Organic Chemistry
Second Edition
David Klein
Chapter 15
Infrared Spectroscopy and Mass Spectrometry
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Klein, Organic Chemistry 2e
15.1 Introduction to Spectroscopy
• Spectroscopy involves an interaction between matter
and light (electromagnetic radiation)
• Light can be thought of as waves of energy or packets
(particles) of energy called photons
• Properties of light waves include wavelength and
frequency
• Is wavelength directly or inversely proportional to
energy? WHY?
• Is frequency directly or inversely proportional to
energy? WHY?
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Klein, Organic Chemistry 2e
15.1 Introduction to Spectroscopy
• There are many wavelengths of light that can not be
observed with your eyes
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Klein, Organic Chemistry 2e
15.1 Introduction to Spectroscopy
• When light interacts with molecules, the effect depends
on the wavelength of light used
• This chapter focuses on IR spectroscopy
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Klein, Organic Chemistry 2e
15.1 Introduction to Spectroscopy
• Matter exhibits particle-like properties
• On the macroscopic scale, matter appears to exhibit
continuous behavior rather than quantum behavior
– Consider the example of an engine powering the rotation of a
tire. The tire should be able to rotate at nearly any rate
• Matter also exhibits wave-like properties as we learned
in section 1.6
• Matter on the molecular scale exhibits quantum
behavior
– A molecule will only rotate or vibrate at certain rates
(energies)
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Klein, Organic Chemistry 2e
15.1 Introduction to Spectroscopy
• For each of the types of molecular motion/energy
below, describe how it is quantized
– Rotation
– Vibration
– Energy of electrons
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Klein, Organic Chemistry 2e
15.1 Introduction to Spectroscopy
• For each different bond, vibrational energy levels are
separated by gaps (quantized)
• If a photon of light strikes the molecule with the exact
amount of energy needed, a molecular vibration will
occur
• Energy is eventually released from the molecule
generally in the form of heat
• Infrared (IR) Light generally causes molecular vibration
• HOW might IR light absorbed give you information
about a molecule’s structure
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• Molecular bonds can vibrate by stretching or by bending
in a number of ways
• This chapter will focus mostly on stretching frequencies
• WHY do objects emit IR light?
• WHY do some objects emit more IR radiation than
others?
• WHERE does that light come from?
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• Some night vision goggles can detect IR light that is
emitted
• IR or thermal imaging is also used to detect breast
cancer
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• The energy necessary to cause vibration depends on the
type of bond
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• An IR spectrophotometer irradiates a sample with all
frequencies of IR light
• The frequencies that are absorbed by the sample tell us
the types of bonds (functional groups) that are present
• How do we measure the frequencies that are absorbed?
• Most commonly, samples are deposited neat on a salt
(NaCl) plate. WHY is salt used?
• Alternatively, the compound may be dissolved in a
solvent or embedded in a KBr pellet
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• In the IR spectrum below, WHAT is % transmittance and
how does it relate to molecular structure?
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• Analyze the units for the wavenumber,
• ν = frequency and c = the speed of light
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• HOW are wavelength and wavenumber different?
• HOW are wavenumbers and energy related?
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Klein, Organic Chemistry 2e
15.2 IR Spectroscopy
• A signal on the IR spectrum has three important
characteristics: wavenumber, intensity, and shape
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• The wavenumber for a stretching vibration depends on
the bond strength and the mass of the atoms bonded
together
• Should bonds between heavier atoms require higher or
lower wavenumber IR light to stretch?
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Rationalize the trends below using the wavenumber
formula
1.
2.
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• The wavenumber formula and empirical observations
allow us to designate regions as representing specific
types of bonds
• Explain the regions above
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• The region above 1500 cm-1 is called the diagnostic
region. WHY?
FINGERPRINT REGION
DIAGNOSTIC REGION
• The region below 1500 cm-1 is called the fingerprint
region. WHY?
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Analyze the diagnostic and fingerprint regions below
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Analyze the diagnostic and fingerprint regions below
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Compare the IR spectra
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Given the formula below and the given IR data, predict
whether a C-H or O-H bond is stronger
• C-H stretch ≈ 3000 cm-1
• O-H stretch ≈ 3400 cm-1
• Practice with conceptual checkpoint 15.1
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Compare the IR stretching wavenumbers below
• Are the differences due to mass or bond strength?
• Which bond is strongest, and WHY?
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Note how the region ≈3000 cm-1 in the IR spectrum can
give information about the functional groups present
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Is it possible that an alkene or alkyne could give an IR
spectra without any signals above 3000 cm-1?
• Predict the wavenumbers that would result (if any)
above 3000 cm-1 for the molecules below
• Practice with conceptual checkpoint 15.2
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Resonance can affect the wavenumber of a stretching
signal
• Consider a carbonyl that has two resonance contributors
• If there were more contributors with C-O single bond
character than C=O double bond character, how would
that affect the wavenumber?
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Klein, Organic Chemistry 2e
15.3 IR Signal Wavenumber
• Use the given
examples to
explain HOW
and WHY the
conjugation
and the –OR
group affect
resonance
and thus the
IR signal?
• Practice with conceptual checkpoint 15.3
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Klein, Organic Chemistry 2e
15.4 IR Signal Strength
• The strength of IR signals can vary
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Klein, Organic Chemistry 2e
15.4 IR Signal Strength
• When a bond undergoes a stretching vibration, its
dipole moment also oscillates
• Recall the formula for dipole moment includes the
distance between the partial charges,
• The oscillating dipole moment creates an electrical field
surrounding the bond
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15-30
Klein, Organic Chemistry 2e
15.4 IR Signal Strength
• The more polar the bond, the greater the opportunity
for interaction between the waves of the electrical field
and the IR radiation
• Greater bond polarity = stronger IR signals
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Klein, Organic Chemistry 2e
15.4 IR Signal Strength
• Note the general strength of
the C=O stretching signal vs.
the C=C stretching signal
• Imagine a symmetrical
molecule with a completely
nonpolar C=C bond: 2,3dimethyl-2-butene
• 2,3-dimethyl-2-butene does
not give an IR signal in the
1500-2000 cm-1 region
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Klein, Organic Chemistry 2e
15.4 IR Signal Strength
• Stronger signals are also observed when there are
multiple bonds of the same type vibrating
• Although C-H bonds are not very polar, they often give
very strong signals, WHY?
• Because sample concentration can affect signal strength,
the Intoxilyzer 5000 can be used to determine blood
alcohol levels be analyzing the strength of C-H bond
stretching in blood samples
• Practice with conceptual checkpoints 15.5 – 15.7
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• Some IR signals are broad, while others are very narrow
• O-H stretching signals are often quite broad
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• When possible, O-H bonds form H-bonds that weaken
the O-H bond strength
• WHY does H-bonding
affect the O-H bond
strength?
• The H-bonds are transient, so the sample will contain
molecules with varying O-H bond strengths
• Why does that cause the O-H stretch signal to be broad?
• The O-H stretch signal will be narrow if a dilute solution
of an alcohol is prepared in a solvent incapable of Hbonding
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15-35
Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• In a sample with an intermediate concentration, both
narrow and broad signals are observed. WHY?
• Explain the cm-1 readings for the
two O-H stretching peaks
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• Consider how broad the O-H stretch is for a carboxylic
acid and how its wavenumber is around 3000 cm-1
rather than 3400 cm-1 for a typical O-H stretch
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• H-bonding is often more pronounced in
carboxylic acids, because they can forms
H-bonding dimers
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15-38
Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• For the molecule below, predict all of the stretching
signals in the diagnostic region
• Practice with conceptual checkpoint 15.9
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• Primary and secondary amines exhibit N-H stretching
signals. WHY not tertiary amines?
• Because N-H bonds are capable of H-bonding, their
stretching signals are often broadened
• Which is generally more polar, an O-H or an N-H bond?
• Do you expect N-H stretches to be strong or weak
signals?
• See example spectra on next slide
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• The appearance of two N-H signals
for the primary amine is NOT simply
the result of each N-H bond giving a
different signal
• Instead, the two N-H bonds vibrate
together in two different ways
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Klein, Organic Chemistry 2e
15.5 IR Signal Shape
• A single molecule can only vibrate symmetrically or
asymmetrically at any given moment, so why do we see
both signals at the same time?
• Similarly, CH2 and CH3 groups can also vibrate as a group
giving rise to multiple signals
• Practice with conceptual checkpoint 15.10
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Klein, Organic Chemistry 2e
15.6 Analyzing an IR Spectrum
• Table 15.2 summarizes some of the key signals that help
us to identify functional groups present in molecules
• Often, the molecular structure can be identified from an
IR spectra
1. Focus on the diagnostic region (above 1500 cm-1)
a)
b)
c)
d)
1600-1850 cm-1 – check for double bonds
2100-2300 cm-1 – check for triple bonds
2700-4000 cm-1 – check for X-H bonds
Analyze wavenumber, intensity, and shape for each signal
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Klein, Organic Chemistry 2e
15.6 Analyzing an IR Spectrum
•
Often, the molecular
structure can be identified
from an IR spectra
2. Focus on the 2700-4000
cm-1 (X-H) region
•
Practice with SkillBuilder
15.1
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15-45
Klein, Organic Chemistry 2e
15.7 Using IR to Distinguish
Between Molecules
• As we have learned in previous chapters, organic
chemists often carry out reactions to convert one
functional group into another
• IR spectroscopy can often be used to determine the
success of such reactions
• For the reaction below, how might IR spectroscopy be
used to analyze the reaction?
• Practice with SkillBuilder 15.2
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Klein, Organic Chemistry 2e
15.7 Using IR to Distinguish
Between Molecules
• For the reactions below, identify the key functional
groups, and describe how IR data could be used to verify
the formation of product
• Is IR analysis qualitative or quantitative?
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15-47
Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• Mass spectrometry is primarily used to determine the
molar mass and formula for a compound
1. A compound is vaporized and then ionized
2. The masses of the ions are detected and graphed
• Can you think of ways to get an organic molecule to
ionize?
• Will the molecule need to absorb energy or release
energy?
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Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• The most common method of ionizing molecules is by
electron impact (EI)
• The sample is bombarded with a beam of high energy
electrons (1600 kcal or 70 eV)
• EI usually causes an electron to be ejected from the
molecule. HOW? WHY?
• What is a radical cation?
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Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• How does the mass of the radical cation compare to the
original molecule?
• If the radical cation remains intact, it is known as the
molecular ion (M+•) or parent ion
• Often, the molecular ion undergoes some type of
fragmentation. WHY?
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Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• The resulting fragments may undergo even further
fragmentation
• The ions are deflected by a magnetic field
• Smaller mass and higher charge fragments are affected
more by the magnetic field. WHY?
• Neutral fragments are not detected. WHY?
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Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• Explain the units on the x and
y axes for the mass spectrum
for methane
• The base peak is the tallest
peak in the spectrum
• For methane the base peak
represents the M+•
• Sometimes, the M+• peak is
not even observed in the
spectrum, WHY?
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15-52
Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• Peaks with a mass of less than M+• represent fragments
• Subsequent H radicals can be fragmented to give the
ions with a mass/charge = 12, 13 and 14
• The presence of a peak representing (M+1) +• will be
explained in section 15.10
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Klein, Organic Chemistry 2e
15.8 Into to Mass Spectrometry
• Mass spec is a relatively sensitive analytical method
• Many organic compounds can be identified
– Pharmaceutical: drug discovery and drug metabolism,
reaction monitoring
– Biotech: amino acid sequencing, analysis of macromolecules
– Clinical: neonatal screening, hemoglobin analysis
– Environmental: drug testing, water quality, food
contamination testing
– Geological: evaluating oil composition
– Forensic: Explosive detection
– Many More
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15-54
Klein, Organic Chemistry 2e
15.9 Analyzing the M+• Peak
• In the mass spec for benzene, the M+•
peak is the base peak
• The M+• peak does not easily fragment
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15-55
Klein, Organic Chemistry 2e
15.9 Analyzing the M+• Peak
• Like most compounds,
the M+• peak for
pentane is NOT the
base peak
• The M+• peak
fragments easily
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15-56
Klein, Organic Chemistry 2e
15.9 Analyzing the M+• Peak
• The first step in analyzing a mass spec is to identify the
M+• peak
– It will tell you the molar mass of the compound
– An odd massed M+• peak MAY indicate an odd number of N
atoms in the molecule
– An even massed M+• peak MAY indicate an even number of N
atoms or zero N atoms in the molecule
• Give an alternative explanation for a M+• peak with an
odd mass
• Practice with conceptual checkpoint 15.19
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15-57
Klein, Organic Chemistry 2e
15.10 Analyzing the (M+1)+• Peak
• Recall that the (M+1)+• peak in
methane was about 1% as
abundant as the M+• peak
• The (M+1)+• peak results from
the presence of 13C in the
sample. HOW?
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15-58
Klein, Organic Chemistry 2e
15.10 Analyzing the (M+1)+• Peak
• For every 100 molecules of decane,
what percentage of them are made of
exclusively 12C atoms?
• Comparing the heights of the (M+1)+•
peak and the M+• peak can allow you to
estimate how many carbons are in the
molecule. HOW?
• The natural abundance of deuterium is
0.015%. Will that affect the mass spec
analysis?
• Practice with SkillBuilder 15.3
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15-59
Klein, Organic Chemistry 2e
15.11 Analyzing the (M+2)+• Peak
• Chlorine has two abundant isotopes
• 35Cl=76% and 37Cl=24%
• Molecules with chlorine
often have strong (M+2)+•
peaks
• WHY is it sometimes
difficult to be absolutely
sure which peak is the
(M)+• peak?
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Klein, Organic Chemistry 2e
15.11 Analyzing the (M+2)+• Peak
•
79Br=51%
and 81Br=49%, so molecules with bromine
often have equally strong (M)+• and (M+2)+• peaks
• Practice with
conceptual
checkpoints
15.23 and
15.24
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15-61
Klein, Organic Chemistry 2e
15.12 Analyzing the Fragments
• A thorough analysis of the molecular fragments can
often yield structural information
• Consider pentane
• Remember, MS only
detects charged
fragments
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Klein, Organic Chemistry 2e
15.12 Analyzing the Fragments
• WHAT type of
fragmenting is
responsible for the
“groupings” of
peaks observed?
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Klein, Organic Chemistry 2e
15.12 Analyzing the Fragments
• In general, fragmentation will be more prevalent when
more stable fragments are produced
• Correlate the relative
stability of the fragments
here with their abundances
on the previous slide
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Klein, Organic Chemistry 2e
15.12 Analyzing the Fragments
• Consider the fragmentation below
• All possible fragmentations are generally observed
under the high energy conditions employed in EI-MS
• If you can predict the most abundant fragments and
match them to the spectra, it can help you in your
identification
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Klein, Organic Chemistry 2e
15.12 Analyzing the Fragments
• Alcohols generally undergo two main types of
fragmentation: alpha cleavage and dehydration
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Klein, Organic Chemistry 2e
15.12 Analyzing the Fragments
• Amines generally undergo alpha cleavage
• Carbonyls generally undergo McLafferty rearrangement
• Practice with conceptual checkpoints
15.25 – 15.28
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Klein, Organic Chemistry 2e
15.13 High Resolution Mass Spec
• High Resolution Mass Spectrometry allows m/z to be
measured with up to 4 decimal places
• Masses are generally not whole number integers
– 1 proton = 1.0073 amu and 1 neutron = 1.0086 amu
• One 12C atom = exactly 12.0000 amu, because the amu
scale is based on the mass of 12C
• All atoms other than 12C will have a mass in amu that
can be measured to 4 decimal places by a highresolution mass spec instrument
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Klein, Organic Chemistry 2e
15.13 High Resolution Mass Spec
• Note the exact masses and natural abundances below
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Klein, Organic Chemistry 2e
15.13 High Resolution Mass Spec
• Why are the values in table 15.5 different from those on
the periodic table?
• Imagine you want to use
high-res MS to distinguish
between the molecules
below
• Why can’t you use low-res?
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15-70
Klein, Organic Chemistry 2e
15.13 High Resolution Mass Spec
• Using the exact masses and natural abundances for each
element, we can see the difference high-res makes
• The molecular ion results from the molecule composed
of the isotopes with the greatest natural abundance
• What if the molecular ion is not observed?
• Practice with conceptual checkpoints 15.19 and 15.30
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15-71
Klein, Organic Chemistry 2e
15.14 High Resolution Mass Spec
• MS is suited for the identification of pure substances
• However, MS instruments are often connected to a gas
chromatograph so mixtures can be analyzed
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Klein, Organic Chemistry 2e
15.14 High Resolution Mass Spec
• GC-MS gives two main forms of information
1. The chromatogram
gives the retention
time
2. The Mass
Spectrogram (low-res
or high-res)
• GC-MS is a great technique for detecting compounds
such as drugs in solutions such as blood or urine
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15-73
Klein, Organic Chemistry 2e
15.15 MS of Large Biomolecules
• To be analyzed by EI mass spec, substances generally
must be vaporized prior to ionization
• Until recently (last 30 years), compounds that
decompose before they vaporize could not be analyzed
• In Electrospray ionization (ESI), a high-voltage needle
sprays a liquid solution of an analyte into a vacuum
causing ionization
• HOW is ESI relevant for analyzing large biomolecules?
• ESI is a “softer” ionizing technique. WHAT does that
mean?
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15-74
Klein, Organic Chemistry 2e
15.16 Degrees of Unsaturation
• Mass spec can often be used to determine the formula
for an organic compound
• IR can often determine the functional groups present
• Careful analysis of a molecule’s formula can yield a list
of possible structures
• Alkanes follow the formula below, because they are
saturated
CnH2n+2
• Verify the formula by drawing some isomers of pentane
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Klein, Organic Chemistry 2e
15.16 Degrees of Unsaturation
• Notice that the general formula for the compound,
CnH2n+2, changes when a double or triple bond is present
• Adding a degree of unsaturation decreases the number
of H atoms by two
• How many degrees of unsaturation are there in
cyclopentane?
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Klein, Organic Chemistry 2e
15.16 Degrees of Unsaturation
• Consider the isomers of C4H6
• How many degrees of unsaturation are there?
• 1 degree of unsaturation = 1 unit on the hydrogen
deficiency index (HDI)
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Klein, Organic Chemistry 2e
15.16 Degrees of Unsaturation
• For the HDI scale, a halogen is treated as if it were a
hydrogen atom
• How many degrees of unsaturation are there in C5H9Br?
• An oxygen does not affect the HDI. WHY?
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Klein, Organic Chemistry 2e
15.16 Degrees of Unsaturation
• For the HDI scale, a nitrogen increases the number of
expected hydrogen atoms by ONE
• How many degrees of unsaturation are there in
C5H8BrN?
• You can also use the formula below
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Klein, Organic Chemistry 2e
15.16 Degrees of Unsaturation
• Calculating the HDI can be very useful. For example, if
HDI=0, the molecule can NOT have any rings, double
bonds, or triple bonds
• Propose a structure for a molecule with the formula
C7H12O. The molecule has the following IR peaks
– A strong peak at 1687 cm-1
– NO IR peaks above 3000 cm-1
• Practice with SkillBuilder 15.4
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Klein, Organic Chemistry 2e
Additional Practice Problems
• Explain why a completely nonpolar bond will not give a
stretching signal in the IR spectra. Would you expect to
see a signal for C-H stretching for a nonpolar molecule?
Why or why not?
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Klein, Organic Chemistry 2e
Additional Practice Problems
• Explain how IR might be used to qualitatively determine
the degree of substitution when ammonia is treated
with excess bromoethane.
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Additional Practice Problems
• How might you use EI GCMS to distinguish between
constitutional isomers?
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Klein, Organic Chemistry 2e
Additional Practice Problems
• Explain how an experiment involving isotopic labeling
might be used to explore the type of fragmentation that
occurs in the MS analysis of organic compounds.
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