB W2 ad 1HAHB four transitions can give rise to an N MR signal corresponding to transitions involving ip of one spin single quantum transitions Ifthere is no scalar coupling the frequencies of the two HA and HB lines are identical and two singlets are observed relaxation can occur across six transitions the four single quantum transitions and the 2 double and zero quantum transitions with the transition rates probabilities W1 W0 and W2 The maximum possible NOE is given by the ratio of cross to total relaxation I QL VVZ WO cross relaxation WIA WIB WO VVZ total spin lattice relaxationl 7 81 T1 Relaxation requires the presence of randomly oscillating magnetic fields In solution that occurs if Brownian motion of the molecules modulates the effective field around a nucleus So we need to closer look at two things molecular motion and interactions which depend on the orientation of the molecule anisotropic relaxation molecular motionfrequency 39 interaction changing field 83 Molecular motion and Relaxation The frequency of any random fields is directly related to molecular motion Brownian motion and internal rotation Brownian motion is described by the rotational diffusion coefficient Dmt not to be mixed up with the translational diffusion or the correlation time the time a molecule on average remains in one orientation For spherical molecules the Stokes Einstein equation can be used D L kT rot 67 8m3n 82 solv As one can see its value depends on molecular size given by the volume r3 temperature T and viscosity of the solvent 1 note that the same symbol is used for the NOE don t get them mixed upD represents an average rate of motion of with a certain distribution of frequencies present wt in the ensemble of molecules determined by the Boltzmann distributionThis distribution is described TC byJW 83 J 0 number of molecules at a certain frequency l spectral density D ltlt 10 large molecule low T rot high viscosity D t 10 1 0 Drot gtgt 10 small molecule high T wiscosity l 00 2 The relaxation rates W will be most efficient when the tumbling at the transition frequencies reaches a maximum One can already see that for very fast or very slow tumbling the overall relaxation at frequency 00 will become very inefficient The double quantum transition W2 at o 2100 will be more important for fast tumbling small molecules WO at o 2 DA DB 0 will become important for small molecules so the sign of H H NOE will be different for different size molecule 84 Interactions leading to relaxation The second ingredient to relaxation is an interaction which gets modulated by the motion Some important interactions are Chemical shielding anisotropy see part 3 of lecturenotes important for tertiary carbons dipolar through space nuclear spin spin coupling part 4 of lecturenotes protons C H N H dipolar nuclear spin electron spin coupling only when paramagnetic molecules are present quadrupolar coupling orientation of non spherical nuclei in electric field Only for I gt 12 nuclei but then almost always dominant All these interactions cause the nucleus to experience a uctuating field in the presence of motion 4 even if only a constant average value is observed in solution All interactions will contribute to single quantum transitions at 00 W 1 only dipolar coupling between A and B will cause double and zero quantum transitions W0 and W2 gt only dipolar coupling will create NOE s as a consequence maximum NOE will only be reached if dipolar interaction dominant and other processes negligible problem I gt 12 nuclei paramagnetic impurities like oxygen 841 Dipolar relaxation and NOE The dipolar coupling between two spins depends only on the distance and the gyromagnetic ratio of the spins DAB 2 yA yB FHA 84 From a lengthy and tedious quantum mechanical deviation one can obtain values for the relevant relaxation rates WIA 2 320DABZ JaA W13 2 320 DABZ JaB W0 2 110 DABZ JaA 13 110 DABZ J0 for AB both 1H 85 W2 2 610 39DABZ JaAaB 610 DABZ J2a for AB both 1H Combining that with 81 and 83 and some more algebra for an homonuclear spin system IA 2 1B leads to 5 4wg z39f 86 10 23wfz39f 4w5 rf max nAB Lets consider two extreme cases very small molecules D gtgt 00 or 2300 ltlt 1 nABm 12 mt very large molecules or viscous solutions D ltlt 00 or 2300 gtgt 1 nABm 1 mt somewhere in between there will be a case with 1 ABMquot 0 ll Now lets discuss the apparently ridiculous result that the maximum NOE does not appear on the distance A B anymore which was the main reason for all the trouble For a two spin system it is not the NOE itself which depends on the distance but the rate at which it builds up which is 0 2 W2 W0 Therefore measurement of NOE buildup will be a more appropriate experiment in most cases Meaning if there where only two protons in the universe irradiating one would always eventually result in the same NOE no matter what the distance But if they are 1 1y apart it will take the age of the universe to build up that NOE Usually more than two isolates spins are present and it is the competing effect of different couplings which will lead to different NOE values Only cross relaxation between A and B will contribute to the NOE between the two numerator of 81 all other relaxation including dipolar interaction with other protons will contribute to the denominator only Thus steady state NOE s will have to be treated with caution 6 example linear four spin system steady state NOE can be deceiving for isolated protons as they don t have a competing relaxation mechanism As B is the only proton near A saturation of B will give almost full NOE on A On the other hand saturation of A has barely any effect on B due to the close proximity of C to B which will contribute to the denominator of 81 Reversely saturation of C will have a large effect on B In practice that means that more than one NOE should be considered before interpreting any result Also NOE s to CH3 groups will be weaker than the NOE from the CH3 group as the methyl protons will relax each other so saturating the methyl group will cause to larger NOE 842 Transient vs steady state NOE W2 W0 which is directly related to protonproton distance it is usually more meaningful value to measure This can be achieved by the transient NOE experiment Instead of a saturation one applies a selective 07 As it is the rate of cross relaxation O O 04 08 12 16 2 24 28 180 pulse on one proton resonance waits TmiX 3 for a time 17m and then excites all protons with a 90quot pulse As long rum is kept short enough the NOE intensities between A and B will be proportional to the cross relaxation rate GAB As shown in the figure the NOE A D grows in at a much smaller rate than A B or A C 85 Two dimensional NOESY The 1D transient NOE experiment requires irradiating one signal at a time using a selective pulse In practice often many distances are required gt two dimensional experiment xrxxrx x XXrxrx t1 I 1 I acquisitionx x xx Sequence is identical to DQF COSY with a mixing time 1 added However a different phase cycle is employed selecting the z magnetization present after the second 90quot pulse term I in the description of the COSY experiment lncremented delay t1 will make z magnetization dependent of chemical shift 900X QA tl 900Xselectz lzA gt gt gt IZA cosQA t1 NOEW2W0 lI JAB exchangekAB 17 H gt IZA cosQA t11 1T11m WZWokAB Tm IchosQ t1 Diagonal peak cross peak A further 90quot x pulse converts the z magnetization after 1 into detectable y magnetization The linear relationship towards 1 is only valid as approximation for short values of 1 otherwise a bi exponential function has to be used case 1 CAR dominated by WO large molecules or chemical exchange negative GAB gt Diagonal and cross peak have the same sign both normally phased positive negative NOE gives positive cross peaks case 11 NOE dominated by W2 small molecules positive GAB gt Diagonal and cross peak have opposite sign phase correction typically makes diagonal positive and cross peaks negative positive NOE gives negative cross peaks The two dimensional NOESY experiment will give cross peaks between protons exhibiting NOE between each other and which are therefore in close spatial proximity For short enough mixing times small molecules 400 800 ms large molecules 100 200 ms g 90 9 1 3 y z 2 AZ NOEW2 1 9 a y 39 y 39 1 4quotO o 6 Z A z z the cross peak intensity is proportional to the cross relaxation rate and thus to rab correlation time 17C 2 16Dmt is usually not known or may be different for different parts of the molecule in case of internal rotation Therefore only relative distances can be determined Usually known fixed distances can be used for calibration geminal CH2 aromatic CHCH or qualitative interpretation For small molecules NOE and chemical exchange will give cross peaks of opposite sign and can be distinguished As sequence is identical to DQF COSY coupling can cause artefacts in spectrum These will be anti phase in nature 86 ROESY experiment If the molecular weight is around 1000 g mol 1 Dmt is of the order of the lamor frequency and W0 and W2 become very similar or equal and NOE s will be very weak or zero no matter how close the distance 5 00217024000414 2 0 NOESY experiments will fail in that case The location of the zero NOE point is given by 1001C g and thus depends both on no and 10 For a given molecule both parameters can be varied to reach a region where NOE 0 vary 1C by changing temperature or solvent viscosity This option is often limited due to sample considerations change 00 Using a different magnetic BOfield In practice this option is usually even more limited and extremely weak magnets would give poor sensitivity and resolution However relaxation taking place during a spin lock will only experience the B1 field as effective field The later alterative is employed in the ROESY experiment otating Frame Qverhauser Effect X y I t1 acquisition Spin lock pulse 10 Instead of generating z quotUL f magnetization the y magnetization will be locked along the y axis by the spin 0 lock pulse During that time the spins o5 NOE only experience the B1 field of the spin 39 lock pulse so the effective lamor 2301 01 1 1o 10 frequencies for transitions will be of sunquot 7IO 3 if DISL YHBi WhiCh is typically Figure 447 Nucleu Ovarhnuaexenhnm mmtMormplogledalmnceim of the rotation correlation time T for NOESY mORMHSY d e 1 39 spectrometer equency o 1 an an arpr 3 5 kHz Therefore the condition DI0t m3 a p cnlculaud an distance of 20 A gtgt of or I we ltlt 1 will be always be fulfilled and NOE s will always be positive ROE s will be observable for molecules where NOESY gives no signal also for large molecules or low temperature ROE peaks and exchange peaks will have opposite sign and can be distinguished as ROE will be dominated by W2 861 ROESY versus TOCSY The ROESY experiment utilizes the same pulse sequence as the TOCSY experiment The only difference are the strength and length of the spin lock pulse TOCSY yBlZn 10 kHz IPSL 15 80 ms pulse is too short for effective ROESY ROESY yBlZn 15 kHz IPSL 400 ms field is too weak for effective TOCSY sign of TOCSY peaks is always same as diagonal positive for ROESY peaks it is opposite to diagonal negative Note the TOCSY spectrum of Gramicidin Experiment 4 of lab has some weak negative peaks due to ROESY 11 88 Examples and applications of NOE experiments 881 Assignment of methyl groups in camphor 39 6 A Camphour NOESY W ppm M 036 39 3 z I 078 391 3936 i l 3 080 g aquot quot39 8 082 I 084 39 w a 086 Q 93 I 0 088 090 a a 092 Wquot a a 36 539 9 094 1 1 24 23 212 21 20 19 13 17 16 ppm 2D NOESY spectrum of Camphor CH38 is the one on the side of the CO group as it shows NOE s to H4A whereas CH39 exhibits NOE s to H6A and H7A Also H7A appears to be the exo hydrogen and H7B the endo 12 882 NOE and conformational analysis NOE s can be used to supplement coupling constants in the measurement of dihedral angles For a pair of Vicinal protons coupling and NOE information is exactly complementary in case of staggered conformations but may be used to identify possible eclipsed conformations H1 H1 H1 H2 H23 H2A H213 H2A lt5HZ lt5HZ lt5HZ lt5HZ gt9HZ NOE1 2A strong weak medium NOE1ZB strong strong strong Combination of coupling constant and NOE for three different conformations In many examples Vicinal coupling constants and NOE s alone do not allow the determination of the conformation In the example below amino acid side chain an additional distance HN HB NOE is needed in order to determine the dihedral angle N Coc CB CR and assign the diastereotopic protons of the prochiral CH2 group x 50 180 60 0 II If n H53 Ha GEE N NH CO NH CO NH a CO H33 52 H52 n Has CH2 H H Ha R H R 31 0432 Hz lt4 lt4 gt 10 310413 Hz lt4 gt10 lt4 NOEs aBZaB3 aBZgtaB3 a62ltth3 NOEs NH32ltNHB3 NHBZNHB3 NH52gtNHB3 13 883 Structure determination of Macromolecules NOE39s and coupling constants can be used to determine three dimensional structures of macromolecules like proteins nucleic acids and carbohydrates Typically the sequence of aminoacids or nuclides is known connectivity but the folding three dimensional structure needs to be established and the procedure usually follows the steps listed below C c ccl N 3E mmu y 14 Some special considerations when working with biological macromolecules molecular weight of such compounds is 5000 30000 gmol short T2 gt broad lines experiments generating anti phase cross peaks through small couplings COSY HMBC may fail heavy overlap of lines due to complexity of molecule gt highest magnetic field possible should be used two dimensional experiments reach their limits usually computer software needed to keep track of assignment data 15 Lecture notes Chem 781 fall 2008 F Holger F O39rsterling September 10 2008 1 Basic Theory 11 Definitions Like all other spectroscopic techniques NMR Nuclear Magnetic Resonance spectroscopy involves the interaction of the material being examined with electromagnetic interaction The latter may be characterized by the corresponding range of wavelength A frequency V and energy E A cV hcE h Plank39s constant c velocity of light The main principle of any spectroscopy in order for a particle to absorb a photon of electromagnetic radiation the particle must first exhibit a uniform periodic motion with a frequency that exactly matches the frequency of the absorbed radiation In terms of quantum mechanics the energy of the absorbed photon should be equal to an energy splitting of oscillating particles AE h v 11 v n 39 of Change of 7 Chang of 39 Nuclear Change Of 11110162 Change of Spin Orientation Con guration Change of Electron Distribution configuration orientation A A A A r Visible and nq 1 nmr esr Microwave Infrared ultraviolet Xray Tray 9 0 g O o o O i I f 9 Hy or 0 0 C 5 A K J 10 2 l 100 10 cmquot 10quot wavenumber IO I I I I I I IO m 100 cm 1 cm l00 um I inn 101nm wavelength l00 pm I I I I 4 3 x 10quot 3 gtlt108 3 X10 3 x 10392 3 x10quot HZ 3 gtlt10l6 frequency 3 x10 1 X 10 l l l I l 1 10393 10quot IO i03 105 joulesmole 10quot energy 1039 I I I I I I Figure adapted from C Banwell Fundamentals of Molecular SpectroscopyThird editionMcGraw Hill London 1983 See table of spectroscopic methods at different frequencies yradiation to radio waves Characteristics of NMR NMR employs radio frequencies 60 900 MHZ it is therefore at the low energy end of spectroscopic methods NQR is still lower looks directly at the nucleus change in nuclear spin state That means we can get an atom by atom view of the molecule F Holger FO39rsterling September 10 2008 l requires external magnetic field Magnet is one of the most prominent parts of the spectrometer 12 Spin angular moment Spin of an elementary particle can be visualized as rotation about its axis resulting in an angular momentum J direction of Jaaccording to right hand rule In the case of a charged particle rotation also results in magnetism with a magnetic moment 39 semi classic description Classical electromagnetism Charged particle with angular momentum I has a magnetic moment Example electron in orbital e2me E 12 e elementary charge E orbital angular momentum me mass of electron gt minus sign because of negative charge by defining the gyromagnetic ratio VI e2me one obtains in E 13 Electron spin with spin angular momentum T gvlf Yef 14 g e 20023 In this case the experimental magnetic moment differs by the g factor of the electron from what would be expected by treating the spin just as a rotating charge This can not be derived from classical electrodynamics nor from the Schrodinger Equation Relativistic Schrodinger equation gives exactly 2 QED adds the remaining 00023 Nuclear spin gN e2mpgt 1 w J 15 with mp mass of proton YN is called g omagnetic ratio of the nucleus and is a constant specific for each nucleus Handout of table of nuclei F Holger Forsterling September 10 2008 2 Since mass of a proton n1p much larger than mass of e by a factor of 1861 y is expected to be a factor 1860 smaller than 11B However neutrons and protons are no true elementary particles but are composed of quarks and determination of gN is not straightforward gp 57 for p and gn 39 for neutron resulting in VP 26752239108 rad T ls 1 and y 1867939108 rad T ls 1 The magnetic moment of the neutron can only be understood by its composite nature The more heavy nuclei are composed of protons and neutrons and their total spin angular and magnetic moments are a net effect of many particles Quantum Mechanics of angular momentum Spin angular momentum of an atomic particle has to be treated on a quantum mechanical basis Analogous to the rotation about a free axis the value of the spin angular momentum can only take multiples of Plancks constant determined by the spin quantum number I h J 2 5110 1 hII 1 16 As opposed to molecular rotation each particular particle has one and only one specific spin quantum number which can be integer or half integer i e the angular momentum and overall magnetic moment are are a constant for each particle Examples for particles with different spin quantum numbers I 12 electron proton 1H neutron In 13C I 1 photon y deuteron 2H 11 of photon is important for selection rules I gt 1 7Li 32 170 52 59C0 72 I 0 12C For a list of most important nuclei see handout with table of nuclei with spin and gyromagnetic ratio Spin of complex nuclei Nuclei of all elements are composed of protons p and neutrons 11 both of which have I 12 F Holger Farsterling September 10 2008 3 The total spin of a nucleus can be determined by employing the shell model of the nucleus The following rules apply protons and neutrons pair up separately with reverse Hund s rule as high spin pairing energy results in orbitals filled up one by one we are looking at nuclei not electrons filled shells even number of p or n have zero angular momentum for each type of particle the unpaired spin 12 combines with its integer orbital angular momentum strong spin orbit coupling with the higher angular momentum state considerably lower in energy p and n contributions combine to total spin of nucleus Shell model of the nucleus Energy levels of modified harmonic osszilator note that this is not the coulomb potential used for electron orbits around nucleus E2n 1l n l with spin orbit coupling jzliJz 1 2 1d32 2 1d2s 2 0 2s12 1 2 1d52 1 1p 1 1 1p12 1 1 1pm 0 1s 7 0 0 1s12 p n Three cases can be distinguished even number of both p and n all spins paired l 0 example 12C 6p 6n 160 8p 8n even p 11 odd 11 p1 l n2 half integer 7Li 3 p1 4n 1 32 170 8p 9n gt 1n in 1d52 state I 52 11B 5p6 11 gt 1 p in 1p32 state I 32 19F 39 l3 1 ENE C 6p 7n gt 1 n 1n1p12statel 12 39 15N 7 p1 8n gt 1 pin 1p2 state I V 31131 F Holger Forsterling September 10 2008 4 odd p odd n I n integer 2H 1p 1 n I 1 14N 7 p 7 n I 1 10B 5 p 5n I 3 6Li 3p 3n I 1 U More useful classification for NMR purposes I 0 useless no NMR signal I 12 sharp lines because of spherical symmetry of nucleus the unpaired proton or neutron is in a s type orbital spherical symmetry zero orbital momentum no quadrupole interaction with electric field gradient of surrounding electrons I gt 12 often broad lines due to quadrupole interactions non spherical symmetry of nucleus can interact with non spherical distribution of electrons Most of this course will be about spin 12 nuclei H 13C 13 Spins in magnetic eld In the presence of a magnetic field also the orientation of the angular momentum vector is restricted by an additional magnetic quantum number n1S such that the z component of the vector is a multiple of in hlmz 1 12 F Holger Farsterling September 10 2008 5 J h n1S 17a n1S l l1l 1l z z Yh ms 171 For a proton I 12 that results in two different allowed orientations n1S 12 and n1S 12 and three for deuterium I 1 n1S 1 0 1 or in general 2 I 1 allowed orientations If the z component is given the Xy components will be undetermined and any orientation with that mZ will be possible Precession 0f the magnetic moment vector The magnetic quantum number n1S restricts the component of gt the angular momentum parallel to the magnetic field JZ and therefore z but not the components perpendicular to the gt 17 0 du field That means that the vectors J and u can take any lt gt LL orientation on a cone corresponding to a certain value of ms From a classical point of view the magnetic field tries to align the magnetic moment parallel to the field However like a spinning top or the earth in the suns gravitational field the nucleus is subject to a force perpendicular to 3 0 and 1 1 7B0 x z 18 As a result the nucleus will not align parallel to the field but precess at an angle to the field This is consistent with the quantum mechanical requirement of restricted values of the z component of the angular momentum but no restriction of the X or y component The angular velocity of the precession is obtained as Q 30 7B0 0r 10 2 19 F Holger Farsterling September 10 2008 6 o is the angular velocity in rad s 1 angle s whereas v is the frequency measured in s 1 revolutions s with o 2 1T 1 as one revolution 2 360 2 2n rad The precession frequency 00 is also called the Lamor frequency of a nucleus Exercise Precession of a nucleus in the Earth magnetic field Beam 2 47 39 10 5 Tand the eld of a typical high eld superconducting magnet 1175 T Earth magnetic eld 40 4710395T 39267522 107 rad T1 s 1 12573 rads or 10 2 kHz Superconducting magnet 40 313 106 rads or 10 50013 MHz eld strengths can be expressed by lamor frequencies Energy of spin states Without external magnetic field the two orientations of the spin states are identical degenerated and have the same energy In the presence of a magnetic field the energies depend on the magnetic quantum number ms Em VhKO n1S 110 For a proton that means that the energy of the parallel orientation is lowered and for the the anti parallel it is raised The transition energy between the two levels AB E W E JQ is E ms 39 12 AE 7130 71600 hVO 111 AE no angular frequency angle s rad s39l ms 12 00 7O frequency rotations s s39l Bo Resonance condition for electromagnetic radiation photons thF th Selection Rule AmS i 1 due to spin 1 of photon Magnetization of an ensemble of spins A macroscopic samples consist of many spins 6022 1023 for 1 mol Due to the difference in energy there is a small excess of the spins aligned with the field m8 12 compared to the F Holger Farsterling September 10 2008 7 number of spins aligned in the opposite direction The probability to find a spin in a certain state is the Boltzmann distribution divided over the total number of states 5 mm Pm 8 k1quot 8 k1quot z lH mSJI BO 112 Z 21 1 2 CT Z is the partition function which is the sum over all states weighted by energy Z Ea m exp EmkT 11221 In the high temperature approximation kT gt ABquot Z equals the total number of states or 21 1 2 for a spin 12 The relative excess in the lower energy level in then given by AN N N NP P yhBO 7 Z T Z T Z W 03 o In the earth magnetic field one gets an excess of 3310 for afield of1175 T 8316 106 Since the spins are not precessing synchronized they ms 2 12 will have random phase all orientations on the cone have the same probability Therefore all components of the magnetic moment perpendicular to 3 0 will cancel and only a static z magnetization aligned with the magnetic field will exist on a macroscopic scale see figure For an ensemble of spin 12 nuclei only the excess spins in the level n1S 12 lower energy will contribute to the overall magnetization l WBUNl M0ANz N2h2 z n M 294 7 NW2 114 121 4n B 2 F Holger Farsterling September 10 2008 8 For the more general case of a nuclei of spin I the macroscopic z magnetization under equilibrium conditions can be obtained as the sum of the contributions from all states 1 I M3 Z a N39 Nyh Z um 114a Any spin 1 quot171 myl Within the high temperature approximation one obtains Total macroscopic magnetization of many spins Curie Law This general equation applies to all materials containing spins Curie Law Note that nuclear magnetism is MUCH smaller than paramagnetism or ferromagnetism arising from unpaired electrons and is normally even weaker than electron induced diamagnetism gt In a magnetic field in equilibrium a constant magnetization of nuclear spins parallel to 3 0 is present longitudinal magnetization lt Observable magnetization M0 will depend on magnetic field B0 we need as strong magnetic field as possible number of spins N so concentration is important square of gyromagnetic ratio y one factor due to Boltzmann distribution one due to dependence of magnetic moment on y inverse temperature T 14 Generating Xy transversal magnetization Static z magnetization by itself does not give rise to a signal in the NMR spectrometer Only magnetization oscillating perpendicular to HO will induce a signal in a receiver coil gt we need a way perturb the equilibrium and generate Xy magnetization Quantum mechanical picture Irradiate with EM radiation of the frequency according to F Holger Farsterling September 10 2008 9 gt hv AEzyhBozhvo 116 n1S 12 will stimulate transitions between energy levels AbsorptionEmission Selection rule for transitions absorption emission of photon with spin 1 The effect of RF radiation on macroscopic magnetization can be best understood in a classical vector picture The coil inside the probe generates an oscillating radio frequency field along x axis B LXO 2 B1 cosut 117 A linear oscillating magnetic field of amplitude 2B1 is equivalent to two counter rotating fields of amplitude B1 If the resonance condition DRF no is fulfilled only the component rotating with the lamor precession of will interact with the effect of the other component on U can be neglected as B1 B0 If n e no the additional B1 field exerts a torque on MZ and will cause net magnetization M to precess r to both HO and 310 w 7 B nutation Since B1 lt BO only a Blfield which is synchronized with IX 1quot the lamor precession will be able to have a significant effect on the orientation on NI in a sense the B1 vector will follow the precession of NI about B0 and keep B1 r M thus exerting maximum torque at all times This is like pushing a pendulum in sync with its frequency even a small push can have a large effect As a result the magnetization will be tipped from its initial orientation along the z axis while precessing about the static BO field The tip angle 6 depends only on the strength of the B1 field and the amount of time the oscillating field is applied F Holger Farsterling September 10 2008 10 Egure 9 Effect of a B1 field rotating in the xy plane with the lamor frequeny o 00 Left Combined precession of the magnetization M about 3 0 and the rotating B 1 field Right Trajectory of the magnetization vector during application of a RF pulse Note that 01 is NOT the frequency of the RF field which is DRF no but the precession of the magnetization about B1 ie the rate of ipping the spins and is related to the strength of BL The result will be the creation of transverse magnetization Mmy as a function of 6 MKy M0 sin 6 M0 sinyB11P 119 A maximum of transverse magnetization and therefore a maximum NMR signal is observed for 6 1T2 90quot since all of MZ has been converted into MW Once the oscillating B1 field has been turned off the magnetization is precessing freely about B0 with 00 YBO and it is this oscillating magnetization detected in the NMR experiment Rotating reference Frame The complex motion of the magnetization vector can be simplified when the observer is rotating with the B1 field at the frequency DRF This compares to us observing each other on the surface on the earth without worrying about the rotation of the earth about its axis F Holger Farsterling September 10 2008 l l 4 b Figure 12 I Larmor precession as seen in a the laboratory reference frame md b a rotating reference frame In this frame of reference the B1 eld appears as a static eld Since we are rotating the precession of the magnetic moments about B0 will be reduced to Q coo u and will vanish if JORFO no on resonance That is equivalent to reducing or removing the effect of the main static magnetic eld B0 0 That results in the following transformations laboratory frame Rotating frame Rotating frame on resonance BEG Blx cosut Bly sinut Blx Blx 90 Q no 9 Q 0 B0 B0 DRFY 0 Be mb2 mfg Deff 11 YB1 From now on x and y axes will refer to the coordinates rotating aboatz with frequency a z 00 Analogy When describing events on earth the coordinate system rotating with the earth is used not the xed stars As a result On resonance 1 no the effect of the B0 eld has been removed since the observer rotates with the spins about Bo Only rotation about HI eld from rf pulse has to be regarded F Holger Forsterling September 10 2008 12 gt Most NMR experiments can be understood on the basis of simple rotations dM a a F 1 X M 120 Therefore the Magnetization will be rotated perpendicular to 3 1 according to the right hand rule For a B1 field along the rotating x axis one obtains MZ M0 lZ cosi11P MX 0 My 2 MO Iy sini11p If 0139171 112 90quot pulse then after pulse M220 MX0andMyMO For 01 171 2 1T 180 pulse MZ MO 12 MX My 2 0 after the pulse That means that a 90quot pulse converts all z magnetization into y magnetization whereas a gure 11 ACT2X pulse creates ma netization alon 180 pulse inverts z magnetization into z g g y magnetization If A is close but not exactly on resonance with no as in the case of different chemical shifts there will be precession about the z axis with the frequency Q no mRF during the pulse However as long as the condition 01 gt Q is fulfilled then this precession will be small Thus 01 21W1 YB1 is a measure of the bandwidth of excitation The stronger B1 the shorter the duration of the pulse 171 and the wider the excitation range for which 91 gtgt Q A typical example In a proton spectrum at 10 002 7 2300 MHz the typical chemical shift range is 10 ppm which corresponds to 2 2 7T 39 3kHz For 4 10 39 Q 27 39 30kHz we obtain a durationfora 90 rotation 90 degree pulse of DU2 14 392 7001 83 Us which is a typical value F Holger Forsterling September 10 2008 l3 With QmZ 7 3 kHz we obtain an angle of 9 of precession during the pulse for signals at the edge of spectrum some important consequences of off resonance effects Finite excitation range Problems can arise in particular in 13C spectra or other hetero nuclei where the chemical shift range is much larger or when multiple pulses have to be ms12 A ix 397 I Pulse H 03111 m2 applied and errors will accumulate Phase errors arising from rotation about z axis during the excitation pulse On the other hand application of very weak pulses low value of B1 long 17P90 will allow for selective excitation water suppression Relation to microscopic quantum mechanical picture The macroscopic magnetization is produced by an ensemble of many individual spins Without going into the quantum mechanical treatment of the effect of an 1f pulse on the spins we can determine from the macroscopic effect that application of radio frequency has two effects change in population and alignment of the exited spins In the case of YBIIP 1T2 90quot pulse the result is Equal population of 06 and 5 spin state MZ 0 Excess magnetization becomes aligned along y axis coherence My MZO If YBIIP 2 1T 180 pulse Populations of spin states are inverted each spin is ipped once M MO no alignment in xy plane all temporary alignment during the pulse is lost 1le My 2 0 F Holger Forsterling September 10 2008 l4 15 Free precession If the rotating radio frequency field B is turned of the magnetization is precessing freely about the main field 3 0 with the lamor frequency 1O YBO When Viewed in the rotating frame this amounts to a static magnetization along the y axis in the on resonance case 1 10 My Moiy and Mx o for a mo 121 In the more general case of a 3 1 field not exactly on resonance with 1O memo free precession occurs in the rotating frame with the offset frequency Q no 1 M Moiycos t y M Moix sinQt x 122 Relaxation of signal The experimentally observed signal is decaying exponentially gt rotating xy magnetization results from NON equilibrium state of sample System will always restore back to equilibrium state magnetization along z axis random phases Three cases have to be distinguished dephasing of spins due to differences in lamor frequency inhomogeneity restoration of equilibrium magnetization spin lattice relaxation Eexchange of energy with lattice random dephasing of spins Spin spin relaxation Signal decay through inhomogeneity In real sample Field is not perfectly homogeneous Spins in different parts of sample will experience a different field gt different 1O F Holger Forsterling September 10 2008 15 result aligned spins will dephase with time and macroscopic xy magnetization will eventually vanish The effects effect of linear vs square and higher order gradients in the field are mentioned in lab lmportan Y v is reversibleif field gradient or orientation of spins is reversed Situation is like different chemical shifts Spins within small subvolumes are still aligned and precessing with one particular frequency Z ab cd B0 tO tgt0 Spinlattice relaxation relaxation of zmagnetization population of energy levels will return to equilibrium levels thermal equilibrium and reestablish z magnetization This is called longitudinal relaxation time constant T1 All processes contributing to longitudinal relaxation T1will also contribute to a corresponding reduction of transverse xy magnetization T2 see below gt buildup of z magnetization on expense of xy magnetization Spinspin relaxation relaxation of xy magnetization Even in perfectly homogeneous BO field xy magnetization will decay Reason interaction of nuclei with lattice AND each other In addition to random transitions to reestablish equilibrium z magnetization with the time constant T1 spin spin interactions chemical exchange spin diffusion F Holger Forsterling September 10 2008 l6 dependence of 00 on orientation can result in dephasing of spins without change in population energy levels Result is random statistical change in phase gt irreversible Entropy The relaxation of xy magnetisation is described by time constant T2 transversal relaxation It is always T1 2 T2 The difference between inhomogeneity 198 PHOTON ECHOES broadening and T2 relaxation can be visualized by 09 39V4 runners on a track D1fferent1nd1v1duals have different r x m 39 A 1 1 8 ii s eed and at some oint all runners will be e uall K I K 39 p p q y quot 1 distributed around the track But if they reverse p 14 K quot 47 iquot direction they will meet all again in the same location v quot39 quot quotquot t C J t D O x l In real life runners Will not have the same speed W 3 k m M throughout the run so they Will be more out of sync M A Q vjgt g x when meeting up than at the start gt T2 dephas1ng 39 K E F Any runner dropping out of the race T1 Wlll also Emar QcoNVIWWJj M 4 1 contribute to the reduction of the signal and T2 2EL K12 t i13533232 lioifhiii i39 32 ef f9f ReProduced by permission The experimentally observed decay of NMR signal is a combined effect of inhomogeneity broadening and transversal relaxation with time constant T with UT 2 yABOZ lT2 where ABO is the maximum spread of field values The line width is given by the decay rate of the signal it is AV 2 2TET2 123 Considering free precession and relaxation the three components of the magnetization present after a 900 pulse along x can be described as follows F Holger Forsterling September 10 2008 l7 t M MOlZ ei Moly cosQ t 6 T2 13924 t 397 Molx sinQ t e T M y M c Bloch Equations of free precession in a magnetic field after a 90 pulse Mzt00 Mxt00 Myt0MD 1y Implications of signal decay by T for experiment Maximizing the homogeneity will improve the resolution of the experiment longer T and therefore smaller Av Sensitivity will also be improved why 7 Acquisition time must be long enough for signal to decay to zero in order to avoid artefacts This is often not possible in 2D experiments gt sine or qsine filter functions have to be applied before processing Extending acquisition time beyond 3 T2 will not accumulate any more signal Implications of finite recovery time T1 one scan 90quot excitation pulse will give maximum signal If many scans are to be accumulated one will have to wait for z magnetization to recover between scans normally 5T1for full recovery With T1 e1 55 for protons this is very time consuming Often it is more advantageous to excite with shorter pulses and use shorter repetition time The optimum pulse angle is called Ernst Angle 6E eXPTRT1 TR 2 aq d1 repetition time 125 for quantitative determinations intensity more accurate integral will only be meaningful if same z magnetization was present before each pulse for each species If multiple scans are added repetition time needs to be 5 T1max for accurate results F Holger Forsterling September 10 2008 18 After inserting sample into magnet it will take a few seconds for magnetization to build up Normally no problem But some low y nuclei 103Rh or isolated spins CHCl3 in degassed deuterated solvent can have T1 of several minutes Exercise a Generalize the Bloch equation for free precession after an arbitrary pulse width 6 b What are the relationships for a sample just inserted into the magnet Further Reading General Textbooks JW Akitt NMR and Chemistry fourth ed Chapter 1 Andrew Derome Modern NMR techniques for Chemistry Research Chapter 41 433 and 44443 Harald Gunther NMR Spectroscopy second ed Chapter 1 Chapter 7 pp221 233 Edwin Becker High Resolution NMR Theory and Chemical Applications Third Ed Chapter 21 29 Jeremy Sanders Brian Hunter Modern NMR Spectroscopy second Ed Ch 11 125 and 127 Frank van de Ven Multidimensional NMR in Liquids Ch 11 14 Nuclear Shell Model G Friedlander et al Nuclear and RadioChemistry 3 01 Ed Ch 10D httpenwikipediaorgwikiShellimodel httpwwwhepphyssotonacukhepwwwstaffDRossphys3002shellpdf F Holger Farsterling September 10 2008 19


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Amaris Trozzo George Washington University

"I made $350 in just two days after posting my first study guide."

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.