1H NMR BRUSH UP PRIMER

Nuclear Magnetic Resonance
            A.  Theory
                        1.  The nuclei of some atoms behave as though they spin on their axis.
                                    a)  Electrons have the same behavior.
                        2.  The ¹H nucleus (a proton) has two nuclear spin states with quantum numbers +1/2 and -1/2.
                        3.  Spinning charged particles create a magnetic field, so the proton has a magnetic moment.
                                    a)  The proton acts like a small bar magnet.
                        4.  The two spin states are identical in energy in the absence of an applied magnetic field.
                                    a)  In absence of an applied magnetic field, the nuclear magnetic moments are randomly oriented.
                        5.  When placed in a magnetic field the magnetic moments align either with or against the field.
                                    a)  The two spin states are no longer of equal energy.
                                                i)  The +1/2 spin state aligns with the applied magnetic field and is of lower energy.
                                                ii)  The −1/2 spin state aligns against the applied magnetic field and is of higher energy.
                                    b)  Example: think of the Earth's magnetic field as an applied field in which a compass is placed.
                                                i)  The compass needle aligns with the magnetic field of the Earth.
                                                ii)  This is of lower energy than if you were to align the needle in the opposite direction.
                        6.  The nucleus in the lower (+1/2) spin state can be elevated to the higher-energy spin state by the
                                    absorption of energy (electromagnetic radiation) equal to the energy difference of the two spin states.
                                    a)  This change from the +1/2 to −1/2 spin state is referred to as "spin flipping".
                                    b)  The amount of energy required for this transition is dependent upon the magnitude of the applied
                                                 magnetic field.
                                    c)  The greater the applied field, the higher the required energy to flip the spin.
                        7.  Many nuclei lack the property of spin.
                                    a)  Nuclei having an even number of protons and an even number of neutrons lack spin.
                                                i)  Pairs of protons and pairs of neutrons in such nuclei have opposed spins which cancel each
                                                   other out, leaving the nucleus with no net spin.     

            B.  NMR Spectroscopy
                        1.  The NMR spectrometer consists of a  powerful magnet.
                        2.  A sample is placed in the magnetic field of the spectrometer.
                        3.  A radio-frequency source irradiates energy through the sample until the frequency of the energy source
                                     is equal to the energy difference of the two spin states.
                                    a)  At this time the two are said to be in resonance with each other.
                        4.  The absorption of energy is detected by a radio-frequency receiver and recorded as a peak on an NMR
                                    spectrum.
                        5.  The fact that radio waves are used to cause the spin flip demonstrates that the energy difference
                                     between the two spin states is of very low energy.
                                    a)  In a very strong magnetic field of 14,100 gauss, the energy difference for the two spin states of a
                                                 proton is only 2.5 x 10^-5 kJ/mole.
                        6.  NMR instruments can operate either with a constant magnetic field and changing the frequency, or
                                    with constant frequency and changing magnetic field.
                                    a)  Either way, the spectra are identical.
                        7.  Not long ago 60 MHz instruments were in common usage, then 100 MHz, 220 MHz, and 360 MHz
                                     instruments were in use; now 400, 500 and 600 MHz instruments are in use.
 

            C.  Chemical Shift
                        1.  If all hydrogen nuclei spin flipped at the same frequency, NMR spectroscopy would have no value in
                                     compound structure determination since all NMR spectra would look the same:  a single   peak at the
                                    same place.

                        2.  Fortunately, the chemical environment of  hydrogen atoms affects the frequency of the energy
                                    absorbed for resonance to take place.
                        3.  In a constant frequency instrument, where the magnetic field is slowly changed to cause the spin flip, it
                                    is observed that the applied magnetic field required to spin flip a proton in an organic compound is
                                     greater than that required for a bare proton.
                                    a)  This is because the electrons in a molecule "shield" the nucleus from the applied magnetic field.
                                                i)  This is referred to as "diamagnetic shielding".
                                    b)  This shielding occurs because the application of the magnetic field results in the electrons moving
                                                and setting up their own magnetic field which opposes the applied field.
                                    c)  The nucleus of the bound proton "feels" less magnetic field than is actually applied.
                                    d)  In order to achieve the spin flip, a greater magnetic field must be applied as compared to the field
                                                required to flip a bare proton.
                        4.  The change in the resonance position of a nucleus (due to the chemical environment of the atom) is
                                    referred to as a chemical shift.
                        5.  A peak at a higher field is said to be "upfield".
                                    a)  A peak at a lower field is said to be "downfield".

                         6.  The chemical shift is measured relative to a substance standard
                                    a)  The standard used is TMS, tetramethylsilane.
                                    b)  The NMR signal from TMS is set at 0 (on the delta scale) since protons in TMS are highly
                                                 shielded due to the low electronegativity of Si.
                                                i)  There is therefore a relatively high electron density around the hydrogen atoms.
                                                ii)  Shielding in TMS is greater than that of most organic compounds, making it an ideal standard.
                                                iii)  The spectrum is adjusted electronically on the chart so that the TMS signal is set to 0.
                        7.  The placement of an NMR signal relative to TMS varies with magnetic field strength.
                                    a)  Because NMR instruments use many different magnetic field strengths, a unitless measure of
                                                chemical shift independent of field strength has been developed.
                                    b)  The delta scale is the ratio of the chemical shift of a given signal to the total radio frequency used:
            Chemical shift (d) =  (position of signal - position of TMS peak)  x 10⁶/   radio frequency of spectrometer
                                    c)  The values in the above equation are in Hz.
                                    d)  Because the difference in the two positions is so small, it is multiplied by 10⁶.
                                                i)  The delta scale is therefore expressed in ppm.

                        8.  The chemical shift yields important information.
                                    a)  By the magnitude of the shift, some details of chemical structure can be deduced.
                                                i)  The text has a more complete listing of chemical shifts; below are some representative values.

                                     b) Highly electronegative elements deshield the protons, resulting in downfield resonances. 

Chemical Shifts of CH₃Z          

CH3Z	CH3F	CH3OH	CH3Cl	CH3Br	CH3I	CH3CH3	CH4	(CH3)4Si
Z	F	O	Cl	Br	I	C	H	Si
Electronegativity * very polarizable	4.0	3.5	3.1	2.8	2.5*	2.5	2.1	1.8
d, ppm	4.26	3.40	3.05	2.68	2.16	0.7	0.23	0.00

 

 Substitution Effects on Chemical Shifts

CHCl3	CH2Cl2	CH3Cl	-CH2Br	-CH2-CH2Br	-C­H2(CH2)2Br
7.27	5.30	3.05	3.30	1.69	1.25

 

                        9.  The chemical shift for benzene is around 7 ppm and that for the protons of alkenes is around 5.
                                    a)  Resonance occurs far downfield than might be expected for hydrocarbons.
                                    b)  The effect is due to anisotropy.
                                                i)  The pi cloud of electrons of these systems circulate due to the induced magnetic field,
                                                            creating their own magnetic field.
                                    c)  In the case of benzene, the induced magnetic field opposes the applied field in the middle of the
                                                 ring, but is with the applied field outside the ring.
                                                i)  The protons of benzene are in the space where the secondary magnetic field generated by
                                                            the pi electrons of the pi cloud circulating deshields the protons.
                                                ii)  A strong clue to the presence of the benzene ring in a compound is a NMR signal at ~7 ppm.

                                    d)  In the case of alkenes, the induced field opposes the applied field in the middle of the double
                                                bond, but are with the applied field in the area of the vinyl hydrogens.

             D.  Splitting Patterns 

                        1.  The NMR spectrum of ethane is a single peak since there is only one chemical environment for H                  
                        2.  The NMR spectrum of 1,1,2-trichloroethane should show how many NMR signals, and what would
                                     you expect their approximate chemical shifts?

                                    Answer:  Two.   

                                    The H at C-1:  ~6 ppm.  It is further downfield than the H’s at C-2 since two Cl are attached to its C.
                                    The H's at C-2:  ~4 ppm           

                                    a)  Two single peaks are NOT observed for the above compound.
                                    b)  A triplet is observed at 5.8 ppm and a doublet is observed at 3.9 ppm. 

                         3.  Multiplets are observed in place of singlets on the NMR spectrum when hydrogen atoms exist on
                                     adjacent carbon atoms.
                                    a)  Multiplets arise from "spin-spin splitting".
                                    b)  The chemical shift is in the center of the multiplet.
                        4.  Splitting occurs because the magnetic field of a proton of a hydrogen atom on carbon 1 adds or
                                     subtracts to the applied field felt by the hydrogen atom on carbon 2.
                                    a)  The "effective field" felt by a proton can be greater, less than, or equal to the applied field,
                                                depending on the orientation of spin on neighboring hydrogens.
                        5.  The two hydrogens on C-2 of 1,1,2-trichloroethane experience two different effective magnetic fields
                                     depending upon if the hydrogen on C-1 is in the spin up   or spin down state.
                        6.  The H at C-1 is near two hydrogens on C-2 and as a result can feel 3 different effective fields.
                                    a)  Both C-2  H-spins can be up, both could be down, or one could be up while the other is down.

                                     b)  Notice that the central peak of the triplet is twice the size of the other two since there are twice
                                          as many spin combinations possible than for the other two.
                                    c)  The hydrogens on the adjacent carbons are said to be "coupled".
                        7.  In simple cases, the observed multiplet is equal to the number of H on adjacent carbon atoms + 1.
                          This is called the "(n+1) rule"
                        8.  The splitting pattern for the ethyl group is quite distinctive.
                                    a)  Predict what the proton NMR spectrum would look like for CH­₃CH₂F, and show the spin
                                                combinations and net spins below each signal. 

                        9.  The following chart shows the type of multiplet observed as a function of adjacent hydrogens, and the
                                     area ratios of the peaks of the multiplet. You may recognize this pattern as Pascal’s Triangle

# of Adjacent [vicinal] Equivalent H's	Number of Peaks	Peak  Description learn the lingo!	Area Ratios
0	1	singlet	1
1	2	doublet	1:1
2	3	triplet	1:2:1
3	4	quartet	1:3:3:1
4	5	quintet	1:4:6:4:1
5	6	sextet	1:5:10:10:5:1
6	7	septet	1:6:15:20:15:6:1

                        10.  Protons of the hydroxyl group on alcohols usually appear as a singlet and are not split due to rapid
                                     intermolecular exchange of the hydroxyl protons. The same is true for amine hydrogens. 

            E.  Coupling Constants

                        1.  The distance between peaks in a multiplet, measured in Hz, is referred to as the coupling constant, J.
                        2.  The coupling constant is independent of the magnetic field strength of the NMR instrument.
                        3.  For the example fill in the blanks:
                                    The peak for H_a appears as a doublet.
                                    The peak for H_c appears as a doublet.  
                                    a)  Because the coupling constants J_ab and J_bc are not equal (J_ab = 3.6 Hz while J_bc = 6.8 Hz) the peak
                                                for H_b is not a quintet but instead a doubled quartet (or quadrupled doublet.)
                                                i)  This makes the multiplet rather messy rather than being a nice and easy-to-decipher peaks. 

                         4. Unequal coupling constants can result in complex spectra which are more difficult to interpret.
                        5. Coupling constants can yield important information about conformations in cyclohexane rings.
                                    a)  The coupling constant for adjacent axial hydrogens in cyclohexanes is 10-13 Hz.
                                    b)  The coupling constant for adjacent equatorial hydrogens in cyclohexanes is 3-5 Hz.   

Jaa' = 11.1 Hz Jae  =  4.3 Hz	Jee' = 2.7 Hz Jae  = 3.0 Hz

 

                         6.  Coupling constant values can be used for the determination of stereostructure in alkenes.
                                    a)  J is always greater for trans H’s than for cis H’s for a pair of isomeric cis and trans alkenes.
                                                i)  J_trans = 11-19 Hz.
                                                ii)  J_cis = 5-14 Hz.
                                    b)  Notice there is overlap in the J values.
                                                i)  If the vicinal coupling constant is < 10, one can infer the alkene is the cis isomer.
                                                ii)  If the vicinal coupling constant is > 14, one can infer the alkene is the trans isomer.
                                                iii)  If the J value is between 10 and 14, then the spectra of both isomers must be available
                                                            to make a conclusion (J_cis is always less than J_trans for isomeric alkenes.)
                                    c)  The NMR spectrum of (E)- and (Z)-3-chloropropenenitrile illustrate how the coupling constants
                                                 can be used to assign structure.  

           

F.   Integration 

                        1.   In proton NMR the area of the peaks for each proton environment is proportional to the number of
                                    protons in that environment.
                        2.  When the NMR spectrometer is set to the integration mode it draws a step-like line whose height is
                                    directly proportional to the number of protons responsible for the signal.
                        3.  Integrations do not actually indicate the precise number of hydrogens at each chemical environment.
                                    a)  They instead yield a ratio of protons at each environment.
                        4.  The example below illustrates how integration yields a ratio of protons at each chemical environment. 

                                     a)  In the above example notice how the integral ratios are calculated.
                        5.  The information given by the chemical shifts, splitting patterns, coupling constants, and the
                                    integration can often lead to the structure of the compound in question.
                                    a)  In addition to this we also have information given by IR (functional groups present), mass
                                                spectrometry (molecular weight and other information), elemental analysis, and
                                                other spectroscopic methods.