Chapter 9

​

Nuclear Magnetic

Resonance and Mass

Spectrometry

Ch. 9 - 1

Created by

Professor William Tam & Dr. Phillis Chang

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

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

Ch. 9 - 4

• Classic methods for organic structure determination

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• Require large quantities of sample and are time consuming

Ch. 9 - 5

• Spectroscopic methods for organic structure determination
1. Mass Spectroscopy (MS)
• Molecular Mass & characteristic fragmentation pattern
2. Infrared Spectroscopy (IR)
• Characteristic functional groups
3. Ultraviolet Spectroscopy (UV)
• Characteristic chromophore
4. Nuclear Magnetic Resonance (NMR)

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

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

Ch. 9 - 8

• General steps for structure elucidation
3. From molecular formula
• Double bond equivalent (DBE)

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• Infrared spectroscopy (IR)
• Identify some specific functional groups
• e.g. C=O, C–O, O–H, COOH, NH₂ … etc.

Ch. 9 - 9

• General steps for structure elucidation
5. UV
• Sometimes useful especially for conjugated systems
• e.g. dienes, aromatics, enones

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• ¹H, ¹³C NMR and other advanced NMR techniques
• Full structure determination

Ch. 9 - 10

• Electromagnetic spectrum

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

Ch. 9 - 12

Ch. 9 - 13

1. The number of signals in the spectrum tells us how many different sets of protons there are in the molecule

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• 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

Ch. 9 - 14

3. The area under the signal tells us about how many protons there are in the set being measured

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• 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

Ch. 9 - 15

• Typical ¹H NMR spectrum
• Chemical Shift (δ)

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• Integration (areas of peaks ⇒ no. of H)

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• Multiplicity (spin-spin splitting) and coupling constant

Ch. 9 - 16

• Typical ¹H NMR spectrum

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

Ch. 9 - 18

Ch. 9 - 19

Ch. 9 - 20

• Normal range of ¹H NMR

Ch. 9 - 21

• Reference compound
• TMS = tetramethylsilane

​

as a reference standard (0 ppm)

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• 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. = 28^oC)

Ch. 9 - 22

• NMR solvent
• Normal NMR solvents should not contain hydrogen
• Common solvents
• CDCl₃
• C₆D₆
• CD₃OD
• CD₃COCD₃ (d₆-acetone)

Ch. 9 - 23

• The 300-MHz ¹H NMR spectrum of 1,4-dimethylbenzene

Ch. 9 - 24

2B. Integration of Signal Areas

�Integral Step Heights

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​

​

​

​

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• 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

Ch. 9 - 25

H^a

H^b

2 H^a

3 H^b

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

Ch. 9 - 27

• Examples

Ch. 9 - 28

Ch. 9 - 29

• Examples

Note: All H^b’s are chemically and magnetically equivalent.

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

Ch. 9 - 31

• Pascal’s Triangle

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​

• For

​

​

​

​

• For

Due to symmetry, H^a and H^b are identical

⇒ a singlet

H^a ≠ H^b

⇒ two doublets

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)

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• Use chemical shift tables or charts to correlate chemical shifts with possible structural environments

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

Ch. 9 - 34

• Example: ¹H NMR (300 MHz) of an unknown compound with molecular formula C₃H₇Br

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)

Ch. 9 - 36

δ (ppm): 3.4 1.8 1.1

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Integral: 2 2 3

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

Ch. 9 - 38

Complete structure:

• 2 H's from integration
• triplet

• 2 H's from integration
• sextet

• 3 H's from integration
• triplet

most upfield signal

most downfield

signal

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

Ch. 9 - 40

Ch. 9 - 41

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

Ch. 9 - 43

5. Detecting the Signal: Fourier �Transform NMR Spectrometers

Ch. 9 - 44

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

Ch. 9 - 46

Ch. 9 - 47

• Deshielding by electronegative groups

• Greater electronegativity
• Deshielding of the proton
• Larger δ

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)

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)

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

Shielded

(δ 2 – 3 ppm)

Ch. 9 - 51

• Aromatic system

Ch. 9 - 52

• e.g.

δ (ppm)

H^a & H^b: 7.9 & 7.4 (deshielded)

H^c & H^d: 0.91 – 1.2 (shielded)

Ch. 9 - 53

• Alkenes

Ch. 9 - 54

• Aldehydes

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. = 28^oC)

7. The Chemical Shift

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

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)

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• This is the origin of the delta scale for the expression of chemical shifts relative to TMS

Ch. 9 - 58

• For example, the chemical shift for benzene protons is 2181 Hz when the instrument is operating at 300 MHz. Therefore

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• The chemical shift of benzene protons in a 60 MHz instrument is 436 Hz:

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• 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)

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

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• Chemically equivalent protons are chemical shift equivalent in ¹H NMR spectra

8. Chemical Shift Equivalent and �Nonequivalent Protons

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

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• Homotopic hydrogens have identical environments and will have the same chemical shift. They are said to be chemical shift equivalent

Ch. 9 - 61

• The six hydrogens of ethane are homotopic and are, therefore, chemical shift equivalent
• Ethane, consequently, gives only one signal in its ¹H NMR spectrum

same compounds

same compounds

Ch. 9 - 62

• If replacing hydrogens by a different atom gives different compounds, the hydrogens are said to be heterotopic

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• Heterotopic atoms have different chemical shifts and are not chemical shift equivalent

Ch. 9 - 63

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

Ch. 9 - 64

• CH₃CH₂Br
• two sets of hydrogens that are heterotopic with respect to each other
• two ¹H NMR signals

Ch. 9 - 65

• Other examples

⇒ 2 ¹H NMR signals

⇒ 4 ¹H NMR signals

Ch. 9 - 66

• Other examples

⇒ 3 ¹H NMR signals

Ch. 9 - 67

• Application to ¹³C NMR spectroscopy
• Examples

⇒ 1 ¹³C NMR signal

⇒ 4 ¹³C NMR signals

Ch. 9 - 68

⇒ 5 ¹³C NMR signals

⇒ 4 ¹³C NMR signals

Ch. 9 - 69

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

Ch. 9 - 70

• Enantiotopic hydrogen atoms have the same chemical shift and give only one ¹H NMR signal:

enantiomer

enantiotopic

Ch. 9 - 71

diastereomers

diastereotopic

Ch. 9 - 72

diastereomers

diastereotopic

Ch. 9 - 73

• 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

Ch. 9 - 74

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

Ch. 9 - 75

9B. Splitting Tree Diagrams and the �Origin of Signal Splitting

• Splitting analysis for a doublet

Ch. 9 - 76

• Splitting analysis for a triplet

Ch. 9 - 77

• Splitting analysis for a quartet

Ch. 9 - 78

• 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

Ch. 9 - 79

9C. Coupling Constants – Recognizing �Splitting Patterns

Ch. 9 - 80

9D. The Dependence of Coupling �Constants on Dihedral Angle

• ³J values are related to the dihedral angle (φ)

Ch. 9 - 81

• Karplus curve

• φ ~0^o or 180^o

⇒ Maximum ³J value

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• φ ~90^o

⇒ ³J ~0 Hz

Ch. 9 - 82

• Karplus curve
• Examples

Ch. 9 - 83

• Karplus curve
• Examples

Ch. 9 - 84

9E. Complicating Features

• The 60 MHz ¹H NMR spectrum of ethyl chloroacetate

Ch. 9 - 85

• The 300 MHz ¹H NMR spectrum of ethyl chloroacetate

Ch. 9 - 86

9F. Analysis of Complex Interactions

Ch. 9 - 87

• The 300 MHz ¹H NMR spectrum of 1-nitropropane

Ch. 9 - 88

• 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

Ch. 9 - 89

• 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)

Ch. 9 - 90

Trick:

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• 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

Ch. 9 - 91

• Deuterium Exchange

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• 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)

Ch. 9 - 92

• 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

Ch. 9 - 93

• Phenols
• Intramolecular H-bonding causes downfield shift

12.1 ppm

Ch. 9 - 94

• 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

Ch. 9 - 95

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)

Ch. 9 - 96

• Note for a 200 MHz NMR (field strength 4.70 Tesla)

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• ¹H NMR ⇒ Frequency = 200 MHz

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• ¹³C NMR ⇒ Frequency = 50 MHz

Ch. 9 - 97

• Example:
• 2-Butanol

Proton-coupled

¹³C NMR spectrum

Ch. 9 - 98

• Example:
• 2-Butanol

Proton-decoupled

¹³C NMR spectrum

Ch. 9 - 99

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

Ch. 9 - 100

• Factors affecting chemical shift
1. Diamagnetic shielding due to bonding electrons
2. Paramagnetic shielding due to low-lying electronic excited state
3. 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)

Ch. 9 - 101

• Electronegative substituents cause downfield shift

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• Increase in relative atomic mass of substituent causes upfield shift

Ch. 9 - 102

• Hybridization of carbon

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• sp² > sp > sp³

123.3 ppm

71.9 ppm

5.7 ppm

Ch. 9 - 103

• Anisotropy effect

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• Shows shifts similar to the effect in ¹H NMR

shows large

upfield shift

Ch. 9 - 104

Ch. 9 - 105

Ch. 9 - 106

(a)

(b)

(c)

Ch. 9 - 107

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)

Ch. 9 - 108

Off-resonance decoupled ¹³C NMR

Broadband proton-decoupled ¹³C NMR

Ch. 9 - 109

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

Ch. 9 - 110

(a)

(b)

(c)

Ch. 9 - 111

• The broadband proton-decoupled ¹³C NMR spectrum of methyl methacrylate

Ch. 9 - 112

• HCOSY
• ¹H–¹H correlation spectroscopy

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• HETCOR
• Heteronuclear correlation spectroscopy

12. Two-Dimensional (2D) NMR

Techniques

Ch. 9 - 113

• HCOSY of 2-chloro-butane

H₂

H₁

H₁

H₃

H₃

H₄

H₄

H₂

Ch. 9 - 114

• HETCOR of 2-chloro-butane

C₁

C₂

C₃

C₄

Ch. 9 - 115

• Partial MS of octane (C₈H₁₈, M = 114)

13. An Introduction to Mass �Spectrometry

Ch. 9 - 116

• The M⁺ peak at 114 is referred to as the parent peak or molecular ion

• 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.

Ch. 9 - 117

• Masses are usually rounded off to whole numbers assuming:

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H = 1, C = 12, N = 14, O = 16, F = 19 etc.

Ch. 9 - 118

• 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

Ch. 9 - 119

• 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

Ch. 9 - 120

15. Depicting the Molecular Ion

Radical cations from ionization

of nonbonding on π electron

Ch. 9 - 121

• Ionization potentials of selected molecules

Ch. 9 - 122

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

Ch. 9 - 123

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:

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ArCH₂⁺ > CH₂=CHCH₂⁺ > 3^o > 2^o > 1^o > CH₃⁺

Ch. 9 - 124

• e.g.

X

• Site of ionization: n > π > σ

non-bonding

Ch. 9 - 125

• As the carbon skeleton becomes more highly branched, the intensity of the molecular ion peak decreases
• Butane vs. isobutane

a

b

Ch. 9 - 126

16B. Fragmentation of Longer Chain �and Branched Alkanes

• Octane vs. isooctane

Ch. 9 - 127

16C. Fragmentation to Form �Resonance-Stabilized Cations

• Alkenes
• Important fragmentation of terminal alkenes
• Allyl carbocation (m/e = 41)

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Ch. 9 - 128

• 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

Ch. 9 - 129

• Alcohols
• Most common fragmentation: - loss of alkyl groups

a

b

Ch. 9 - 130

• Carbon–carbon bonds next to the carbonyl group of an aldehyde or ketone break readily because resonance-stabilized ions called acylium ions are produced

Ch. 9 - 131

• Aldehydes
• M⁺ peak usually observed but may be fairly weak

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• Common fragmentation pattern
• α-cleavage

Ch. 9 - 132

• Ketones
• α-cleavage

a

a

b

b

Ch. 9 - 133

• 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

Ch. 9 - 134

• Aromatic hydrocarbons
• very intense M⁺ peaks
• characteristic fragmentation pattern (when an alkyl group attached to the benzene ring): - tropylium cation

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Ch. 9 - 135

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

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• May lose H₂O by 1,2- or 1,4-elimination

Ch. 9 - 136

Ch. 9 - 137

• Cycloalkenes show a characteristic fragmentation pattern which corresponds to a reverse Diels-Alder reaction

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​

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• e.g.

Ch. 9 - 138

• Aromatic hydrocarbons
• e.g.

Ch. 9 - 139

• Ketones
• McLafferty rearrangement

Ch. 9 - 140

Ch. 9 - 141

• Characteristic of McLafferty rearrangement
1. No alkyl migrations to C=O, only H migrates

X

Ch. 9 - 142

• Characteristic of McLafferty rearrangement
2. 2^o is preferred over 1^o

Ch. 9 - 143

17. How To Determine Molecular

Formulas and Molecular Weights

Using Mass Spectrometry

17A. Isotopic Peaks & the Molecular Ion

Ch. 9 - 144

• 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

Ch. 9 - 145

• The M + 1 peak can be used to determine the number of carbons in a molecule

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• The M + 2 peak can indicate whether bromine or chlorine is present

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• The isotopic peaks, in general, give us one method for determining molecular formulas

Ch. 9 - 146

• Example
• Consider 100 molecules of CH₄

C¹²: 100 C¹³: 1.11

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H¹: 100 H²: 0.016

Ch. 9 - 147

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

Ch. 9 - 148

≈

100

1.17

m/z

relative ion abundance

M

M +1

Ch. 9 - 149

17B. How To Determine the Molecular

Formula

Ch. 9 - 150

• 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)

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• For our unknown, M is even. The compound must have an even number of nitrogen atoms

Ch. 9 - 151

• 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

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• For our unknown

Ch. 9 - 152

• 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

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• The molecular formula can now be established by determining the number of hydrogen atoms and adding the appropriate number of oxygen atoms, if necessary

Ch. 9 - 153

• 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”

Ch. 9 - 154

• For a molecule composed of C and H only

H = 72 – (4 x 12) = 24

but C₄H₂₄ is impossible

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• 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

Ch. 9 - 155

17C. High-Resolution Mass Spectrometry

Ch. 9 - 156

• Example 1

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• O₂, N₂H₄ and CH₃OH all have M.W. of 32 (by MS), but accurate masses are different
• O₂ = 2(15.9949) = 31.9898

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• N₂H₄ = 2(14.0031) + 4(1.00783) = 32.0375

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• CH₄O = 12.00000 + 4(1.00783) + 15.9949 = 32.0262

Ch. 9 - 157

• Example 2

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• Both C₃H₈O and C₂H₄O₂ have M.W. of 60 (by MS), but accurate masses are different

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• C₃H₈O = 60.05754

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• C₂H₄O₂ = 60.02112

Ch. 9 - 158

18. Mass Spectrometer Instrument �Designs

Ch. 9 - 159

19. GC/MS Analysis

Ch. 9 - 160

🕭 END OF CHAPTER 9 🕭