ORGANIC CHEMISTRY CH 12 BY MCMURRY 20233

© 2016 Cengage Learning. All Rights Reserved.
John E. McMurry
www.cengage.com/chemistry/mcmurry
Chapter 12
Structure Determination: Mass
Spectrometry and Infrared
Spectroscopy

Learning Objectives
(12.1)
 Mass spectrometry of small molecules:
Magnetic-sector instruments
(12.2)
 Interpreting mass spectra
(12.3)
 Mass spectrometry of some common functional
groups
(12.4)
 Mass spectrometry in biological chemistry: Time-
of-flight (TOF) instruments

Learning Objectives
(12.5)
 Spectroscopy and the electromagnetic spectrum
(12.6)
 Infrared spectroscopy
(12.7)
 Interpreting infrared spectroscopy
(12.8)
 Infrared spectra of some common functional
groups

Mass Spectrometry of Small Molecules:
Magnetic-Sector Instruments
 Mass spectrometry (MS) determines molecular
weight by measuring the mass of a molecule
 Components of a mass spectrometer:
 Ionization source - Electrical charge assigned to
sample molecules
 Mass analyzer - Ions are separated based on
their mass-to-charge ratio
 Detector - Separated ions are observed and
counted

Electron-Ionization, Magnetic-
Sector Mass Spectrometer
 Small amount of sample undergoes vaporization
at the ionization source to form cation radicals
 Amount of energy transferred causes
fragmentation of most cation radicals into
positive and neutral pieces

Electron-Ionization, Magnetic-
Sector Mass Spectrometer
 Fragments pass through a strong magnetic field
in a curved pipe that segregates them according
to their mass-to-charge ratio
 Positive fragments are sorted into a detector
and are recorded as peaks at the various m/z
ratios
 Mass of the ion is the m/z value

Figure 12.1 - The electron-ionization,
magnetic-sector mass spectrometer

Quadrupole Mass Analyzer
 Comprises four iron rods arranged parallel to
the direction of the ion beam
 Specific oscillating electrostatic field is created
in the space between the four rods
 Only the corresponding m/z value is able to pass
through and reach the detector
 Other values are deflected and crash into the
rods or the walls of the instrument

Figure 12.2 - The Quadrupole
Mass Analyzer

Representing the Mass
Spectrum
 Plot mass of ions (m/z) (x-axis) versus the
intensity of the signal (roughly corresponding to
the number of ions) (y-axis)
 Tallest peak is base peak (Intensity of 100%)
 Peak that corresponds to the unfragmented
radical cation is parent peak or molecular ion
(M+)

Interpreting Mass Spectra
 Provides the molecular weight from the mass of
the molecular ion
 Double-focusing mass spectrometers have a
high accuracy rate
 In compounds that do not exhibit molecular ions,
soft ionization methods are used

Other Mass Spectral Features
 Mass spectrum provides the molecular
fingerprint of a compound
 The way molecular ions break down, can
produce characteristic fragments that help in
identification
 Interprets molecular fragmentation pattern,
assisting in the derivation of structural
information

Mass Spectral Fragmentation of
Hexane
 Hexane (m/z = 86 for parent) has peaks at m/z =
71, 57, 43, 29

Worked Example
 The male sex hormone testosterone contains
only C, H, and O and has a mass of 288.2089
amu, as determined by high-resolution mass
spectrometry
 Determine the possible molecular formula of
testosterone

Worked Example
 Solution:
 Assume that hydrogen contributes 0.2089 to the
mass of 288.2089
 Dividing 0.2089 by 0.00783 ( difference between
the atomic weight of one H atom and 1) gives
26.67
 Approximate number of H in testosterone
 Determine the maximum number of carbons by
dividing 288 by 12
 List reasonable molecular formulas containing
C,H, and O that contain 20-30 hydrogens and
whose mass is 288

Worked Example
 The possible formula for testosterone is C19H28O2

Mass Spectrometry of Some
Common Functional Groups
 Alcohols
 Fragment through alpha cleavage and
dehydration

 Amines
 Nitrogen rule of mass spectrometry
 A compound with an odd number of nitrogen atoms
has an odd-numbered molecular weight
 Amines undergo -cleavage, generating alkyl
radicals and a resonance-stabilized, nitrogen-
containing cation

 Halides
 Elements comprising
two common isotopes
possess a distinctive
appearance as a mass
spectra

Fragmentation of Carbonyl
Compounds
 A C–H that is three atoms away leads to an
internal transfer of a proton to the C=O called
the McLafferty rearrangement
 Carbonyl compounds can also undergo -
cleavage

Worked Example
 List the masses of the parent ion and of several
fragments that can be found in the mass
spectrum of the following molecule

Worked Example
 Solution:
 The molecule is 2-Methyl-2-pentanol
 It produces fragments resulting from dehydration and
alpha cleavage
 Peaks may appear at M+=102(molecular ion), 87, 84, 59

Mass Spectroscopy in Biological
Chemistry: Time-of-Flight (TOF)
Instruments
 Most biochemical analyses by MS use soft
ionization methods that charge molecules with
minimal fragmentation
 Electrospray ionization (ESI)
 High voltage is passed through the solution sample
 Sample molecule gain one or more protons from the
volatile solvent, which evaporates quickly
 Matrix-assisted laser desorption ionization (MALDI)
 Sample is absorbed onto a suitable matrix compound
 Upon brief exposure to laser light, energy is transferred
from the matrix compound to the sample molecule

Figure 12.15 - MALDI–TOF Mass
Spectrum of Chicken Egg-White
Lysozyme

Spectroscopy and the
Electromagnetic Spectrum
 Waves are classified by frequency or
wavelength ranges

 Electromagnetic radiation seems to have dual
behavior
 Possesses the properties of a photon
 Behaves as an energy wave

 Speed of the wave
 The unit of electromagnetic energy is called
quanta
–1
Wavelength × Frequency = Speed
λ (m)× ν(s ) = c (m / s)
c c
λ = or ν =
v λ
= h =
hc
 


 Considering the plank equation and multiplying ɛ
by Avogadro’s number NA:
-4
A
-5
N hc 1.20 × 10 kJ/mol
E = =
λ λ(m)
or
2.86 × 10 kcal/mol
E=
λ(m)

Absorption Spectrum
 Organic compound exposed to electromagnetic
radiation can absorb energy of only certain
wavelengths (unit of energy)
 Transmits energy of other wavelengths
 Changing wavelengths to determine which are
absorbed and which are transmitted produces
an absorption spectrum
 In infrared radiation, absorbed energy causes
bonds to stretch and bend more vigorously
 In ultraviolet radiation, absorbed energy causes
electrons to jump to a higher-energy orbital

Worked Example
 Calculate the energy in kJ/mol for a gamma ray
with λ = 5.0×10-11m
 Solution:
-4
1.20×10 kj/ mol
E =
λ(in m)
-4
-11
1.20×10 kJ / mol
E =
5.0×10
6
E = 2.4×10 kJ / mol for a gamma ray

Infrared Spectroscopy
 IR region has lower energy than visible light
(below red - produces heating as with a heat
lamp)
 Wavenumber: 1 1
( )
( )
cm
cm





Infrared Energy Modes
 Molecules possess a certain amount of energy
that causes them to vibrate
 Molecule absorbs energy upon electromagnetic
radiation only if the radiation frequency and the
vibration frequency match

Interpreting Infrared Spectra
 IR spectrum interpretation is difficult as the
arrangement of organic molecules is complex
 Disadvantage - Generally used only in pure
samples of fairly small molecules
 Advantage - Provides a unique identification of
compounds
 Fingerprint region - 1500cm-1 to 400 cm-1 (approx)
 Complete interpretation of the IR spectrum is not
necessary to gain useful structural information
 IR absorption bands are similar among
compounds

Table 12.1 - Characteristic IR Absorptions of
Some Functional Groups

Figure 12.20 - IR Spectra of Hexane,
1-Hexene, and 1-Hexyne

Regions of the Infrared
Spectrum
 Region from 4000 to 2500 cm-1 can be divided
into areas characterized by:
 Single-bond stretching motions
 Triple-bond stretching motions
 Absorption by double bonds
 Fingerprint portion of the IR spectrum

Harmonic Oscillator
 System comprising two atoms connected by a
bond
 In a vibrating bond, the vibrating energy is in a
constant state of change from kinetic to potential
energy and vice versa
 Equation for natural frequency of vibration for a
bond -
OSC
h
OSC
E  
1 K
v =
2Πc μ

Worked Example
 Using IR spectroscopy, distinguish between the
following isomers:
 CH3CH2OH and CH3OCH3
 Solution:
 CH3CH2OH is a strong hydroxyl bond at 3400–
3640 cm-1
 CH3OCH3 does not possess a band in the region
3400–3640 cm-1

Infrared Spectra of Some
 Alkanes
 No functional groups
 C–H and C–C bonds are responsible for
absorption
 C–H bond absorption ranges from 2850 to 2960
cm-1
 C–C bonds show bands between 800 to 1300
cm-1

 Alkenes
 Vinylic =C–H bonds are responsible for
absorption from 3020 to 3011cm-1
 Alkene C=C bonds are responsible for absorption
close to 1650cm-1
 Alkenes possess =C–H out-of-plane bending
absorptions in the 700 to 1000 cm-1 range

 Alkynes
 C≡C stretching absorption exhibited at 2100 to
2260 cm-1
 Similar bonds in 3-hexyne show no absorption
 Terminal alkynes such as 1-hexyne possess
≡C–H stretching absorption at 3300 cm-1

Aromatic Compounds
 Weak C–H stretch at 3030 cm1
 Weak absorptions at 1660 to 2000 cm1 range
 Medium-intensity absorptions at 1450 to 1600
cm1

Aromatic Compounds
 Alcohols
 O–H 3400 to 3650 cm1
 Usually broad and intense
 Amines
 N–H 3300 to 3500 cm1
 Sharper and less intense than an O–H

Carbonyl Compounds
 Strong, sharp C=O peak in the range of 1670 to
1780 cm1
 Exact absorption is characteristic of type of
carbonyl compound
 Principles of resonance, inductive electronic
effects, and hydrogen bonding provides a better
understanding of IR radiation frequencies

Carbonyl Compounds
 Aldehydes
 1730 cm1 in saturated aldehydes
 1705 cm1 in aldehydes next to double bond or
aromatic ring
 Low absorbance frequency is due to the
resonance delocalization of electron density from
the C=C into the carbonyl

Ketones
 Saturated open-chain ketones and six-
membered cyclic ketones absorb at 1715cm-1
 Five-membered ketones absorb at 1750cm-1
 Stiffening of C=O bond due to ring strain
 Four members absorb at 1780cm-1

Carbonyl Compounds
 Esters
 Saturated esters absorb at 1735 cm-1
 Esters possess two strong absorbances within
the range of 1300 to 1000 cm-1
 Esters adjacent to an aromatic ring or a double
bond absorb at 1715 cm-1

Worked Example
 Identify the possible location of IR absorptions in
the compound below

Worked Example
 Solution:
 The compound possesses nitrile and ketone
groups as well as a carbon–carbon double bond
 Nitrile absorption occurs at 2210–2260 cm-1
 Ketone exhibits an absorption bond at 1690 cm-1
 Double bond absorption occurs at 1640–1680 cm-1

Summary
 Mass spectrometry determines the molecular
weight and formula of a molecule
 Infrared (IR) spectroscopy identifies functional
groups in a molecule
 Electromagnetic radiation is used in infrared
spectroscopy
 Organic molecules absorb a certain frequency
from electromagnetic radiation

ORGANIC CHEMISTRY CH 12 BY MCMURRY 20233

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