Nuclear Magnetic Resonance and Mass Spectrometryhome.konkuk.ac.kr/~parkyong/Classes/ch9.pdf · Professor William Tam & Dr. Phillis Chang Chapter 9 Nuclear Magnetic Resonance and Mass

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Created byProfessor William Tam & Dr. Phillis Chang

Chapter 9

Nuclear MagneticResonance and Mass

Spectrometry

Copyright © 2014 by John Wiley & Sons, Inc. All rights reserved.

Classic methods for organic structure determination● Require large quantities of

sample and are time consuming

© 2014 by John Wiley & Sons, Inc. All rights reserved.

2

Spectroscopic methods for organic structure determinationa) 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)


General steps for structure elucidation 1. Elemental analysis

● Empirical formula● e.g. C2H4O

2. Mass spectrometry● Molecular weight● Molecular formula● e.g. C4H8O2, C6H12O3 … etc.● Characteristic fragmentation

pattern for certain functional groups


3

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.


General steps for structure elucidation 5. UV (Ultraviolet) Spectroscopy

● Sometimes useful especially for conjugated systems

● e.g. dienes, aromatics, enones

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


4

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


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


5


Typical 1H NMR spectrum● Chemical Shift ()

● Integration (areas of peaks no. of H)

● Multiplicity (spin-spin splitting) and coupling constant


6

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


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


7


Normal range of 1H NMR

15 -10 ppm

"upfield" (more shielded)

"downfield" (deshielded)

(high field strength)

(low field strength)


8

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)

MeSi MeMeMe


NMR solvent● Normal NMR solvents should not

contain hydrogen● Common solvents

CDCl3 C6D6

CD3OD

CD3COCD3 (d6-acetone)


9

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


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


10

O

Ha HaHb

HbHbR

HaHb

2 Ha 3 Hb


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, its multiplicity is n + 1


11

Examples

Hb C C ClHaHb

Hb Ha

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

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

(1)

Cl C C ClHbHa

Cl Hb

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

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

(2)



12

Examples

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

Hb C C BrHaHb

Hb

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

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

(3)

HbHb

Hb


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


13

Pascal’s Triangle

● For

● For

Br C C BrHbHa

Cl Cl

Due to symmetry, Ha

and Hb are identical a singlet

Cl C C BrHbHa

Cl Br

Ha ≠ Hb

two doublets


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


14

Three distinct signals at ~ 3.4, 1.8and 1.1 ppm 3.4 ppm: likely to be near an

electronegative group (Br)


(ppm): 3.4 1.8 1.1

Integral: 2 2 3


15

(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


Complete structure:

BrCH2

CH2

CH3

• 2 H's from integration

• triplet


• sextet


• triplet

most upfield signalmost downfieldsignal


16

4. Nuclear Spin:The Origin of the Signal

The magnetic field associated with a spinning

proton

The spinning proton

resembles a tiny bar magnet



17


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


18

5. Detecting the Signal: Fourier Transform NMR Spectrometers


19



20



21



22

6. The Chemical Shift


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

6A. PPM and the Scale

23

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)



24


Protons absorb at different NMR frequencies depending on the electron density around them and the effects of local induced magnetic fields

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

7. Shielding & Deshielding of Protons


25

15 -10 ppm

"upfield" (more shielded)

"downfield" (deshielded)

(high field strength)

(low field strength)



26


27

Protons of Hydrogen Atoms in AlkylC–H Groups● The chemical shift for hydrogens of

unsubstituted alkanes is typically in the range of 0.8–1.8

Protons of Hydrogens Near Electro-negative Groups● The chemical shift of hydrogens bonded

to a carbon bearing oxygen or a halogen is typically in the range of 3.1–4.0


28

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

(ppm) 4.26 3.40 3.05 2.68 2.16 0.23

● Greater electronegativity Deshielding of the proton Larger


Electronegativity

29

● 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)© 2014 by John Wiley & Sons, Inc. All rights reserved.

● In fact, protons of terminal alkynes absorb between 2.0 and 3.0, and the order is

(higher frequency)

sp2 < sp < sp3 (lower frequency)


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Shielding and deshielding by circulation of electrons


31

● e.g.

Hd Hc

Hb

Ha

(ppm)Ha & Hb: 7.9 & 7.4 (deshielded)Hc & Hd: 0.91 – 1.2 (shielded)


● Aldehydes

OR

H

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


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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


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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HC CHH

HH

H

Ethane

HC CHH

HH

Br

HC CHH

BrH

H

HC CHH

HBr

HHCC HH

HBr

H

HCC HH

BrH

H

HCC HH

HH

Br

The six hydrogens of ethane are homotopicand are, therefore, chemical shift equivalent

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

sam

e co

mpo

unds

same com

pounds


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


34

HC CHH

HH

Br

HC CHH

HCl

BrBrCC HH

HCl

H

BrCC HH

HCl

H

BrCC HH

ClH

H

BrCC HH

HH

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 CH3group gives only one 1H NMR signal


HC CHH

HH

Br

CH3CH2Br● two sets of hydrogens that are

heterotopic with respect to each other

● two 1H NMR signals


35

Other examples

(1) C CH

H

CH3

CH3 2 1H NMR signals

(2) H

CH3H

H

H CH3

4 1H NMR signals


Other examples

(3) H3CCH3

H H

H

H

H H

3 1H NMR signals


36

Application to 13C NMR spectroscopy● Examples

(1) H3C CH3 1 13C NMR signal

(2)

CH3

CH3 4 13C NMR signals


(3)

OHHO

5 13C NMR signals

(4)HO

HO 4 13C NMR signals


37

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


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


38

CH3

H OHH3C

HaHb

dias

tere

omer

s

diastereotopic

CH3

H OHH3C

GHb

CH3

H OHH3C

HaG

chiralitycenter


HbBrHa

H

dias

tere

omer

s

diastereotopic

HbBrG

H

GBr

Ha

H


39


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

9. Signal Splitting:Spin–Spin Coupling

Ha Hb3J or vicinal coupling


40

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


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

Splitting analysis for a doublet

C CHaHb


41

C C

H H

C C

H HA A

upfielddownfield

Bo

THE CHEMICAL SHIFT OF PROTON HA IS AFFECTED BY THE SPIN OF ITS NEIGHBORS

50 % ofmolecules

50 % ofmolecules

At any given time about half of the molecules in solution willhave spin +1/2 and the other half will have spin -1/2.

aligned with Bo opposed to Bo

neighbor aligned neighbor opposed

+1/2 -1/2

42

Splitting analysis for a triplet

CHb

CHaHb

C C CHaHb Hb


43

Splitting analysis for a quartet

Hb CHb

CHaHb


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


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9C. Coupling Constants – Recognizing Splitting Patterns

X CHa

CHb

HbHbHa


9D. The Dependence of Coupling Constants on Dihedral Angle

3J values are related to the dihedral angle ()

H

H


45

Karplus curve

~0o or 180o

Maximum 3J value

~90o

3J ~0 Hz


Karplus curve● Examples

Hb

Ha

Hb

Ha

= 180º

Ja,b = 10-14 Hz

(axial, axial)

Hb

Ha

Hb

Ha

= 60º

Ja,b = 4-5 Hz

(equatorial, equatorial)


46

Karplus curve● Examples

Hb

Ha

Hb

Ha

= 60º

Ja,b = 4-5 Hz

(equatorial, axial)


9E. Complicating Features The 60 MHz 1H NMR spectrum of ethyl

chloroacetate


47

The 300 MHz 1H NMR spectrum of ethyl chloroacetate


9F. Analysis of Complex Interactions


48

The 300 MHz 1H NMR spectrum of1-nitropropane


Protons of alcohols (ROH) and amines (RNH2) 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


49

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.


+ 50 °C

- 30 °C

50

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


51

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


Phenols● Intramolecular H-bonding causes

downfield shift

OH

O12.1 ppm


52



53


Unlike 1H with natural abundance ~99.98%, only 1.1% of carbon, namely 13C, is NMR active

11. Carbon-13 NMR Spectroscopy11A. Interpretation of 13C NMR

Spectra


54

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 requiring a lot more scans for a good spectrum)


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

● 1H NMR Frequency = 200 MHz

● 13C NMR Frequency = 50 MHz


55

12C is not NMR-active I = 0

however…. 13C does have spin, I = 1/2 (odd mass)

1. Natural abundance of 13C is small (1.08% of all C)

2. Magnetic moment of 13C is small

13C signals are weaker than 1H because:

13C NMR

PULSED FT-NMR IS REQUIRED

The chemical shift range is larger than for protons

0 - 200 ppm

56

13C NMR

For a given field strength 13C has its resonance at adifferent (lower) frequency than 1H.

1H

13C1.41 T 60 MHz2.35 T 100 MHz7.05 T 300 MHz

1.41 T 15.1 MHz2.35 T 25.0 MHz7.05 T 75.0 MHz

Divide the hydrogenfrequency by 4 (approximately)

for carbon-13

CH3 C CH2 CH3

H

OH

Example:● 2-Butanol

Proton-coupled13C NMR spectrum


57

CH3 C CH2 CH3

H

OH

Example:● 2-Butanol

Proton-decoupled13C NMR spectrum


AldehydesKetones

Acids AmidesEsters Anhydrides

Aromatic ringcarbons

Unsaturated carbon - sp2

Alkyne carbons - sp

Saturated carbon - sp3

electronegativity effects

Saturated carbon - sp3

no electronegativity effects

C=O

C=O

C=CC C

200 150 100 50 0

200 150 100 50 0

8 - 30

15 - 55

20 - 60

40 - 80

35 - 80

25 - 65

65 - 90

100 - 150

110 - 175

155 - 185

185 - 220

Correlation chart for 13C Chemical Shifts (ppm)

C-O

C-Cl

C-Br

R3CH R4C

R-CH2-R

R-CH3

RANGE

/

58

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


Electronegative substituents cause downfield shift


59



60



61

Hybridization of carbon

● sp2 > sp > sp3

H2C CH2 HC CH H3C CH3

e.g.

123.3 ppm 71.9 ppm 5.7 ppm


Anisotropy effect

● Shows shifts similar to the effect in 1H NMR

C C

e.g.

C

shows largeupfield shift


62



63

BROMOCYCLOHEXANE

Cl CH2 CH CH3

OH

(a) (b) (c)

1-Chloro-2-propanol

(a)(b)

(c)


64

Cl

Cla

a

b

b

c

c

1,2-DICHLOROBENZENE

65

Because of its low natural abundance (0.0108) thereis a low probability of finding two 13C atoms next toeach other in a single molecule.

However, 13C does couple to hydrogen atoms (I = 1/2)

13C - 13C coupling NO!

13C - 1H coupling YES!

Spectra are determined by many molecules contributingto the spectrum, each having only one 13C atom.

13C NMR (cont)

not probable

very common

Homonuclear: Spin-spin splitting patterns where the interaction occurs between two

atoms that are of the same type of nucleus.

3JH-H = 1-20 Hz

Heteronuclear: Spin-spin splitting patterns where the interaction occurs between two atoms that are of different types of nuclei.

1JC-H = 100-250 Hz

Spin-Spin Coupling (scalar)

H

C C

H

H

C C

H

66

The effect of attached protons on 13C resonances

n+1 = 4 n+1 = 3 n+1 = 2 n+1 = 1

C13

3 protons 2 protons 1 proton 0 protons

H

H

H

C13

H

H

C13

H C13

Methylcarbon

Methylenecarbon

Methinecarbon

Quaternarycarbon

( n+1 rule applies )

COUPLING TO ATTACHED PROTONS

(J’s are large ~ 100 - 200 Hz)

ETHYL PHENYLACETATE

13C coupledto the hydrogens

67

ETHYL PHENYLACETATE

13C coupledto the hydrogens

13C decoupledfrom the hydrogens

in some casesthe peaks of the multiplets willoverlap

this is aneasier spectrumto interpret

The NMR Spectrum - 13C

1H Coupled (undecoupled)

1H Decoupled

68

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


11D. 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


69

Cl CH2 CH CH3

OH

(a) (b) (c)

1-Chloro-2-propanol

(a)(b) (c)



70

HCOSY● 1H–1H correlation spectroscopy

HETCOR● Heteronuclear correlation

spectroscopy

12. Two-Dimensional (2D) NMR Techniques


12A. 1H–1H COSY Cross-Peak Correlations

HCOSY of 2-chloro-butane

H2

H1

H1

H3

H3

H4

H4

H2


71

12B. 1H–13C Heteronuclear Corre-lationCross-Peak Correlations

HETCOR of 2-chloro-butane

H1

H2

H3

H4

C1

C2

C3 C

4



72



73

74

75

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-

M© 2014 by John Wiley & Sons, Inc. All rights reserved.

EI

76

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.


77

Partial MS of octane (C8H18, M = 114)

13. An Introduction to Mass Spectrometry

114

85

71

57M+

29 (CH3CH2)

14 (CH2)



78

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]

(85)

fragmentation

-CH3CH2 (29)

[C5H11]

(71)-CH3CH2CH2 (29+14)


Fragmentation

79

80

81

15. Depicting the Molecular Ion

CH3CH2 CH3

H3C OH H3C N CH3

CH3

H2C CHCH2CH3

Methanol Trimethylamine 1-Butene

Radical cations from ionizationof nonbonding on electron


82

CompoundIonization

Potential (eV)CH3(CH2)3NH2 8.7C6H6 (benzene) 9.2C2H4 10.5CH3OH 10.8C2H6 11.5CH4 12.7

Ionization potentials of selected molecules


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 fragmentation of the more stable carbocation:ArCH2

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

+


83

e.g.

R CH

R+ CH3+

R +CH3+X

● Site of ionization: n > >

non-bonding


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)

aCH3+

(43)a

b CH2CH3+(29)

b

CH3+(43)


84

16B. Fragmentation of Longer Chain and Branched Alkanes

Octane vs. isooctane

M+(114)

(85)

(71)

(57)

(43)

M+(114)

+

+

+

+

+(57)


2,2,4-trimethylpentane

• Branched alkanes have small or absent M+

85

16C. Fragmentation to Form Resonance-Stabilized Cations

Alkenes● Important fragmentation of terminal

alkenes Allyl carbocation (m/e = 41)

R

(41)

R +


Alkene Fragmentation– Fairly prominent M+

– Fragment ions of CnH2n+ and CnH2n-1+

– Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic cations

R - R

[CH2=CHCH2 CH2 CH3 ] CH2 =CHCH2+ + • CH2 CH3

+•

86

Carbon–carbon bonds next to an atom with an unshared electron pair usually break readily because the resulting carbocation is resonance stabilized

Ethers● Cleavage (to ether oxygen) C–C bonds

O

O

(m/e = 59)

+ OCH3


n-butyl ether

87

Alcohols● Most common fragmentation: loss of

alkyl groups

OH

M+(74)

a

OH OHCH3CH2 +

b

(m/e = 45)

b


2-pentanol

88

3-methyl-2-butanol

89

Hydrogen atom transfer

Aldehydes● M+ peak usually observed but may

be fairly weak

● Common fragmentation pattern -cleavage

RR H

OH C OR

C OH

+

+(m/e = 29)

acylium ion


90

hexanal

Ketones● -cleavage

O a

a

b

b


91

92

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


Benzene derivatives produce a variety of resonance stabilized ions.

CH2CH2CH2CH3 CH2CH2 CH2 CH2

CH2

m/e = 91

93

+

+

Tropylium ion

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

+●


94

1,2-elimination:OH

+ H2O

M (M - 18)


1,4-elimination

2-pentanol

95

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

e.g.

retro Diels-Alder+

+


96

H

CH2

H

H(m/e = 92)

+

McLafferty Rearrangement


Mc Lafferty Rearrangement

Mc Lafferty Rearrangement

97

Ketones● McLafferty rearrangement

O

HOH OH

OHOH

OH

+

(m/e = 86)

(m/e = 58)

+

1st McL. Rearr.

2nd McL. Rearr.


98

2-octanone displays both cleavage and McLafferty

CH3CO+

resulting from cleavage.

CH3CH2CH2CH2CH2CH2CO+

resulting from cleavage.

99

OH H

OH OH

(m/e = 86)2º radical

observedi

i

OH

1º radical

OH

(m/e = 114)NOT observed

ii

ii


Characteristics of McLaffertyrearrangements1. No alkyl migrations to C=O, only H

migrates

OH

O

O

R

R

R

H

HX


100

Characteristics of McLaffertyrearrangements2. 2o is preferred over 1o

OH H

iiiOH

2º radical

OH

1º radical

not


101

17. Isotopes in Mass Spectra


13C and 12CAbout 1.1% of all carbon atoms are the 13C isotope

About 98.9% of the methane molecules in the sample will contain 12C, and the other 1.1% will contain 13C


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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© 2014 by John Wiley & Sons, Inc. All rights reserved.

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

+●

+●


103

≈

100

1.17m/z

rela

tive

ion

abun

danc

eM

M +1

+●

+●


104


105


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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


Example 2

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

C3H8O = 60.05754

C2H4O2 = 60.02112


107

Low Resolution Mass

108

High Resolution Mass

109

110

111

112

113

114

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Identify which one of the following isomers of C6H14 has the C-13 NMR below.

A) CH3CH2CH2CH2CH2CH3 C) (CH3)2CHCH(CH3)2

B) CH3CH2CH2CH(CH3)2 D) CH3CH2C(CH3)3

A) CH3CH2CH2CH2CH2CH3 C) (CH3)2CHCH(CH3)2 B) CH3CH2CH2CH(CH3)2 D) CH3CH2C(CH3)3

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120

m/e = 72 ?

72

O

Nuclear Magnetic Resonance and Mass Spectrometryhome.konkuk.ac.kr/~parkyong/Classes/ch9.pdf · Professor William Tam & Dr. Phillis Chang Chapter 9 Nuclear Magnetic Resonance and Mass

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