Organic Chemistry Nmr Notes 3z6l5z

Chapter 9 Nuclear Magnetic Resonance and Mass Spectrometry Created by Professor William Tam & Dr. Phillis Chang Ch. 9 - 1

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 ed 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

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 Ch. 9 - 3

  Classic

methods for organic structure determination ●  Require large quantities of sample and are time consuming

Ch. 9 - 4

 

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) Ch. 9 - 5

 

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 - 6

 

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 - 7

 

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 - 8

 

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 - 9

 

Electromagnetic spectrum

cosmic & γ-rays λ:

X-rays

0.1nm

ultraviolet

200nm

X-Ray Crystallography

visible

400nm

infrared

800nm IR

UV

microwave

radiowave

50µm

NMR

1Å = 10-10m 1nm = 10-9m 1µm = 10-6m

Ch. 9 - 10

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 - 11

Ch. 9 - 12

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 - 13

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 - 14

 

Typical 1H NMR spectrum ●  Chemical Shift (δ) ●  Integration (areas of peaks ⇒ no. of H) ●  Multiplicity (spin-spin splitting) and coupling constant

Ch. 9 - 15

 

Typical 1H NMR spectrum

Record as:

1

H NMR (300 MHz, CDCl3):

δ 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) chemical shift (δ) in ppm

coupling constant no. of H (integration) multiplicity in Hz

Ch. 9 - 16

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 - 17

Ch. 9 - 18

Ch. 9 - 19

 

Normal range of 1H NMR

"upfield" (more shielded) "downfield" (deshielded) 15 (low field strength)

δ ppm

-10 (high field strength)

Ch. 9 - 20

 

Reference compound ●  TMS = tetramethylsilane Me

Me Si Me Me

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) Ch. 9 - 21

 

NMR solvent ●  Normal NMR solvents should not contain hydrogen ●  Common solvents  

CDCl3

 

C6D6

 

CD3OD

 

CD3COCD3 (d6-acetone) Ch. 9 - 22

 

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

Ch. 9 - 23

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 - 24

Ha R

Ha

O Hb

Hb Hb a

2H

H

b

b

3H

H

a Ch. 9 - 25

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 - 26

 

Examples

(1)

Hb Ha

Hb

C

C

a

H : multiplicity = 3 + 1 = 4 (a quartet) Cl

Hb Ha

(2) Ha Hb Cl

C

C

Cl Hb

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

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

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

Ch. 9 - 27

Ch. 9 - 28

 

Examples

(3)

Hb Ha

Hb

C Hb Hb

C

Br

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

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

H

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

Ch. 9 - 29

 

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) doublet (d) triplet (t) quartet (q) quintet sextet

1 11 121 1331 14641 1 5 10 10 5 1

Ch. 9 - 30

 

Pascal’s Triangle Ha Hb

●  For

Br

C

C

Br

Cl Cl

Ha Hb

●  For

Cl

C

C

Cl Br

Br

Due to symmetry, Ha and Hb are identical ⇒ a singlet Ha ≠ Hb ⇒ two doublets Ch. 9 - 31

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 Ch. 9 - 32

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. 

the fragments to make a molecule in a fashion that is consistent with the data Ch. 9 - 33

 

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

Ch. 9 - 34

 

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 - 35

δ (ppm):

3.4

1.8

1.1

Integral:

2

2

3 Ch. 9 - 36

δ (ppm):

3.4

Multiplicity: triplet 2 H's on adjacent C

1.8

1.1

sextet

triplet

5 H's on adjacent C

2 H's on adjacent C

Ch. 9 - 37

Complete structure: most downfield signal Br •  2 H's from integration •  triplet

CH2

most upfield signal CH2

CH3

•  2 H's from integration •  sextet

•  3 H's from integration •  triplet

Ch. 9 - 38

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 - 39

Ch. 9 - 40

Ch. 9 - 41

  1H:

Spin quantum number (I) 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 - 42

5.  Detecting the Signal: Fourier Transform NMR Spectrometers

Ch. 9 - 43

Ch. 9 - 44

6.  Shielding & Deshielding of Protons  

   

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 "upfield" (more shielded) "downfield" (deshielded) 15 (low field strength)

δ ppm

-10 (high field strength)

Ch. 9 - 45

Ch. 9 - 46

  Deshielding

by electronegative groups

CH3X X=

F

Electro4.0 negativity

OH

Cl

Br

I

H

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

Ch. 9 - 47

  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 (lower 2 3 sp < sp < sp frequency) frequency) Ch. 9 - 48

●  In fact, protons of terminal alkynes absorb between δ 2.0 and δ 3.0, and the order is (higher (lower 2 3 sp < sp < sp frequency) frequency)

Ch. 9 - 49

●  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)

Ch. 9 - 50

●  Aromatic system Shielded region

Deshielded region

Ch. 9 - 51

●  e.g. Hd

Hc

H

b

Ha

δ (ppm) a

b

c

d

H & H : 7.9 & 7.4 (deshielded) H & H : 0.91 – 1.2 (shielded)

Ch. 9 - 52

●  Alkenes

H Deshielded (δ 4.5 – 7 ppm)

Ch. 9 - 53

●  Aldehydes

R H

O

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

Ch. 9 - 54

7.  The Chemical Shift  

Reference compound ●  TMS = tetramethylsilane

Me Me

Si Me Me

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)

Ch. 9 - 55

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 - 56

 

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) Ch. 9 - 57

 

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

 

300 x 106 Hz

= 7.27 ppm

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

 

2181 Hz x 106

436 Hz x 106 60 x 106 Hz

= 7.27 ppm

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 - 58

8.  Chemical Shift Equivalent and Nonequivalent Protons   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 Ch. 9 - 59

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 - 60

Br H

same compounds

Br

H

C

C

H

H

H

H

C

C

H

H

H

H

C

C

Br H

   

H

H

H

H

H

H

C

C

H

H

H

H

Ethane H

H

Br

C

C

H

H

H

H

C

C

H

H

H

H

C

C

H

Br

H

Br

H

same compounds

H

H

The six hydrogens of ethane are homotopic and are, therefore, chemical shift equivalent Ethane, consequently, gives only one Ch. 9 - 61 signal in its 1H NMR spectrum

  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 - 62

same Cl Br compounds H C C H ⇒ these 3 H H H’s of the H Br CH3 group H Cl C C H are homotopic H H ⇒ the CH3 H Br group gives H C C H H Br only one 1H Cl H NMR signal H C C Cl H

These 2 H’s are also H Br homotopic C C H to each H H other

H

H

H

Br

C

C

H

Cl

different compounds ⇒ heterotopic

H

Ch. 9 - 63

H

H

Br

C

C

H

H

H

  CH3CH2Br

●  two sets of hydrogens that are heterotopic with respect to each other ●  two 1H NMR signals Ch. 9 - 64

  Other

examples

H (1)

CH3 C

C

H

CH3

H (2)

⇒ 2 1H NMR signals

CH3

H

H H

⇒ 4 1H NMR signals

CH3 Ch. 9 - 65

  Other H (3)

examples H

H CH3

H3C H

H

H

⇒ 3 1H NMR signals

Ch. 9 - 66

  Application

to 13C NMR spectroscopy ●  Examples

(1)

H3C

CH 3

⇒1

13C

NMR signal

CH3 (2)

CH3

⇒4

13C

NMR signals

Ch. 9 - 67

(3) HO

(4)

⇒5

13C

NMR signals

⇒4

13C

NMR signals

OH

HO OH

Ch. 9 - 68

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 - 69

  Enantiotopic

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

enantiotopic H H3C

H3C

G Br

H

enantiomer

Br G H3C

H Br Ch. 9 - 70

H

OH

H3C b

H

Ha

OH

H3C Hb

G

CH3

CH3

diastereotopic

H

OH

H3C G

Ha

diastereomers

chirality centre

H

CH3

Ch. 9 - 71

Br

Ha Br

Hb H

Hb

Ha

H

diastereotopic

Br

diastereomers

G

G H Ch. 9 - 72

9.  Signal Splitting: Spin–Spin Coupling   Vicinal

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

Hb

3

J or vicinal coupling Ch. 9 - 73

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 - 74

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

analysis for a doublet

Hb Ha C

C

Ch. 9 - 75

  Splitting

analysis for a triplet

Hb H a C

C

Hb Hb H a Hb C

C

C

Ch. 9 - 76

  Splitting

analysis for a quartet

H Hb

b

C

H

a

C

Hb

Ch. 9 - 77

 

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) doublet (d) triplet (t) quartet (q) quintet sextet

1 11 121 1331 14641 1 5 10 10 5 1

Ch. 9 - 78

9C. Coupling Constants – Recognizing Splitting Patterns Ha Hb X

C

C

Hb

Ha Hb

Ch. 9 - 79

9D. The Dependence of Coupling Constants on Dihedral Angle   3J

values are related to the dihedral angle (φ) H

φ H

Ch. 9 - 80

  Karplus

curve

 

φ ~0o or 180o ⇒ Maximum 3J value

 

φ ~90o ⇒ 3J ~0 Hz

Ch. 9 - 81

  Karplus

curve ●  Examples

b

H

H Ha (axial, axial) H

b H a

(equatorial, equatorial)

b a

H

b

a

H φ = 180º Ja,b = 10-14 Hz

H

φ = 60º Ja,b = 4-5 Hz

Ch. 9 - 82

  Karplus

curve ●  Examples

b

H

Ha (equatorial, axial) Hb

Ha

φ = 60º Ja,b = 4-5 Hz

Ch. 9 - 83

9E. Complicating Features   The

60 MHz 1H NMR spectrum of ethyl chloroacetate

Ch. 9 - 84

  The

300 MHz 1H NMR spectrum of ethyl chloroacetate

Ch. 9 - 85

9F. Analysis of Complex Interactions

Ch. 9 - 86

  The

300 MHz 1H NMR spectrum of 1nitropropane

Ch. 9 - 87

10.  Proton NMR Spectra and Rate Processes  

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

in high dilution (free OH): R

O

H

δ = ~0.5-1.0 ppm

in conc. solution (H-bonded): − R R δ H O O H R δ+ H O proton more deshielded

Ch. 9 - 88

  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 - 89

Trick: ●  Run NMR in d6-DMSO where Hbonding with DMSO’s oxygen prevents H’s from exchanging and we may be able to see the coupling

Ch. 9 - 90

  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) R

O

H

D2O

R

O

D

+

HOD Ch. 9 - 91

  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 - 92

  Phenols

●  Intramolecular H-bonding causes downfield shift O

12.1 ppm H O

Ch. 9 - 93

11.  Carbon-13 NMR Spectroscopy 11A. Interpretation of 13C NMR Spectra   Unlike 1H

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

Ch. 9 - 94

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 - 95

  Note

for a 200 MHz NMR (field strength 4.70 Tesla) ●  1H NMR ⇒ Frequency = 200 MHz ●  13C NMR ⇒ Frequency = 50 MHz

Ch. 9 - 96

H

  Example:

●  2-Butanol

CH3

C

CH2

CH3

OH

Proton-coupled 13C NMR spectrum

Ch. 9 - 97

H

  Example:

●  2-Butanol

CH3

C

CH2

CH3

OH

Proton-decoupled 13C NMR spectrum

Ch. 9 - 98

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 - 99

 

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) Ch. 9 - 100

 

Electronegative substituents cause downfield shift

 

Increase in relative atomic mass of substituent causes upfield shift

X

Electronegativity Atomic Mass

13

C NMR: CH3X

Cl

2.8

35.5

23.9 ppm

Br

2.7

79.9

9.0 ppm

I

2.2

126.9

-21.7 ppm Ch. 9 - 101

 

Hybridization of carbon ●  sp2 > sp > sp3 e.g. H2C

CH2

123.3 ppm

HC

CH

71.9 ppm

H3C

CH3

5.7 ppm

Ch. 9 - 102

 

Anisotropy effect ●  Shows shifts similar to the effect in 1H NMR e.g. C

C

C

shows large upfield shift Ch. 9 - 103

Ch. 9 - 104

Ch. 9 - 105

Cl

(a) CH2

(b) CH

(c) CH3

OH 1-Chloro-2-propanol

(c) (b) (a)

Ch. 9 - 106

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 - 107

9

Off-resonance decoupled 13C NMR

2

N

1

8

3

N

7 6

4 5

O

Broadband proton-decoupled

13C

NMR

Ch. 9 - 108

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 - 109

Cl

(a) CH2

(b) CH

(c) CH3

(b)

(a)

(c)

OH 1-Chloro-2-propanol

Ch. 9 - 110

  The

broadband proton-decoupled 13C NMR spectrum of methyl methacrylate

Ch. 9 - 111

12.  Two-Dimensional (2D) NMR Techniques   HCOSY

●  1H–1H correlation spectroscopy   HETCOR

●  Heteronuclear correlation spectroscopy

Ch. 9 - 112

  HCOSY

of 2-chloro-butane H2

H3

H1

H4

H4 H1 H3

H2

Ch. 9 - 113

  HETCOR

of 2-chloro-butane C3

C1

C4

C2

H4 H1

H3

H2

Ch. 9 - 114

13.  An Introduction to Mass Spectrometry   Partial

MS of octane (C8H18, M = 114) 14 (CH2) 29 (CH3CH2)

57

85 71

M+ 114

Ch. 9 - 115

  The

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

C8H18

e 70 eV

[C8H18] + 2 e(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. Ch. 9 - 116

  Masses

are usually rounded off to whole numbers assuming: H = 1, C = 12, N = 14, O = 16, F = 19 etc.

[C8H18]

fragmentation

(M , 114) -CH3CH2 (29) +

-CH3CH2CH2 (29+14) Molecular ion (parent peak)

[C6H13]

Daughter ions

(85) [C5H11] (71) Ch. 9 - 117

14.  Formation of Ions: Electron Impact Ionization  

 

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 M

70 eV e-

M Ch. 9 - 118

  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

- m1º

A

- m2º

B

- m3º

C

etc.

mº = neutral fragment radical

Ch. 9 - 119

15.  Depicting the Molecular Ion CH 3CH 2 CH 3

Radical cations from ionization of nonbonding on π electron

H3C

OH

H3C

N

CH3

H2C

CHCH2CH3

CH3 Methanol

Trimethylamine

1-Butene Ch. 9 - 120

  Ionization

molecules

potentials of selected

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 Ch. 9 - 121

16.  Fragmentation 1. 

2. 

3. 

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 We use single-barbed arrows to depict mechanisms involving single electron movements The relative ion abundances, as indicated by peak intensities, are very important

Ch. 9 - 122

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

Ch. 9 - 123

  e.g. R+ R

CH

X

●  Site of ionization:

R

+

+

CH3

+

CH3

n > π > σ

non-bonding Ch. 9 - 124

  As

the carbon skeleton becomes more highly branched, the intensity of the molecular ion peak decreases   Butane vs. isobutane 70eV e-

70eV e

a b+ M (58)

M+(58)

a

b

(43)

(29)

(43)

+ CH3

+ CH2CH3

+ CH3 Ch. 9 - 125

16B. Fragmentation of Longer Chain and Branched Alkanes   Octane

vs. isooctane (85) + (71) + (57) +

M+(114)

(43) +

+ M+(114)

(57) Ch. 9 - 126

16C. Fragmentation to Form Resonance-Stabilized Cations   Alkenes

●  Important fragmentation of terminal alkenes   Allyl carbocation (m/e = 41) R R

+ (41) Ch. 9 - 127

  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

CH3

+

O

O

(m/e = 59) Ch. 9 - 128

  Alcohols

●  Most common fragmentation: - loss of alkyl groups a

CH3 +

OH (m/e = 59)

a OH b

M+(74)

b

OH

CH3CH2 +

OH

OH

(m/e = 45) Ch. 9 - 129

  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 - 130

  Aldehydes

●  M+ peak usually observed but may be fairly weak ●  Common fragmentation pattern   α-cleavage H

+ R C O acylium ion

R

+ H C O (m/e = 29)

O R

H

Ch. 9 - 131

  Ketones

●  α-cleavage O

a b

O

+

(m/e = 71)

a b

O

+

(m/e = 99)

Ch. 9 - 132

  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 - 133

  Aromatic

hydrocarbons ●  very intense M+ peaks ●  characteristic fragmentation pattern (when an alkyl group attached to the benzene ring): tropylium cation CH3

CH3 +

CH2 rearrangement

benzyl cation

tropylium cation (m/e = 91)

Ch. 9 - 134

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 1,4elimination

Ch. 9 - 135

1,2-elimination:

OH

+ (M - 18)

M

1,4-elimination: M

H2O

OH

H

(M - 18)

OH + CH3CH2

+

H2O Ch. 9 - 136

  Cycloalkenes

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

+

  e.g.

+ Ch. 9 - 137

  Aromatic

hydrocarbons

●  e.g.

H McLafferty Rearrangement CH2 H H (m/e = 92)

+

Ch. 9 - 138

  Ketones

●  McLafferty rearrangement H OH

O

1st McL. Rearr.

OH

H

O

OH

(m/e = 86)

2nd McL. Rearr. OH

(m/e = 58)

+

+ Ch. 9 - 139

OH

i H

ii

O

2º radical

H

i ii

OH

1º radical

OH

(m/e = 86) observed

OH

(m/e = 114) NOT observed

Ch. 9 - 140

 

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

O

H

O

R

R

X

R

O

H

Ch. 9 - 141

 

Characteristic of McLafferty rearrangement 2.  2o is preferred over 1o H H O ii i

OH 2º radical

not

OH 1º radical Ch. 9 - 142

17.  How To Determine Molecular Formulas and Molecular Weights Using Mass Spectrometry 17A. Isotopic Peaks & the Molecular Ion

Ch. 9 - 143

  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:

C, H, N

M + 2 Elements:

O, S, Br, Cl Ch. 9 - 144

  The

+ ●

  The

+ ●

M + 1 peak can be used to determine the number of carbons in a molecule 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 - 145

  Example

●  Consider 100 molecules of CH4 1

H 1

H

12

C

H 1

H

H

1

1

H

1

13

C

H

H H

1

H

1

1

12

C

2

H

H1

1

M + 1 = 17

M : 16

C12: 100

C13: 1.11

H1: 100

H2: 0.016 Ch. 9 - 146

H1 1

H

12

C

H1

H1 1

H

H

1

H1

13

C

H

1

H

1

C12

H2

H1

H1 M + 1 = 17

M : 16

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 - 147

relative ion abundance

M

+ ●

100

+ ●

≈ m/z

M +1 1.17 Ch. 9 - 148

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

Ch. 9 - 149

  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 - 150

  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 - 151

+ ●

 

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 - 152

  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 - 153

  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 - 154

17C. High-Resolution Mass Spectrometry

Ch. 9 - 155

  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 - 156

  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 - 157

18.  Mass Spectrometer Instrument Designs

Ch. 9 - 158

19. GC/MS Analysis

Ch. 9 - 159

 END OF CHAPTER 9 

Ch. 9 - 160