Download - Organic Chemistry NMR Notes
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Created by Professor William Tam & Dr. Phillis Chang
Ch. 9 - 1
Chapter 9
Nuclear Magnetic Resonance and Mass
Spectrometry
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About The Authors
These PowerPoint Lecture Slides were created and prepared by Professor William Tam and his wife Dr. Phillis Chang. Professor William Tam received his B.Sc. at the University of Hong Kong in 1990 and his Ph.D. at the University of Toronto (Canada) in 1995. He was an NSERC postdoctoral fellow at the Imperial College (UK) and at Harvard University (USA). He joined the Department of Chemistry at the University of Guelph (Ontario, Canada) in 1998 and is currently a Full Professor and Associate Chair in the department. Professor Tam has received several awards in research and teaching, and according to Essential Science Indicators, he is currently ranked as the Top 1% most cited Chemists worldwide. He has published four books and over 80 scientific papers in top international journals such as J. Am. Chem. Soc., Angew. Chem., Org. Lett., and J. Org. Chem. Dr. Phillis Chang received her B.Sc. at New York University (USA) in 1994, her M.Sc. and Ph.D. in 1997 and 2001 at the University of Guelph (Canada). She lives in Guelph with her husband, William, and their son, Matthew.
Ch. 9 - 2
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Ch. 9 - 3
1. Introduction
Classic methods for organic structure determination ● Boiling point ● Refractive index ● Solubility tests ● Functional group tests ● Derivative preparation ● Sodium fusion (to identify N, Cl, Br, I &
S) ● Mixture melting point ● Combustion analysis ● Degradation
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Ch. 9 - 4
Classic methods for organic structure determination
● Require large quantities of sample and are time consuming
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Ch. 9 - 5
Spectroscopic methods for organic structure determination a) Mass Spectroscopy (MS) ● Molecular Mass & characteristic
fragmentation pattern b) Infrared Spectroscopy (IR) ● Characteristic functional groups
c) Ultraviolet Spectroscopy (UV) ● Characteristic chromophore
d) Nuclear Magnetic Resonance (NMR)
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Ch. 9 - 6
Spectroscopic methods for organic structure determination ● Combination of these
spectroscopic techniques provides a rapid, accurate and powerful tool for Identification and Structure Elucidation of organic compounds
● Rapid ● Effective in mg and microgram
quantities
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Ch. 9 - 7
General steps for structure elucidation 1. Elemental analysis ● Empirical formula ● e.g. C2H4O
2. Mass spectroscopy ● Molecular weight ● Molecular formula ● e.g. C4H8O2, C6H12O3 … etc. ● Characteristic fragmentation
pattern for certain functional groups
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Ch. 9 - 8
General steps for structure elucidation 3. From molecular formula ● Double bond equivalent (DBE)
4. Infrared spectroscopy (IR) ● Identify some specific
functional groups ● e.g. C=O, C–O, O–H, COOH,
NH2 … etc.
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Ch. 9 - 9
General steps for structure elucidation 5. UV ● Sometimes useful especially
for conjugated systems ● e.g. dienes, aromatics, enones
6. 1H, 13C NMR and other advanced NMR techniques ● Full structure determination
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Ch. 9 - 10
Electromagnetic spectrum
cosmic & γ-rays X-rays ultraviolet visible infrared micro-
waveradio-wave
1Å = 10-10m1nm = 10-9m1µm = 10-6m
λ: 0.1nm 200nm 400nm 800nm 50µm
X-RayCrystallography
UV IR NMR
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Ch. 9 - 11
2. Nuclear Magnetic Resonance (NMR) Spectroscopy
A graph that shows the characteristic energy absorption frequencies and intensities for a sample in a magnetic field is called a nuclear magnetic resonance (NMR) spectrum
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Ch. 9 - 12
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Ch. 9 - 13
1. The number of signals in the spectrum tells us how many different sets of protons there are in the molecule
2. The position of the signals in the spectrum along the x-axis tells us about the magnetic environment of each set of protons arising largely from the electron density in their environment
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Ch. 9 - 14
3. The area under the signal tells us about how many protons there are in the set being measured
4. The multiplicity (or splitting pattern) of each signal tells us about the number of protons on atoms adjacent to the one whose signal is being measured
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Ch. 9 - 15
Typical 1H NMR spectrum ● Chemical Shift (δ)
● Integration (areas of peaks ⇒ no. of H)
● Multiplicity (spin-spin splitting) and coupling constant
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Ch. 9 - 16
Typical 1H NMR spectrum
Record as: 1H NMR (300 MHz, CDCl3):
4.35 (2H, t, J = 7.2 Hz, Hc)2.05 (2H, sextet, J = 7.2 Hz, Hb)1.02 (3H, t, J = 7.2 Hz, Ha)
δ
chemicalshift (δ) in ppm no. of H
(integration) multiplicity
couplingconstantin Hz
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Ch. 9 - 17
2A. Chemical Shift The position of a signal along the x-axis of
an NMR spectrum is called its chemical shift
The chemical shift of each signal gives information about the structural environment of the nuclei producing that signal
Counting the number of signals in a 1H NMR spectrum indicates, at a first approximation, the number of distinct proton environments in a molecule
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Ch. 9 - 18
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Ch. 9 - 19
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Ch. 9 - 20
Normal range of 1H NMR
15 -10δ ppm
"upfield" (more shielded)"downfield" (deshielded)
(high field strength)
(low field strength)
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Ch. 9 - 21
Reference compound ● TMS = tetramethylsilane
as a reference standard (0 ppm)
● Reasons for the choice of TMS as reference Resonance position at higher field
than other organic compounds Unreactive and stable, not toxic Volatile and easily removed
(B.P. = 28oC)
Me
Si MeMe
Me
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Ch. 9 - 22
NMR solvent ● Normal NMR solvents should not
contain hydrogen ● Common solvents
CDCl3
C6D6
CD3OD
CD3COCD3 (d6-acetone)
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Ch. 9 - 23
The 300-MHz 1H NMR spectrum of 1,4-dimethylbenzene
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Ch. 9 - 24
2B. Integration of Signal Areas
Integral Step Heights
The area under each signal in a 1H NMR spectrum is proportional to the number of hydrogen atoms producing that signal
It is signal area (integration), not signal height, that gives information about the number of hydrogen atoms
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Ch. 9 - 25
O
Ha HaHb
HbHbR
Ha Hb
2 Ha 3 Hb
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Ch. 9 - 26
2C. Coupling (Signal Splitting)
Coupling is caused by the magnetic effect of nonequivalent hydrogen atoms that are within 2 or 3 bonds of the hydrogens producing the signal
The n+1 rule ● Rule of Multiplicity:
If a proton (or a set of magnetically equivalent nuclei) has n neighbors of magnetically equivalent protons. It’s multiplicity is n + 1
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Ch. 9 - 27
Examples
Hb C C Cl
HaHb
Hb Ha
Ha: multiplicity = 3 + 1 = 4 (a quartet)
Hb: multiplicity = 2 + 1 = 3 (a triplet)
(1)
Cl C C Cl
HbHa
Cl Hb
Ha: multiplicity = 2 + 1 = 3 (a triplet)
Hb: multiplicity = 1 + 1 = 2 (a doublet)
(2)
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Ch. 9 - 28
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Ch. 9 - 29
Examples
Note: All Hb’s are chemically and magnetically equivalent.
Hb C C Br
HaHb
Hb
Ha: multiplicity = 6 + 1 = 7 (a septet)
Hb: multiplicity = 1 + 1 = 2 (a doublet)
(3)
HbHb
Hb
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Ch. 9 - 30
Pascal’s Triangle ● Use to predict relative intensity of
various peaks in multiplet ● Given by the coefficient of
binomial expansion (a + b)n
singlet (s) 1 doublet (d) 1 1 triplet (t) 1 2 1 quartet (q) 1 3 3 1 quintet 1 4 6 4 1 sextet 1 5 10 10 5 1
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Ch. 9 - 31
Pascal’s Triangle
● For
● For
Br C C Br
HbHa
Cl Cl
Due to symmetry, Ha and Hb are identical ⇒ a singlet
Cl C C Br
HbHa
Cl Br
Ha ≠ Hb ⇒ two doublets
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Ch. 9 - 32
3. How to Interpret Proton NMR Spectra
1. Count the number of signals to determine how many distinct proton environments are in the molecule (neglecting, for the time being, the possibility of overlapping signals)
2. Use chemical shift tables or charts to correlate chemical shifts with possible structural environments
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Ch. 9 - 33
3. Determine the relative area of each signal, as compared with the area of other signals, as an indication of the relative number of protons producing the signal
4. Interpret the splitting pattern for each signal to determine how many hydrogen atoms are present on carbon atoms adjacent to those producing the signal and sketch possible molecular fragments
5. Join the fragments to make a molecule in a fashion that is consistent with the data
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Ch. 9 - 34
Example: 1H NMR (300 MHz) of an unknown compound with molecular formula C3H7Br
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Ch. 9 - 35
Three distinct signals at ~ δ3.4, 1.8 and 1.1 ppm ⇒ δ3.4 ppm: likely to be near an
electronegative group (Br)
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Ch. 9 - 36
δ (ppm): 3.4 1.8 1.1 Integral: 2 2 3
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Ch. 9 - 37
δ (ppm): 3.4 1.8 1.1
Multiplicity: triplet sextet triplet
2 H's on adjacent C
5 H's on adjacent C
2 H's on adjacent C
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Ch. 9 - 38
Complete structure:
BrCH2
CH2
CH3
• 2 H's from integration
• triplet
• 2 H's from integration
• sextet
• 3 H's from integration
• triplet
most upfield signal most downfield signal
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Ch. 9 - 39
4. Nuclear Spin: The Origin of the Signal
The magnetic field associated with a spinning
proton
The spinning proton
resembles a tiny bar magnet
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Ch. 9 - 40
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Ch. 9 - 41
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Ch. 9 - 42
Spin quantum number (I)
1H: I = ½ (two spin states: +½ or -½) ⇒ (similar for 13C, 19F, 31P)
12C, 16O, 32S: I = 0 ⇒ These nuclei do not give an NMR
spectrum
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Ch. 9 - 43
5. Detecting the Signal: Fourier Transform NMR Spectrometers
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Ch. 9 - 44
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Ch. 9 - 45
All protons do not absorb energy at the same frequency in a given external magnetic field
Lower chemical shift values correspond with lower frequency
Higher chemical shift values correspond with higher frequency
6. Shielding & Deshielding of Protons
15 -10δ ppm
"upfield" (more shielded)"downfield" (deshielded)
(high field strength)
(low field strength)
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Ch. 9 - 46
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Ch. 9 - 47
Deshielding by electronegative groups
CH3X
X = F OH Cl Br I H
Electro-negativity 4.0 3.5 3.1 2.8 2.5 2.1
δ (ppm) 4.26 3.40 3.05 2.68 2.16 0.23
● Greater electronegativity Deshielding of the proton Larger δ
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Ch. 9 - 48
Shielding and deshielding by circulation of π electrons ● If we were to consider only the
relative electronegativities of carbon in its three hybridization states, we might expect the following order of protons attached to each type of carbon:
(higher frequency) sp < sp2 < sp3 (lower
frequency)
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Ch. 9 - 49
● In fact, protons of terminal alkynes absorb between δ 2.0 and δ 3.0, and the order is
(higher frequency) sp2 < sp < sp3 (lower
frequency)
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Ch. 9 - 50
● This upfield shift (lower frequency) of the absorption of protons of terminal alkynes is a result of shielding produced by the circulating π electrons of the triple bond
H
Shielded (δ 2 – 3 ppm)
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Ch. 9 - 51
● Aromatic system
Shielded region
Deshielded region
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Ch. 9 - 52
● e.g.
HdHc
Hb
Ha
δ (ppm)
Ha & Hb: 7.9 & 7.4 (deshielded)
Hc & Hd: 0.91 – 1.2 (shielded)
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Ch. 9 - 53
● Alkenes
Deshielded (δ 4.5 – 7 ppm)
H
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Ch. 9 - 54
● Aldehydes
OR
H
Electronegativity effect + Anisotropy effect ⇒ δ = 8.5 – 10 ppm (deshielded)
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Ch. 9 - 55
Reference compound ● TMS = tetramethylsilane
as a reference standard (0 ppm)
● Reasons for the choice of TMS as reference Resonance position at higher field
than other organic compounds Unreactive and stable, not toxic Volatile and easily removed
(B.P. = 28oC)
Me
Si MeMe
Me
7. The Chemical Shift
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Ch. 9 - 56
7A. PPM and the δ Scale
The chemical shift of a proton, when expressed in hertz (Hz), is proportional to the strength of the external magnetic field
Since spectrometers with different magnetic field strengths are commonly used, it is desirable to express chemical shifts in a form that is independent of the strength of the external field
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Ch. 9 - 57
Since chemical shifts are always very small (typically 5000 Hz) compared with the total field strength (commonly the equivalent of 60, 300, or 600 million hertz), it is convenient to express these fractions in units of parts per million (ppm)
This is the origin of the delta scale for the expression of chemical shifts relative to TMS
δ =(observed shift from TMS in hertz) x 106
(operating frequency of the instrument in hertz)
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Ch. 9 - 58
For example, the chemical shift for benzene protons is 2181 Hz when the instrument is operating at 300 MHz. Therefore
The chemical shift of benzene protons in a 60 MHz instrument is 436 Hz:
Thus, the chemical shift expressed in ppm is the same whether measured with an instrument operating at 300 or 60 MHz (or any other field strength)
δ =2181 Hz x 106
300 x 106 Hz= 7.27 ppm
δ =436 Hz x 106
60 x 106 Hz= 7.27 ppm
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Ch. 9 - 59
Two or more protons that are in identical environments have the same chemical shift and, therefore, give only one 1H NMR signal
Chemically equivalent protons are chemical shift equivalent in 1H NMR spectra
8. Chemical Shift Equivalent and Nonequivalent Protons
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Ch. 9 - 60
8A. Homotopic and Heterotopic Atoms
If replacing the hydrogens by a different atom gives the same compound, the hydrogens are said to be homotopic
Homotopic hydrogens have identical environments and will have the same chemical shift. They are said to be chemical shift equivalent
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Ch. 9 - 61
H
C CH
H
H
H
H
Ethane
H
C CH
H
H
H
Br
H
C CH
H
Br
H
H
H
C CH
H
H
Br
HH
CC H
H
H
Br
H
H
CC H
H
Br
H
H
H
CC H
H
H
H
Br
The six hydrogens of ethane are homotopic and are, therefore, chemical shift equivalent
Ethane, consequently, gives only one signal in its 1H NMR spectrum
sam
e co
mp
ou
nd
s sam
e com
po
un
ds
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Ch. 9 - 62
If replacing hydrogens by a different atom gives different compounds, the hydrogens are said to be heterotopic
Heterotopic atoms have different chemical shifts and are not chemical shift equivalent
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Ch. 9 - 63
H
C CH
H
H
H
Br
H
C CH
H
H
Cl
BrBr
CC H
H
H
Cl
H
Br
CC H
H
H
Cl
H
Br
CC H
H
Cl
H
H
Br
CC H
H
H
H
Cl
These 2 H’s are also homotopic to each other
different compounds ⇒ heterotopic
same compounds ⇒ these 3 H’s of the CH3 group are homotopic ⇒ the CH3 group gives only one 1H NMR signal
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Ch. 9 - 64
H
C CH
H
H
H
Br
CH3CH2Br ● two sets of hydrogens that are
heterotopic with respect to each other ● two 1H NMR signals
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Ch. 9 - 65
Other examples
(1) C C
H
H
CH3
CH3⇒ 2 1H NMR signals
(2) H
CH3H
H
H CH3
⇒ 4 1H NMR signals
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Other examples
(3) H3CCH3
H H
H
H
H H
⇒ 3 1H NMR signals
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Application to 13C NMR spectroscopy ● Examples
(1) H3C CH3 ⇒ 1 13C NMR signal
(2)
CH3
CH3⇒ 4 13C NMR signals
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(3)
OHHO
⇒ 5 13C NMR signals
(4)
OH
HO⇒ 4 13C NMR signals
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8B. Enantiotopic and Diastereotopic Hydrogen Atoms
If replacement of each of two hydrogen atoms by the same group yields compounds that are enantiomers, the two hydrogen atoms are said to be enantiotopic
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Enantiotopic hydrogen atoms have the same chemical shift and give only one 1H NMR signal:
H3C Br
H H
H3C Br
H G
H3C Br
G H
enantiomer
enantiotopic
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CH3
H OHH3C
HaHb
dias
tere
omer
s
diastereotopic
CH3
H OHH3C
GHb
CH3
H OHH3C
HaG
chirality centre
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HbBr
Ha
H
dias
tere
omer
s
diastereotopic
HbBr
G
H
GBr
Ha
H
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Vicinal coupling is coupling between hydrogen atoms on adjacent carbons (vicinal hydrogens), where separation between the hydrogens is by three σ bonds
9. Signal Splitting: Spin–Spin Coupling
Ha Hb
3J or vicinal coupling
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9A. Vicinal Coupling
Vicinal coupling between heterotopic protons generally follows the n + 1 rule. Exceptions to the n + 1 rule can occur when diastereotopic hydrogens or conformationally restricted systems are involved
Signal splitting is not observed for protons that are homotopic (chemical shift equivalent) or enantiotopic
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9B. Splitting Tree Diagrams and the Origin of Signal Splitting
Splitting analysis for a doublet
C C
HaHb
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Splitting analysis for a triplet
C
HbC
HaHb
C C C
HaHb Hb
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Splitting analysis for a quartet
Hb C
HbC
HaHb
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Pascal’s Triangle ● Use to predict relative intensity of
various peaks in multiplet ● Given by the coefficient of
binomial expansion (a + b)n
singlet (s) 1 doublet (d) 1 1 triplet (t) 1 2 1 quartet (q) 1 3 3 1 quintet 1 4 6 4 1 sextet 1 5 10 10 5 1
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9C. Coupling Constants – Recognizing Splitting Patterns
X C
HaC
HbHb
HbHa
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9D. The Dependence of Coupling Constants on Dihedral Angle
3J values are related to the dihedral angle (φ)
H
H
φ
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Karplus curve
φ ~0o or 180o
⇒ Maximum 3J value
φ ~90o
⇒ 3J ~0 Hz
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Karplus curve ● Examples
Hb
Ha
Hb
Ha
φ = 180ºJa,b = 10-14 Hz
(axial, axial)
Hb
Ha
Hb
Ha
φ = 60ºJa,b = 4-5 Hz
(equatorial, equatorial)
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Karplus curve ● Examples
Hb
Ha
Hb
Ha
φ = 60ºJa,b = 4-5 Hz
(equatorial, axial)
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9E. Complicating Features
The 60 MHz 1H NMR spectrum of ethyl chloroacetate
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The 300 MHz 1H NMR spectrum of ethyl chloroacetate
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9F. Analysis of Complex Interactions
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The 300 MHz 1H NMR spectrum of 1-nitropropane
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Protons of alcohols (ROH) and amines may appear over a wide range from 0.5 – 5.0 ppm ● Hydrogen-bonding is the reason for this
range
10. Proton NMR Spectra and Rate Processes
in high dilution (free OH):
δ = ~0.5-1.0 ppm
in conc. solution (H-bonded):
H O
R
HO
R
H OR
δ−
δ+
proton more deshielded
R O H
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Why don’t we see coupling with the O–H proton, e.g. –CH2–OH (triplet?) ● Because the acidic protons are
exchangeable about 105 protons per second (residence time 10-5 sec), but the NMR experiment requires a time of 10-2 – 10-3 sec. to “take” a spectrum, usually we just see an average (thus, OH protons are usually a broad singlet)
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Trick:
● Run NMR in d6-DMSO where H-bonding with DMSO’s oxygen prevents H’s from exchanging and we may be able to see the coupling
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Deuterium Exchange
● To determine which signal in the NMR spectrum is the OH proton, shake the NMR sample with a drop of D2O and whichever peak disappears that is the OH peak (note: a new peak of HOD appears)
D2O+ HODR O H R O D
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Phenols ● Phenol protons appear downfield
at 4-7 ppm ● They are more “acidic” - more H+
character ● More dilute solutions - peak
appears upfield: towards 4 ppm
OH O H
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Phenols ● Intramolecular H-bonding causes
downfield shift
O
HO
12.1 ppm
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Unlike 1H with natural abundance ~99.98%, only 1.1% of carbon, namely 13C, is NMR active
11. Carbon-13 NMR Spectroscopy
11A. Interpretation of 13C NMR Spectra
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11B. One Peak for Each Magnetically Distinct Carbon Atom
13C NMR spectra have only become commonplace more recently with the introduction of the Fourier Transform (FT) technique, where averaging of many scans is possible (note 13C spectra are 6000 times weaker than 1H spectra, thus require a lot more scans for a good spectrum)
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Note for a 200 MHz NMR (field strength 4.70 Tesla)
● 1H NMR ⇒ Frequency = 200 MHz
● 13C NMR ⇒ Frequency = 50 MHz
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CH3 C CH2 CH3
H
OH
Example: ● 2-Butanol
Proton-coupled 13C NMR spectrum
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CH3 C CH2 CH3
H
OH
Example: ● 2-Butanol
Proton-decoupled 13C NMR spectrum
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11C. 13C Chemical Shifts
Decreased electron density around an atom deshields the atom from the magnetic field and causes its signal to occur further downfield (higher ppm, to the left) in the NMR spectrum
Relatively higher electron density around an atom shields the atom from the magnetic field and causes the signal to occur upfield (lower ppm, to the right) in the NMR spectrum
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Factors affecting chemical shift i. Diamagnetic shielding due to bonding
electrons ii. Paramagnetic shielding due to low-lying
electronic excited state iii. Magnetic Anisotropy – through space
due to the near-by group (especially π electrons)
In 1H NMR, (i) and (iii) most significant; in 13C NMR, (ii) most significant (since chemical shift range >> 1H NMR)
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Electronegative substituents cause downfield shift
Increase in relative atomic mass of substituent causes upfield shift
X
Cl
Br
I
Electronegativity
2.8
2.7
2.2
Atomic Mass
35.5
79.9
126.9
13C NMR: CH3X
23.9 ppm
9.0 ppm
-21.7 ppm
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Hybridization of carbon
● sp2 > sp > sp3
H2C CH2 HC CH H3C CH3
e.g.
123.3 ppm 71.9 ppm 5.7 ppm
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Anisotropy effect
● Shows shifts similar to the effect in 1H NMR
C C
e.g.
C
shows large upfield shift
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Cl CH2 CH CH3
OH
(a) (b) (c)
1-Chloro-2-propanol
(a) (b)
(c)
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11D. Off-Resonance Decoupled Spectra NMR spectrometers can differentiate among carbon
atoms on the basis of the number of hydrogen atoms that are attached to each carbon
In an off-resonance decoupled 13C NMR spectrum, each carbon signal is split into a multiplet of peaks, depending on how many hydrogens are attached to that carbon. An n + 1 rule applies, where n is the number of hydrogens on the carbon in question. Thus, a carbon with no hydrogens produces a singlet (n = 0), a carbon with one hydrogen produces a doublet (two peaks), a carbon with two hydrogens produces a triplet (three peaks), and a methyl group carbon produces a quartet (four peaks)
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Off-resonance decoupled 13C NMR
N
N
O
1
2
3 45
6
789
Broadband proton-decoupled 13C NMR
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11E. DEPT 13C Spectra
DEPT 13C NMR spectra indicate how many hydrogen atoms are bonded to each carbon, while also providing the chemical shift information contained in a broadband proton-decoupled 13C NMR spectrum. The carbon signals in a DEPT spectrum are classified as CH3, CH2, CH, or C accordingly
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Cl CH2 CH CH3
OH
(a) (b) (c)
1-Chloro-2-propanol
(a) (b) (c)
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The broadband proton-decoupled 13C NMR spectrum of methyl methacrylate
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HCOSY ● 1H–1H correlation spectroscopy
HETCOR ● Heteronuclear correlation
spectroscopy
12. Two-Dimensional (2D) NMR Techniques
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HCOSY of 2-chloro-butane
H2
H1
H1
H3
H3
H4
H4
H2
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HETCOR of 2-chloro-butane
H1
H2
H3
H4
C1
C2
C3 C
4
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Partial MS of octane (C8H18, M = 114)
13. An Introduction to Mass Spectrometry
11485
71
57M+
29 (CH3CH2)14 (CH2)
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The M+ peak at 114 is referred to as the parent peak or molecular ion
C8H18e-
70 eV+ 2 e-[C8H18]
(M+)
The largest or most abundant peak is called the base peak and is assigned an intensity of 100%, other peaks are then fractions of that e.g. 114(M+,40), 85(80), 71(60), 57(100) etc.
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Masses are usually rounded off to whole numbers assuming:
H = 1, C = 12, N = 14, O = 16, F = 19 etc.
Molecular ion (parent peak)
Daughter ions [C8H18]
(M+, 114)[C6H13]
(85)
fragmentation
-CH3CH2 (29)
[C5H11](71)
-CH3CH2CH2 (29+14)
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In the mass spectrometer, a molecule in the gaseous phase under low pressure is bombarded with a beam of high-energy electrons (70 eV or ~ 1600 kcal/mol)
This beam can dislodge an electron from a molecule to give a radical cation which is called the molecular ion, M+ or more accurately
14. Formation of Ions: Electron Impact Ionization
M 70 eV e-M
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This molecular ion has considerable surplus energy so it can fly apart or fragment to give specific ions which may be diagnostic for a particular compound
M A B C- m1º - m2º - m3º
mº = neutral fragment radical
etc.
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15. Depicting the Molecular Ion
CH3CH2 CH3
H3C OH H3C N CH3
CH3
H2C CHCH2CH3
Methanol Trimethylamine 1-Butene
Radical cations from ionization of nonbonding on π electron
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Compound
Ionization Potential (eV)
CH3(CH2)3NH2 8.7
C6H6 (benzene) 9.2
C2H4 10.5
CH3OH 10.8
C2H6 11.5
CH4 12.7
Ionization potentials of selected molecules
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16. Fragmentation
1. The reactions that take place in a mass spectrometer are unimolecular, that is, they do not involve collisions between molecules or ions. This is true because the pressure is kept so low (10-6 torr) that reactions involving bimolecular collisions do not occur
2. We use single-barbed arrows to depict mechanisms involving single electron movements
3. The relative ion abundances, as indicated by peak intensities, are very important
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16A. Fragmentation by Cleavage at a Single Bond
When a molecular ion fragments, it will yield a neutral radical (not detected) and a carbocation (detected) with an even number of electrons
The fragmentation will be dictated to some extent by the fragmention of the more stable carbocation:
ArCH2+ > CH2=CHCH2
+ > 3o > 2o > 1o > CH3+
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e.g.
R CH
R+ CH3+
R +CH3+X
● Site of ionization: n > π > σ
non-bonding
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As the carbon skeleton becomes more highly branched, the intensity of the molecular ion peak decreases
Butane vs. isobutane
70eVe-
M+(58)
70eVe-
M+(58)
a CH3+
(43)a
b CH2CH3+(29)
b
CH3+(43)
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16B. Fragmentation of Longer Chain and Branched Alkanes
Octane vs. isooctane
M+(114)
(85)
(71)
(57)
(43)M+(114)
+
+
+
+
+(57)
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16C. Fragmentation to Form Resonance-Stabilized Cations
Alkenes ● Important fragmentation of
terminal alkenes Allyl carbocation (m/e = 41)
R
(41)
R +
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Carbon–carbon bonds next to an atom with an unshared electron pair usually break readily because the resulting carbocation is resonance stabilized
Ethers ● Cleavage α (to ether oxygen) C–C bonds
O
O
(m/e = 59)
+ OCH3
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Alcohols ● Most common fragmentation: - loss
of alkyl groups
OH
M+(74)
CH3+OHOH
a(m/e = 59)
a
OH OHCH3CH2 +b
(m/e = 45)
b
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Carbon–carbon bonds next to the carbonyl group of an aldehyde or ketone break readily because resonance-stabilized ions called acylium ions are produced
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Aldehydes ● M+ peak usually observed but may
be fairly weak
● Common fragmentation pattern α-cleavage
RR H
OH C OR
C OH
+
+(m/e = 29)
acylium ion
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Ketones ● α-cleavage
O a
a
b
b
O+
(m/e = 71)
O+
(m/e = 99)
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Alkyl-substituted benzenes ionize by loss of a π electron and undergo loss of a hydrogen atom or methyl group to yield the relatively stable tropylium ion (see Section 14.7C). This fragmentation gives a prominent peak (sometimes the base peak) at m/z 91
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Aromatic hydrocarbons ● very intense M+ peaks ● characteristic fragmentation
pattern (when an alkyl group attached to the benzene ring): - tropylium cation
CH3CH2
CH3
(m/e = 91)tropylium cation
rearrangement+
benzyl cation
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16D. Fragmentation by Cleavage of Two Bonds
Alcohols frequently show a prominent peak at M - 18. This corresponds to the loss of a molecule of water
● May lose H2O by 1,2- or 1,4-elimination
+●
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1,2-elimination: OH+ H2O
M (M - 18)
1,4-elimination:OH H OH
+ H2O
+ CH3CH2
M
(M - 18)
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Cycloalkenes show a characteristic fragmentation pattern which corresponds to a reverse Diels-Alder reaction
e.g.
retro Diels-Alder+
+
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H
CH2
HH
(m/e = 92)
+
McLafferty Rearrangement
Aromatic hydrocarbons ● e.g.
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Ketones ● McLafferty rearrangement
OH
OH OH
OHOH
OH
+
(m/e = 86)
(m/e = 58)
+
1st McL. Rearr.
2nd McL. Rearr.
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OH H
OH OH
(m/e = 86)2º radical
observedi
i
OH
1º radical
OH
(m/e = 114)NOT observed
ii
ii
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Characteristic of McLafferty rearrangement 1. No alkyl migrations to C=O, only H
migrates
OH
O
O
R
R
R
H
HX
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Characteristic of McLafferty rearrangement 2. 2o is preferred over 1o
OH H
iiiOH
2º radical
OH
1º radical
not
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17. How To Determine Molecular Formulas and Molecular Weights Using Mass Spectrometry
17A. Isotopic Peaks & the Molecular Ion
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The presence of isotopes of carbon, hydrogen, and nitrogen in a compound gives rise to a small M + 1 peak
The presence of oxygen, sulfur, chlorine, or bromine in a compound gives rise to an M + 2 peak
M + 1 Elements:
M + 2 Elements:
C, H, N
O, S, Br, Cl
+●
+●
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The M + 1 peak can be used to determine the number of carbons in a molecule
The M + 2 peak can indicate whether bromine or chlorine is present
The isotopic peaks, in general, give us one method for determining molecular formulas
+●
+●
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Example ● Consider 100 molecules of CH4
M : 16
H1
C12H1 H1
H1
H1
C13H1 H1
H1
H1
C12H1 H2
H1
M + 1 = 17
C12: 100 C13: 1.11 H1: 100 H2: 0.016
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M : 16
H1
C12H1 H1
H1
H1
C13H1 H1
H1
H1
C12H1 H2
H1
M + 1 = 17
1.11 molecules contain a 13C atom
4x0.016 = 0.064 molecules contain a 2H atom
Intensity of M + 1 peak: 1.11+0.064=1.174% of the M peak
+●+●
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≈
100
1.17
m/z
rela
tive
ion
abun
danc
e M
M +1
+●
+●
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17B. How To Determine the Molecular Formula
m/z Intensity
(% of M )
72 73.0/73 x 100 = 100
73 3.3/73 x 100 = 4.5
74 0.2/73 x 100 = 0.3
+●
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Is M odd or even? According to the nitrogen rule, if it is even, then the compound must contain an even number of nitrogen atoms (zero is an even number)
● For our unknown, M is even. The compound must have an even number of nitrogen atoms
+●
+●
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The relative abundance of the M +1 peak indicates the number of carbon atoms. Number of C atoms = relative abundance of (M +1)/1.1
● For our unknown
Number of C atoms =4.5
1.1~ 4
+●
+●
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The relative abundance of the M +2 peak indicates the presence (or absence) of S (4.4%), Cl (33%), or Br (98%) ● For our unknown M +2 = 0.3%; thus,
we can assume that S, Cl, and Br are absent
The molecular formula can now be established by determining the number of hydrogen atoms and adding the appropriate number of oxygen atoms, if necessary
+●
+●
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Since M is m/z 72 ⇒ molecular weight = 72
As determined using the relative abundance of M +1 peak, number of carbons present is 4
Using the “nitrogen rule”, this unknown must have an even number of N. Since M.W. = 72, and there are 4 C present, (12 x 4 = 48), adding 2 “N” will be greater than the M.W. of the unknown. Thus, this unknown contains zero “N”
+●
+●
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For a molecule composed of C and H only
H = 72 – (4 x 12) = 24
but C4H24 is impossible
For a molecule composed of C, H and O
H = 72 – (4 x 12) – 16 = 8
and thus our unknown has the molecular formula C4H8O
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17C. High-Resolution Mass Spectrometry
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Example 1
● O2, N2H4 and CH3OH all have M.W. of 32 (by MS), but accurate masses are different O2 = 2(15.9949) = 31.9898
N2H4 = 2(14.0031) + 4(1.00783) = 32.0375
CH4O = 12.00000 + 4(1.00783) + 15.9949 = 32.0262
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Ch. 9 - 157
Example 2
● Both C3H8O and C2H4O2 have M.W. of 60 (by MS), but accurate masses are different
C3H8O = 60.05754
C2H4O2 = 60.02112
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18. Mass Spectrometer Instrument Designs
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19. GC/MS Analysis
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END OF CHAPTER 9