created by professor william tam & dr. phillis chang ch. 9 - 1 chapter 9 nuclear magnetic...

Created byProfessor William Tam & Dr. Phillis

Chang Ch. 9 - 1

Chapter 9

Nuclear MagneticResonance and Mass

Spectrometry

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 -

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

●Require large quantities of sample and are time consuming

Ch. 9 - 5

Spectroscopic methods for organic structure determinationa) Mass Spectroscopy (MS)

● Molecular Mass & characteristic fragmentation pattern

b) Infrared Spectroscopy (IR)● Characteristic functional

groupsc) Ultraviolet Spectroscopy (UV)

● Characteristic chromophored) Nuclear Magnetic Resonance

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)

4. 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

6. 1H, 13C NMR and other advanced NMR techniques● Full structure determination

Ch. 9 - 10

Electromagnetic spectrum

cosmic & -rays

X-rays ultraviolet visible infrared micro-wave

radio-wave

1Å = 10-10m

1nm = 10-9m

1m = 10-6m

0.1nm 200nm 400nm 800nm 50m

X-RayCrystallography

UV IR NMR

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

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

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

Ch. 9 - 15

Typical 1H NMR spectrum● Chemical Shift ()

● Integration (areas of peaks no. of H)

● Multiplicity (spin-spin splitting) and coupling constant

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)

chemicalshift () in ppm

no. of H(integration) multiplicity

couplingconstantin Hz

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

15 -10 ppm

"upfield" (more shielded)

"downfield" (deshielded)

(high field strength)

(low field strength)

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)

Si MeMe

Ch. 9 - 22

NMR solvent● Normal NMR solvents should

not contain hydrogen● Common solvents

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

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

2 Ha 3 Hb

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

Hb C C Cl

Ha: multiplicity = 3 + 1 = 4 (a quartet)

Hb: multiplicity = 2 + 1 = 3 (a triplet)

Cl C C Cl

Ha: multiplicity = 2 + 1 = 3 (a triplet)

Hb: multiplicity = 1 + 1 = 2 (a doublet)

Ch. 9 - 28

Ch. 9 - 29

Examples

Note: All Hb’s are chemically and magnetically equivalent.

Hb C C Br

Ha: multiplicity = 6 + 1 = 7 (a septet)

Hb: multiplicity = 1 + 1 = 2 (a doublet)

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) 1doublet (d) 1 1triplet (t) 1 2 1quartet (q) 1 3 3 1quintet 1 4 6 4 1sextet 1 5 10 10 5 1

Ch. 9 - 31

Pascal’s Triangle

● For

Br C C Br

Due to symmetry, Ha and Hb are identical a singlet

Cl C C Br

Ha ≠ Hb

two doublets

Ch. 9 - 32

3. How to Interpret Proton NMRSpectra

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

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 ~ d3.4, 1.8 and 1.1 ppm d3.4 ppm: likely to be near an

electronegative group (Br)

Ch. 9 - 36

d (ppm): 3.4 1.8 1.1

Integral: 2 2 3

Ch. 9 - 37

d (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

• sextet

• triplet

most upfield signalmost downfieldsignal

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

15 -10 ppm

"upfield" (more shielded)

"downfield" (deshielded)

(high field strength)

(low field strength)

Ch. 9 - 46

Ch. 9 - 47

Deshielding by electronegative groupsCH3X

X = F OH Cl Br I H

Electro-negativity

4.0 3.5 3.1 2.8 2.5 2.1

d (ppm)4.26

●Greater electronegativity Deshielding of the proton Larger d

Ch. 9 - 48

Shielding and deshielding by circulation of p 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 d 2.0 and d 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 p electrons of the triple bond

Shielded( 2 – 3 ppm)

Ch. 9 - 51

●Aromatic system

Shielded region

Deshielded region

Ch. 9 - 52

●e.g.

d (ppm)

Ha & Hb: 7.9 & 7.4 (deshielded)

Hc & Hd: 0.91 – 1.2 (shielded)

Ch. 9 - 53

●Alkenes

Deshielded( 4.5 – 7 ppm)

Ch. 9 - 54

●Aldehydes

Electronegativity effect + Anisotropy effect = 8.5 – 10 ppm (deshielded)

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)

Si MeMe

7. The Chemical Shift

Ch. 9 - 56

7A.PPM and the d 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)

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

(observed shift from TMS in hertz) x 106

(operating frequency of the instrument in hertz)

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)

=2181 Hz x 106

300 x 106 Hz= 7.27 ppm

=436 Hz x 106

60 x 106 Hz= 7.27 ppm

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

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

Ch. 9 - 61

Ethane

The six hydrogens of ethane are homotopic and are, therefore, chemical shift equivalent

Ethane, consequently, gives only one signal in its 1H NMR spectrum

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

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

(1) C C

2 1H NMR signals

4 1H NMR signals

Ch. 9 - 66

Other examples

(3) H3CCH3

3 1H NMR signals

Ch. 9 - 67

Application to 13C NMR spectroscopy●Examples

(1) H3C CH3 1 13C NMR signal

CH3 4 13C NMR signals

Ch. 9 - 68

5 13C NMR signals

HO 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:

H3C Br

enantiomer

enantiotopic

Ch. 9 - 71

H OHH3C

diastereotopic

H OHH3C

chiralitycentre

Ch. 9 - 72

diastereotopic

Ch. 9 - 73

Vicinal coupling is coupling between hydrogen atoms on adjacent carbons (vicinal hydrogens), where separation between the hydrogens is by three s bonds

9. Signal Splitting:Spin–Spin Coupling

3J or vicinal 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

HaHb Hb

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) 1doublet (d) 1 1triplet (t) 1 2 1quartet (q) 1 3 3 1quintet 1 4 6 4 1sextet 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 ()

Ch. 9 - 81

Karplus curve

~0o or 180o

Maximum

3J value

3J ~0 Hz

Ch. 9 - 82

Karplus curve●Examples

= 180º

Ja,b = 10-14 Hz

(axial, axial)

= 60º

Ja,b = 4-5 Hz

(equatorial, equatorial)

Ch. 9 - 83

Karplus curve●Examples

= 60º

Ja,b = 4-5 Hz

(equatorial, axial)

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

in high dilution (free OH):

= ~0.5-1.0 ppm

in conc. solution (H-bonded):

proton more deshielded

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:

●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

●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

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

OH O H

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)

●1H NMR Frequency = 200 MHz

●13C NMR Frequency = 50 MHz

Ch. 9 - 97

CH3 C CH2 CH3

Example:●2-Butanol

Proton-coupled13C NMR spectrum

Ch. 9 - 98

CH3 C CH2 CH3

Example:●2-Butanol

Proton-decoupled13C 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 shifti. 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 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

Increase in relative atomic mass of substituent causes upfield shift

Electronegativity

Atomic Mass

13C NMR: CH3X

23.9 ppm

9.0 ppm

-21.7 ppm

Ch. 9 - 102

Hybridization of carbon

● sp2 > sp > sp3

H2C CH2 HC CH H3C CH3

123.3 ppm 71.9 ppm 5.7 ppm

Ch. 9 - 103

Anisotropy effect

● Shows shifts similar to the effect in 1H NMR

shows largeupfield shift

Ch. 9 - 104

Ch. 9 - 105

Ch. 9 - 106

Cl CH2 CH CH3

(a) (b) (c)

1-Chloro-2-propanol

(a)(b)

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

Cl CH2 CH CH3

(a) (b) (c)

1-Chloro-2-propanol

(a)(b)(c)

Ch. 9 - 111

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

Ch. 9 - 112

HCOSY●1H–1H correlation spectroscopy

HETCOR●Heteronuclear correlation

spectroscopy

12.Two-Dimensional (2D) NMR Techniques

Ch. 9 - 113

HCOSY of 2-chloro-butane

Ch. 9 - 114

HETCOR of 2-chloro-butane

Ch. 9 - 115

Partial MS of octane (C8H18, M = 114)

13.An Introduction to Mass Spectrometry

29 (CH3CH2)

14 (CH2)

Ch. 9 - 116

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

C8H18e-

70 eV+ 2 e-[C8H18]

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:

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

Molecular ion (parent peak)

Daughterions[C8H18]

(M+, 114)

[C6H13]

fragmentation

-CH3CH2 (29)

[C5H11]

(71)-CH3CH2CH2 (29+14)

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

M70 eV e-

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

M A B C- m1º - m2º - m3º

mº = neutral fragment radical

Ch. 9 - 120

15.Depicting the Molecular Ion

CH3CH2 CH3

H3C OH H3C N CH3

H2C CHCH2CH3

Methanol Trimethylamine 1-Butene

Radical cations from ionization

of nonbonding on p electron

Ch. 9 - 121

CompoundIonization

Potential (eV)

CH3(CH2)3NH2 8.7

C6H6 (benzene)

C2H4 10.5

CH3OH 10.8

C2H6 11.5

CH4 12.7

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:

ArCH2+ > CH2=CHCH2

+ > 3o > 2o > 1o >

Ch. 9 - 124

R+ CH3+

R +CH3+X

●Site of ionization: n > p > s

non-bonding

Ch. 9 - 125

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)

b CH2CH3+(29)

CH3+(43)

Ch. 9 - 126

16B.Fragmentation of Longer Chain and Branched Alkanes

Octane vs. isooctane

M+(114)

Ch. 9 - 127

16C.Fragmentation to Form Resonance-Stabilized Cations

Alkenes●Important fragmentation of

terminal alkenes Allyl carbocation (m/e = 41)

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 a (to ether oxygen) C–C

bondsO

(m/e = 59)

+ OCH3

Ch. 9 - 129

Alcohols●Most common fragmentation: -

loss of alkyl groups

M+(74)

CH3+OHOH

a(m/e = 59)

OH OHCH3CH2 +

(m/e = 45)

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

●Common fragmentation pattern a-cleavage

OH C OR

+(m/e = 29)

acylium ion

Ch. 9 - 132

Ketones●a-cleavage

(m/e = 71)

(m/e = 99)

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

CH3CH2

(m/e = 91)tropylium cation

rearrangement+

benzyl cation

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

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

Ch. 9 - 136

1,2-elimination:OH

M (M - 18)

1,4-elimination:OH H OH

+ CH3CH2

(M - 18)

Ch. 9 - 137

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

retro Diels-Alder+

Ch. 9 - 138

H(m/e = 92)

McLafferty Rearrangement

Aromatic hydrocarbons●e.g.

Ch. 9 - 139

Ketones●McLafferty rearrangement

HOH OH

(m/e = 86)

(m/e = 58)

1st McL. Rearr.

2nd McL. Rearr.

Ch. 9 - 140

(m/e = 86)2º radical

observedi

1º radical

(m/e = 114)NOT observed

Ch. 9 - 141

Characteristic of McLafferty rearrangement1. No alkyl migrations to C=O,

only H migrates

Ch. 9 - 142

Characteristic of McLafferty rearrangement2. 2o is preferred over 1o

2º radical

1º radical

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 peakM + 1 Elements:

M + 2 Elements:

C, H, N

O, S, Br, Cl

Ch. 9 - 145

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

Ch. 9 - 146

Example●Consider 100 molecules of CH4

M : 16

C12H1 H1

C13H1 H1

C12H1 H2

M + 1 = 17

C12: 100 C13: 1.11

H1: 100 H2: 0.016

Ch. 9 - 147

M : 16

C12H1 H1

C13H1 H1

C12H1 H2

M + 1 = 17

1.11 molecules contain a 13C

4x0.016 = 0.064 molecules contain a 2H

Intensity of M + 1 peak:1.11+0.064=1.174% of the M

Ch. 9 - 148

tive ion a

bundance

Ch. 9 - 149

17B.How To Determine the Molecular Formula

m/zIntensity(% 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

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)

●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

●For our unknown

Number of C atoms =4.5

1.1~ 4

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

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

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

● 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

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

Ch. 9 - 158

18. Mass Spectrometer Instrument Designs

Ch. 9 - 159

19. GC/MS Analysis

Ch. 9 - 160

END OF CHAPTER 9

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