Ch. 9 - 1

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Chapter 9

• Nuclear Magnetic
• Resonance and Mass
• Spectrometry

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
• Degradation

3

• Classic methods for organic structure
  determination.
• These methods require large quantities of sample
  and are time consuming.

4

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

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.
• Effective in mg and microgram quantities.

6

• General steps for structure elucidation.
• Elemental analysis
• Empirical formula
• e.g. C2H4O
• Mass spectroscopy
• Molecular weight
• Molecular formula
• e.g. C4H8O2, C6H12O3 etc.
• Characteristic fragmentation pattern for certain
  functional groups.

7

• General steps for structure elucidation.
• From molecular formula
• Double bond equivalent (DBE)
• Infrared spectroscopy (IR)
• Identify some specific functional groups
• e.g. CO, CO, OH, COOH, NH2 etc.

8

• General steps for structure elucidation.
• UV
• Sometimes useful especially for conjugated
  systems.
• e.g. dienes, aromatics, enones
• 1H, 13C NMR and other advanced NMR techniques.
• Full structure determination

9

• Electromagnetic spectrum

10

1. Nuclear Magnetic Resonance(NMR) Spectroscopy

• A graph that shows the characteristic energy
  absorption frequencies and intensities for a
  sample in a magnetic field is called a nuclear
  magnetic resonance (NMR) spectrum.

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• Typical 1H (Proton) NMR spectrum

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.

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1. The area under the signal tells us about how many
   protons there are in the set being measured.
2. The multiplicity (or splitting pattern) of each
   signal tells us about the number of protons on
   atoms adjacent to the one whose signal is being
   measured.

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

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• Typical 1H NMR spectrum

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2A. Chemical Shift

• The position of a signal along the x-axis of an
  NMR spectrum is called its chemical shift.
• The chemical shift of each signal gives
  information about the structural environment of
  the nuclei producing that signal.
• Counting the number of signals in a 1H NMR
  spectrum indicates, at a first approximation, the
  number of distinct proton environments in a
  molecule.

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• 1H Chemical Shifts

18

• 13C Chemical Shifts

19

• Normal range of 1H NMR

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• Reference compound
• TMS tetramethylsilane
• as a reference standard (0 ppm).
• Reasons for the choice of TMS
• Resonance position at higher field than other
  organic compounds.
• Unreactive and stable, not toxic
• Volatile and easily removed
• (B.P. 28oC)

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

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• The 300-MHz 1H NMR spectrum of 1,4-dimethylbenzene
  .
• Peaks are in blue, integration in black.

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2B. Integration of Signal Areas Integral Step
Heights

• The area under each signal in a 1H NMR spectrum
  is proportional to the number of hydrogen atoms
  producing that signal.
• It is signal area (integration), not signal
  height, that gives information about the number
  of hydrogen atoms.

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3 Hb
2 Ha
Splitting of protons on adjacent atoms.
Ha
Hb
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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 n1 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.

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• Examples

27

• Example of splitting

28

• Examples

Note All Hbs are chemically and magnetically
equivalent.
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• Pascals Triangle
• Used to predict relative intensity of various
  peaks in multiplet.
• Given by the coefficient of binomial expansion (a
  b)n.

singlet (s) 1 doublet (d) 1 1 triplet (t) 1 2
1 quartet (q) 1 3 3 1 quintet 1 4 6 4 1 sextet 1
5 10 10 5 1
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• Pascals Triangle
• For
• For

Due to symmetry, Ha and Hb are identical ? a
singlet
Ha ? Hb ? two doublets
31

1. 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.

32

1. 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.
2. 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.
3. Join the fragments to make a molecule in a
   fashion that is consistent with the data.

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

34

• Three distinct signals at d3.4, 1.8 and 1.1
  ppm.
• ? d3.4 ppm likely to be near an electronegative
  group (Br).

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d (ppm) 3.4 1.8 1.1 Integral 2 2 3
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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
37
Complete structure
most upfield signal
most downfield signal

• 2 H's from integration
• triplet

• 2 H's from integration
• sextet

• 3 H's from integration
• triplet

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1. Nuclear SpinThe Origin of the Signal

The magnetic field associated with a spinning
proton
The spinning proton resembles a tiny bar magnet
39

• Alignment of nuclei in presence and absence of an
  applied magnetic field
• random with or against

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Difference in energy is proportional to applied
field strength.
41

• 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
42

1. Detecting the Signal Fourier Transform NMR
   Spectrometers

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1. 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.

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• 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
d (ppm) 4.26 3.40 3.05 2.68 2.16 0.23

• Greater electronegativity
• Deshielding of the proton
• Larger d

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• 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)
(lower frequency)
sp lt sp2 lt sp3
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• In fact, protons of terminal alkynes absorb
  between d 2.0 and d 3.0, and the order is

(higher frequency)
(lower frequency)
sp2 lt sp lt sp3
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• 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 (anisotropy).

Shielded (d 2 3 ppm)
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• Aromatic system

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• e.g.

d (ppm) Ha Hb 7.9 7.4 (deshielded) Hc Hd
0.91 1.2 (shielded)
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• Alkenes

Anisotropic effect
53

• Aldehydes

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1. The Chemical Shift

• 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).

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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.

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• 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.

57

• 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).

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1. 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.

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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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same compounds
same compounds

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

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• If replacing hydrogens by a different atom gives
  different compounds, the hydrogens are said to be
  heterotopic.
• Heterotopic atoms have different chemical shifts
  and are not chemical shift equivalent.

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same compounds ? these 3 Hs of the CH3 group are
homotopic ? the CH3 group gives only one 1H NMR
signal
These 2 Hs are also homotopic to each other
different compounds ? heterotopic
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• CH3CH2Br
• two sets of hydrogens that are heterotopic with
  respect to each other.
• two 1H NMR signals.

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• Other examples

? 2 1H NMR signals
? 4 1H NMR signals
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• Other examples

? 3 1H NMR signals
66

• Application to 13C NMR spectroscopy
• Examples

? 1 13C NMR signal
? 4 13C NMR signals
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? 5 13C NMR signals
? 4 13C NMR signals
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8B. Enantiotopic and Diastereotopic Hydrogen
Atoms

• If replacement of each of two hydrogen atoms by
  the same group yields compounds that are
  enantiomers, the two hydrogen atoms are said to
  be enantiotopic.

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

enantiotopic
enantiomer
70
diastereomers
diastereotopic
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diastereomers
diastereotopic
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1. Signal SplittingSpinSpin Coupling

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

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9A. Vicinal Coupling

• Vicinal coupling between heterotopic protons
  generally follows the n 1 rule. Exceptions to
  the n 1 rule can occur when diastereotopic
  hydrogens or conformationally restricted systems
  are involved.
• Signal splitting is not observed for protons that
  are homotopic (chemical shift equivalent) or
  enantiotopic.

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

• Splitting analysis for a doublet.

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• Splitting analysis for a triplet

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• Splitting analysis for a quartet

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• Pascals Triangle
• Use to predict relative intensity of various
  peaks in multiplet.
• Given by the coefficient of binomial expansion (a
  b)n.

singlet (s) 1 doublet (d) 1 1 triplet (t) 1 2
1 quartet (q) 1 3 3 1 quintet 1 4 6 4 1 sextet 1
5 10 10 5 1
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9C. Coupling Constants Recognizing Splitting
Patterns
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9E. Complicating Features

• The 60 MHz 1H NMR spectrum of ethyl chloroacetate.

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• The 300 MHz 1H NMR spectrum of ethyl
  chloroacetate.

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9F. Analysis of Complex Interactions
82

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

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1. 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.

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• Why dont we see coupling with the OH proton,
  e.g. CH2OH (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).

85

• Trick
• Run the spectrum at lower temperaturewhichslows
  the H exchangeand coupling is observed,
• or
• Run NMR in d6-DMSO where H-bonding with DMSOs
  oxygen prevents Hs from exchanging and we may be
  able to see the coupling.

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• Deuterium Exchange
• To determine which signal in the NMR spectrum is
  the OH proton, shake the NMR sample with a drop
  of D2O and whichever peak disappears that is the
  OH peak (note a new peak of HOD appears).

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• Phenols
• Phenol protons appear downfield at 4-7 ppm.
• They are more acidic - more H character.
• More dilute solutions - peak appears upfield
  towards 4 ppm.

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• Phenols
• Intramolecular H-bonding causes downfield shift.

12.1 ppm
89

1. 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.

90
11B. One Peak for Each Magnetically Distinct
Carbon Atom

• 13C NMR spectra have only become commonplace more
  recently with the introduction of the Fourier
  Transform (FT) technique, where averaging of many
  scans is possible (note 13C spectra are 6000
  times weaker than 1H spectra, thus require a lot
  more scans for a good spectrum).

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• Note for a 200 MHz NMR (field strength 4.70
  Tesla)
• 1H NMR ? Frequency 200 MHz
• 13C NMR ? Frequency 50 MHz

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• Example
• 2-Butanol

Proton-coupled 13C NMR spectrum
93

• Example
• 2-Butanol

Proton-decoupled 13C NMR spectrum
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11C. 13C Chemical Shifts

• Decreased electron density around an atom
  deshields the atom from the magnetic field and
  causes its signal to occur further downfield
  (higher ppm, to the left) in the NMR spectrum.
• Relatively higher electron density around an atom
  shields the atom from the magnetic field and
  causes the signal to occur upfield (lower ppm, to
  the right) in the NMR spectrum.

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

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• Electronegative substituents cause downfield
  shift.
• Increase in relative atomic mass of substituent
  causes upfield shift.

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• Hybridization of carbon and 13C NMR
• sp2 gt sp gt sp3

123.3 ppm
71.9 ppm
5.7 ppm
98

• Anisotropy effect and 13C NMR
• Shows shifts similar to the effect in 1H NMR.

shows large upfield shift
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13C NMR Chemical Shifts
101
13C NMR spectrum
(c)
(a)
(b)
102
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. A carbon with no hydrogens produces
  a singlet (n 0), a carbon with one hydrogen
  produces a doublet (two peaks), a carbon with two
  hydrogens produces a triplet (three peaks), and a
  methyl group carbon produces a quartet (four
  peaks).

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Off-resonance decoupled 13C NMR
Broadband proton-decoupled 13C NMR
104
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.

105
(c)
(a)
(b)
106

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

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1. An Introduction to Mass Spectrometry

• Partial MS of octane (C8H18, M 114)

108

• 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.

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

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1. 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.

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• This molecular ion has considerable surplus
  energy so it can fly apart or fragment to give
  specific ions which may be diagnostic for a
  particular compound.

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1. Depicting the Molecular Ion

Radical cations from ionization of nonbonding on
p electron
113

• Ionization potentials of selected molecules

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
114

1. Fragmentation

1. The reactions that take place in a mass
   spectrometer are unimolecular, that is, they do
   not involve collisions between molecules or ions.
   This is true because the pressure is kept so low
   (10-6 torr) that reactions. involving bimolecular
   collisions do not occur
2. We use single-barbed arrows to depict mechanisms
   involving single electron movements.
3. The relative ion abundances, as indicated by peak
   intensities, are very important.

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16A. Fragmentation by Cleavage at a Single Bond

• When a molecular ion fragments, it will yield a
  neutral radical (not detected) and a carbocation
  (detected) with an even number of electrons.
• The fragmentation will be dictated to some extent
  by the fragmention of the more stable
  carbocation
• ArCH2 gt CH2CHCH2 gt 3o gt 2o gt 1o gt CH3

116

• e.g.

X

• Site of ionization n gt p gt s

non-bonding
117

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

a
b
118
16B. Fragmentation of Longer Chain and Branched
Alkanes

• Octane vs. isooctane

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16C. Fragmentation to Form Resonance-Stabilized
Cations

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

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• Carboncarbon 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) CC bonds

121

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

a
b
122

• Carboncarbon bonds next to the carbonyl group of
  an aldehyde or ketone break readily because
  resonance-stabilized ions called acylium ions are
  produced via an a-cleavage.

123

• Aldehydes
• M peak usually observed but may be fairly weak.
• Common fragmentation pattern.
• a-cleavage

124

• Ketones
• a-cleavage

a
a
b
b
125

• Alkyl-substituted benzenes ionize by loss of a p
  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.

126

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

127
16D. Fragmentation by Cleavage of Two Bonds

• Alcohols frequently show a prominent peak at M -
  18. This corresponds to the loss of a molecule of
  water.
• May lose H2O by 1,2- or 1,4-elimination.

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129

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

130

• Aromatic hydrocarbons
• e.g.

131

• Ketones
• McLafferty rearrangement

132
McLafferty rearrangement
133

• Characteristic of McLafferty rearrangement.
• No alkyl migrations to CO, only H migrates.

X
134

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

135

• How To Determine Molecular
• Formulas and Molecular Weights
• Using Mass Spectrometry

17A. Isotopic Peaks the Molecular Ion
136

• 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.

137

• 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.

138

• Example
• Consider 100 molecules of CH4

C12 100 C13 1.11 H1 100 H2 0.016
139
1.11 molecules contain a 13C atom
4x0.016 0.064 molecules contain a 2H atom
Intensity of M 1 peak 1.110.0641.174 of
the M peak
140
M
100
relative ion abundance
M 1

1.17
m/z
141
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
142

• 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.

143

• 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

144

• 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.

145

• 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.

146

• 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.

147
17C. High-Resolution Mass Spectrometry
148

• 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

149

• Example 2
• Both C3H8O and C2H4O2 have M.W. of 60 (by MS),
  but accurate masses are different
• C3H8O 60.05754
• C2H4O2 60.02112

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1. Mass Spectrometer Instrument Designs

151
19. GC/MS Analysis
152
? END OF CHAPTER 9 ?