Unit 10a – Measurement and
Uncertainty
Topic 11
Pre-Lesson 1
To Be Done On Your Own!
Mass spectrometry
 Used to determine relative atomic and molecular
masses. The fragmentation pattern can be used as a
fingerprint technique to identify unknown
substances or for evidence for the arrangements of
atoms in a molecule.
Mass Spectrometry
For review…
The Mass Spectrometer
Relative atomic masses (among other things we will
discuss when we get to organic chemistry) can be
determined using this instrument.
How it works (5 basic steps):
1. Vaporization: if the sample is not already as gas,
the sample is heated to this point.
How it works (5 basic steps):
2. Ionization: sample is bombarded with a stream of
high energy electons. In practice, the instrument is
set so that only ions with a single positive charge are
formed (M+).
How it works (5 basic steps):
3. Acceleration: resulting unipositive ions pass
through slits in parallel plates under the influence
of an electric field.
How it works (5 basic steps):
4. Deflection: ions are then passed over an external magnetic
field. The magnetic field causes the ions to be deflected, and
the amount of deflections is proportional to the charge/mass
ratio. Ions with smaller masses are deflected more than
heavier ions. Ions with higher charges are deflected more as
they interact more effectively with the magnetic field.
Heavier
particles
lighter
particles
How it works (5 basic steps):
5. Detection: positive ions of a particular mass/charge
ratio are detected and a signal is sent to a recorder.
The strength of the signal is a measure of the
number of ions with that charge/mass ratio that are
detected.
Example: Find the relative atomic mass (Ar) of naturally occurring
lead from the data below. Record your answer to the nearest tenth.
Mass Spectrum of Pb
6
5.2
5
relative abundance
4
3
2.4
2.2
Isotopic
mass
Relative
abundance
% relative
abundance
204
0.2
2
206
2.4
24
207
2.2
22
208
5.2
52
2
𝐴𝑟 =
1
2 × 204 + 24 × 206 + 22 × 207 + 52 × 208
100
0.2
0
203
204
205
206
207
208
209
mass/charge
Figure: The Mass Spectrum of Naturally Occurring Lead
𝐴𝑟 =207.2
Determining the molecular mass
of a compound
 Can also use to determine relative molecular mass of a
compound (Mr)
 If empirical formula is known, can be used to determine
molecular formula.
Fragmentation Patterns
 Ionization process involves an e- from an electron gun
hitting the incident species and removing an electron:
 X(g) + e- → X+(g) + 2e This collision can be so energetic that it causes the molecule to
break up into different fragments.
Fragmentation Patterns
 The largest mass peak corresponds to a parent ion passing through
the instrument unscathed, but other ions produced as a result of
this break up are also detected.
Fragmentation Patterns
 The fragmentation pattern can provide useful evidence for the
structure of the compound.
Fragmentation Patterns
 A chemist pieces together the fragments to form a
picture of the complete molecule, just as the
archaeologist finds clues about the past from pieces of
artifacts discovered on the ground.
Example: ethanol
100
relative abundance
31
45
15
29
46
0
0
30
mass/charge
60
Example: ethanol
Example: ethanol
Note: This fragmentation will
yield either CH3+ and CH2OH or
CH3 and CH2OH+, yielding peaks
at both 15 and 31
Example: ethanol
100
31
relative abundance
CH2OH+
C2H5O+
45
CH3+
15
C2H5+
29
46
C2H5OH+
0
0
30
mass/charge
60
So…
 The highest mass fragment represents the Mr of the compound.
 Fragments provide clues about structure because certain numbers
correspond to particular groups.
Fragments you will be expected to recognize:
Mr
loss of…
15
CH3+
17
OH+
29
C2H5+ or CHO+
31
CH3O+
45
COOH+
Lesson 1: Advanced Mass
Spectrometry
Tuesday, December 8
Understandings 11.3
 The degree of unsaturation or index of hydrogen deficiency (IHD)
can be used to determine from a molecular formula the number of
rings or multiple bonds in a molecule.
 Mass spectrometry (MS), proton nuclear magnetic resonance
spectroscopy (1H NMR), and infrared spectroscopy (IR) are
techniques that can be used to help identify and to determine the
structure of compounds.
Guidance
 The electromagnetic spectrum (EMS) is given in the data booklet in
section 3. The regions employed for each technique should be
understood.
 The operating principles are not required for any of these methods.
Applications and Skills
 Determination of the IHD from a molecular formula.
 Deduction of information about the structural features of a
compound from percentage composition data, MS, 1H NMR, or IR.
Guidance
 The data booklet contains characteristic ranges for IR absorptions
(section 26), 1H NMR data (section 27), specific MS fragments (section
28), and the formula to determine IHD. For 1H NMR, only the ability to
deduce the number of different hydrogen (proton) environments and
the relative numbers of hydrogen atoms in each environment is
required. Integration traces should be covered but splitting patterns
are not required.
Analytical Techniques
 Qualitative analysis: the detection of the presence but
not the quantity of a substance in a mixture; for example,
forbidden substances in an athlete’s blood
 Quantitative analysis: the measurement of the quantity
of a particular substance in a mixture; for example, the
alcohol levels in a driver’s breath
 Structural analysis: a description of how the atoms are
arranged in molecular structures; for example, the
determination of the structure of a naturally occurring or
artificial product.
Tools We Will Study
 Infrared spectroscopy is used to identify the bonds in a molecule.
 Mass spectrometry is used to determine relative atomic and
molecular masses. The fragmentation pattern can be used as a
fingerprint technique to identify unknown substances or for
evidence for the arrangements of atoms in a molecule.
 Nuclear magnetic resonance spectroscopy is used to show the
chemical environment of certain isotopes (hydrogen, carbon,
phosphorus, and fluorine) in a molecule and so gives vital
structural information.
Mass Spectrometry - REVIEW
 Remember that in mass spectrometry, we take a substance and
ionize it, giving it a charge, and then accelerate it and run it through
a magnet that changes the pathway of that particle according to its
mass to charge ratio
 Previously, we looked at mass spectrometry readouts for isotopes
to get relative abundance and calculate the average atomic mass,
as seen in the POGIL activity
 However, when an electron from an electron gun hits a molecule
instead of atoms, the collision can be so energetic that it causes
the molecule to break up into different fragments.
 These fragments can help us determine the formula of a
compound!
Fragmentation Pattern
 The fragmentation pattern can provide useful evidence for the
structure of the compound
 Let’s look at an example together
Discussion: What do each of the peaks below
refer to? Hint: Look at the masses.
Fragmentation Patterns
 The parent ion can break up into smaller ions in a mass
spectrometer.
 A compound is characterized by this fragmentation pattern.
 ONLY CHARGED SPECIES CAN BE DETECTED BY THE MASS SPEC!
Using Mass Spec
 Fully identifying a substance using mass spectrometry is a complex
process
 To help you, the IB gives you the formula for some common organic
mass fragments in TABLE 28 of the data booklet to help you; you
are not expected to memorize these but be able to recognize them
 Don’t forget to include the positive charge on the ions detected by
the mass spectrometer when identifying different fragments.
Mass Spec Fragments
Let’s Practice Together
Fragments
Molecular
Formula
Step 1: Molecular Formula
 First let’s calculate the molecular formula to be able to figure out
fragment pieces
Step 2: Identify Fragments
 There are many different ways we can arrange the C, H, and O in
this compound.
 We need to start trying to identify the fragments from their relative
masses.
Answer
A (the spectrum on the left) corresponds to CH3CH2CHO
B (the spectrum on the right) corresponds to CH3COCH3
Similarities
Both have a molecular ion corresponding to 58.
Differences
A has peaks corresponding to 29 (CH3CH2+), 28 (loss of CH3CH2) and 57
(CH3CH2CO+).
B has a peak corresponding to 43 (loss of CH3).
Lesson 2: Index of Hydrogen
Deficiency and Intro To IR
Tuesday, December 8
Hydrocarbons - Review
 A hydrocarbon is a compound which contains only hydrogen and
carbon
 Hydrocarbons can be classified as saturated or unsaturated
 Saturated hydrocarbons have only single bonds between the
carbons; we can think of them as being “saturated” with as many
hydrogen atoms as possible can bond to the carbons
 Unsaturated hydrocarbons contain one or more multiple bonds
between the carbons; since carbon can only make four bonds when
it contains a multiple bond it can bond with fewer hydrogen atoms
Definition
 The degree of unsaturation or index of hydrogen deficiency (IHD) is
a measure of how many molecules of H2 would be needed in
theory to convert the molecule to the corresponding saturated,
non-cyclic molecule.
 If a compound has a hydrogen deficiency of 1, that means one
molecule of hydrogen could be added to the molecule if multiple
bonds (or rings!) are broken
Let’s Practice
Spectroscopy is the main method we have of
probing into the atom and the molecule.
EM Spectrum
 Wavelength (λ): the distance between successive
crests or troughs
 Frequency (ν): the number of waves which pass a
point every second.
Type of EM radiation
Radio waves (low energy)
Microwaves
Infrared (IR)
Visible (ROYGBIV)
Ultraviolet (UV)
X-rays
Gamma rays (high energy)
Typical v (s-1)
3 x 106
3 x 1010
3 x 1012
3 x 1015
3 x 1016
3 x 1018
> 3 x 1022
Note: f x  = c = 3.0 x 108 ms-1
Typical  (m)
102
10-2
10-4
10-7
10-8
10-10
< 10-14
Thus, f = c/ 
Energy of electromagnetic radiation is carried in discrete
packets of energy called photons or quanta.
E = hv
 E = energy of a single photon of radiation
 h = 6.63 x 10-34 Js (Plank’s constant)
 v = frequency of the radiation
Example: Calculate the energy of a photon of visible light with a
frequency of 3.0 x 1014 s-1. Express in kJ mol-1.
E = hv
E = (6.63 x 10-34 Js)(3.0 x 1014 s-1)
E = 1.989 x 10-19 J
1.989  1019 J 6.02  1023 photons
1kJ


photon
mol
1000J
Wavenumber
 In IR spectroscopy, the frequency of radiation is often
measured as number of waves per centimeter (cm-1), also
called wavenumber.
 Example: Calculate the wavenumber in cm-1 for an IR wave with
a frequency of 3 x 1013 s-1.
Info From Different EM Regions
(You need to know the underlined!)
 Radio waves can be absorbed by certain nuclei, causing
them to reverse their spin. They are used in NMR and can
give information about the environment of certain atoms.
 Microwaves cause molecules to increase their rotational
energy. This can give information about bond lengths. It is
not necessary to know the details at this level.
 Infrared radiation is absorbed by certain bonds causing
them to stretch or bend. This gives information about the
bonds in a molecule.
Info From Different EM Regions
(You need to know the underlined!)
 Visible light and ultraviolet light can produce
electronic transitions and give information about
the electronic energy levels within the atom or
molecule.
 X rays are produced when electrons make
transitions between inner energy levels. They have
wavelengths of the same order of magnitude as the
inter-atomic distances in crystals and produce
diffraction patterns which provide direct evidence
of molecular and crystal structure.
Absorption and Emission spectra
 Absorption: black lines
where light is absorbed by
sample.
 Emission: bands of colored
lines where light is
transmitted from excited
sample.
Infrared (IR) Spectroscopy
Chemical Bonds As Springs
 Bond Frequency:
 A chemical bond can be thought of as a
spring.
 Each bond vibrates and bends at a natural
frequency which depends on the bond
strength and the masses of the atoms.
 Light atoms, for example, vibrate at higher
frequencies than heavier atoms and
multiple bonds vibrate at higher
frequencies than single bonds.
Bond Vibrations
 Simple diatomic molecules such as HCl, HBr, and HI, can only
vibrate when the bond stretches
 In more complex molecules, different types of vibration can occur,
such as bending, so that a complex range of frequencies is present
Diatomic
More
Complex
Using IR to Excite Molecules
 Energy needed to excite the bonds in a molecule to make them
vibrate with greater amplitude occurs in the IR region.
 IR radiation can cause a bond to stretch or bend.
 A bond will only interact with the electromagnetic infrared
radiation, however, if it is polar. The presence of separate areas of
partial positive and negative charge in a molecule allows the
electric field component of the electromagnetic wave to excite the
vibrational energy of the molecule
Energy Absorbed
 The change in the vibrational energy produces a corresponding
change in the dipole moment of the molecule.
 The intensity of the absorption depends on the polarity of the
bond.
 Symmetrical non-polar bonds in N≡N and O=O do not absorb
radiation, as they cannot interact with an electric field.
Polyatomic Atoms
 Monatomic atoms can only stretch in one direction but for
polyatomic atoms, it is more appropriate to think of the whole
molecule bending and stretching
 Any stretches where a non-polar molecule remains symmetrical
(i.e. there is NO net dipole) however do not show up in the IR
spect.
 Let’s look at some examples on the next slides!
Stretching and bending
a) Bending (symmetrical)
b) Stretching (asymmetrical)
c) Stretching (symmetrical)
Polar Molecules
 All stretches show up on the IR because there is always a dipole
moment
Non-Polar Linear Molecules
 Any stretches that retain the symmetry do not show up
Vibrations of H2O, SO2 & CO2
Molecule
H2O
Asymmetrical
stretching
Symmetrical
stretching
Symmetrical
bending
-
-
-
O
O
O
+H
H +
+H
IR active
+
SO2
O -
O
CO2
O
+
C
IR active
IR active
+
+
S
O -
- O
IR active
-
O
-
O
H +
IR active
- O
IR active
-
+H
S
S
-
H +
+
C
O IR active
-
O
IR inactive
-
O
+
C
IR active
-
O
Double-beam IR spectrometer
 One beam passes through
sample
 Second beam passes through
reference
 Purpose of reference: to eliminate
absorptions caused by CO2 and H2O
vapor in air, or absorptions from the
bonds in the solvent used.
 Baseline = 100% transmittance
Matching wavenumbers with bonds
“fingerprint
region”
lots of overlap, so
not very useful
very strong
broad and strong
broad and strong
Usually sharper
than OH
IR spectrum of ethanol, CH3CH2OH
IR spectrum of ethyl ethanoate, CH3COOCH2CH3
C-H
C=O
“fingerprint region”
Reference Numbers
Lesson 3: Infrared Spectroscopy
Practice
Friday, December 11th
Hydrogen Bonds
 Hydrogen bonds can be detected by a broadening of the
absorptions. For example, hydrogen bonding between hydroxyl
groups changes the O–H vibration; it makes the absorption much
broader and shifts it to a lower frequency.
 The broad O-H region is one of the most recognizable parts of the
IR spectrum
Propanone
 The baseline at the top corresponds to 100% transmittance and the
key features are the troughs which occur at the natural frequencies
of the bonds present in the molecule.
 The absorption at just below 1800 cm–1 shows the presence of the
C= O bond and the absorption near 3000 cm–1 is due to the
presence of the C– H bond. The more polar C= O bond produces the
more intense absorption.
Ethanol
 The presence of the C–H bond can again been seen near 3000 cm–1
in the spectrum of ethanol.
 The broad peak at just below 3400 cm–1 shows the presence of
hydrogen bonding which is due to the hydroxyl (OH) group.
Lesson 4: Introduction to NMR
Monday, December 14
Nuclear Magnetic Resonance (NMR)
Spectroscopy
 Used to show the chemical environment of certain isotopes
(hydrogen, carbon, phosphorous and fluorine) in a molecule
and so gives vital structural information.
Odd Nuclei
 The nuclei of atoms with an odd number of protons such as 1H, 13C,
19F, and 31P, spin and behave like tiny bar magnets.
 If placed in an external magnetic eld, some of these nuclei will line
up with an applied field and, if they have sufficient energy, some
will line up against it.
 This arrangement leads to two nuclear energy levels; the energy
needed for the nuclei to reverse their spin and change their
orientation in a magnetic field can be provided by radio waves.
NMR
• In practice, a sample is placed in an electromagnet. The field strength is varied until the
radio waves have the exact frequency needed to make the nuclei flip over and spin in the
opposite direction.
• This is called resonance and can be detected electronically and recorded in the form of a
spectrum
• Resonance is when the radio waves have the exact frequency needed to make the nuclei
flip over and spin in the opposite direction
Results
The Chemical Shift
 As electrons shield the nucleus from the full effects of the
external magnetic field, differences in electron distribution
produce different energy separations between the two spin
energy levels.
 Nuclei in different chemical environments produce
different signals in the spectrum.
The Chemical Shift
 Proton or 1H NMR is particularly useful.
 Hydrogen nuclei are in all organic molecules.
 Act as spies and give information about their position in a
molecule.
The TMS Standard
 If we are to measure
shift, we need a
standard, or point of
reference from which to
measure.
 Tetramethylsilane (TMS)
is the perfect standard.
The TMS Standard
 All 12 H’s are in identical chemical environments, so one
signal is recorded.
The TMS Standard
 Because Si is less electronegative than C, TMS absorbs
radio waves in a different region from that absorbed by
H attached only to C.
 This ensures that the signal does not overlap with any
signals under investigation
The TMS Standard
The TMS Standard
 TMS is also inert.
 TMS is soluble in most organic solvents.
 TMS can be easily removed from the sample because it has a
low boiling point.
 The position of the NMR signal relative to this standard is
called the chemical shift of the proton. Hydrogen nuclei in
particular environments have characteristic chemical shifts.
 A more complete list is given in section 27 of the IB data
booklet.
Ethanal
 Let’s see if we can interpret the 1H NMR spectrum for ethanal
1. What do each of the peaks represent?
2. What do the relative heights mean?
IB Tips!
 Avoid losing marks through carelessness. The number of peaks
does not simply give the number of different chemical
environments – it gives the number of different chemical
environments in which hydrogen atoms are located.
Wait? Why are these
spiky?
MORE ON THAT IN
THE HL LEVEL
TOMORROW!!!
Lesson 5: Advanced NMR
Tuesday, December 15
Warm-Up
21.1 Advanced Identification
 Structural identification of compounds involves several different
analytical techniques, including IR, 1H NMR, and MS.
 In a high-resolution 1H NMR spectrum, single peaks present in low
resolution can split into further clusters of peaks.
 The structural technique of single crystal X-ray crystallography can
be used to identify the bond lengths and bond angles of crystalline
compounds.
Guidance
 The operating principles are not required for any of these methods.
 High resolution 1H NMR should be covered.
 The precise details of single crystal X-ray crystallography need not
be known in detail, but students should be aware of the existence
of this structural technique in the wider context of structural
identification of both inorganic and organic compounds.
21.1 Advanced Identification
 Explanation of the use of tetramethylsilane (TMS) as the reference
standard.
 Deduction of the structure of a compound given information from
a range of analytical characterization techniques (X-ray
crystallography, IR, 1H NMR, and MS).
Guidance
 Students should be able to interpret the following from 1H NMR
spectra: number of peaks, area under each peak, chemical shift, and
splitting patterns. Treatment of spin–spin coupling constants will not
be assessed but students should be familiar with singlets, doublets,
triplets, and quartets.
Advanced 1H NMR
 Recall the sneak peak from yesterday’s lesson where we looked at
a 1H NMR spectrum that was weirdly spiky
 What are those spikes? What do they represent?
 The NMR spectrum of an organic compound does not generally
consist of a series of single peaks shown in the low-resolution
spectra presented earlier. Instead, a sensitive, high-resolution NMR
machine reveals a hidden structure, with the single peaks split or
resolved into a group of smaller parts.
Spin-Spin Coupling
 The splitting of the peaks occurs as the effective magnetic field,
experienced by particular nuclei, is modified by the magnetic field
produced by neighboring protons.
 This effect is known as spin–spin coupling.
Here the magnetic field experienced by the protons in the methyl group,
for example, depends on the spin of the proton attached to the carbon
atom of the carbonyl group (CHO). The local magnetic field is increased
when the magnetic field of the CHO proton is aligned with the external field
and decreased when aligned against it. As the energy separation between
the two spin states of a proton depends on the local magnetic field, this
results in two possible values for the energy difference between the two
nuclear energy levels for the CH protons
3
Ethanal
 Instead of one signal corresponding to one energy difference, ∆E,
two signals corresponding to ∆Ea and ∆En are produced. Each line
corresponds to a different spin of the neighboring proton. As they
are both equally likely, the lines are of equal intensity
Ethanal
 In a similar way, the low-resolution peak corresponding to the CHO
proton is split due to the different magnetic fields produced by the
combinations of spin for the three protons of the neighboring
methyl group. As there are two possible orientations for each
proton, a total of 23 combinations are possible, resulting in four
different local magnetic fields. This produces four signals with
relative intensities 1, 3, 3, 1 – as shown in the table below.
Splitting Patterns
 The splitting patterns produced from different numbers of
neighboring protons can be deduced from Pascal’s triangle and are
summarized in the table below.
Short Cut!
 If a proton has n protons as nearest neighbors, its NMR peak is split
into a group of (n + 1) peaks.
 Thus, for CH3CH2F, we would expect the — CH2— proton signal to
be split into a quartet as it has three protons as nearest neighbors.
Additional Points
 Protons bonded to the same atom do not interact with one
another as they are equivalent and behave as a group
 Protons on non-adjacent carbon atoms do not generally interact
with one another
 The O—H single peak in ethanol does not split unless the sample is
pure as rapid exchange of the protons between ethanol molecules
averages out the different possible spins.
Let’s Practice
 Predict the splitting pattern produced by a neighboring – CH2–
group.
Lesson 6: X-Ray Diffraction
Thursday, December 17
X-Ray Diffraction – Warm-Up
 Imagine you could not see your hand directly. How could you you
use a flashlight and a white wall to find out the shape of your hand?
Observing Atoms
 The most direct way to perceive an object is to shine light on it and
then observe the light that is scattered from it.
 The difficulty with examining individual atoms and molecules in this
way is that the wavelength of visible light is too long for light to
interact effectively with matter on this scale.
 The wave could go completely above or below the atom without
actually interacting with it
 Interatomic distances are of the order of 10–9 m which corresponds
to the wavelength of X rays
Diffraction Patterns
 When X rays pass through a crystalline solid they are scattered in an
orderly way by their interaction with electrons in the substance.
The scattered waves interact with each other to cause a diffraction
pattern
Interference
 (NOTE: Ask someone in IB Physics to help if you want to learn more
about waves!)
 In places where the waves are in phase, with the peaks still aligned,
the waves interfere constructively (i.e. resulting wave gets bigger)
 Destructive interference occurs at places where the waves are out
of phase by 180°, with the peak of one wave aligned with the
trough of another
X-Ray Crystallography
 When X rays shine on a crystal, they are reflected in consecutive
planes.
 The scattered waves interfere as they travel different distances as
they pass through a crystal, and so are at different phases,
dependent on their wavelength, when they hit the detector or
screen.
 The diffraction pattern depends in a complex way on the
relationship between the angle of incidence (θ), the wavelength of
the incident X rays (λ), and the distance between the atoms and
their relative orientations (d)
 Monochromatic X rays are used to ensure a simple
correspondence between the diffraction pattern and the crystal
structure. Similarly, the sample must be in the solid state as only
orderly structures give ordered diffraction patterns that can be
interpreted.
Electron Density
A map of the electron density in a
solid can be determined directly
from the X-ray diffraction pattern.
One of the first applications of X-ray
crystallography was to study
inorganic ionic substances which
have regular crystal structures, but
it is now also applied to organic
compounds. The electron density
map of the organic molecule
anthracene is shown on the right.
Contour lines connect points with
the same electron density.
Hydrogen
 The identity of the atoms can be determined from the electron
density map as the pattern in electron densities are related to an
element’s electron configuration.
 It should be noted that hydrogen atoms, with only one electron,
are not visible as their electrons densities are too low.