Download Mass Spec, IR - faculty at Chemeketa

12. Structure Determination:
Mass Spectrometry and
Infrared Spectroscopy
Determining the Structure of an
Organic Compound
 The analysis of the outcome of a reaction requires
that we know the full structure of the products as well
as the reactants
 In the 19th and early 20th centuries, structures were
determined by synthesis and chemical degradation
that related compounds to each other
 Physical methods now permit structures to be
determined directly. We will examine:




mass spectrometry (MS)
infrared (IR) spectroscopy
nuclear magnetic resonance spectroscopy (NMR)
ultraviolet-visible spectroscopy (VIS)
2
Why this Chapter?
 Finding structures of new molecules
synthesized is critical
 To get a good idea of the range of structural
techniques available and how they should be
used
3
12.1 Mass Spectrometry of Small
Molecules:Magnetic-Sector Instruments
Bonds in cation radicals
begin to break (fragment)
Sample
vaporized and
subjected to
bombardment
by electrons
that remove an
electron
creating a
cation radical
Mass to charge (m/z)
ratio is measured
Molecular ion (M+)
shows molecular weight
4
The Mass Spectrum
 Plot mass of ions (m/z) (x-axis) versus the intensity of the signal
(roughly corresponding to the number of ions) (y-axis)
 Tallest peak is base peak (100%)
 Other peaks listed as the % of that peak
 Peak that corresponds to the unfragmented radical cation is
parent peak or molecular ion (M+)
Propane
MW = M+ = 44
H H H
H C C C H
H H H
5
12.2 Interpreting Mass Spectra
 Molecular weight from the mass of the molecular ion
 Double-focusing instruments provide high-resolution
“exact mass”
 0.0001 atomic mass units – distinguishing specific
atoms
 Example MW “72” is ambiguous: C5H12 and C4H8O but:
 C5H12 72.0939 amu exact mass C4H8O 72.0575 amu
exact mass
 Result from fractional mass differences of atoms 16O =
15.99491, 12C = 12.0000, 1H = 1.00783
 Instruments include computation of formulas for each
peak
 If parent ion not present due to electron bombardment causing
breakdown, “softer” methods such as chemical ionization are used
6
M+ peak Example:
 Possible formulas can be determined from the mass. (M+)
Predict possible molecular formulas containing C, H, and
maybe O if :
M+ = 86
M+ = 156
7
M+ peak Example:
 Possible formulas can be predicted from the mass. (M+)
Predict possible molecular formulas containing C, H, and
maybe O if :
M+ = 86
C6H14
C5H10O
C4H6O2
C3H2O3
M+ = 156
C11H24
C12H12
C11H8O
C10H20O
C10H4O2
C9H16O2
C8H12O3
C7H8O4
C6H4O5
8
M+ peak :
 Peaks above the molecular weight appear as a result
of naturally occurring heavier isotopes in the sample

(M+1) from 1.1% 13C in nature

(M and M+2) in 75.8%/24.2% ratio = 35Cl and 37Cl
(M and M+2) in 50.7%/49.3% ratio = 79Br and 81Br

Propane
MW = M+ = 44
H H H
H C C C H
H H H
M+1 = 45
From 1.1% 13C
9
Click on image to enlarge
M+ peak: Halides
M+ and M+2 in
75.8%:24.2%
(~ 3:1) ratio
= 35Cl and 37Cl
M+ and M+2 in
50.7%:49.3%
(~ 1:1) ratio
= 79Br and 81Br
CH3Cl
Br
10
Mass-Spec Fragmentation Patterns
 Molecular ions break into characteristic fragments that can be
identifed
 Serves as a “fingerprint” for comparison with known materials in
analysis (used in forensics)
 Positive charge goes to fragments that best can stabilize it
11
Mass-Spec Fragmentation Patterns
M -15 = 57
CH3=15
MW=72;
M+ peak not seen
12
Mass Spec: Fragmentation of Hexane
Hexane (m/z = 86 for parent) has peaks at
m/z = 71, 57, 43, 29
71=Loss of CH3 (15)
57=Loss of CH3CH2 (29)
13
CH3
Learning Check:
MW=98
MW=98
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.
14
CH3
Solution:
MW=98
MW=98
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.
98-69=29 loss of (CH3CH2)
98-83=15 (loss of CH3)
M+ =98
15
CH3
Learning Check:
MW=98
MW=98
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.
16
CH3
Solution:
MW=98
MW=98
Identify the following MS as that of methylcyclohexane or ethylcyclopentane.
98-83=15 (loss of CH3)
M+ =98
17
12.3 Mass Spec: Common Functional Groups
Alcohols:
 Alcohols undergo -cleavage (at bond next to the C-OH)
as well as loss of H-OH (18) to give C=C
Loss of 18
18
Mass Spec: Alcohols
CH3
H3C
CH
19
Mass Spec: Amines
 Nitrogen Rule: Amines with odd # of N’s have Odd M+
 Amines undergo -cleavage, generating radicals
20
Mass Spec: Carbonyl Compounds
 A C-H that is three atoms away leads to an internal transfer of a
proton to the C=O, called the McLafferty rearrangement
Loss of 28
•Carbonyl compounds can also undergo  cleavage
21
12.4 Mass Spectrometry in Biological
Chemistry: Time-of-Flight (TOF) Instruments
 Most biochemical analyses by MS use:
- electrospray ionization (ESI)
- Matrix-assisted laser desorption ionization (MALDI)
• Linked to a time-of-flight mass analyzer
MALDI-TOF MS of chicken
eg-white lysozyme
22
12.5 Spectroscopy and the
Electromagnetic Spectrum
 Radiant energy is proportional to its frequency (cycles/s = Hz)
as a wave (Amplitude is its height)
 Different types are classified by frequency or wavelength ranges
23
Electromagnetic Spectrum
High frequency (u)
Short wavelength (l)
= High energy
Low frequency (u)
Long wavelength (l)
= Low energy
24
Absorption Spectra
 Organic compound exposed to electromagnetic radiation, can
absorb energy of certain wavelengths.
 Changing wavelengths to determine which are absorbed and
which are transmitted produces an absorption spectrum
 Energy absorbed is shown as dips in spectrum
High frequency (u)
Short wavelength (l)
= High energy
Infrared Absorption of Ethyl Alcohol CH3CH2OH
Low frequency (u)
Long wavelength (l)
= Low energy
25
12.6 Infrared Spectroscopy
High frequency = High E
Low frequency = Low E
 IR region lower energy than visible light (< red – produces heating as
with a heat lamp)
 2.5  106 m to 2.5  105 m region used by organic chemists for
structural analysis
 IR energy in a spectrum is usually measured as wavenumber (cm-1),
the inverse of wavelength and proportional to frequency
 Specific IR absorbed by organic molecule related to its structure
26
Infrared Spectroscopy
 IR energy absorption corresponds to atomic movements, such as
vibrations and rotations from bending and stretching of bonds
between groups of atoms
 Energy is characteristic of the bonding of atoms in a functional group
 Bond stretching dominates higher energy modes
 Light objects connected to heavy objects vibrate fastest: C-H, N-H,
O-H
 For two heavy atoms, stronger bond requires more energy: C
C  N > C=C, C=O, C=N > C-C, C-O, C-N, C-halogen
 C,
27
12.7 Interpreting Infrared Spectra
 Most functional groups absorb at a characteristic energy
 4000-2500 cm-1 N-H,
 2000-1500 cm-1 double bonds
C-H, O-H (stretching)


3300-3600 N-H, O-H
3000 C-H
C=C, C=O, C=N (stretching)
 C=O 1680-1750
 C=C 1640-1680 cm-1
Below 1500 cm-1 “fingerprint”
Single bonds C-C, C-O, C-N,
C-X (vibrations)
2500-2000 cm-1
CC and C  N
(stretching)
28
Infrared Spectra: Functional Grps
 Characteristic higher energy IR absorptions used to confirm the
existence of the presence of functional groups
29
12.8 IR Spectra: Functional Grps
Alkane
-C-H
C-C
Alkene
Alkyne
30
IR: Aromatic Compounds
(Subsituted benzene “teeth”)
C≡C
31
IR: Alcohols and Amines
O-H broadens with Hydrogen bonding
CH3CH2OH
C-O
Amines similar to OH
N-H broadens with Hydrogen bonding
32
IR: Alcohols: O-H stretch
Gas phase
(no H-bonding)
CCl4 sln (0.25M)
(some H-bonding)
Liquid Film
(Lots of H-bonding)
33
IR: Alcohols
C-O
34
IR: Amines
Primary and secondary
amines exhibit a
characteristic broad IR N-H
stretching absorption between
3250 and 3500 cm-1.
Primary amines show two
strong peaks in this range,
whereas secondary amines
show only one.
Primary amines also show a
band near 1600 cm-1 due to a
scissoring motion of the NH2
group.
Tertiary amines do not show
any of these signals since
they do not have a hydrogen
bound to nitrogen.
Amines similar to OH
N-H broadens with Hydrogen bonding
35
IR: Amines Examples
36
IR: Carbonyls: C=O Aldehydes
 Carbonyls in general:
 Strong, sharp C=O peak 1670 to 1780 cm1
 Conjugation lowers stretching frequency
37
IR: C=O: Aldehydes
38
IR: C=O: Ketones
Conjugation with a double bond or
benzene ring lowers the stretching
frequency by 30 to 40 cm-1.
39
IR: C=O: Ketones
Ring strain increases frequency:
•
Incorporation of the carbonyl group in
a small ring (5, 4 or 3-membered),
raises the stretching frequency.
40
IR: C=O: Esters
 1735 cm1 in saturated esters
 Electron donating O increased the frequency
 1715 cm1 in esters next to aromatic ring or a double bond
 Conjugation decreases the frequency
41
IR: Carboxylic Acids
42
Learning Check:
32. Which of the following represents cyclohexane and which cyclohexene?
43
Solution:
32. Which of the following represents cyclohexane and which cyclohexene?
=C- H
-C-H
-C=C-
44
Learning Check:
41.Propose a structure for the following unknown hydrocarbon:
45
Solution:
68-53 =15
loss of CH3
41.Propose a structure for the following unknown hydrocarbon:
M-1=67
C5H8
2 deg of unsat
M=68
≡C- H
-C≡C-C-H
46
Learning Check:
42. Propose a structure for the following unknown hydrocarbon:
47
Solution:
42. Propose a structure for the following unknown hydrocarbon:
70-55 =15
loss of CH3
C5H10
M=70
=C-H
-C-H
-C=C-
48
In mass spectrometry, what term is used to describe the ion
that results from the ejection of one electron from a
molecule?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
base peak
parent peak
fragment
analyte
none of these
1
2
5
What quantity is detected by mass spectrometry?
1.
2.
3.
4.
5.
the energy of a
molecule
the number of electrons
ejected from a molecule
the number of ions of a
particular mass to
charge ratio
the number of electrons
needed to ionize a
molecule
the number of hydrogen
atoms in a molecule
20%
1
20%
2
20%
20%
3
4
20%
5
High-resolution mass spectrometry enables one
to determine the molecular formula of a
molecule.
50%
50%
True
2. False
1.
1
2
High-resolution mass spectrometry would allow one to
distinguish between the following molecules.
50%
Br
50%
Br
True
2. False
1.
1
2
Which of the given C6H12O isomers would be expected
to produce an m/z peak at M+-18?
25%
25%
25%
25%
2.
1.
O
O
4.
3.
OH
O
1
2
3
4
Which of the following reasons explains why some
mass spectrum peaks are larger than others?
1.
2.
3.
4.
5.
The larger peaks
represent fragments that
are more stable.
The larger peaks
represent fragments that
are less stable.
Cations tend to give larger
peaks than radical cations.
Radical cations tend to
give larger peaks than
cations.
None of these
20%
1
20%
2
20%
20%
3
4
20%
5
Which of the following compounds is most likely to
undergo a McLafferty rearrangement?
20%
1.
20%
20%
20%
3
4
20%
2.
O
NH2
3.
O
4.
5.
OH
OH
1
2
5
Frequency is commonly reported in units of:
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
Joules
nm
amu
Hz
m/z
1
2
5
The higher the wavenumber of a molecular vibration, the lower
the energy of the infrared radiation needed to stimulate it.
50%
50%
True
2. False
1.
1
2
Which type of electromagnetic radiation
possesses the highest energy?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
IR light
UV light
microwaves
visible light
FM radio waves
1
2
5
Infrared spectroscopy is based on _____
excitation.
25%
25%
25%
25%
electronic
2. rotational
3. nuclear
4. vibrational
1.
1
2
3
4
What functional groups are most likely present in a compound
whose IR spectrum shows absorbances at 2217 and 1648 cm-1?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
aldehyde and alkene
alkene and alcohol
alcohol and nitrile
alkyne and ketone
nitrile and alkene
1
2
5
(R)-2-pentanol and (S)-2-pentanol give
identical IR spectra.
50%
50%
True
2. False
1.
1
2
Which of the following compounds will have its
carbonyl absorb at the lowest frequency in IR
spectroscopy?
20%
1.
20%
20%
20%
3
4
20%
2.
O
O
O
3.
O
H
5.
4.
O
O
O
1
2
5
What approximate frequency range is considered
the fingerprint region in infrared spectroscopy?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
4000 – 3500 cm-1
3500 – 3000 cm-1
3000 – 2000 cm-1
2000 – 1500 cm-1
1500 – 500 cm-1
1
2
5
(E)-2-Butene and (Z)-2-butene give
identical IR spectra.
True
2. False
1.
50%
1
50%
2