SPECTROSCOPIC IDENTIFICATION
OF ORGANIC COMPOUNDS
11.3/21.1
THANK YOU
Many notes taken/adapted from Mr. Michaelides
http://teachers.wrdsb.ca/amichael/sch-4wi-home-page/
Some slides taken from InThinking – Dr. Geoffrey Neuss
11.3
Understandings:
• 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 compounds and to determine their structure.
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.
21.1
Understandings:
• 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.
Applications and skills:
• Explanation of the use of tetramethylsilane (TMS) as the reference standard.
of a compound given information from a range of analytical characterization techniques (X-ray
• Deduction of the structure
1
crystallography, IR, H NMR and MS)
MAIN PURPOSES OF SPECTROSCOPIC ANALYSIS
•Identify and characterize unknown
substances
•Determine composition of a mixture
•Identify impurities
INSTRUMENTS FOR STRUCTURAL ANALYSIS
•Infrared spectroscopy (IR)
•Used to identify the bonds in a molecule
•Mass spectrometry
•Used to determine relative atomic and molecular
masses
•Fragmentation pattern can be used as a fingerprint to
identify unknown substances
•Can be used to evidence for the arrangements of atoms
in a molecule
INSTRUMENTS FOR STRUCTURAL ANALYSIS
•Nuclear magnetic resonance spectroscopy (NMR)
•Used to show the chemical environment of certain isotopes
(hydrogen, carbon, phosphorus, and fluorine) in a molecule
•Gives vital structural information
•X-ray crystallography
•Map of electron density can be determined directly
•Identity of atoms can be determined from the electron density
map
•No one method is definitive
•Combination of techniques can provide strong evidence
for the structure
THE DEGREE OF UNSATURATION/IHD
DEGREE OF UNSATURATION /
INDEX OF HYDROGEN
DEFICIENCY (IHD)
•Provides a clue to the structure of
the molecule once the formula is
known
•Measure of how many H2 molecules
would be needed in theory to
convert the molecule to the
corresponding saturated, non-cyclic
molecule
•The degree of unsaturation is used
to calculate the number of rings
and pi bonds present in a structure
Degree of Unsaturation
(IHD)
Type of Bond Present
1
Double Bond
1
Ring Structure
2
Triple Bond
(or could be one double and one
cyclic ring)
4
Benzene Ring
(3 double bonds and 1 ring)
I hate having to remember
extra formulas so the easiest
method is to use the general
formula of an alkane – unless
you have a nitrogen present
and then the other formula is
easier
Based on the number of
Carbons determine the number
of Hs that should be there
(CnH2n+2)
Compare the formulas treating
halogens as hydrogens and
ignoring oxygens - each H2
missing is one degree of
hydrogen deficiency.
Molecule Saturated non-cyclic
Index of hydrogen deficiency
target (CnH2n+2)
(IDH)
C2H4
C2H6
The molecule is missing 2H or 1 H 2,
therefore the IDH is:
1
C4H4
C2H5OH
C2H5Cl
C2H4O
C3H7N
MASS SPECTROMETRY
MASS
SPECTROMETER
IONIZATION AND FRAGMENTATION
•Involves an electron from an electron gun
hitting the incident species and removing an
electron
X(g) + e- → X+(g) + 2e•Collision can be so energetic that it causes the
molecule to break up into different fragments
•Largest peak in the mass spectrum
corresponds to a parent ion passing through
the instrument unbroken
• But other ions produced are also detected
FRAGMENTATION OF ETHANOL
Fragment
Mr
[C2H5OH]+
46
[C2H5O]+
45
[CH2OH]+
Loss of
H
**[CH2OH]+ 31
CH3
(Mr=15)
**[CH3]+
15
CH2OH
[C2H5]+
29
OH
** when fragmentation occurs, one of the products keeps
the positive charge (the cation is what will show up on the
mass spectrum)
[C2H5O]+
[CH3
]+
[C2H5]+
ANALYSING MASS SPECTRUM
Mass Lost
Fragment lost
•Can be a complex process
15
CH3
17
OH
•Make use of mass differences to identity
pieces which have fallen off
18
H 2O
•Section 28 of data booklet gives details
28
CH2=CH2, C=O
29
CH3CH2, CHO
31
CH3O
45
COOH
MASS SPECTRUM EXAMPLE
A molecule with an empirical formula
CH2O has the following simplified mass
spectrum. Deduce the molecular formula
and give a possible structure of the
compound
READING A MASS SPECTRUM - WARNING
What should I be careful about when
reading a spectrum?
Not all possible fragments are formed
Unexpected and unusual fragments can
also occur
 i.e. Only look at the peaks that you are asked
to look at ☺
ELECTROMAGNETIC SPECTRUM
CHARACTERISTIC
FREQUENCIES
Table 26 of data
booklet: Characteristic
ranges for infrared
absorption due to
stretching vibrations in
organic molecules.
WHAT IS IT!
One can tell the difference between alcohols, aldehydes and carboxylic
acids by comparison of their spectra.
O-H STRETCH
ALCOHOL
C=O STRETCH
ALDEHYDE
O-H STRETCH
AND
C=O STRETCH
CARBOXYLIC ACID
PRACTICE SOME MORE SPECTRA
http://undergrad-ed.chemistry.ohio-state.edu/anim_spectra/index.html
1 H-NMR
SPECTROSCOPY
INTRO TO NMR
NMR SPECTROSCOPY
INTERPRETATION OF SPECTRA
•NMR spectra provide information about the structure of organic molecules from the...
• number of different signals in the spectrum
• position of the signals (chemical shift)
• intensity of the signals
• splitting pattern of the signals
OBTAINING SPECTRA
•a liquid sample is placed in a tube which spins in a magnetic field
•solids are dissolved in solvents which won’t affect the spectrum - CCl4, CDCl3
•TMS, tetramethylsilane, (CH3)4Si, is added to provide a reference signal
•when the spectrum has been run, it can be integrated to find the relative peak areas
•spectrometers are now linked to computers to analyse data and store information
HIGH-RESOLUTION 1H-NMR
•low resolution NMR gives 1 peak for each environmentally different group of protons
•high resolution gives more complex signals - doublets, triplets, quartets, multiplets
•the signal produced indicates the number of protons on adjacent carbon atoms
The splitting pattern depends on the number of hydrogen atoms on adjacent atoms and it’s due to spin-spin coupling
HIGH RESOLUTION SPECTRUM OF 1-BROMOPROPANE
The broad peaks
are split into
sharper signals
1H-NMR
spectroscopy - summary
An NMR spectrum provides several types of information :
number of signal groups
chemical shift
peak area (integration)
multiplicity
tells you
tells you
tells you
tells you
the number of different proton environments
the general environment of the protons
the number of protons in each environment
how many protons are on adjacent atoms
In many cases this information is sufficient to deduce the structure of an organic
molecule but other forms of spectroscopy are used in conjunction with NMR.
Number of Signal Groups
When is a hydrogen chemically different?
1
2
3
4
CHEMICAL SHIFT
•each proton type is said to be chemically shifted relative to a standard (usually TMS)
Chemical Shift
•the chemical shift is the difference between the field strength at which it absorbs and the field
strength at which TMS protons absorb
•the delta (d) scale is widely used as a means of reporting chemical shifts
•the TMS peak is assigned a value of ZERO (d = 0.00 ppm)
Approximate
chemical shifts
The actual values
depend on the
environment
Chemical Shift
CHEMICAL SHIFTS – TABLE IN DATABOOKLET
Integration
Measure the
distance between
the top and bottom
lines.
Integration
Compare the
heights from each
signal and make
them into a simple
ratio.
HOW TO WORK OUT THE SIMPLE RATIOS
• Measure how much each integration line rises as it goes of a set of signals
• Compare the relative values and work out the simple ratio between them
• In the above spectrum the rises are in the ratio... 1:2:3
IMPORTANT: It doesn’t provide the actual number of H’s in
each environment, just the ratio
Splitting Patterns
Splitting patterns - multiplicity
O adjacent H’s
There is no effect
1 adjacent H
can be aligned either with a or against b the field
there are only two equally probable possibilities
the signal is split into 2 peaks of equal intensity
Splitting Patterns
2 adjacent H’s
more possible combinations
get 3 peaks in the ratio 1 : 2 : 1
3 adjacent H’s
even more possible combinations
get 4 peaks in the ratio 1 : 3 : 3 : 1
Splitting patterns - multiplicity
Splitting Patterns
4 adjacent H’s
gives 5 peaks in the ratio 1 : 4 : 6 : 4 : 1
Splitting Patterns
OH bonds and splitting patterns
• The signal due to the hydroxyl (OH) hydrogen is a singlet ... there is no splitting
• H’s on OH groups do not couple with adjacent hydrogen atoms
Splitting Patterns
Arises because the H on the OH, rapidly exchanges with protons on other molecules and is not
attached to any particular oxygen long enough to register a splitting signal.
OH hydrogens are always seen as a singlet
... there is no splitting
This is a quartet despite the fact that there
are 4 H’s on adjacent atoms - the H on the
OH doesn’t couple
1H-NMR
spectroscopy
HOW TO WORK OUT AN NMR SPECTRUM
1. Get the formula of the compound
2. Draw out the structure
3. Go to each atom in turn and ask the ‘census’ questions
4. Work out what the spectrum would look like ... signals due to H’s nearer electronegative atoms (Cl, Br, O)
are shifted downfield to higher d values
THE BASIC “CENSUS”
Ask each hydrogen atom to...
- describe the position of the atom on which it lives
- say how many hydrogen atoms live on that atom
- say how many chemically different hydrogen atoms live
on adjacent atoms
BUT, REMEMBER THAT
H atoms on OH groups - ONLY PRODUCE ONE PEAK
- DON’T COUNT AS A NEIGHBOUR
1H-NMR
spectroscopy
“CENSUS” QUESTIONS
- describe where each hydrogen lives
- say how many hydrogens live on that atom
- say how many chemically different hydrogen
atoms live on adjacent atoms
1
2
3
1-BROMOPROPANE
ATOM
UNIQUE DESCRIPTION OF THE POSITION OF THE
HYDROGEN ATOMS
H’S ON THE ATOM
CHEMICALLY DIFFERENT H’S ON
ADJACENT ATOMS
SIGNAL SPLIT
INTO
1
On an end carbon, two away from the
carbon with the bromine atom on it
3
2
2+1 = 3
2
On a carbon atom second from the
end and one away from the carbon
with the bromine atom
2
3+2 = 5
5+1 = 6
3
On an end carbon atom which also
has the bromine atom on it
2
2
2+1 = 3
1H-NMR
spectroscopy: 1-bromopropane
CHEMICAL SHIFTS
1
2
3
3 environments = 3 signals
Triplet
Sextet
Triplet
d = 3.4
d = 1.9
d = 1.0
TMS
Signal for H’s on carbon 3 is
shifted furthest downfield
from TMS due to proximity of
the electronegative halogen
5
4
3
2
1
0
d
1H-NMR
spectroscopy: 1-bromopropane
INTEGRATION
1
2
3
Area ratio from relative
heights of integration lines =
2:2:3
Carbon 1
Carbon 2
Carbon 3
2
3
TMS
3
2
2
2
5
4
3
2
1
0
d
1H-NMR
spectroscopy: 1-bromopropane
SPLITTING
1
1
2
3
SPLITTING PATTERN
Carbon 1
TMS
Chemically different
hydrogen atoms on adjacent
atoms = 2
2+1 =3
The signal will be a
TRIPLET
5
4
3
2
1
0
d
1H-NMR
spectroscopy: 1-bromopropane
SPLITTING
2
1
2
3
SPLITTING PATTERN
Carbon 2
TMS
Chemically different
hydrogen atoms on adjacent
atoms = 5
5+1 =6
The signal will be a
SEXTET
5
4
3
2
1
0
d
1H-NMR
spectroscopy: 1-bromopropane
SPLITTING
3
1
2
3
SPLITTING PATTERN
Carbon 3
TMS
Chemically different
hydrogen atoms on adjacent
atoms = 2
2+1 =3
The signal will be a
TRIPLET
5
4
3
2
1
0
d
The signal is shifted furthest away (downfield) from TMS as the hydrogen atoms
are nearest the electronegative bromine atom.
1H-NMR
spectroscopy: 1-bromopropane
SPLITTING
3
1
2
2
1
3
3 environments = 3 signals
1 Triplet
2 Sextet
3 Triplet
d = 1.0
d = 1.9
d = 3.4
3 H’s
2 H’s
2 H’s
TMS
Signal for H’s on carbon 3 is
shifted furthest downfield
from TMS due to proximity of
the electronegative halogen
5
4
3
2
1
0
d
1H-NMR
1
spectroscopy: 1-bromopropane
2
3
TMS
SUMMARY
Peaks
Shift
Splitting
Integration
4
3
2
1
0
d
Three different signals as there are three chemically different protons.
Signals are shifted away from TMS signal, are nearer to the halogen.
Signals include a triplet (d = 1.0) sextet (d = 1.8) triplet (d = 3.4)
The integration lines show that the ratio of protons is 2:2:3
The signals due to the protons attached to carbon ...
C1 triplet
C2 sextet
C3 triplet
(d = 1.0) coupled to the two protons on carbon C2
(d = 1.8) coupled to five protons on carbons C1 and C3
(d = 3.4) coupled to the two protons on carbon C2
( 2+1 = 3 )
( 5+1 = 6 )
( 2+1 = 3 )
TETRAMETHYLSILANE – TMS
(PROVIDES THE REFERENCE SIGNAL)
•non-toxic liquid - SAFE TO USE
•inert - DOESN’T REACT WITH COMPOUND BEING ANALYSED
The molecule contains four methyl
groups attached to a silicon atom in a
tetrahedral arrangement. All the
hydrogen atoms are chemically
equivalent.
•all the hydrogen atoms are chemically equivalent - PRODUCES A SINGLE
PEAK
•twelve hydrogens so it produces an intense peak - DON’T NEED TO USE
MUCH
•signal is outside the range shown by most protons - WON’T OBSCURE
MAIN SIGNALS
•given the chemical shift of d = 0
•the position of all other signals is measured relative to TMS
USES Structural
OF NMR
SPECTROSCOPY
determination
 Gives information on chemical environments of all the protons in the molecule
 Other forms of NMR such as 13C also provide detailed structural information such
as distinguishing cis and trans isomers.
Medicinal uses
 1H-NMR is used in body scanning as protons in water, lipids, carbohydrates, etc,
give different signals (MRI or magnetic resonance imaging).
 MRI can be used to diagnose and monitor conditions, such as cancer, multiple
sclerosis and hydrocephalous.
 Useful in the medical field because radio waves are harmless.
1. Name the compound
C2H3Br3
2. Name the compound
C2H4O2
3. Name the compound
C4H8O2
4. Name the compound
C3H6O
5. Name the compound
C3H6O
6. Name the compound
C4H8O
7. Name the compound
C8H10
X-RAY CRYSTALLOGRAPHY