Welcome to Infrared Spectroscopy By H.T.Wong
IR & NMR to provide information about the structure of a molecule. • Modern instrumental technique to understand the structures of molecules • IR-vibrations of atoms • NMR-magnetic properties of atomic nuclei • Different functional groups have characteristic absorption wavenumbers.
Notes on IR Spectroscopy • ABSORPTION OF INFRARED RADIATION • The atoms in molecules are not static, but vibrate about their equilibrium positions. When molecules absorb infrared radiation, the absorbed energy causes an increase in the amplitude of the vibration of the bonded atoms.
The molecule is in an excited vibrational state. The absorbed energy is then dissipated as heat when the molecule returns to the ground state.
A particular type of bond between a certain pair of atoms will have a characteristic frequency of vibration. The natural frequency of vibration lies in the infrared region (1.20 1013- 1.20 1014 Hz) of the electromagnetic spectrum. Absorption of infrared radiation causes the atoms in molecules to vibrate more vigorously. • IR spectrum is obtained by measuring the absorption of IR radiation of a sample at different frequencies. The absorption is usually quoted in wavenumber (cm-1) which is the reciprocal of the wavelength (1/l). Since frequency () and wavelength (l) are related by the equation, c=l (where c is the speed of light), wavenumber will be a direct measure of frequency.
Relationship between Frequency, Wavelength and Wavenumber • c = • where c = velocity of light (ms-1) • = frequency (Hz or s-1) • = wavelength (m) • In IR spectroscopy, the frequency of radiation is often expressed as wavenumber (cm-1) : • wavenumber = 1/
Frequency(Hz)Wavelength-(m)Wavenumber- (cm-1 ) 1.2 1013 2.50 10-5 400 4.5 1013 6.67 10-6 1500 1.2 1014 2.50 10-6 4000 • Example:The wavenumber of a vibration frequency with the value 4.5 1013 Hz can be calculated as follows: • Since 4.5 1013 Hz 4.5 1013 s-1 • wavenumber= 4.5 1013 / 3 108 102 • =1500 cm-1
Photo 4: Apparatus for Handling Samples for Infrared Spectroscopy
FEATURES OF AN INFRARED SPECTRUM • An infrared spectrum is a plot of per cent transmittance (or absorbance) against wavenumber (frequency or wavelength). A typical infrared spectrum is shown below.
A 100 per cent transmittance in the spectrum implies no absorption of IR radiation. When a compound absorbs IR radiation, the intensity of transmitted radiation decreases. This results in a decrease of per cent transmittance and hence a dip in the spectrum. The dip is often called an absorption peak or absorption band. • Different types of groups of atoms (C-H, O-H, N-H, etc…) absorb infrared radiation at different characteristic wavenumbers.
INTERPRETATION OF IR SPECTRA • In general, the IR spectrum can be split into four regions for interpretation: • 4000 2500 cm-1: Absorption of single bonds formed by hydrogen and other elements e.g. CH, OH, NH • 2500 2000 cm-1: Absorption of triple bonds e.g. C≡C, C≡N • 2000 1500 cm-1: Absorption of double bonds e.g. C=C, C=O • 1500 400 cm-1: This region often consists of many different, complicated bands. This part of the spectrum is unique to each compound and is often called the fingerprint region. It is rarely used for identification of particular functional groups.
The region between 4000 and 1500 cm-1 of an infrared spectrum is often used for the identification of various functional groups. • Different functional groups have characteristic absorption wavenumbers. A correlation table providing information is shown below. It could help to interpret an infrared spectrum.
Table 1: Correlation Table of Functional Groups and IR Absorption Wavenumbers • Characteristic Range • Bond Wavenumber (cm-1) • C=C Alkenes 1610 to 1680 • C=O Aldehydes, ketones, acids, esters 1680 to 1750 • C≡C Alkynes 2070 to 2250 • C≡N Nitriles 2200 to 2280 • OH Acids (hydrogen-bonded) 2500 to 3300 • CH Alkanes, alkenes, arenes 2840 to 3095 • OH Alcohols, phenols(hydrogen-bonded) 3230 to 3670 • NH Primary amine 3350 to 3500
If the infrared spectrum of an unknown compound shows an absorption peak in the region of 3200 to 3700 cm-1, it is reasonable to assume that the compound may contain either OH or NH group. • The following examples show how to make use of the characteristic absorption wavenumbers to identify the functional groups present in organic compounds. Fig.1 shows a portion of the IR spectrum of ethanol.
The peak at 2950 cm-1 corresponds to the absorption of CH group and that at 3340 cm-1 corresponds to the absorption of OH group. Hence, a OH group is present in the ethanol.
The peak at 1710 cm-1 corresponds to the absorption of C=O group and that at 3100 cm-1 corresponds to the absorption of OH group. Hence, both the C=O and OH groups are present in the methanoic acid. • It should be noted that the absorption of the OH group in alcohols and carboxylic acids does not usually appear as a sharp peak. Instead, a broad band is observed because the vibration of OH group is complicated by the hydrogen bonding.
A SUMMARY OF STRATEGY FOR INDENTIFICATION OF FUNCTIONAL GROUPS • 1. Look at the high-wavenumber end of the spectrum (above 1500 cm-1) and concentrate initially on the major absorption peaks. • 2. Classify the major absorption peaks in the spectrum as follows: • (a) X-H absorption (4000 - 2500 cm-1) • (b) triple bond absorption (2500 - 2000 cm-1) • (c) double bond absorption (2000 - 1500 cm-1)
3. For each absorption peak, short-list the possibilities using a correlation table or chart. (Note: Do not expect to be able to assign every absorption peak in the spectrum.) • 4. Place as much reliance on negative evidence as on positive evidence, e.g. if there is no absorption peak in 1600-1800 cm-1 region, it is likely that a carbonyl group is absent in the test sample.
LIMITATIONS OF THE USE OF IR SPECTROSCOPY IN THE IDENTIFICATION OF ORGANIC COMPOUNDS • 1. Some infrared absorptions have very close wavenumbers and the peaks coalesce. • 2. NOT all vibrations give rise to strong absorption peaks. • 3. NOT all absorption peaks in a spectrum can be associated with a particular bond or part of the molecule.
4. Intermolecular interactions in molecules complicate infrared spectra. • 5. Infrared spectroscopy can only provide limited structural information of molecule
Characteristic Absorptionsof Carbonyl Groups • Wavenumber / cm-1 • RCOCl Acyl chloride1815 - 1790 • RCOOR’ Aliphatic ester 1750 - 1730 • RCHO Aliphatic aldehyde 1740 - 1720 • RCOOH Aliphatic acid1725 - 1700 • RCOR’ Aliphatic ketone 1725 - 1700 • ArCHO Aromatic aldehyde1715 - 1695 • ArCOR Aromatic ketone 1700 - 1680 • ArCOAr Diaromatic ketone 1670 - 1650 • RCONH2 Aliphatic amide 1680 - 1640 • (Source: Modern Chemical Techniques, The Royal Society of Chemistry, Cambridge, 1992.)
Wavenumber (cm-1) 2970 1640 Bond in Hex-1-ene CH C=C
Wavenumber (cm-1) 3300 2900 to 2950 2200 Bond in hex-1-yne CH * CH CC
Wavenumber (cm-1) 2980 2830, 2720 1730 Bond in propanal Aliphatic CH CH of CHO C=O
Wavenumber (cm-1) 2880 to 2980 1720 Bond in 3-methylbutan-2-one CH C=O
Wavenumber (cm-1) 3080 1720 1635 Bond in hex-5-en-2-one CH C=O C=C
Wavenumber (cm-1) 3330 2940, 2820 1030 Bond in methanol OH CH CO
Wavenumber (cm-1) 3380 3050 1220 Bond in phenol OH Aromatic CH CO
Wavenumber (cm-1) 3100 1710 Bond in methnoic acid OH C=O
Wavenumber (cm-1) 2900 to 2950 1740 Bond in butyl ethanoate CH C=O
Wavenumber (cm-1) 3300 to 3400 2850 to 2950 Bond in butylamine NH CH
Wavenumber (cm-1) 3000 2250 propanenitrile CH CN
a) C=C 1610-1680 O-H 3230-3670 b) CC 2070-2250 c) N-H 3350-3500 a) C=O 1680-1750 b) C=C 1610-1680 c) no 1. Explain how you can distinguish between each isomer in the following pairs of compounds by infrared absorption peaks.(a) H2C=CHCH2OH and (CH3)2C=O (b) HCCCH2CH2CH3 and H2C=CHCH2CH=CH2 (c) CH3CH2CH2NH2 and (CH3)3N
No C=O (A) & (B) impossible (C) is the unknown 2. An unknown compound with molecular formula C3H6O shows no absorption peak in the region of 1700 to 1750 cm1 of its infrared spectrum. Which one of the following is likely to be the unknown compound? propanone; propanal; prop-2-en-1-ol (A) (B) (C)
3. An organic compound Z, with relative molecular mass below 100, has the following composition by mass.C 66.7%, H 11.1% and O 22.2%.(a) Determine the molecular formula of Z.(b) A portion of the infra-red (IR) spectrum of Z is shown below:
Using the IR spectrum and the result from (a), deduce two possible structures of Z, each belonging to a different homologous series. • a) C H O • no of mol: 66.7/12 11.1/1 22.2/16 • 5.56 11.1 1.39 • simplest ratio: 4 8 1 • empirial formula is C4H8O • let the molecular formula be (C4H8O)n • rel. mol. mass: (4x12+8x1+16) x n < 100 • n=1 • molecular formula is C4H8O
Wavenumber Bond • 2950 C-H • 1720 C=O • CH3CH2CH2CHO • CH3CH2COCH3
4. The infrared spectra in Figure 1 and 2 represent an ester and an alkyne. Identify the peaks marked with “ * ” on each spectrum and hence determine which spectrum represents which compound.Which of the following is an ester or alkyne? Figure 1:
Figure 1: 2920cm-1 : C-H 1720cm-1 : C=O an ester Figure 2: 2920cm-1 : C-H 2180cm-1 : CC an alkyne
5. An alkene A (C5H10), on ozonolysis, gave two different compounds, B and C with molecular formulae C3H6O and C2H4O respectively. Compound B has an infrared spectrum as shown in Figure 3 below:
When B and C were separately treated with acidified potassium dichromate(VI) solution, B did not react but C gave D with molecular formula C2H4O2. D gave effervescence when mixed with sodium hydrogencarbonate.Deduce possible structures for A, B, C and D. • A : 2-methylbut-2-ene • B : propanone • C : ethanal • D : ethanoic acid