Chapter 13: Nuclear Magnetic Resonance Spectroscopy
Chapter 13: Nuclear Magnetic Resonance Spectroscopy Coverage: 1. The nucleus in an NMR experiment. 2. Shielding/deshielding of the nucleus and the chemical shift. 3. Spin-spin interactions, N+1 rule, and the coupling constant J. 4. 1H NMR spectroscopy and integration 5. Unknown analysis, Degrees of Unsaturation 6. Fourier Transform NMR Spectroscopy
End of Chapter Problems:13.28, 13.29, 13.31, 13.32,13.37, 13.40, 13.46, 13.48, 13.51. Goals: 1. 2. 3. 4. 5 6
Know terms: chemical shift, shielding/deshielding, TMS, J. Be able to convert between chemical shift in ppm to Hz relative to TMS (vice versa). Be able to apply N+1 rule to predict the spectrum of simple structures. Be able to calculate degree of unsaturation from formula and know what results means. Be able to identify the number of chemical equivalent hydrogen or carbon atoms in a molecule. Complete assigned homework on spectroscopy. See syllabus.
Nuclear Magnetic Resonance Spectroscopy (NMR) • Provides a map of the C and H framework of the molecule. • Physical Basis: The 13C and 1H nuclei both possess spin. The nucleus can be irradiated with radiowaves to cause a spin transition. The frequency of the this transition depends on the electronic environment of the nucleus. By measuring the frequency, we can learn about the way in which the nucleus is bonded to other atoms in the molecule. Do all nuclei possess spin?
No.
Only those with an odd atomic number or odd atomic mass: 13C 6
12C 6
1H 1
16O 8
14N 7
32S 16
15N 7
31 15 P
NMR active
NMR inactive
A nucleus with spin behaves much like a small magnet. It possesses a magnetic moment which has both magnitude and direction. Consider a collection of 1H nuclei with randomly oriented magnetic moments.
β spin state B0 α spin state No applied magnetic field Randomly oriented B0 is the applied magnetic field. The α spin state is lower energy. There is an excess of α spin states. β Irradiate
E α
radiowaves
β
α
Recall E = hν. By irradiating with the proper frequency, the α spin state is “flipped” to the β spin state with absorption of energy. The nucleus is in “resonance”
The energy difference, ∆E, between and the two spin states depends on the strength of the applied field, H0. h Planck’s constant γ Nuclear constant
∆E = γ h B0 2π
The strengths of magnets will vary from one institution to another, and will be measured in units of gauss, or alternatively, in terms of the frequency (MHz) at which hydrogen nuclei absorb with that particular magnet. Magnet Strength (gauss) 14,092 23,486 70,458 117,430 1 MHz = 1 million Hertz
Frequency at which 1H nuclei absorb (MHz) 60 100 300 (UAF) 500
1.
Shielding of Nuclei by Electrons
Not all 13C nuclei absorb at the same frequency. Different nuclei absorb at different frequencies, depending on their electronic environments. Consider the following alkyl halide. CH3CH2Br
Two types of 13C
The electrons that surround the nuclei tend to shield the nuclei from the applied field, B0. This weakens B0 in the vicinity of the nucleus and causes The nuclei to absorb at lower energy and lower frequency. The methylene carbon (CH2) is less shielded due the close proximity the electronegative bromine atom, which withdraws electrons away from it.
CH3CH2Br More shielded, absorbs at lower frequency
Less shielded, absorbs at higher frequency
Thus, two different absorptions are observed in the 13C NMR spectrum of this molecule.
Deshielded High ν
Shielded Low ν CH3CH2Br
Intensity Reference compound (TMS) defined as 0.0 ppm
Chemical Shift, δ (ppm)
2. Chemical Shift - variations in frequencies of NMR absorptions due to shielding /deshielding by the electrons. Reference Compound: Tetramethylsilane (TMS) CH3 H3C Si
CH3
CH3
The methyl carbons are highly shielded and absorb at low frequency. Most other carbons absorb at higher frequencies.
Chemical Shift = Absorption Frequency relative to TMS (Hz) Spectrometer Frequency (MHz) 1 Hz MHz
is _ 1_Hz__ or 1 part per million (ppm) 1 million Hz
δ(delta) scale with units of ppm. TMS assigned a value of 0.0 ppm
Carbon Chemical Shifts
Mode of Operation – under typical operating conditions, each type of 13C nucleus yields a single, sharp line. This mode of operation is called the spin-decoupled spectrum, for reasons that will be apparent later.
O CH3CCH2CH3 a
b
b c
d
c
a
d
1H
NMR Spectroscopy
• The same fundamental concepts apply to the 1H nucleus as with the 13C nucleus. • The chemical shift range is 0-15 ppm for 1H • However, there are three additional considerations. 3. Spin-spin coupling 4. Integration of Peak Areas 5. Coupling constant
Hydrogen Chemical Shifts
3. Spin-Spin Coupling Consider the partial structure:
Ha
Hb
C
C
• Ha and Hb have different electronic environments and therefore different absorption frequencies. • Ha and Hb are on adjacent carbons and therefore interact through a process spin-spin coupling.
Consider the signal of Ha. If Hb were not present, Ha would yield a single line in its NMR spectrum. However, Hb interacts with Ha and causes it to split into two lines. This happens because Hb possesses spin, with its own small magnetic moment. This magnetic moment of Hb interacts with Ha in two possible ways, as shown on the next slide.
Signal of Ha Ha
Hb aligned with B0 Reinforces B0
Ha is shifted to higher ν
Ha yields a single line without Hb
Hb aligned against B0 Diminishes B0
Ha is shifted to lower ν
Conclusion: The signal of Ha is split into two signals depending on the alignment of Hb with the applied field.
N + 1 Rule: a hydrogen nucleus with N adjacent equivalent hydrogen nuclei will be split into N + 1 lines. Consider the 1H NMR spectrum of ethyl bromide. CH3CH2Br The CH2 hydrogens are split into 3 + 1 = 4 lines The CH3 hydrogens are split into 2 + 1 = 3 lines
a b CH3CH2Br b
a
The number of lines is referred to as the multiplicity Number of Lines
Multiplicity
1 2 3 4 5 6
singlet doublet triplet quartet quintet sextet
Area Ratios 1 1 1 1 1 1
2 3
4 5
1 1 3
1
6
4
10 10
1 5
Consider Isopropyl bromide. Predicts its 1H NMR spectrum. a b a CH3CHCH3 Hydrogen Chemical Shift Multiplicity Br
a b
0.9 3.5
doublet septet
1
a b
a
CH3CHCH3 Br
Intensity
a b
4
3
2
1
0 ppm
4. The Coupling Constant, J - the separation between peaks in a multiplet measured in units of Hz. Jba a b CH3CH2Br
Jba
3 .5 0
3 .4 5
Jba
3 .4 0
CH2 b
3 .3 5
3 .3 . 30
3 .2 5
Jba = Jab = 7.3 Hz Jab Jab CH3 a
1 ..77 5
1 .7 0
1 .6 5
1 .6 0
1 .5 5
1 .5 0
• J is independent of magnetic field strength • Mutually coupled hydrogen nuclei have the same J value. The magnitude of J gives information about structure. H H
H C C
C C
H
J = 7 Hz H
H
J = 15 Hz H
C C
C C H
J = 10 Hz
J = 2 Hz
5. Integration - the area under a peak is proportional to the number of of hydrogens that give rise to that peak.
a b CH3CH2Br
2.0 b
3.0 a
Integration values will only yield only relative numbers. If you know the formula of the compound, then you can calculate the actual hydrogens for each peak.
Unknown Analysis Given: Molecular Formula Steps to follow: 1. 2. 3.
4.
Calculate Degree of Unsaturation (DU). Examine IR spectrum to determine functionality of molecule. Examine 13C spectrum to give the number of different types of carbon. Check to see if it agrees with formula or not. a. If yes, then each carbon is unique electronic environment b. If no, then some carbons are equivalent to each other. a. Examine 1H NMR spectrum to determine number of different types of hydrogen. b. Correlate chemical shifts with those in table. c. Look at integration. Does it sum to total in the formula. It should! Clues integral 3 Probably a CH3 integral 2 Probably a CH2
http://www.chem.ucla.edu/~webspectra/ Work problems 5,6,7,11,12,16,17,19 for practive
C4H8O
C4H8O
C4H8O
C4H10O
C4H10O
C4H10O
C8H8O2
C8H8O2
C7H8O2
C7H8O2
C7H8O2
The Basic One-Pulse Experiment If we place a single nucleus with a spin into a magnetic field, B0, it will precess at a characteristic frequency, termed the Larmor frequency. z y α B0
x β
If we consider all nuclei, then we have many vectors, and we can consider the bulk Magnitization, M, of the sample, which is the sum of the individual spins. M z α B0
y x
β
Nα > Nβ Nα / Nβ = e-∆E/kT where k is Boltmann constant and T is temp (K). ∆E = hγB0/2π or ν = γB0/2π γ- magnetogyric ratio (nuclear constant) Apply a pulse of electromagnetic radiation, B1, at an angle perpendicular to the applied field, B0. All of the individual magnetic moments focus and become phase coherent. M tips away from the z axis and precesses at the Larmor frequency. α z
B0
α
z
M y
α
MZ
M y
x
β
x
β
MXY
The tip angle, α, depends on the power and duration of the pulse. We are interested in the MXY component of M since it the signal in this plane That gives rise to the NMR signal. The Rotating Frame of Reference M is rotating in the xy plane at the Larmor frequency. For protons in our NMR Spectrometer, this 300 MHz. This is too fast for our purposes, and we will get dizzy. Suppose we allow the x and y axes to rotate along with M and that we are riding on the xy plane along with M. To us, M is not moving at all, at least not in the xy plane. This is the rotating frame of reference and the axes are now defined as x’, y’ and z’. z’
α
z’
M y’
α x’
β
y’ x’
β
M
900 pulse
Now consider what happens to M after the pulse is turned off. It relaxes back to the original Boltzmann distribution. Looking along the x’ axis, the signal decays with time to x’=0. This is called the Free Induction Decay or FID. Intensity x’=0
If M is moving slightly faster than the rotating frame, i.e. it has a slightly different Larmor frequency, the signal will oscillate at a characteristic frequency. Intensity x’=0
If this FID is mathematically transformed by a process called Fourier Transformation, then a spectrum will be obtained.
Intensity FT
x’=0
3.90 4.00
Time, sec
3.95 3.85
3.90 3.80
3.85 3.75
3.703.80
Frequency, Hz
3.65 3.75
3.60 3.70
End of Chapter Problems:13.28, 13.29, 13.31, 13.32,13.37, 13.40, 13.46, 13.48, 13.51. Goals: 1. 2. 3. 4. 5 6
Know terms: chemical shift, shielding/deshielding, TMS, J. Be able to convert between chemical shift in ppm to Hz relative to TMS (vice versa). Be able to apply N+1 rule to predict the spectrum of simple structures. Be able to calculate degree of unsaturation from formula and know what results means. Be able to identify the number of chemical equivalent hydrogen or carbon atoms in a molecule. Complete assigned homework on spectroscopy. See syllabus.
Nuclear Magnetic Resonance Spectroscopy (NMR) • Provides a map of the C and H framework of the molecule. • Physical Basis: The 13C and 1H nuclei both possess spin. The nucleus can be irradiated with radiowaves to cause a spin transition. The frequency of the this transition depends on the electronic environment of the nucleus. By measuring the frequency, we can learn about the way in which the nucleus is bonded to other atoms in the molecule. Do all nuclei possess spin?
No.
Only those with an odd atomic number or odd atomic mass: 13C 6
12C 6
1H 1
16O 8
14N 7
32S 16
15N 7
31 15 P
NMR active
NMR inactive
A nucleus with spin behaves much like a small magnet. It possesses a magnetic moment which has both magnitude and direction. Consider a collection of 1H nuclei with randomly oriented magnetic moments.
β spin state B0 α spin state No applied magnetic field Randomly oriented B0 is the applied magnetic field. The α spin state is lower energy. There is an excess of α spin states. β Irradiate
E α
radiowaves
β
α
Recall E = hν. By irradiating with the proper frequency, the α spin state is “flipped” to the β spin state with absorption of energy. The nucleus is in “resonance”
The energy difference, ∆E, between and the two spin states depends on the strength of the applied field, H0. h Planck’s constant γ Nuclear constant
∆E = γ h B0 2π
The strengths of magnets will vary from one institution to another, and will be measured in units of gauss, or alternatively, in terms of the frequency (MHz) at which hydrogen nuclei absorb with that particular magnet. Magnet Strength (gauss) 14,092 23,486 70,458 117,430 1 MHz = 1 million Hertz
Frequency at which 1H nuclei absorb (MHz) 60 100 300 (UAF) 500
1.
Shielding of Nuclei by Electrons
Not all 13C nuclei absorb at the same frequency. Different nuclei absorb at different frequencies, depending on their electronic environments. Consider the following alkyl halide. CH3CH2Br
Two types of 13C
The electrons that surround the nuclei tend to shield the nuclei from the applied field, B0. This weakens B0 in the vicinity of the nucleus and causes The nuclei to absorb at lower energy and lower frequency. The methylene carbon (CH2) is less shielded due the close proximity the electronegative bromine atom, which withdraws electrons away from it.
CH3CH2Br More shielded, absorbs at lower frequency
Less shielded, absorbs at higher frequency
Thus, two different absorptions are observed in the 13C NMR spectrum of this molecule.
Deshielded High ν
Shielded Low ν CH3CH2Br
Intensity Reference compound (TMS) defined as 0.0 ppm
Chemical Shift, δ (ppm)
2. Chemical Shift - variations in frequencies of NMR absorptions due to shielding /deshielding by the electrons. Reference Compound: Tetramethylsilane (TMS) CH3 H3C Si
CH3
CH3
The methyl carbons are highly shielded and absorb at low frequency. Most other carbons absorb at higher frequencies.
Chemical Shift = Absorption Frequency relative to TMS (Hz) Spectrometer Frequency (MHz) 1 Hz MHz
is _ 1_Hz__ or 1 part per million (ppm) 1 million Hz
δ(delta) scale with units of ppm. TMS assigned a value of 0.0 ppm
Carbon Chemical Shifts
Mode of Operation – under typical operating conditions, each type of 13C nucleus yields a single, sharp line. This mode of operation is called the spin-decoupled spectrum, for reasons that will be apparent later.
O CH3CCH2CH3 a
b
b c
d
c
a
d
1H
NMR Spectroscopy
• The same fundamental concepts apply to the 1H nucleus as with the 13C nucleus. • The chemical shift range is 0-15 ppm for 1H • However, there are three additional considerations. 3. Spin-spin coupling 4. Integration of Peak Areas 5. Coupling constant
Hydrogen Chemical Shifts
3. Spin-Spin Coupling Consider the partial structure:
Ha
Hb
C
C
• Ha and Hb have different electronic environments and therefore different absorption frequencies. • Ha and Hb are on adjacent carbons and therefore interact through a process spin-spin coupling.
Consider the signal of Ha. If Hb were not present, Ha would yield a single line in its NMR spectrum. However, Hb interacts with Ha and causes it to split into two lines. This happens because Hb possesses spin, with its own small magnetic moment. This magnetic moment of Hb interacts with Ha in two possible ways, as shown on the next slide.
Signal of Ha Ha
Hb aligned with B0 Reinforces B0
Ha is shifted to higher ν
Ha yields a single line without Hb
Hb aligned against B0 Diminishes B0
Ha is shifted to lower ν
Conclusion: The signal of Ha is split into two signals depending on the alignment of Hb with the applied field.
N + 1 Rule: a hydrogen nucleus with N adjacent equivalent hydrogen nuclei will be split into N + 1 lines. Consider the 1H NMR spectrum of ethyl bromide. CH3CH2Br The CH2 hydrogens are split into 3 + 1 = 4 lines The CH3 hydrogens are split into 2 + 1 = 3 lines
a b CH3CH2Br b
a
The number of lines is referred to as the multiplicity Number of Lines
Multiplicity
1 2 3 4 5 6
singlet doublet triplet quartet quintet sextet
Area Ratios 1 1 1 1 1 1
2 3
4 5
1 1 3
1
6
4
10 10
1 5
Consider Isopropyl bromide. Predicts its 1H NMR spectrum. a b a CH3CHCH3 Hydrogen Chemical Shift Multiplicity Br
a b
0.9 3.5
doublet septet
1
a b
a
CH3CHCH3 Br
Intensity
a b
4
3
2
1
0 ppm
4. The Coupling Constant, J - the separation between peaks in a multiplet measured in units of Hz. Jba a b CH3CH2Br
Jba
3 .5 0
3 .4 5
Jba
3 .4 0
CH2 b
3 .3 5
3 .3 . 30
3 .2 5
Jba = Jab = 7.3 Hz Jab Jab CH3 a
1 ..77 5
1 .7 0
1 .6 5
1 .6 0
1 .5 5
1 .5 0
• J is independent of magnetic field strength • Mutually coupled hydrogen nuclei have the same J value. The magnitude of J gives information about structure. H H
H C C
C C
H
J = 7 Hz H
H
J = 15 Hz H
C C
C C H
J = 10 Hz
J = 2 Hz
5. Integration - the area under a peak is proportional to the number of of hydrogens that give rise to that peak.
a b CH3CH2Br
2.0 b
3.0 a
Integration values will only yield only relative numbers. If you know the formula of the compound, then you can calculate the actual hydrogens for each peak.
Unknown Analysis Given: Molecular Formula Steps to follow: 1. 2. 3.
4.
Calculate Degree of Unsaturation (DU). Examine IR spectrum to determine functionality of molecule. Examine 13C spectrum to give the number of different types of carbon. Check to see if it agrees with formula or not. a. If yes, then each carbon is unique electronic environment b. If no, then some carbons are equivalent to each other. a. Examine 1H NMR spectrum to determine number of different types of hydrogen. b. Correlate chemical shifts with those in table. c. Look at integration. Does it sum to total in the formula. It should! Clues integral 3 Probably a CH3 integral 2 Probably a CH2
http://www.chem.ucla.edu/~webspectra/ Work problems 5,6,7,11,12,16,17,19 for practive
C4H8O
C4H8O
C4H8O
C4H10O
C4H10O
C4H10O
C8H8O2
C8H8O2
C7H8O2
C7H8O2
C7H8O2
The Basic One-Pulse Experiment If we place a single nucleus with a spin into a magnetic field, B0, it will precess at a characteristic frequency, termed the Larmor frequency. z y α B0
x β
If we consider all nuclei, then we have many vectors, and we can consider the bulk Magnitization, M, of the sample, which is the sum of the individual spins. M z α B0
y x
β
Nα > Nβ Nα / Nβ = e-∆E/kT where k is Boltmann constant and T is temp (K). ∆E = hγB0/2π or ν = γB0/2π γ- magnetogyric ratio (nuclear constant) Apply a pulse of electromagnetic radiation, B1, at an angle perpendicular to the applied field, B0. All of the individual magnetic moments focus and become phase coherent. M tips away from the z axis and precesses at the Larmor frequency. α z
B0
α
z
M y
α
MZ
M y
x
β
x
β
MXY
The tip angle, α, depends on the power and duration of the pulse. We are interested in the MXY component of M since it the signal in this plane That gives rise to the NMR signal. The Rotating Frame of Reference M is rotating in the xy plane at the Larmor frequency. For protons in our NMR Spectrometer, this 300 MHz. This is too fast for our purposes, and we will get dizzy. Suppose we allow the x and y axes to rotate along with M and that we are riding on the xy plane along with M. To us, M is not moving at all, at least not in the xy plane. This is the rotating frame of reference and the axes are now defined as x’, y’ and z’. z’
α
z’
M y’
α x’
β
y’ x’
β
M
900 pulse
Now consider what happens to M after the pulse is turned off. It relaxes back to the original Boltzmann distribution. Looking along the x’ axis, the signal decays with time to x’=0. This is called the Free Induction Decay or FID. Intensity x’=0
If M is moving slightly faster than the rotating frame, i.e. it has a slightly different Larmor frequency, the signal will oscillate at a characteristic frequency. Intensity x’=0
If this FID is mathematically transformed by a process called Fourier Transformation, then a spectrum will be obtained.
Intensity FT
x’=0
3.90 4.00
Time, sec
3.95 3.85
3.90 3.80
3.85 3.75
3.703.80
Frequency, Hz
3.65 3.75
3.60 3.70
