NMR Spectroscopy. for 1 st B.Tech INTRODUCTION Lecture -1 Indian Institute of Technology, Dhanbad

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1 NMR Spectroscopy for 1 st B.Tech Lecture -1 Indian Institute of Technology, Dhanbad by Dr. R P John & Dr. C. Halder INTRODUCTION Nucleus of any atom has protons and neutrons Both Proton and Neutron has spin quantum number m s =±½ Depending upon their pairing, the Total angular momentum of the nucleus (Nuclear spin) I, will be different If the spins are paired, no net angular momentum or I = 0 When there is an unpaired spin, it imparts a net nuclear magnetic moment to the nucleus For even mass number nuclei, the spin is an integer. 2 D, For odd mass number nuclei, the spin is a half integer (1/2, 3/2,...). 1 H, 13 C, 31 P, 19 F, etc. All nuclei with an even number of protons and neutrons have zero spin in their ground state. 16 O, 12 C, 32 S etc. 1

2 NMR is possible with all nuclei except I = 0. But I =½ has the simplest physics. Allowed orientation of the nuclear magnetic moment vector in a magnetic field is given by nuclear spin angular momentum quantum number, m I. m I can take values I, (I-1),..-(I-1),-I and there are a total of (2I 1) components in each case. Example: If I= ½, m I can take two values, ½, -½, If I =1, m I can take three values -1, 0, 1. These components are normally degenerate but degeneracy may lifted and (2I 1) different energy level result if an external magnetic field is applied. Nuclear Spin A spinning charge, such as the nucleus of 1 H or 13 C, generates a magnetic field. The magnetic field generated by a nucleus of spin 1/2 is opposite in direction from that generated by a nucleus of spin 1/2. 2

3 The distribution of nuclear spins is random in the absence of an external magnetic field. An external magnetic field causes nuclear magnetic moments to align parallel and antiparallel to applied field. H 0 3

4 There is a slight excess of nuclear magnetic moments aligned parallel to the applied field H 0 Energy Differences Between Nuclear Spin States E -1/2 = 1/2(g N N B) -1/2, - spin DE = g N N B Potential energy of interaction of a nucleus With external magnetic field B E = -(g N N m I )B increasing field strength 1/2, - spin E 1/2 = -1/2(g N N B) no energy difference in absence of magnetic field proportional to strength of external magnetic field 4

5 For Proton: Value of g N is found to be , substituting the values of m I and g N along with N, we get, E 1/2 = -(g N N )(1/2)B =- ( JT -1 )(5.5854)(1/2) B = - ( JT -1 ) B E -1/2 = -(g N N )(-1/2)B = ( JT -1 )(5.5854)(1/2) B = ( JT -1 ) B The energy difference between two levels is DE g N N B = ( JT -1 ) B In terms of frequency ν=de/h = g N N B/h= ( JT -1 B)/( Js) = s -1 T -1 B For the field strength of 1.0 T, the frequency of radiation requires for the reversal of the spin of proton is ν= s -1 T -1 (1T) = s -1 = 42.6 MHz The frequency lies in the radiofrequency region of the electromagnetic spectrum ( Hz) Value of nuclear magneton: N = eh 4 m p ( C)( Js = = JT ( Kg) Q1. What magnetic field strength is required for proton magnetic resonance at 100 MHz. Given the factor g N for proton Ans: Since DE = g N N B, we have B= DE/g N N = (hdν) ( JT 1 )(5.585) ( s 1 )( J S) = = T ( JT 1 )(5.585) 5

6 Q2. Calculate the resonance frequency of a proton in Mega Hertz in a magnetic field of 9.4 Tesla, Given N for H = JT -1 and g N =5.585, h = Js. Ans: Since ν = DE/h = g N N B/h, we have = ( JT T) ( Js) = Hz = MHz The energy difference between the two nuclear levels can also be expresses in the unit of angular Frequency ω expresses as (radians/sec) Since Angular frequency = 2πν = 2π (g N N B)/h = {(2πg N N )/h}. B = γ. B Where = 2πg N N /h = gyromagnetic ration or the magnetogyric ratio. The spinning nucleus in an applied magnetic field would precess with a frequency = B, is the Larmor precession frequency. 6

7 Q3. Calculate the value of gyromagnetic ratio γ for a proton. (Given N for H = JT -1 and g N =5.585, h = Js.) Ans: T -1 s Calculate the precessional frequency of a proton in a field of 1.5 T. (Given N for H = JT -1 and g N =5.585, h = Js.) Ans: T -1 s -1 For magnetic field with strengths 1-20 Tesla, the energy of transition is 0.1cal/mole [Earth magnetic field is 0.1mT] For an infrared transition it is 1-10kcal/mole Irradiation with Radio frequency in the range MHz result in transition depending upon the nucleus under study 7

8 1 H NMR The Spectrum An NMR spectrum is a plot of the intensity of a peak against its chemical shift, measured in parts per million (ppm). 8

9 Shielding by electrons & Chemical Shift The electrons surrounding a nucleus generate a field that oppose the applied field. Thus the proton feel less magnetic field. Hence, higher magnetic field need to be applied to achieve resonance The higher the electron density around a proton the higher the field applied The lower the electron density around a proton the lower the field applied The extent of field applied for resonance to occur is measured in ppm For a 100MHz instrument, 42ppm means 4,200Hz For most organic compounds, resonance appear between 0-12ppm A standard sample is used to measure the position of the resonance signals Tetramethyl silane, Si(CH 3 ) 4 is the standard sample used for NMR measurement The resonance signals positions are identified using Chemical Shift values ( ) expressed in ppm 17 NMR absorptions generally appear as sharp peaks. Increasing chemical shift is plotted from left to right. Most protons absorb between 0-10 ppm. The terms upfield and downfield describe the relative location of peaks. Upfield means to the right. Downfield means to the left. NMR absorptions are measured relative to the position of a reference peak at 0 ppm on the scale due to tetramethylsilane (TMS). TMS is a volatile inert compound that gives a single peak upfield from typical NMR absorptions. 9

10 19 Q.5 In a 60 MHz instrument CH 3 Br display a signal 162 Hz downfield from TMS. Calculate the chemical shift (δ in ppm) of CH 3 Br proton. Ans: So, δ = 162 Hz/( ) Hz = 2.70 ppm Q6. In a 100 MHz NMR instrument shift of the protons of X is 270 Hz downfield from TMS. Calculate the shift of X proton from TMS in a 60 MHz NMR instrument. Ans: δ = 270 Hz/( ) Hz = 2.70 ppm Again δ = Shift from TMS (in Hz)/(Spectrometer frequency in MHz) So, 2.70 ppm = Shift from TMS (in Hz)/ 60 MHz Shift from TMS (in Hz) = Hz = 162 Hz 10

11 1 H NMR spectrum provide information about a compound s structure: a.number of signals b.position of signals c.intensity of signals. d.spin-spin splitting of signals. Number of Signals The number of NMR signals equals the number of different types of protons in a compound. Protons in different environments give different NMR signals. Equivalent protons give the same NMR signal. 11

12 1 H NMR Position of Signals In the vicinity of the nucleus, the magnetic field generated by the circulating electron decreases the external magnetic field that the proton feels. 12

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14 Qns: The solvents used for NMR measurements are always deuterated. Why? [CDCl 3, D 2 O, C 6 D 6, CD 2 Cl 2, (CD 3 ) 2 CO etc.] Ans: Deuterium has different Mag. Moment and spin hence its signals does not appear in spectrometers tuned to proton. The more an electronegative an element attached to C the less the electron density around the proton attached to it The protons thus are called de-shielded and resonance appear at higher values Compound (CH 3 ) 4 C (CH 3 ) 3 N (CH 3 ) 2 O CH 3 F δ Compound (CH 3 ) 4 Si (CH 3 ) 3 P (CH 3 ) 2 S CH 3 Cl δ Cpd. / Sub. X=Cl X=Br X=I X=OR X=SR CH 3 X CH 2 X CHX

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17 Curious case of Aromatic and unsaturated compounds The -electrons above and below the ring plane circulate in response to the applied field, generating an induced field that oppose the applied field at the center of the ring and support the at the edge of the ring Hence, the Protons at the periphery of the ring give signals at higher values 34 17

18 Intensity of signals. The ratio of integrals to one another gives the ratio of absorbing protons in a spectrum. Note that this gives a ratio, and not the absolute number, of absorbing protons. 18

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