organic chemistry(Infrared)

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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



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,

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holder.

www.ochem4free.com 5 July 2005




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





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.





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.





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





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





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.





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.





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.





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)





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





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.





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





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.





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)





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





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





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.





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.





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



Cycloheptene 4-Methylaniline p-Dichlorobenzene

Ethynylbenzene 3-Methylphenol Iodobenzene

(Note: aniline is aminobenzene (PhNH2) and phenol ishydroxybenzene (PhOH).)

a)

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1 0 0

0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650

Wavenumber (cm–1

)

b)





T

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1 0 0

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4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650

Wavenumber (cm–1

)

c)

T

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1 0 0

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Wavenumber (cm–1

)

d)

T

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ta

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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





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





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





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



Benzonitrile Heptanenitrile Cyclohexanol

Cyclohexanone Methanoic acid Cyclohexanecarboxaldehyde

a)

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0( % )





4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650

Wavenumber (cm–1

)

b)

Tr

a

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s

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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

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a

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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.





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





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



trans-3-Hexene cis-2-Pentene 3-Hexyne





1,7-Octadiyne 4-Methylnitrobenzene 3-Buten-2-ol

a)

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1 0 0

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Wavenumber (cm–1

)

b)T

r

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1 0 0

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Wavenumber (cm–1

)

c)T

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1 0 0

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Wavenumber (cm–1

)

d)





T

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1 0 0

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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.





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

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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.





T

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a

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s

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a

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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

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1 0 0

0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650

Wavenumber (cm–1

)

b)





T

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1 0 0

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Wavenumber (cm–1

)

c)T

r

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n

s

mi

t

t

a

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1 0 0

0( % ) 4000 3000 2500 2000 1500 1300 1100 1000 900 800 700 650

Wavenumber (cm–1

)

d)T

r

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it

t

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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.





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.





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.





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.





••

•

•••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





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





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.





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.





++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





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.

organic chemistry(Infrared)

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