What can you remember from
last lesson?
Aims:
- To understand the basis of IR spectroscopy
- To interpret IR spectra
- To combine IR and MS data to identify
molecules
INFRA RED SPECTROSCOPY
Molecules can absorb Infra-red radiation
The absorbed energy makes covalent bonds vibrate
BENDING AND STRETCHING IN
WATER MOLECULES
SYMMETRIC STRETCHING
ASYMMETRIC STRETCHING
BENDING
Symmetrical
stretching
Rocking
Antisymmetrical
stretching
Wagging
Scissoring
Twisting
For a molecule to absorb IR photons there must be a CHANGE IN
DIPOLE in the bond stretch or vibration
e.g. H2O – only the stretching (top row) involve a change in dipole
Symmetrical
stretching
Rocking
Antisymmetrical
stretching
Scissoring
Wagging
Twisting
Molecules such as O2 and N2 do not appear on an IR spectrum.
Symmetrical
stretching
Rocking
Antisymmetrical
stretching
Scissoring
Wagging
Twisting
INFRA RED SPECTROSCOPY
Each bond vibrates and therefore absorbs infra-red
radiation.
The exact frequency depends on:
- The bond strength
- The bond length
- The mass of each atom in the bond
The Infra-red Spectrophotometer
• a beam of infra red radiation is passed through the sample
• a similar beam is passed through the reference cell
• the frequency of radiation is varied
• bonds vibrating with a similar frequency absorb the radiation
• the amount of radiation absorbed by the sample is compared with the
reference
• the results are collected, stored and plotted
The Infra-red Spectrophotometer
A bond will absorb radiation of a frequency similar to its vibration(s)
normal vibration
vibration having absorbed energy
The spectrum
Vertical axis Absorbance
Horizontal axis
the stronger the absorbance the larger the peak
Frequency
wavenumber (waves per centimetre) / cm-1
INFRA RED SPECTRA - USES
IDENTIFICATION OF
PARTICULAR BONDS
IN A MOLECULE
The presence of bonds
such as O-H and C=O
within a molecule can be
confirmed because they
have characteristic peaks
in identifiable parts of the
spectrum.
INFRA RED SPECTRA - USES
IDENTIFICATION OF
PARTICULAR BONDS
IN A MOLECULE
The presence of bonds such as O-H
and C=O within a molecule can be
confirmed because they have
characteristic peaks in identifiable
parts of the spectrum.
IDENTIFICATION OF
COMPOUNDS BY
DIRECT COMPARISON
OF SPECTRA
The only way to
completely identify a
compound using IR is to
compare its spectrum with
a known sample. The part
of the spectrum known as
the ‘Fingerprint Region’ is
unique to each compound.
Uses of IR spectroscopy
- Forensic science
- Breathalyzers
- Quality control in perfume production
- Drug analysis
Watch the video clip and answer the questions
1)How can a pair of atoms joined by a covalent bond be described?
2)What frequency is needed to increase the vibration of molecules?
3)How can we identify different functional groups?
4)Why is a blank needed?
5)How can an unknown sample be identified?
http://www.rsc.org/learnchemistry/resource/res00001041/spectroscop
y-videos#!cmpid=CMP00001771
FINGERPRINT REGION
•
•
•
•
organic molecules have a lot of C-C and C-H bonds within their structure
spectra obtained will have peaks in the 1400 cm-1 to 800 cm-1 range
this is referred to as the “fingerprint” region
the pattern obtained is characteristic of a particular compound the frequency
of any absorption is also affected by adjoining atoms or groups.
Interpreting spectra
• You must be able to identify the following
peaks.
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes, ketones and carboxylic acids 1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
O-H in alcohols
3200-3550 (broad)
• You will have a data sheet to help.
IR SPECTRUM OF A CARBONYL COMPOUND
• carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm-1
• this is due to the presence of the C=O bond
IR SPECTRUM OF AN ALCOHOL
• alcohols show a broad absorption between 3200 and 3600 cm-1
• this is due to the presence of the O-H bond
IR SPECTRUM OF A CARBOXYLIC ACID
•
•
•
•
carboxylic acids show a broad absorption between 3200 and 3600 cm-1
this is due to the presence of the O-H bond
they also show a strong absorption around 1700 cm-1
this is due to the presence of the C=O bond
IR SPECTRUM OF AN ESTER
• esters show a strong absorption between 1750 cm-1 and 1730 cm-1
• this is due to the presence of the C=O bond
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
CARBOXYLIC ACID
AND
C=O STRETCH
Identify the group next to the yellow box using the key below.
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes, ketones and carboxylic acids 1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
O-H in alcohols
3200-3550 (broad)
Identify the groups next to the yellow boxes using the key below.
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes, ketones and carboxylic acids 1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
O-H in alcohols
3200-3550 (broad)
C-O in alcohols, esters & carboxylic acids
1000-1300
C=O in aldehydes, ketones and carboxylic acids 1640-1750
C-H in alkanes/alkenes/aldehydes
2850-3100
O-H in carboxylic acids
2500-3300 (very broad)
O-H in alcohols
3200-3550 (broad)
Working backwards
Which peaks would be present in the
spectrum of….?
a) A ketone
b) A secondary alcohol
c) Alkene
Exam question
A compound was analysed by IR spectroscopy
and Mass spectrometry. It gave strong
absorbance at 1740 cm-1 and peaks at m/z values
of 15, 29 and 72. The compound had the
following composition by mass: C, 66.63%; H,
11.18%; O, 22.19%.
Use this information to identify two possible
structures of this compound.
Exam question
A compound was analysed by IR spectroscopy and Mass spectrometry. It gave
strong absorbance at 1740 cm-1 and peaks at m/z values of 15, 29 and 72. The
compound had the following composition by mass: C, 66.63%; H, 11.18%; O,
22.19%.
Use this information to identify two possible structures of this compound.
1) Calculate the empirical formula
2) Use IR spectrum to identify the functional group(s)
3) Use Mass spectrum to identify the parent ion (and therefore the molecular
formula)
4) Use Mass spectrum to identify the fragments
5) Suggest the two possible identities of the molecule
Exam question
A compound was analysed by IR spectroscopy and Mass spectrometry. It gave
strong absorbance at 1740 cm-1 and peaks at m/z values of 15, 29 and 72. The
compound had the following composition by mass: C, 66.63%; H, 11.18%; O,
22.19%.
Use this information to identify two possible structures of this compound.
1) Calculate the empirical formula: C4H8O
2) Use IR spectrum to identify the functional group(s) : C=O
3) Use Mass spectrum to identify the parent ion (and therefore the molecular
formula) 72, so C4H10O
4) Use Mass spectrum to identify the fragments
: 15 = CH3+ 29 = C2H5+
5) Suggest the two possible identities of the molecule: Butanal or butanone
Complete Q3 on page 180
Prep: Complete the exam
questions for next lesson