ORGANIC CHEMISTRY Y. ORGANIC CHEMISTRY - … CHEMISTRY Y. ORGANIC CHEMISTRY
organic chemistry(Infrared)
-
Upload
sahilvermasv92 -
Category
Documents
-
view
213 -
download
0
Transcript of organic chemistry(Infrared)
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 1/42
Richard F. Daley and Sally J. Daleywww.ochem4free.com
Organic
ChemistryChapter 9
Infrared Spectroscopyand
Mass Spectrometry9.1 Electromagnetic Radiation and Spectroscopy 431
9.2 Molecular Vibrations in Infrared Spectroscopy 4349.3 Introduction to Interpreting Infrared Spectra 436
9.4 Hydrogen Bonded to sp 3 Hybrid Atoms 439
9.5 Hydrogen Bonded to sp 2 and sp Hybrid Atoms 443
9.6 Carbon—Heteroatom Bonds 448
9.7 Other Bonds 452
9.8 Interpreting Infrared Spectra, Part 2 456
9.9 Mass Spectrometry 459
9.10 The Molecular Ion 463
Key Ideas from Chapter 9 467
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 2/42
Organic Chemistry - Ch 9 430 Daley & Daley
Copyright 1996-2005 by Richard F. Daley & Sally J. DaleyAll Rights Reserved.
No part of this publication may be reproduced, stored in a retrieval system, or
transmitted in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the copyright
holder.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 3/42
Organic Chemistry - Ch 9 431 Daley & Daley
Chapter 9
Infrared Spectroscopy and MassSpectrometry
Chapter Outline
9.1 Electromagnetic Radiation and SpectroscopyBackground on the basis of molecular spectroscopy
9.2 Molecular Vibrations in Infrared SpectroscopyThe molecular motions that absorb energy in an infrared
spectrum
9.3 Introduction to Interpreting Infrared SpectraHow to “read” an infrared spectrum
9.4 Hydrogen Attached to sp 3 Hybrid AtomsThe infrared absorptions for various hydrogens to sp 3 hybridized
carbon, nitrogen, and oxygen bonds
9.5 Hydrogen Attached to sp 2 and sp Hybrid AtomsThe infrared absorptions for various hydrogens to sp 2 and sp
hybridized carbon, nitrogen, and oxygen bonds
9.6 Carbon—Heteroatom BondsThe infrared absorptions for various hydrogens to sp 2 and sp
hybridized carbon, nitrogen, and oxygen bonds
9.7 Other BondsThe infrared absorptions for various bonds not already covered in
the last three sections
9.8 Interpreting Infrared Spectra, Part 2Using an infrared spectrum to find information about an
unknown compound
9.9 Mass Spectrometry An introduction to the techniques of mass spectrometry
9.10 The Molecular IonUsing the molecular ion in interpreting a mass spectrum
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 4/42
Organic Chemistry - Ch 9 432 Daley & Daley
Objectives
✔ Understand what types of molecular motions absorb infrared
energy
✔ Know the wavenumbers for the absorptions of common types of
bonds in an infrared spectrum
✔ Be able to interpret significant peaks in the functional group
region of an infrared spectrum
✔ Confirm, where appropriate, functional group absorptions with
peaks in the fingerprint region of the spectrum
✔ Understand how mass spectrometry works and what information
you can gather from the molecular ion
What a wondrous pair of spectacles
is here!
—Shakespeare
O rganic chemists frequently obtain many new compounds
through laboratory syntheses and extraction from
biological or environmental sources. They then need to either identify
or confirm the structure of these new compounds. Before the 1950s,they had only three general methods available to obtain this
information. One, they could perform an elemental analysis on the
compound. Two, they could perform chemical transformations, or
reactions, using the compound they wanted to identify, then observe
the results. Many of these transformations were simple chemical teststhat provided information rapidly—future chapters present many of
these reactions, as they are still useful. Three, they could take
physical measurements of the compounds, such as the refractive
index, melting or boiling point, and density. Then they compared these
measurements with the known values of previously studied
compounds.Now chemists have instruments to help identify the structure
of compounds. These instruments supplement, and at times replace,
the chemical tests and physical property measurements. Although
many methods of instrumental analysis are available, the most widely
used techniques are those of molecular spectroscopy. Infrared (IR),
nuclear magnetic resonance (NMR), and mass spectrometry (MS) are
the most prominent of the molecular spectroscopy techniques.
Subjecting a molecule to these three techniques gives sufficient
The basis of molecular
spectroscopy involves
exposing the molecule
to an energy source,
then observing how the
molecule responds to
that energy.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 5/42
Organic Chemistry - Ch 9 433 Daley & Daley
information to allow chemists to draw accurate conclusions about the
identity and structure of most molecules.
This chapter considers infrared spectroscopy and mass
spectrometry. By using IR spectroscopy, chemists obtain informationabout the vibrational motions of a molecule. Each type of molecule has
its own characteristic set of stretching and bending motions. These
motions respond to infrared electromagnetic radiation in specific
ways. Chemists use this information to identify the functional groups
contained in a molecule.Mass spectrometry helps determine the molecular formula and
major groups of a molecule. Fundamentally, mass spectrometry is
different from IR and NMR in that it is not based on the absorption of
electromagnetic energy. MS shows how a molecule fragments when a
beam of energetic electrons bombards it. The spectrometer then
analyzes these fragments and plots them as the relative numbers of the various fragments versus the mass of the fragments. The chemist
then examines the spectrum to obtain the molecular weight of the
molecule as well as information about its structure. Chapter 10 covers
NMR, which helps determine the organic skeleton of the molecule.
9.1 Electromagnetic Radiation and Spectroscopy
Before beginning the study of infrared spectroscopy, you needsome background information about electromagnetic radiation and
spectroscopy. Electromagnetic radiation exists in a broad range of
energies. Visible light, infrared light, ultraviolet light, microwaves,and radio frequencies are different energy subdivisions of
electromagnetic radiation. Electromagnetic radiation consists of
discreet parcels of energy called photons. Photons have dual natures,as they possess properties of both particles and waves. Although
photons at rest have no mass, each does have an amount of energy
called a quantum.
The following equation expresses the relationship of the energy
of a photon and its wavelength.
E = hν
In this equation, E represents the energy of the photon, h is Planck's
constant (6.62 x 10–34 joule/second), and ν is the frequency of the
photon in hertz (or cycles/second). Frequency refers to the number of
oscillations that a particular electromagnetic radiation wave makes
per second. An oscillation is the full cycle of a wave from the crest of
that wave through its trough to the crest of its next wave. Wavelength
is the distance between two crests of a wave and is symbolized as λ.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 6/42
Organic Chemistry - Ch 9 434 Daley & Daley
This equation measures the energy of a photon of a particular
wavelength in joules.
The relationship between the frequency of a photon and its
wavelength is given by the following equation:
c = νλ
Where c is the velocity, or speed, of the electromagnetic radiation. The
speed of the electromagnetic radiation is the same velocity as visible
light in a vacuum, which is 3.0 x 108 m/s, and is a constant. Thus, the
shorter the wavelength of light, the higher the frequency and the
higher the energy of the photon. Radio waves are very low energy,
visible light is higher energy, and x-rays and γ-rays are much higher.
Figure 9.1 shows the range of frequencies in the electromagnetic
spectrum.
1081010101210141016
Energy
Frequency ( ) in Hz
rays X rays Ultraviolet Infrared MicrowaveRadio wave
1020 1018
Visible
Violet Red
Wavelength ( ) in meters
10010-210-410-610-810-1010-12
Figure 9.1. The electromagnetic spectrum.
Recall from Chapter 1 that electrons in atoms and molecules
have different energy levels called atomic and molecular orbitals.
Chapter 1 also discusses the electrons that inhabit these orbitals.When an atom absorbs electromagnetic radiation of a given energy, an
electron moves from a lower energy level to a higher one; that is, it
moves from its normal orbital to a higher energy orbital. Chemists callthis higher energy level an excited state. The excited state of an
electron is a higher energy level than its ground state, the energy
level that an electron normally inhabits. The electron returns to its
ground state when it loses the energy absorbed from the
electromagnetic radiation.
The excited state is a
higher energy level
than the normal, or
ground, state.
The transition of an atom from its normal state to its excited
state requires the input of an exact amount of energy. If the photon
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 7/42
Organic Chemistry - Ch 9 435 Daley & Daley
involved has either too little energy or too much energy, then no
transition occurs. Scientists say that the energy states of atoms are
quantized because atoms require a specific amount of energy to move
an electron from its normal state to its excited state. The following equation gives the difference in energy between the ground state ( E0)
and the excited state ( E1) of an atom.
∆ E = E1 – E0 = hν
An atom absorbs energy only when the energy of the photon of
electromagnetic radiation, hν, equals ∆ E.
Molecules consist of several different energy levels: thus, they
absorb energy at several different regions of the electromagnetic
spectrum. For example, transitions that involve the nonbonding and π
electrons in a molecule absorb radiation in the ultraviolet and visibleregions of the spectrum. Vibrational energy changes take place in the
infrared region while rotational energy changes occur in themicrowave region. The electromagnetic spectrum is an arbitrary
division of the different types of electromagnetic radiation. Table 9.1
summarizes the regions of significance to organic chemists.
Region of
Spectrum Molecular
Change Energy
(kcal mole–1) Frequency
(Hertz) Wavelength
(cm)
Ultraviolet
Visible
Electronic
transitions
50 to 100 1014 to 1016 10–6 to 10–4
Infrared Molecular
vibrations
5 1012 to 1014 10–4 to 10–2
Microwave Molecular
rotations
3 x 10–3 1010 1
Radio-
frequency
Orientation of
the spin of
nucleus
10–7 107 to 109 105 to 106
Table 9.1. Selected regions of the electromagnetic spectrum used in spectroscopy.
-
Because molecules absorb electromagnetic radiation in specific
ways, chemists have developed instruments, called spectrometers, to
measure this information. Chemists use the general term spectroscopy
to describe this method. Spectrometers consist mainly of four parts: asource of electromagnetic radiation, a sample chamber, a detector for
the radiation, and data handling system for the spectrum. Typical
data handling systems range from a plotter to a sophisticated
computer system. Figure 9.2 shows schematically how a spectrometer
works. As the instrument directs electromagnetic radiation at the
molecule sample, it varies the frequency of the radiation. The detectorthen compares that intensity with the intensity of the radiation at the
source. When the sample absorbs energy, the detector senses a
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 8/42
Organic Chemistry - Ch 9 436 Daley & Daley
decrease in the intensity. It then plots the relationship between the
intensity and the wavelength as a spectrum. An infrared spectrum is
plotted as transmittance (the y-axis) versus wavelength (the x-axis). A spectrum is a plot of
the intensity of
electromagneticradiation reaching the
detector versus the
wavelength of the
radiation.
Trasnsmittance is the
intensity of the
infrared radiation at a
given wavelength of the
sample in the
instrument divided by
the intensity with no
sample present.
Figure 9.2. Schematic diagram of a spectrometer.
Modern infrared spectrometers have a schematic that lookssimilar to Figure 9.2. To obtain a spectrum, first run the instrument
with no sample in the beam. This blank spectrum measures spectrum
of the source of the infrared radiation and any background absorptions
in the atmosphere as it passes through the spectrometer. Next, place asample in the instrument and obtain a second spectrum. The
spectrometer subtracts the first spectrum from the second, thus,eliminating the effects of atmospheric components. Placing pure
solvent in the instrument when obtaining the blank allows the
spectrometer to subtract the spectrum of the solvent from a solution.
9.2 Molecular Vibrations in Infrared Spectroscopy
Because the atoms within a molecule move constantly, theycause bond distortion. Atoms move, or vibrate, in two general ways.
They stretch, and they bend. An infrared spectrum detects changes in
these molecular vibrations. Stretching vibrations produce
changes in the bond length. Bending vibrations cause changes in
the bond angle. Because there are a number of specific stretching or
bending vibrations, these are often called stretching or bending modes.Figure 9.3 shows two stretching modes, and Figure 9.4 shows four
bending modes observed in the infrared.
Molecular vibrations
are the motions of
atoms in molecules.
Stretching and
bending vibrations are
the molecular
vibrations observed in
the IR spectrum.
Stretching:
AntisymmetricSymmetric
C C
Figure 9.3. Stretching
modes measured in the
infrared.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 9/42
Organic Chemistry - Ch 9 437 Daley & Daley
WaggingTwisting
Out-of-Plane
In-PlaneRockingScissoring
Bending:
C C
C C
Figure 9.4. Bending modesmeasured in the infrared.
When a beam of light strikes a bond with a dipole moment, the
oscillating electric field component of light causes the bond to vibrate
by alternately stretching and compressing. The bond’s vibration is
analogous to a spring that either stretches or compresses when an
external force is applied. As the bond vibrates in response to thechanging polarity of the electric field, it absorbs energy, as shown in
Figure 9.5. The frequencies at which these bonds absorb energy are in
the infrared region of the spectrum.
+
-
+
-
+
-
-
+
-
+
-
+
Figure 9.5. A bond with a dipole moment is either stretched or compressed in the
presence of the oscillating electric field component of light.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 10/42
Organic Chemistry - Ch 9 438 Daley & Daley
Molecules have many vibrational energy states. The set of
vibrations belonging to each molecule is unique for that particular
molecule. The molecules that make up a homologous series also have a
unique set of vibrations. This uniqueness is the basis of infraredspectroscopy. By understanding the meaning of the various peaks, you
can obtain information about a molecule's functional groups.
Molecules absorb infrared electromagnetic radiation within the
wavelength range of 2.5 to 16 µm (a µm is 10–6 meters). Most infrared
spectra are reported in wavenumbers, so this range corresponds to
4000 to 625 cm–1. An advantage of using wavenumbers is that they
are directly proportional to energy, whereas wavelengths are inversely
proportional to energy. Thus, a wavelength of 2.5 µm corresponds to
4000 cm–1 and is the high-energy end of the scale.
Wavenumbers are the
reciprocal of the
wavelength in
centimeters and are
reported as cm –1.
A typical infrared spectrum consists of a series of absorption
peaks. These peaks vary in shape and intensity in relation to thevibrational energy states of the particular molecule that you areanalyzing. Intensity refers to the size of a peak. The greater the
amount of radiation absorbed, the larger the size of the peak. The
intensity of an absorption peak, or band, corresponds to the magnitude
of the change in the dipole moment caused by the vibration of the
molecule. The larger the change in the dipole moment, the more
intense the absorption. For example, the stretching in a carbonylgroup (C=O) increases an already large dipole moment. This
stretching produces a peak of greater intensity than a peak obtained
from the stretching of a C=C double bond because there is a relatively
small change in the dipole moment of a C=C double bond. Because of
these variations in intensity, the interpretation of an infraredspectrum requires a more careful observation of relative intensitiesthan just simply reading numbers from tables.
9.3 Introduction to Interpreting Infrared Spectra
Each infrared spectrum consists of two parts. The first part
covers the area from 4000 cm–1 to about 1500 cm–1 and gives
information about the functional groups contained in the molecule. Of the two parts, this one is usually the easier to interpret. The second
part, called the fingerprint region, covers the region fromapproximately 1500 cm–1 to 625 cm–1. Table 9.2 summarizes the
characteristic infrared absorption frequencies. The types of bondspresent in a molecule determine the frequencies at which the
functional groups absorb the energy. Figure 9.6 summarizes Table 9.2
by diagramming these frequencies by bond type.
The fingerprint region gives information
about the molecular
structure of the
molecule.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 11/42
Organic Chemistry - Ch 9 439 Daley & Daley
Bond
Type
Functional Group Stretching
(cm–1)
Bending
(cm–1)
C—H Alkane 2970–2850 (s) 1470-1350 (s)
C—H Alkene 3080-3020 (m) 1000-675 (s)C—H Alkyne 3300 (s)
C—H Aldehyde 2900 (m)
2700 (m)C—H Aromatic 3100-3000 (v) 800-675 (s)
O—H Alcohols (H-bonded) 3500-3300 (s, broad) 1620-1590 (v)
O—H Alcohols (No H-bonds) 3650-3590 (v)
O—H Acid 3000-2500 (s, broad) 1655-1510 (v)
N—H Amine 3500-3300 (m)N—H Amide 3500-3350 (v)
C—C Alkane 1200-800(w)
C C Alkene 1680-1620 (v)C C Aromatic 1600-1450 (v)
C C Alkyne 2260-2100 (v)
C—O Alcohol, ether, ester 1300-1000 (s)
C O Ketone 1725-1705 (s)
C O Aldehyde 1740-1720 (s)
C O Aryl ketone 1700-1680 (s)
C O Ester 1750-1735 (s)
C O Acid 1725-1700 (s)
C O Amide 1690-1650 (s)C—N Alkyl amine 1220-1050 (w)
C—N Aryl amine 1360-1250 (s)
C N Nitrile 2260-2100 (v)NO2 Nitro 1580-1560 (s)
1400-1380 (s)
(s) - Strong - the most intense, or largest, peaks in the spectrum.
(m) - Medium - peaks that are 40 to 70% of the intensity of the strong peaks.
(w) - Weak - peaks that are less than 40% of the intensity of the strong peaks.
(v) - Variable - variable intensity peaks. Usually medium to weak.
Table 9.2. Characteristic infrared absorption frequencies.
BendC HC
Stretch, C NC C
O-H, N-H, C-H Stretch N-H Bend
C-H Bend
C-O, C-N Bend
C C C N Stretch,
65070080090010001100130015002000250030004000
100
0
Wavenumber (cm-1)
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 12/42
Organic Chemistry - Ch 9 440 Daley & Daley
Figure 9.6. Regions where different types of bonds absorb energy in the infrared.
To see how the structural features of individual molecules
affect infrared spectra, compare the spectra of hexane (Figure 9.7a)
and hexene (Figure 9.7b). The two spectra look different. For example,
the peak just above 3000 cm–1 present in the spectrum of hexene isabsent in the hexane spectrum. This peak is characteristic of the C—H
stretching of an sp 2 hybridized carbon. The similar peaks just below
3000 cm–1 in both spectra are characteristic of the C—H stretching in
an sp 3 hybridized carbon. The next significant difference is the peak at
1640 cm–1 in hexene. This peak is characteristic of C=C double bond
stretching. The peaks between 1500 and 1300 cm–1, present in both
spectra, are the result of C—H bending on an sp 3 hybridized carbon.
The last difference is the peaks at 990 and 910 cm–1 of the hexene
spectrum. These peaks are characteristic of the bending of the C—Hbonds on a double bond.
0
100
4 00 0 3 00 0 2 50 0 20 00 150 0 130 0 110 0 100 0 90 0 8 0 0 70 0 65 0
Wavenumber (cm-1)
(a)
0
100
4 00 0 3 00 0 2 50 0 20 00 150 0 130 0 110 0 100 0 90 0 8 0 0 70 0 65 0
Wavenumber (cm
-1
) (b)
Figure 9.7. Infrared spectrum of (a) hexane and (b) 1-hexene.
If you are a beginner, and you try to interpret every peak of an
IR spectrum, you may find the analysis very complex and very
difficult. A typical IR spectrum simply contains too much information
to allow easy analysis. Frequently, even experienced chemists require
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 13/42
Organic Chemistry - Ch 9 441 Daley & Daley
help analyzing spectra, so they use extensive tables similar to Table
9.2 and Appendix B. Thus, the goal of this book is to help you
recognize the typical peak patterns and what the main peaks indicate.
This recognition allows you to rapidly identify individual functionalgroups. When identifying the functional groups, examine in detail the
4000 to 1500 cm–1 region of the spectrum. Refer to the fingerprint
region only as necessary to confirm the functional groups that you
found in the 4000 to 1500 cm–1 region.
9.4 Hydrogen Attached to sp 3 Hybrid Atoms
The appearance of each peak in a spectrum is typical of all the
compounds in that particular class of compounds. For example, the
infrared spectra of hexane and hexene in Figure 9.7 both show similarpeaks for sp 3 C—H bonds just below 3000 cm–1. However, even though
the spectral peaks for several compounds have a similar appearance,
none are identical. Even two spectra of the same compound may look
somewhat different on different instruments using different
concentrations of compound. The important considerations whenlooking at a spectrum are the position and general appearance of the
peaks, not the details of their specific shapes. The spectral fragments
shown in this and the following sections are from actual spectra.
Three common types of hydrogen to sp 3 hybrid atom bonds in
organic chemistry are hydrogen to oxygen bonds, hydrogen to nitrogen
bonds, and hydrogen to carbon bonds. This section discusses the
appearance of the IR peaks that arise from the stretching of these
three kinds of bonds.First, examine the three types of hydrogen to oxygen, H—O,
bonds: alcohols, phenols, and carboxylic acids. Both alcohols and
phenols produce a strong broad peak around 3500 - 3300 cm–1. The
first spectral fragment in Figure 9.8 shows an H—O stretch peak of an
alcohol involved in hydrogen bonding. On rare occasions, the H—O
bond is not hydrogen bonded but is a free bond. A free H—O bondoccurs at about 3550 cm–1. The free bond peak is neither as strong nor
as broad as the hydrogen bonded peak and occurs only when the
alcohol is either very dilute or too sterically crowded for hydrogen
bonding to occur. The center spectral fragment in Figure 9.8 shows thefree H—O group as it typically appears when mixed with some of the
hydrogen bonded H—O group. The appearance of a phenolic H—Ogroup is nearly identical with the appearance of the alcohols. The
third spectral fragment in Figure 9.8 shows the H—O bond of a
carboxylic acid. This peak is usually broad and its specific location is
hard to pin down. When you see a very broad peak, generally covering
the region between 3500 and 2500 cm–1, interpret it as a carboxylic
acid.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 14/42
Organic Chemistry - Ch 9 442 Daley & Daley
Typical OH Peaks
2 5 0 030004 0 0 0
100
0
Hydrogen bonded
Alcohol
250030004 0 0 0
Wavenumber (cm-1
)
Nonhydrogen bonded
Alcohol
250030004000
Carboxylic acid
Figure 9.8. Comparison of different types of OH groups in the IR.
Next, examine the two functional groups with hydrogen tonitrogen, H—N, bonds: amines and amides. Each of these groups
contains either one or two H—N bonds. The primary amide and amine
peaks have similar appearances and positions, as do the secondary
amide and amine peaks. Primary amides and amines produce two
peaks, but the secondary amides and amines produce only one. Figure
9.9 shows these four types of H—N bonds. The positions for thesebonds are in the 3500 cm–1 to 3300 cm–1 region of the spectrum. This
region is similar to the region for the H—O bonds. Generally, the
peaks for H—N stretching are weaker than the peaks for H—O
stretching. The major difference between amines and amides is the
strong peak for the carbonyl stretching at about 1700 cm–1. Section
9.6 discusses carbonyls in more detail. This difference illustrates animportant concept for interpreting infrared spectra. When identifying
a peak at one location, you must confirm your identification by using
any other peaks indicating that functional group.
Typical NH Peaks
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 15/42
Organic Chemistry - Ch 9 443 Daley & Daley
2 5 0 03 0 0 04 0 0 0
100
0
Primary amine
250030004000
Wavenumber (cm-1
)
Secondary amine
2 5 0 03 0 0 04 0 0 0
Primary amide
250030004000
Secondary amide
Figure 9.9. Comparison of different types of NH bonds in the IR.
The last type of sp 3 bond is the H—C bond in saturatedhydrocarbon groups. Separating the methyl (CH3), methylene (CH2),
and methine (CH) groups in this region is usually very difficult. Whenyou see a broad, often multiple, peak between 3000 and 2850 cm–1,
you need only note that the molecule has some type of sp 3 hybrid H—
C bonds. In addition, H—C bending appears in the range of 1470 to
1350 cm–1 as a set of two or more peaks. The lower wavenumber peak
increases in intensity and often splits into two or three peaks with
increasing branching. Figure 9.10 shows these two regions.
Typical sp 3 CH Peaks
Wavenumber (cm-1
)
2 5 0 03 0 0 04 0 0 0
1 0 0
0
C—H Stretching C—H Bending
1 1 0 01 3 0 01 5 0 0
Figure 9.10. Typical sp 3 C—H bending and stretching regions. Although
there are differences between the peaks for CH3, CH2, and CH groups,
the signals overlap so much that they are difficult to separate.
Exercise 9.1
Choosing from these molecules, assign a molecular structure to each of
the following IR spectra.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 16/42
Organic Chemistry - Ch 9 444 Daley & Daley
Phenol Butanoic acid Triethyl amine
Cyclohexanol Decanamide N -Methylcyclohexanamine
Note: Do not let yourself become confused by all the peaks in thesespectra. Concentrate only on the sections covered so far. As you add
more fragments, more peaks will be meaningful to you.
a)
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
b)T
r
a
n
s
m
i
t
ta
n
c
e
1 0 0
0( % )
4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
c)
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
d)
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 17/42
Organic Chemistry - Ch 9 445 Daley & Daley
T
r
a
n
sm
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Sample solution
c) This is a spectrum of decanamide. The peaks at 3300 cm–1 and 3200
cm–1 indicate the NH2 group, and the peak for the carbonyl stretching
at 1700 cm–1 confirms the presence of the amide group. The peaksbetween 3000 and 2850 cm–1 are typical of sp 3 C—H stretching.
9.5 Hydrogen Attached to sp 2 and sp Hybrid
Atoms
Three common types of sp 2 hybrid hydrogen to carbon, H—C,
bonds are the H—C bonds in alkenes, aromatic compounds, and
aldehydes. This section examines the peaks made by these bonds.Bond bending is more important than bond stretching when
interpreting alkenes and aromatic compounds. The locations of the
H—C bending peaks between 1000 and 650 cm–1 establishes the
substitution patterns of the compound. Both alkenes and aromatic
compounds have a medium intensity peak between 3100 and 3000 cm–
1. Usually a peak closer to 3000 cm–1 indicates an alkene, and a peakcloser to 3100 cm–1 indicates an aromatic compound. Many aromatic
compounds also have a second peak about 3000 cm–1. However, this
peak is not always present, so use it to confirm an aromatic compound,
but do not depend on it as a diagnostic for aromatics. Figure 9.11
shows typical spectra of alkenes and aromatic compounds.
Typical sp 2 CH Stretching Peaks
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 18/42
Organic Chemistry - Ch 9 446 Daley & Daley
1 0 0
0
4000 3000 2500
Wavenumber (cm-1
)
250030004000 4000 3000 2500
Alkene Aromatic Aldehyde
Figure 9.11. The general appearance of the peaks resulting from the H—C stretching
of sp 2 hybrid CH bonds.
An aldehyde shows two peaks: one at 2900 cm–1 and another at
2700 cm–1. When identifying an aldehyde group, rely mostly on the2700 cm–1 peak, as no other common functional groups have a peak at
this location. Plus, the peak at 2900 cm–1 often hides in the saturated
C—H stretching region. The third spectral fragment in Figure 9.11
shows the appearance of an aldehyde group.
Peaks in the region from 1000 cm–1 to 650 cm–1 indicate the
positions of substituents in a compound with a double bond or on anaromatic compound. The sp 2 carbon hydrogen bond bending causes
these peaks. Figure 9.12 shows the most common types of double
bonds. A monosubstituted double bond (RCH=CH2) has two peaks: one
at approximately 990 cm–1 and the other at 910 cm–1. A trans disubstituted double bond has a single peak at 950 cm–1, and the cis
disubstituted double bond has a broad peak at 720 cm–1. A
trisubstituted double bond has a peak at 890 cm–1 and 820 cm–1. The
trisubstituted example in Figure 9.12 shows the two peaks with
different intensities, but the relative intensity varies. Sometimes the
peaks are of equal intensity, and sometimes the 820 cm–1 peak is
stronger, depending on the details of the structure of the compound.
Typical sp 2 CH Bending Peaks
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 19/42
Organic Chemistry - Ch 9 447 Daley & Daley
Wavenumber (cm-1
)
65070080090010001100
1 0 0
0
Wavenumber (cm-1
)
65070080090010001100
1 0 0
0
monosubstituted trans disubstituted
Wavenumber (cm-1
)
65070080090010001100
1 0 0
0
Wavenumber (cm-1
)
65070080090010001100
1 0 0
0
cis disubstituted trisubstituted
Figure 9.12. Double bond patterns for H—C bending of various types of double bonds.
The 900 cm–1 to 675 cm–1 region of the spectrum contains
peaks that indicate substitutions on aromatic rings. Figure 9.13 shows
the four basic substitution patterns on an aromatic compound. The
first spectrum shows a monosubstituted compound with peaks at 730
and 690 cm–1. Compare this spectrum with the spectrum of the meta
substitution. Frequently, confusion occurs over these two, because
both spectra have two peaks at about the same position. To avoid thisconfusion, note that a monosubstituted aromatic ring has about 40
cm–1 between its two peaks, and the meta has about 75 cm–1 between
its two peaks. Although the positions of the peaks vary, the
separations between the pairs stay the same. People also confuse the
ortho and para patterns, unless they remember that the ortho appearsin the range of 730 to 690 cm–1, and the para is in the range of 830 to
780 cm–1.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 20/42
Organic Chemistry - Ch 9 448 Daley & Daley
X
ortho
meta
para
Relative positions of ortho, meta, and para substituents on an aromatic ring.
Typical Aromatic Substitution Patterns
Wavenumber (cm-1
)
650700800900
1 0 0
0
Wavenumber (cm
-1)
650700800900
1 0 0
0
Monosubstituted Ortho disubstituted
Wavenumber (cm-1
)
650700800900
1 0 0
0
Wavenumber (cm-1
)
650700800900
1 0 0
0
Meta disubstituted Para disubstituted
Figure 9.13. Substitution patterns for aromatic rings.
The location of the H—C stretching for a triple bond is at
approximately 3300 cm–1, which is nearly the same location as the
H—O peaks of alcohols and phenols. You can avoid confusing the twoby remembering that the OH groups have a broad peak and the CH
peak is sharp. Figure 9.14 shows the H—C triple bond peak.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 21/42
Organic Chemistry - Ch 9 449 Daley & Daley
Typical Alkyne CH Stretch
Wavenumber (cm-1
)
250030004000
1 0 0
0
Alkyne
Figure 9.14. The H—C bond of a terminal alkyne.
Note that its position is nearly the same as that of the OH peaks of alcohols and phenols, but its
appearance is very different.
Exercise 9.2
Choosing from these molecules, assign a molecular structure to each of
the following IR spectra.
Cycloheptene 4-Methylaniline p-Dichlorobenzene
Ethynylbenzene 3-Methylphenol Iodobenzene
(Note: aniline is aminobenzene (PhNH2) and phenol ishydroxybenzene (PhOH).)
a)
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
b)
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 22/42
Organic Chemistry - Ch 9 450 Daley & Daley
T
r
a
n
sm
i
t
t
a
n
c
e
1 0 0
0( % )
4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
c)
T
r
a
ns
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
d)
T
r
a
n
s
m
i
t
ta
n
c
e
1 0 0
0( %)
4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Sample solution
c) This spectrum is of ethynyl benzene. The diagnostic peaks are the
terminal alkyne at 3300 cm–1 and the pattern matching of the mono
substituted aromatic ring at 800 to 700 cm–1.
9.6 Carbon—Heteroatom Bonds
This section discusses the three categories of carbon to
heteroatom bonds: carbon to oxygen, carbon to nitrogen, and carbon to
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 23/42
Organic Chemistry - Ch 9 451 Daley & Daley
halogen. All three are part of various functional groups that occur in
many types of organic compounds. First, examine the C—O single
bonds contained in alcohols, phenols, ethers, and esters. Because the
peaks for each of these types of bonds are so similar, they are difficultto differentiate among. Because they occur only in the fingerprint
region, they are difficult to distinguish from the peaks of other bonds
that also appear there. The important C—O peaks are near 1055 cm–1
for primary alkyl groups, 1110 cm–1 for secondary alkyl groups, 1175
cm–1 for tertiary alkyl groups and 1250 cm–1 for phenyl and carbonyl
groups. These peaks are usually among the strongest peaks in theinfrared spectra.
Most carbonyl group peaks occur near 1740 cm–1. As this
region has few potentially confusing absorptions, a peak here
definitely means a carbonyl group. However, identifying the specific
type of carbonyl group from this region of the spectrum is a challengeunless you have a high-resolution spectrum. Fortunately, each
carbonyl group has other absorptions that clearly tell what type of
carbonyl is present. For example, in addition to the carbonyl
absorption near 1740 cm–1, an aldehyde has a peak at 2700 cm–1.
Table 9.3 lists these correlations, and Figure 9.15 shows two carbonyl
absorptions.
Typical Carbonyl groups
15002000
1 0 0
0
Wavenumber (cm-1
) Wavenumber (cm
-1)
1 0 0
0
15002000
Typical AnhydrideCarbonyl
Figure 9.15. Carbonyl group absorptions. The spectrum on the left is a typical
carbonyl group. The difficulty of this scale is to see the differences in location of most
carbonyls. The spectrum on the right is an acid anhydride group. It shows twocarbonyl absorptions.
Type of Carbonyl
Group
Peak position
(cm–1)
Correlating peaks
(cm–1)
Aldehyde 1725 Carbonyl C—H 2900, 2700
Ketone 1710 NoneCarboxylic acid 1725 Acid O—H 3000-2500
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 24/42
Organic Chemistry - Ch 9 452 Daley & Daley
Type of Carbonyl
Group
Peak position
(cm–1)
Correlating peaks
(cm–1)
Ester 1740 Two bands of C—O stretch:
1250-11751175-1000
Amide 1640 1o N, 3450, 3225
2o N, 3335 Acyl halide 1820 C—halogen, 1000-900
Acid anhydride 1820, 1750 C—O stretch, 1250-1000
Table 9.3. Positions of the various types of carbonyl groups and correlating peaks for
identification of carbonyl-containing functional groups.
In many molecules, the carbonyl group is adjacent to either a
double or triple bond or to a phenyl group. Chemists call these
molecules conjugated carbonyl compounds. This bonding affects thelocation of the peak by moving it about 40 to 50 cm–1 away from the
position listed in Table 9.3. For example, cyclohexanecarboxylic acid
has a strong carbonyl peak at 1720 cm–1, whereas benzoic acid has astrong carbonyl peak at 1685 cm–1.
A conjugatedcompound has double
or triple bonds
alternating with single
bonds. Conjugation is
discussed in Chapter
16, which begins on
page 000.
C
O
OHC
O
OH
Cyclohexanecarboxylic acid Benzoic acid
1720 cm-1 1685 cm-1
Frequently, carbonyl groups also produce a weak absorptionnear 3450 cm–1. This small peak, called a carbonyl overtone band,
occurs at exactly twice the wavenumber of the actual carbonyl group
peak. Be careful not to confuse the carbonyl overtone band with H—O
and H—N peaks or with sp hybridized C—H bonds. These absorptions
are much stronger in comparison to an overtone band.
Carbon to nitrogen single bonds form amines, anilines, and
amides. The peaks for these bonds appear at nearly the same position
as the carbon to oxygen single bond. The primary alkyl bond tonitrogen emerges at 1065 cm–1, the secondary bond at 1150 cm–1, and
the tertiary bond at 1235 cm–1. The amide peak occurs in nearly the
same region as the amines. The anilines appear at 1280 cm–1.
Carbon to nitrogen double bonds are much less common in
organic molecules than carbon to nitrogen triple bond. You can
recognize nitriles by the medium intensity peak at 2250 cm–1. The
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 25/42
Organic Chemistry - Ch 9 453 Daley & Daley
only other common functional group in this region is the alkyne triple
bond discussed in the Section 9.7. Figure 9.16 shows the nitrile peak.
Typical Nitrile Peak
Wavenumber (cm-1
)
20002500
1 0 0
0
Nitrile
Figure 9.16. The position of the nitrile group.
Halogen absorptions occur between 1250 and 550 cm–1.
However, because there are numerous absorptions within this range,
identifying the specific halogen present is difficult. The most useful
absorptions are the C—Cl stretch at 725 cm–1 and the C—Br stretchat 645 cm–1. Aromatic chlorines appear about 1090 cm–1, and
aromatic bromines appear about 1030 cm–1. When you suspect
halogens, use mass spectrometry, not infrared spectroscopy, for
identification.
Exercise 9.3
Choosing from these molecules, assign a molecular structure to each of
the following IR spectra.
Benzonitrile Heptanenitrile Cyclohexanol
Cyclohexanone Methanoic acid Cyclohexanecarboxaldehyde
a)
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % )
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 26/42
Organic Chemistry - Ch 9 454 Daley & Daley
4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
b)
Tr
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % )
4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
c)
0
1 0 0
e
c
n
a
t
t
i
m
s
n
a
Tr
( %) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
d)
Tr
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % )
4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Sample solution
c) This spectrum is either benzonitrile or heptanenitrile. The peak at
about 2200 cm–1 readily identifies the compound as a nitrile. Thepeaks at 3000-2950 cm–1 identify it as an alkane. Thus, it must be
heptanenitrile.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 27/42
Organic Chemistry - Ch 9 455 Daley & Daley
9.7 Other Bonds
There are a variety of additional bonds that do not fit easilyinto any of the categories discussed in Sections 9.4, 9.5, and 9.6. These
bonds include the carbon to carbon double and triple bonds and the
nitro groups. The double bonds of both the alkenes and the arenes
appear within the same range in the infrared. For the alkenes this
range is 1690 to 1540 cm–1, with most of the simple alkenes appearing
as a medium intensity peak around 1640 cm–1. Arenes generally havetwo bands: one weak to medium peak appearing at 1600 cm–1 and a
stronger peak at 1500 cm–1. Figure 9.17 shows typical peaks for these
two absorptions.
Typical sp 2 Carbon—Carbon Stretching Peaks
Wavenumber (cm-1
)
15002000
1 0 0
0
Wavenumber (cm-1
)
120015002000
0
1 0 0
Alkene Aromatic
Figure 9.17. The bond absorptions for both the alkene and the aromatic C=C bond
appear at similar locations.
The only two common absorptions in the 2500 to 2000 cm–1
range are the carbon to nitrogen triple bond of the nitriles (discussed
in Section 9.6) and the carbon to carbon triple bond of the alkynes. A
monosubstituted alkyne C C H)(R appears as a medium peak
at 2130 cm–1, and a disubstituted alkyne C C R)(R appears at
2220 cm–1. The disubstituted alkyne is usually a very weak peak. It is
so weak, in fact, that it is often nonexistent. Figure 9.18 illustrates thealkyne carbon—carbon stretching peaks.
Typical sp Carbon-Carbon Stretching Peak
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 28/42
Organic Chemistry - Ch 9 456 Daley & Daley
20002500
1 0 0
0
Wavenumber (cm-1
)
20002500
Terminal Internal Alkyne Alkyne
Figure 9.18. The alkyne bond appears in nearly the same region as the nitrile. They
are the only two common absorptions that appear in the 2500 to 2000 cm–1 range of
the infrared. Most of the time, the internal alkyne is very weak.
The last functional group in this section is the nitro group
(NO2). Usually, the nitro group bonds to an aromatic ring. Although
nitro groups do form alkyl nitrates, they are uncommon. The aromatic
nitro groups occur at 1500 and 1380 cm–1. They also have a peakcommonly found at 740 cm–1 that is easy to confuse with the aromatic
substitution patterns.
Typical Nitro Group Peaks
Wavenumber (cm-1
)
2000 1500 1300
1 0 0
0
Nitro group
Figure 9.19. Two strong absorptions, one at 1500 and the other at 1380 cm–1,
characterize the aromatic nitro group.
Exercise 9.4
Choosing from these molecules, assign a molecular structure to each of
the following IR spectra.
trans-3-Hexene cis-2-Pentene 3-Hexyne
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 29/42
Organic Chemistry - Ch 9 457 Daley & Daley
1,7-Octadiyne 4-Methylnitrobenzene 3-Buten-2-ol
a)
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
b)T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
c)T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
d)
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 30/42
Organic Chemistry - Ch 9 458 Daley & Daley
T
r
a
n
sm
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Sample solution
d) This spectrum is of cis-2-pentene. The peaks just above 3000 cm–1
and at about 1650 cm–1
indicate the double bond. The peak at about700 cm–1 indicates the cis double bonds.
9.8 Interpreting Infrared Spectra, Part 2
There are so many possible peaks in an infrared spectrum that
interpreting them all is sometimes very challenging. The purpose of
this section is to reduce that challenge. As noted earlier, an infrared
spectrum consists of two sections: the functional group region from4000 to 1500 cm–1 and the fingerprint region from 1500 to 625 cm–1.
The functional group region breaks down further into three categories.1) In the 3700 to 2700 cm–1 region, bonds involving hydrogen appear.
2) Between 2500 and 2000 cm–1 is the triple bond and nitrile region. 3)
From 1800 to 1500 cm–1 is the double bond and carbonyl group region.
When interpreting a spectrum, first identify all the peaks of atleast medium intensity in the functional group region. Then, if
necessary, confirm your identifications in the fingerprint region.
Identifying all the peaks in the fingerprint region is usually difficult.
Table 9.4 lists some important correlations between the functional
group region and the fingerprint region for alcohols and amines.
Functional group Functional group
region (cm–1)
Fingerprint region (cm–1)
(C—O or C—N)1o 2o 3o Ar
Alcohol (O—H) 3350 (s, broad) 1055 1110 1175 1250
Amine (RN—H2)(R2N—H)
3400,3200 (m)3200 (m)
1065 1150 1235 1280
Table 9.4. Correlations between the functional group and fingerprint regions of the
infrared spectrum for alcohols and amines.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 31/42
Organic Chemistry - Ch 9 459 Daley & Daley
The best way to learn how to interpret infrared spectra is to
work through a couple of examples. Consider the spectrum in Figure
9.20. Look first at the functional group region. In that region there is abroad weak peak at 3400 cm–1, a group of peaks between 3000 and
2700 cm–1 and a strong peak about 1700 cm–1. You must identify
these peaks to pin down the compound’s class.
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Figure 9.20. The first infrared spectrum example analyzed in this section.
The peak at 3400 cm–1 is similar in appearance to the OH peak
of an alcohol but is much too weak to be one. However, the 3400 cm–1
peak may be an overtone band for the strong 1700 cm–1 peak which
indicates a carbonyl group. The complex region between 3000 and
2850 cm–1 matches the range of an alkyl group.
Table 9.3 shows that an aldehyde has a pair of peaks at 2900and 2700 cm–1. Remember that the alkane region (3000 to 2850 cm–1)
often obscures the 2900 cm–1 peak, so check the 2700 cm–1 region. The
2700 cm–1 peak is clear. Because there is little else at 2700 cm–1, the
compound is an aldehyde. Without further data, you cannot determine
that the actual spectrum is of propanal.
See Table 9.3 on page
000.
Now consider the second spectrum as shown in Figure 9.21. It
contains a large broad peak about 3350 cm–1, a small peak around
3100 cm–1, two peaks just below 3000 cm–1 and another about 1580
cm–1. The 3350 cm–1 peak is almost identical in appearance and
location to that of an OH group. The small peak at 3100 cm–1 suggests
an aromatic compound, and the peak at 1580 cm–1 confirms it. The
peaks just below 3000 cm–1 indicate that the compound contains alkylgroups.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 32/42
Organic Chemistry - Ch 9 460 Daley & Daley
T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Figure 9.21. The infrared spectrum for the second solved exercise analyzed in this
section.
At this point you have enough information to know that the
compound is either an alkyl-substituted phenol or a phenylsubstituted alcohol. To determine which, examine the fingerprint
region. The strong broad peak just above 1200 cm–1 and the lack of
any strong peaks between 1200 and 1000 cm–1 indicate that thecompound is a phenol. The strong peaks around 800 and 700 cm–1 are
characteristic of a meta substituted aromatic ring. Thus, this spectrum
is of a meta-alkylphenol, although the exact alkyl group is difficult to
determine. The spectrum is of 3-tert-butylphenol.
Exercise 9.5
Determine the functional groups and make structural generalizationsfor the compounds shown in the following spectra.
a)T
r
a
n
s
m
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
b)
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 33/42
Organic Chemistry - Ch 9 461 Daley & Daley
T
r
a
n
sm
i
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
c)T
r
a
n
s
mi
t
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
d)T
r
a
n
s
m
it
t
a
n
c
e
1 0 0
0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650
Wavenumber (cm–1
)
Sample solution
d) This compound has only alkyl groups above 1500 cm–1. The only
other obvious peak is at 1100 cm–1. This peak indicates a C—O bond,so the compound is an ether.
9.9 Mass Spectrometry
Mass spectrometry (MS) is different from infrared spectroscopy
(IR) in that it does not analyze the selective absorption of
electromagnetic radiation by the various energy levels of a molecule.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 34/42
Organic Chemistry - Ch 9 462 Daley & Daley
Chemists use mass spectra in addition to IR and NMR spectra
(Chapter 10) to gain information about the structure of molecules. To
produce a mass spectrum, the spectrometer bombards the molecule
with a beam of high-energy electrons. As this beam of electronscollides with the molecule, it breaks the molecule into a series of ions.
The spectrometer then records the masses of these ions on the
spectrum.
As high-energy electrons from the mass spectrometer collide
with a molecule, an electron is knocked out of the molecule. This
process produces a radical-cation called the molecular ion.
A radical-cation is a
species with a positive
charge and an
unpaired electron.
+A+BA B• •e 2e
The molecular ion is
the molecule minus an
electron. The molecular
ion is sometimes calledthe parent ion.
Because the ionization process has a large excess of energy the
molecular ion is unstable, so it breaks into a number of other species:other radical-cations, carbocations, radicals, and various uncharged
species.
A B• B+A •
A B• •A + B
The high-energy electrons can knock any electron from the molecule;
thus, fragmenting the molecule in many possible ways. Within each
type of molecule, however, certain electrons knock out more readily
than others. The fragmentation pattern of fragments from a particularmolecule is characteristic for that molecule. The pattern also contains
fragments that are common with other molecules that contain the
same functional groups.
Figure 9.22 shows a schematic diagram of a mass spectrometer.
To use a mass spectrometer, inject a sample into the injection
chamber. Vapors from this sample then “leak” in a small stream fromthe injection chamber into the high vacuum chamber. Because the
instrument is so sensitive, it requires only a very small amount of
sample for a spectrum. As this stream of sample, or neutral molecules,
passes along the chamber, a beam of high-energy electrons strikes it,
changing it and fragmenting it into positive ions and neutral particles.
Next, a series of negatively charged focusing plates accelerate andfocus, or direct, these positive ions into the magnetic field. Any
radicals or uncharged particles are unaffected by the magnetic field
and are unobserved by the spectrometer.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 35/42
Organic Chemistry - Ch 9 463 Daley & Daley
T o Va c uum
Re c ord e r
Ampl i fi e r
Io n Col l e c t or
Ma gne t Pol e
Ma gne t Pol e
Ne u t r a l
Mol e c ul e s
Pos i i ve I ons
De t e c t or Sl i t s
Re pe l l e r Pl a t e
Pum p
E l e c t ro n Be a m Fi l a m e nt
Foc us i ng Pl a t e s
Figure 9.22. Schematic diagram of a simple mass spectrometer.
In response to the magnetic field, the positively charged ions
move in a curved pathway. For a given magnetic field, the radius of
the curved path of each particular ion is proportional to the mass-to-
charge (m/z) ratio of that ion. The larger the mass-to-charge ratio, the
larger the radius of the pathway. The smaller the mass-to-charge
ratio, the smaller the radius of the pathway. The charge of each ion isusually one, so the factor of consideration for each ion is its mass, or
size. Beyond the magnetic field in the mass spectrometer is a collector
slit. The collector slit allows only the ions with a pathway of a certainradius to pass through it. By varying either the magnetic field
strength or the accelerating potential of the focusing plates, the radius
of the path the ions follow changes. Whether the magnetic fieldstrength of the accelerating potential is changed depends on the
particular instrument. This process selectively focuses the ions of the
various m/z ratios on the ion detector slit allowing the detector
(usually called the ion collector) to count each differently sized ion.
The amplifier then amplifies that count and feeds that signal into the
recorder, which plots the number of ions, or ion current, vs. m/z of the
ions. The abscissa of the plot is the mass-to-charge ratio, and the
ordinate is the relative intensity, or the number of ions, in the sample.
Figure 9.23 is the mass spectrum of acetone (2-propanone). Ananalysis of the mass spectrum of acetone shows the processes thatoccur in mass spectrometry.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 36/42
Organic Chemistry - Ch 9 464 Daley & Daley
ytisnetn
m/z
1 8 01 6 01 4 01 2 01 0 08 06 04 02 0
1 0 0
8 0
6 0
2 0
0
I
4 0
CH3CCH3
O
Figure 9.23. The mass spectrum of acetone.
In the first step of the process, the beam of high-energy
electrons knocks an electron from the molecule. With acetone that
electron is usually one of the nonbonding electrons from the oxygen.
The result is a radical cation, or the molecular ion. The peakrepresenting the molecular ion is important in that it gives the
molecular weight of the molecule.
••
••
•••
O
CH3H3C
–e
O
CH3H3C
Because the radical-cation is unstable, it fragments, or breaks apart,into a series of smaller ions. The molecular ion of acetone is present in
the spectrum as a peak with about 50% of the intensity, or height, of
the highest peak and a m/z of 58. Keep in mind that not all molecules
show a molecular ion on their spectrum. In some cases the molecular
ion is so unstable that it completely fragments before it reaches the
collector slit. Each line represents an individual fragment ion.
The most abundant ion gives the peak with the highest
intensity in the spectrum. This peak is called the base peak. The
mass spectrometer always assigns the base peak the value of 100 and
all other peaks their value proportional to the base peak. For acetone
the base peak has a m/z of 43. When you compare this with themolecular ion, which has a m/z of 58, you see a loss of 15 mass units.
A methyl group has a mass of 15. Thus, the major fragmentation
pathway in acetone is the loss of a methyl group.
The base peak is the
most abundant ion in
the MS.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 37/42
Organic Chemistry - Ch 9 465 Daley & Daley
••
•
•••C
O
CH3H3C
C OH3C + CH3
This fragmentation forms an acylium ion, shown in its most stable
resonance form, and a methyl radical. The methyl radical is not
positively charged, so it does not appear in the spectrum. The other
peaks shown in the spectrum are minor peaks that result from avariety of more complex fragmentation pathways.
Exercise 9.6
One of the following mass spectra is of 2-pentanone and the other is of
3-pentanone. Match the appropriate spectrum with the appropriatecompound. Justify your choices.
0
2 0
6 0
8 0
1 0 0
2 0 4 0 6 0 8 0 1 0 0 1 2 0 1 4 0 1 6 0 1 8 0
m/ z
I
nt
ensi
ty
4 0
0
2 0
6 0
8 0
1 0 0
2 0 4 0 6 0 8 0 1 0 0 1 2 0 1 4 0 1 6 0 1 8 0
m/ z
I
nt
ensi
ty
4 0
9.10 The Molecular Ion
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 38/42
Organic Chemistry - Ch 9 466 Daley & Daley
From an MS, chemists determine the structure of the
fragments represented by the peaks, then assemble these fragments
into the structure of the molecule. However, this task is complex and
beyond the scope of this book. For you, the beginning organic chemist,the most useful information that you can gather from an MS is from
the peaks representing the molecular ion and its isotopes. You can
learn the molecular weight of the compound, and you can tell if the
compound contains certain elements, such as nitrogen or a halogen.
Notice that the mass spectrum in Figure 9.23 and Exercise 9.6each have a small peak to the right of the peak representing the
molecular ion. These peaks, which occur at m/z 59 and m/z 87,
respectively, arise from the presence of isotopes of hydrogen, carbon,
and oxygen. Most elements have more than one isotope, with the
various isotopes present in varying amounts. The heavier isotopes give
rise to the small peaks at a higher mass than the molecular ion peak.
A peak that is one mass unit heavier than the molecular ion is called
the M+1 peak. A peak that is two mass units heavier is called the M+2peak. Table 9.5 contains a list of elements common to organic
compounds, their isotopic masses, and the ratio of each isotope in a
naturally occurring sample of that element. Thus, you can see how
each isotope contributes to the M+1 and M+2 peaks.
Most Common Isotopes Element Mass % Mass % Mass %
H 1 100.0 2 0.016 --- ---
C 12 98.9 13 1.1 --- ---
N 14 99.6 15 0.4 --- ---
O 16 99.8 17 0.04 18 0.2
S 32 95.0 33 0.8 34 4.2
F 19 100.0 --- --- --- ---
Cl 35 75.5 --- --- 37 24.5
Br 79 50.5 --- --- 81 49.5
I 127 100.0 --- --- --- ---
Table 9.5. Natural isotopic abundance of some common elements in organic
molecules.
Isotope peaks give you information about the molecular
formula of the compound. They help you determine what elements are
present and in what ratio. To determine the exact ratio, however, you
must precisely determine the relative intensity of the molecular ion
peak and the M+1 and M+2 peaks. As you see in Table 9.5, the variouselements occur in specific ratios, so by understanding these ratios and
the patterns of their occurrences, you can gain this information.
Compounds with a number of carbon, hydrogen, nitrogen, or
oxygen atoms have isotope peaks at one, two, or more mass units
above the mass calculated from the molecular formula and the masses
of the most common isotopes. The intensity of the M+1 peak increases
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 39/42
Organic Chemistry - Ch 9 467 Daley & Daley
with increasing numbers of carbon, hydrogen, and nitrogen atoms.
The intensity of the M+2 peak increases with increasing numbers of
oxygen atoms. The M+2 peak also increases with increasing numbers
of carbons and hydrogens because of the higher probability of having molecules with both 13C and 2H.
The M+2 peak gives you the information needed to identify a
molecule that contains chlorine, bromine, or sulfur. Here is how it
works. Chlorine has two isotopes, 35Cl and 37Cl, with a ratio of nearly
3:1. In a sample of a molecule containing chlorine, three-fourths of themolecules contain 35Cl and one-fourth contain 37Cl. Every molecularfragment in the mass spectrum that contains chlorine actually
appears as a pair of peaks separated by two mass units as shown in
the spectrum of methyl chloride in Figure 9.24. The molecular ion,
which is CH335Cl, appears at m/z 50. The isotope peak, which is
CH3
37Cl, appears at m/z 52. The peak at m/z 50 is roughly three
times the size of the peak at m/z 52, reflecting the 3:1 abundance of
the two isotopes in nature. This pattern is typical for the peaks in an
MS of a molecule that contains a single chlorine atom.
ytisnetn
m/z
1 8 01 6 01 4 01 2 01 0 08 06 04 02 0
1 0 0
8 0
6 0
2 0
0
I
4 0
CH3Cl
Figure 9.24. The mass spectrum of methyl chloride.
Bromine, like chlorine, has two isotopes (79Br and 81Br). The
bromine isotopes give rise to a different pattern of peaks in an MS,
however, because 79Br and 81Br occur in a nearly 1:1 ratio. As aresult, the mass spectrum of methyl bromide in Figure 9.25 shows
every bromine-containing ion as a pair of peaks of about the same
intensity separated by two mass units. The peak for CH379Br is atm/z 94, and the peak for CH3
81Br is at m/z 96.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 40/42
Organic Chemistry - Ch 9 468 Daley & Daley
ytisnetn
m/z
1 8 01 6 01 4 01 2 01 0 08 06 04 02 0
1 0 0
8 0
6 0
2 0
0
I
4 0
CH3Br
Figure 9.25. The mass spectrum of methyl bromide.
Exercise 9.7
There are no peaks at m/z 35 and 37 in the MS of methyl chloride
(Figure 9.24) that would correspond to the Cl⊕ ion, but the Br⊕ ion
does appear at m/z 79 and 81 in the MS of methyl bromide (Figure
9.25). Explain.
To make the best use of an MS, you must be able to readily
recognize the molecular ion peak. The following generalities may help
you. The molecular ion is, with the exception of the M+1 and M+2peaks, the highest mass peak in the spectrum. That is, the molecular
ion peak appears furthermost to the right. However, molecules with
bromine, chlorine, and sulfur produce peaks that are potentiallyconfusing because each of these elements gives a more intense M+2
peak in relation to the molecular ion peak than do other elements.
With sulfur-containing molecules, the M+2 peak is less obvious thanwith bromine or chlorine.
For all organic compounds, except those with an odd number of
nitrogen atoms, the molecular ion peak has an even mass. Thus, if you
have an even mass peak, at the appropriate position, with most of the
lower mass peaks being odd masses, the even mass peak is the
molecular ion. Conversely, if you have an odd mass peak with most of the lower mass peaks being even, an odd number of nitrogens are
present in the molecule.
The molecular ions in alcohols and amines are not generally
very stable. Frequently, they undergo a reaction and lose a small
molecule. In the case of alcohols, the small molecule is water. Thus,
the mass spectra for these alcohols and amines may not show anyvisible molecular ion peak. For example, the mass spectrum of tert-
butyl alcohol (see Figure 9.26) shows no molecular ion peak at all, andthe mass spectrum of 1-butanamine (see Figure 9.27) shows only a
very weak molecular ion peak.
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 41/42
Organic Chemistry - Ch 9 469 Daley & Daley
++H2O+(CH3)3COH • CH2(CH3)2C •
ytisnetn
m/z
1 8 01 6 01 4 01 2 01 0 08 06 04 02 0
1 0 0
8 0
6 0
2 0
0
I
4 0 Missing the
parent ion
(CH3)3COH
Figure 9.26. The MS of tert-butyl alcohol. The molecular ion is expected at m/z 74
but is absent from this spectrum.
ytisnetn
m/z
1 8 01 6 01 4 01 2 01 0 08 06 04 02 0
1 0 0
8 0
6 0
2 0
0
I
4 0 Parent ion
CH3CH2CH2CH2NH2
Figure 9.27. The MS of 1-butanamine. The molecular ion is very weak.
Key Ideas from Chapter 9
❑ Molecular spectroscopy involves irradiating a molecule with
electromagnetic energy, then observing how that moleculeresponds to the energy.
❑ A molecule must absorb a specific amount, a quantum, of
energy to move from its ground state to an excited state.
❑ The energy absorbed by a molecule excites its molecular
vibrations. The energy absorbed give rise to the infrared
www.ochem4free.com 5 July 2005
7/22/2019 organic chemistry(Infrared)
http://slidepdf.com/reader/full/organic-chemistryinfrared 42/42
Organic Chemistry - Ch 9 470 Daley & Daley
spectrum. The molecular vibrations are the stretching and
bending of the bonds of the molecule.
❑ Small differences in molecular structure, which occurs inhybridization and functional groups, produce very different
infrared spectra.
❑ Each infrared spectrum includes two regions: the functional
group region and the fingerprint region. When interpreting a
spectrum, analyze all the peaks in the functional group region.Then analyze the peaks in the fingerprint region as required to
confirm your analysis of the functional group region.
❏ A mass spectrum is the record of the relative abundance of a
series of ions that form when a molecule collides with a high-energy electron.
❏ The collision between a molecule and an electron causes the
molecule to lose an electron; thus, becoming a radical-cation.
The radical-cation then fragments into a series of cations and
free radicals. Within a sample of a given compound, thenumber of molecules that follow a particular fragmentation
pathway depends on the stability of the ions and radicals
produced.
❏ Because the mass spectrometer records the exact masses of the
various ions, the spectrum shows peaks for the differentisotopes of the various elements in a molecule. This is
particularly important for the identification of Br, Cl, and S in
an organic compound.