Infrared Spectroscopy s1

Infrared Spectroscopy

Atoms and molecules interact with electromagnetic radiation (EMR) in a wide variety of ways. They might absorb and/or emit EMR. Absorption of EMR stimulates different types of motion in atoms and/or molecules. The patterns of absorption (wavelengths absorbed and to what extent) and/or emission (wavelengths emitted and their respective intensities) are called ‘spectra’.

Spectroscopy is an analytical technique which helps determine structure. It destroys little or no sample.

There are different types of spectroscopy:

• Infrared (IR) spectroscopy measures the bond vibration frequencies in a molecule and is used to determine the functional group. • Mass spectrometry (MS) fragments the molecule and measures the masses. • Nuclear magnetic resonance (NMR) spectroscopy detects signals from hydrogen atoms and can be used to distinguish isomers. • Ultraviolet (UV) spectroscopy uses electron transitions to determine bonding patterns. • Examples: X rays, microwaves, radio waves, visible light, IR, and UV. • Frequency and wavelength are inversely proportional. • c = ln, where c is the speed of light. • Energy per photon = hn, where h is Planck’s constant.

The Spectrum and Molecular Effects

The infrared region of electromagnetic spectrum extends from 0.8 μm (800 nm) to 1000 μm (1mm) and is subdivided into near infrared

(0.8-2 μm), middle infrared (2-15 μm), and far infrared (15-1000 μm)

The fundamental region, just below red in the visible region of wavelengths usually 2.5-15 μm is the region that provides the greatest information for elucidation of molecular structure and most IR spectrophotometers are limited to measurements in this region.

Absorption of IR radiation causes changes in the vibration energy of the molecule. The molecular vibrations are classified into stretching and bending vibration. The former involves changes in the length of the bond, as if the atoms are connected by a spring vibrating with a particular frequency.

2 Covalent bonds vibrate at only certain allowable frequencies

Poly atomic molecules may have symmetrical or asymmetrical stretching vibrations e.g. H2O

Nonlinear molecule with n atoms usually has 3n - 6 fundamental vibrational modes.

Not all the vibrations of a molecule give rise to IR absorption because certain vibrations are inactive (symmetrical molecules) or only weakly active and because some of the absorption bands involving similar and even different group overlap.

3 SPECTRA

An IR spectrum is a plot of the light transmitted (%) to wave length or wave number (4000-400cm-1).

IR spectra can be obtained from material as neat liquids (homogenous liquid containing no solvents), as gases, as solids or in solution. A tube holding the solvent can be placed in the reference beam.

Different kinds of atoms have different masses. Different kinds of bonds have different strength. Therefore different combination of atomic masses and bond energies give system that can vibrate at different frequencies when the molecules absorb electromagnetic energy at 1595cm-1 when it bends.

Different vibration motions of the atoms in the same molecule can also lead to absorption at different wave numbers.

Interpretation of the IR spectra:

Interpretation of the absorption spectra is considerably simplified by charts or tables which correlates the frequency at which a band occurs with molecular structure.

5 The region between 4000-1500 cm-1 is probably easier to interpret than that between 1500-650 cm-1 , because of the coupling together of many C-C stretching and C-H modes, the individual bands cannot be assigned to individual bonds; rather the whole region is a characteristic of specific molecule. This part of the spectrum is called fingerprint region.

In organic molecules there are usually many C-C and C-H bonds and the frequencies from the vibration of atoms may couple together to give complicated spectra. In the IR of n-hexane the strongest absorption band are found at 2900 cm-1 and 1450 cm-1 resulting from the stretching and bending respectively. All molecules that contain alkyl group show these absorption bands, so these are found in almost all IR spectra of organic

6 compounds. In addition symmetrical bending of a methyl group lead to absorption at approximately 1350 cm-1 . The triple bond is stronger than the double C=C bond, the former is harder to stretch which leads to absorption at higher wave number than C=C stretching. The C=C is in turn responsible for the absorption at higher wave number than the C-C bond. The triple bond absorbs at 2100-2300cm-1 and the double bond absorb near 1650 cm-1. Infrared spectra are regularly used by organic chemist to help identify functional groups.

An Alkane IR Spectrum

An Alkene IR Spectrum

7 An Alkyne IR Spectrum

8 • O-H and N-H Stretching: Both of these occur around 3300 cm-1, but they look different. • Alcohol O-H, broad with rounded tip.

• Secondary amine (R2NH), broad with one sharp spike.

• Primary amine (RNH2), broad with two sharp spikes.

• No signal for a tertiary amine (R3N)

An Alcohol IR Spectrum

9 An Amine IR Spectrum

Carbonyl Stretching

• The C=O bond of simple ketones, aldehydes, and carboxylic acids absorb around 1710 cm-1. • Usually, it’s the strongest IR signal. • Carboxylic acids will have O-H also. • Aldehydes have two C-H signals around 2700 and 2800 cm-1.

A Ketone IR Spectrum

10 An Aldehyde IR Spectrum

O-H Stretch of a Carboxylic Acid

11 This O-H absorbs broadly, 2500-3500 cm-1, due to strong hydrogen bonding.

Variations in C=O Absorption

• Conjugation of C=O with C=C lowers the stretching frequency to ~1680 cm-1. • The C=O group of an amide absorbs at an even lower frequency, 1640-1680 cm-1. • The C=O of an ester absorbs at a higher frequency, ~1730-1740 cm-1. • Carbonyl groups in small rings (5 C’s or less) absorb at an even higher frequency. An Amide IR Spectrum

12 Carbon - Nitrogen Stretching

• C - N absorbs around 1200 cm-1. • C = N absorbs around 1660 cm-1 and is much stronger than the C = C absorption in the same region.

• C º N absorbs strongly just above 2200 cm-1. The alkyne C º C signal is much weaker and is just below 2200 cm-1.

A Nitrile IR Spectrum

Aromatic Compounds

13 *Weak C–H stretch at 3030 cm-1

Weak absorptions 1660 - 2000 cm-1 range

*Medium-intensity absorptions 1450 to 1600 cm-1

Monosubstituted benzene can be recognized by strong bands at 710- 690 cm-1 and at770-730. Meta substituted benzene usually shows two intermediate bands 710-690 cm-1 and 800-750 cm-1 . Ortho and para substituted benzene show very strong and strong band each at 770-735 cm-1 and medium band at 840-800cm-1

14 Conjugated system

Conjugated system absorbs at lower frequencies than nonconjugated systems. An alkenes absorbs at 1670-1640cm-1 whereas alkenes conjugated with carbonyl absorbs or another double bonds absorbs near 1600 cm-1 . The reason for shift can be seen in terms of resonance forms

- + CH CH CH 2 CH CH CH2 2 CH CH 2

The absorption of the carbonyl group proves to be of great value. The absorption is strong and easy to locate and identify. The carbonyl absorption depends very much on its environment i.e. a carbonyl group in a six membered ring have frequencies of approximately the same value as the corresponding open chain compounds, whereas each incremental decrease in the ring size raise the frequency by approximately 30 cm-1

15 O O O O

1715cm-1 1715cm-1 1745cm-1 1775cm-1

When a carbon-carbon double bond is in conjugation with a carbonyl this conjugation lowers the observed carbonyl frequency by about 30 cm-1 compared to that of nonconjugated carbonyl.

O O CH3 CH3

C CH C CH3 CH CH2 C CH3 CH3 CH3 -1 1715cm-1 1685cm

The number of absorption bands in an IR spectrum may be of value as a mean of identification. Two compounds are one and the same, if their spectra agree in fall respect, i.e. in position and relative intensities of the bands. Identification tests in the B.P. monographs of many drug substances and their formulations are based upon a comparison of the sample spectrum with authentic spectrum published in the B.P. if the spectra are not identical and have been obtained by examination of the sample in the solid state, further test must be carried out. The pharmacopoeia describes the following additional tests:

a) Recrystallise the sample from the solvent specified and repeat the determination. b) Dissolve the sample in a suitable solvent and measure the absorption spectrum against the solvent as blank. Compare with the

16 authentic substance under the same conditions. This procedure is necessary because various crystalline forms of the same substance may give rise to different spectra e.g. cortisone acetate has been obtained in five different crystalline forms. The preparation of disc of a hydrochloric salt of an organic base should be carried out using diluents with the same halide to prevent exchange of halide and distortion of the spectrum that may occur if a different halide is used. Thus a hydrochloride salt of abase should be examined in sodium chloride disc and a hydrobromide salt in potassium bromide disc. c) Very often, it is necessary to identify a completely unknown substance or substances in perhaps a single tablet. All the resources of the analyst must be brought to bear on the problem including chromatography and ultraviolet absorption. As a first step, separation of the active principle from inert tablet base will be necessary and partition between immiscible solvents under controlled condition of pH is very useful for this purpose. It may be possible to use the residue from an appropriate extract directly for IR anslysis, from the results of which the presence or absence of functional group may be inferred. An indication may thus be obtained of the type of compound for which a search should be made in the reference file. The residue may be a mixture and require further treatment. In this connection gas chromatography combined with IR analysis is a useful technique for the isolation and identification of pure components.

17 Qualitative analysis:

The principles underlying quantitative ultraviolet spectrophotometer apply to IR work.

Moreover IR absorption spectra possess an advantage over those in the ultraviolet region in greater number of bands present. It may often be possible to select a fairly strong band for each component in a mixture such that little or no interference occur one with another.

A calibration curve of absorbance against concentration may be constructed. The method of calculation is as shown in the Figure. The band a b c is the recorded absorption of component A and d e f is the absorption caused by solvent and other component. Draw line a g c connecting the two minima a and c or between two suitable wave lengths on each side of the band. The point g is obtained by dropping a line perpendicular to the zero transmittance line to meet a c at g. The

18 absorbance (log I0 /It) is calculated from the distances shown in the Fig.

The problems encountered in the use of the IR for quantitative analysis are:

19 - The sample cell must be transparent to light of this wave length range. The most suitable for this region of spectrum of sodium chloride and potassium bromide. However, the water solubility of both KBr and NaCl makes it generally impossible to determine IR spectra in the presences of water with out dissolving or fogging the sample cell. - The concentrations used in IR work are about 10% solutions because all solvents have some absorption in one part or another of the IR region. This means that very short path lengths of 0.025- 0.1mm must be used in many assays. Path length errors of this magnitude produce a correspondingly large error in calculated concentrations of the unknown sample. - Another problem caused by the material used for sample cell is that of light scattering. This cause the detector to read a greater absorption than is actually present. This is because the walls are not as smooth as the quartz cell in ultraviolet spectrophotometers.

20  Problems on infra-red Spectroscopy:

1. An organic compound with molecular mass C8H18 shows the following characteristic absorption bonds:

(i) 2925 cm-1 (st), (ii) 1465 cm-1, (iii) 1380 cm-1 and (iv) 720 cm-1. Determine the structure of the compound.

2. An organic liquid with molecular formula C7H8 burns with a sooty flame. It shows the following absorption bands in its Infra-red spectrum:(i) 3060 cm-1 (ii) 3040 cm-1 (iii) 2918, 2870 cm-1 (iv) 1500, 1450 cm-1 (v) 750 cm-1. Deduce the structure of the compound.

3. The analytical data and the molecular mass determination gave C8H8O as the molecular formula of the compound. The compound burns with a sooty flame and gave an oxime with hydroxylamine hydrochloride. Following absorption bands appear in its Infra-red spectrum : (i) 2825 cm-1, (ii) 2717 cm-1, (iii) 3060 cm-1 and (iv) 1700 cm-1 (s) and 830 cm-1. Deduce the structure of the compound.

4. An organic compound dissolves in sodium hydroxide to form a yellow coloured solution. It gives brisk effervescence with sodium bicarbonate solution. Its infra-red spectrum exhibits the following absorption bands: (i) 3060-3110 cm-1, (ii) 3000-2520 cm-1 (b), (iii) 1602, 1510, 1450 cm-1 and (iv) 1620, 1375 cm-1(s) and 830 cm-1. . Deduce the structure of the compound

21 6. An organic compound with molecular formula C8H6 decolourises bromine in carbon tetrachloride and gives a white precipitate with ammonical silver nitrate solution. Give the probable structure of the compound. Its infra-red spectrum gives a band at 2150- 2200 cm-1 and the region near 3300 cm-1 is transparent.

7. How will you distinguish between: (i) p—CH3CO—C6H4——OCH3 and (ii) CH3CO—C6H4—NO2?

9. An aromatic organic compound decolourises bromine in carbon tetra - chloride and exhibits the following absorption position.

(i) 3090 cm-1 (ii) 3040 and 3000 cm-1 (iii) 2820 and 2750 cm-1 (iv) 1685 cm-1(s) (v) 1630 cm-1 (vi) 1580 and 1450 cm-1 (vii) 750 cm-1.

 Short Questions

1. What is the range of infra-red radiations?

2. What happens when a substance is irradiated with infrared radiations?

3. Can you distinguish a pair of enantiomers by infra-red spectroscopy?

4. Name the various types of bending vibrations.

5. Compared to the number of bonds in a molecule, there are generally more number of peaks in the infra-red spectrum. Explain.

6. How hydrogen bonding change the position of absorption in the Infra- red spectroscopy?

7. Can you distinguish the type of hydrogen bonding by Infra-red spectroscopy?

8. Discuss the positions of absorption of a particular band in a substance in all the three states.

11. Infra-red spectroscopy is the best technique to establish the identity of organic compounds. Explain.

22 12. Some of the fundamental vibrations are infra-red active while others are not. Explain.

13. What happens to O-H str position when 10 ml of carbon tetrachloride is added to 2 ml of ethyl alcohol?

14. What do you mean by Finger print region?

15. In acetylene, —C C—H str appears at about 3300 cm-1. How will you distinguish it from an O—H str in alcohol?

16. How will you show that the compound under investigation is not aromatic? Use Infra-red technique.

17. How will you distinguish between CH3CH2CHO and CH3COCH3?

18. Which of the following diatomic molecules do not absorbs in the Infra-red region.

HCI, C1Br, N2, H2, O2

19. Why is methanol a good solvent for UV but not for IR spectroscopy?

20. What is the effect of hybridization of carbon on the stretching frequency of C—H bonds?

21. cis-1,2-dichloroethene is Infra-red active while trans-1,2- dichloroethene is IR inactive. Explain.

22. How will you note the progress of the oxidation of 2-Propanol to Propanone in Infra-red spectroscopy?

23. How will you distinguish between cis and trans-cinnamic acid?

24. Can you detect an anhydride by Infra-red spectroscopy?

25. How will you distinguish between CH3CONH2 and CH3CH2NH2?

26. Infra-red absorption due to carbonyl stretching occurs at higher frequencies than stretching of C = C bond. Explain.

27. An aliphatic aldehyde containing unconjugated double bond shows C = C and C = O stretching at ~1645 cm-1 and 1720-1740 cm-1 respectively. Explain why crotonaldehyde shows the corresponding absorptions at 1700 cm-1 and 1630 cm-1.

23 Specific type of Absorption Bond Type of bond Appearance bond peak

C─H 1260 cm−1 strong

1380 cm−1 weak methyl 2870 cm−1 medium to strong

2960 cm−1 medium to strong alkyl 1470 cm−1 strong

methylene 2850 cm−1 medium to strong

2925 cm−1 medium to strong

methine 2890 cm−1 weak

−1 vinyl C═CH2 900 cm strong

2975 cm−1 medium

3080 cm−1 medium

24 C═CH 3020 cm−1 medium

900 cm−1 strong monosubstituted alkenes 990 cm−1 strong

cis-disubstituted 670–700 strong alkenes cm−1

trans- disubstituted 965 cm−1 strong alkenes

trisubstituted 800–840 strong to medium alkenes cm−1

benzene/sub. 3070 cm−1 weak benzene

700–750 strong cm−1 monosubstituted benzene 690–710 strong cm−1 aromatic ortho-disub. 750 cm−1 strong benzene

750–800 strong cm−1 meta-disub. benzene 860–900 strong cm−1

25 para-disub. 800–860 strong benzene cm−1

alkynes any 3300 cm−1 medium

2720 cm−1 aldehydes any medium 2820 cm−1

C─C monosub. 1645 cm−1 medium alkenes

1,1-disub. 1655 cm−1 medium alkenes

cis-1,2-disub. acyclic C─C 1660 cm−1 medium alkenes

trans-1,2-disub. 1675 cm−1 medium alkenes

trisub., tetrasub. 1670 cm−1 weak alkenes

dienes 1600 cm−1 strong conjugated C─C 1650 cm−1 strong

with benzene 1625 cm−1 strong ring

with C═O 1600 cm−1 strong

26 1640–1680 C═C (both sp2) any medium cm−1

1450 cm−1

1500 cm−1 weak to strong aromatic C═C any (usually 3 or 4) 1580 cm−1

1600 cm−1

2100–2140 terminal alkynes weak cm−1 C≡C 2190–2260 very weak (often disubst. alkynes cm−1 indisinguishable)

saturated aliph./cyclic 6- 1720 cm−1 membered

α,β-unsaturated 1685 cm−1 C═O aldehyde/ketone aromatic 1685 cm−1 ketones

cyclic 5- 1750 cm−1 membered

27 cyclic 4- 1775 cm−1 membered

influence of aldehydes 1725 cm−1 conjugation (as with ketones)

saturated 1710 cm−1 carboxylic acids

unsat./aromatic 1680–1690 carb. acids cm−1

influenced by esters and conjugation and 1735 cm−1 lactones ring size (as with ketones)

1760 cm−1 carboxylic anhydrides acids/derivates 1820 cm−1

acyl halides 1800 cm−1

associated amides 1650 cm−1 amides

carboxylates 1550–1610 (salts) cm−1

amino acid 1550–1610 zwitterions cm−1

28 low 3610–3670 concentration cm−1 alcohols, phenols high 3200–3400 broad concentration cm−1 O─H low 3500–3560 concentration cm−1 carboxylic acids high 3000 cm−1 broad concentration

3400–3500 strong cm−1 primary amines any 1560– strong 1640 cm−1 N─H secondary any >3000 cm−1 weak to medium amines

2400–3200 multiple broad ammonium ions any cm−1 peaks

C─O alcohols 1040–1060 primary strong, broad cm−1

secondary ~1100 cm−1 strong

tertiary 1150–1200 medium cm−1

29 phenols any 1200 cm−1

aliphatic 1120 cm−1

ethers 1220–1260 aromatic cm−1

1250–1300 carboxylic acids any cm−1

two bands (distinct from 1100–1300 esters any ketones, which cm−1 do not possess a C─O bond)

1020–1220 aliphatic amines any often overlapped cm−1

similar 1615–1700 C═N any conjugation cm−1 effects to C═O

unconjugated 2250 cm−1 medium C─N C≡N (nitriles) conjugated 2230 cm−1 medium

R─N─C 2165–2110 any (isocyanides) cm−1

2140–1990 R─N═C═S any cm−1

30 1000–1100 ordinary cm−1 fluoroalkanes 1100–1200 two strong, broad trifluromethyl cm−1 bands

C─X 540–760 chloroalkanes any weak to medium cm−1

500–600 bromoalkanes any medium to strong cm−1

iodoalkanes any 500 cm−1 medium to strong

1540 cm−1 stronger aliphatic 1380 cm−1 weaker N─O nitro compounds

1520, 1350 lower if aromatic cm−1 conjugated