Laboratory Report Materials Physics Laboratory The Infra-Red Spectroscopy of Polymers Yufei Chang • Group H
Abstract The aim of this experiment is to to determine the types of polymers using infrared spectroscopy. Samples were placed in infrared spectrometer and hit by infrared beam. Background and sample spectra were produced by computer. With the help of a numerical listing of wavenumbe range. One or several types of polymers could be deduced. Some other techniques could be employed for determining the exact structure.
Introduction Infrared Spectroscopy is the study of the interaction of infrared light with matter. Light is composed of electric and magnetic waves. These two waves are in planes perpendicular to each other, and the light wave moves through space in a plane perpendicular to the planes containing the electric and magnetic waves. It’s the electric part of light, called electric vector, that interacts with molecules. The wavenumber of a light wave is defined as the reciprocal of the wavelength. W=1/λ, where λ is wavelength measured in cm, then W is reported as cm⁻¹. Wavenumbers are the units typically used in infrared spectroscopy to denote different kinds of light. As E=hf, the wavenumber could directly proportional to energy as follow E=hcW , where h is the Planck’s constant, c is the speed of light and W is the wavenumber. Thus high wavenumber light has more energy than low wavenumber light. In this experiment the rang of we used in spectrometer is 4000∼400 cm⁻¹. When infrared light is passed through a sample of an organic compound, some of the frequencies are absorbed , cause the vibration of the matter. While other frequencies are transmitted through the sample without being absorbed. ❷A new unit is introduced here as transmittance T(%) which is defined as I/I₀. Where I is the sample spectrum and I₀ is the background spectrum. The spectrum we plot here is transmittance against wavenumber. For a single molecule (not all kinds of, discussed later) , it could have more than one wavenumber. This due to different vibration types. In this experiment we mainly concerned are CH2 or CH3 , both have stretching and bending vibration. For CH₃ ,when it is under stretching vibration, it has asymmetric (2960±10 cm⁻¹ ) and symmetric type (2870±10 cm⁻¹ ). When it is under bending vibration, it has asymmetric (1460±10 cm⁻¹ ) and symmetric umbrella type (1375±10 cm⁻¹ ). For Ch₂ the situation is almost the same except the wavenumbers.
Experimental There are two parts in this experiment. First is obtaining the spectra of four unknown polymers. What we measured this time are from sample 2 to sample 5. These are samples are vinyl polymers with a simple C-C backbone with hydrogen atom or other functional groups.
Part two is the identification of a common polymer. One article form the range of everyday article made form polymers should be selected. This time we chose ruler as our sample. Then compare the spectrum with those obtained in part 1 and identify some absorption bands.
Results Fig 1
Fig 3
Fig 4
Fig 5
Sample 2 From the fig 1. Big band in spectrum indicates high infrared absorption. For example, the peak with wavenumber 2839.32 cm-1 has intensity approximately 28.540, namely, only 28.540% of infrared beam is transmitted through the sample while 71.460% of beam is been absorbed. According to Fig.1, peaks just below 3000 cm-1 might be due to C–H stretching. Peak at 2950.53 cm-1 could be CH3 asymmetric stretch, the peak at 2918.69 cm-1 is likely to be CH2 asymmetric stretch and that at 2866.86 cm-1 is CH3 symmetric stretch. Moving to the right, peaks at about 1400 cm-1 are suggested to be either C–H bending (either due to CH2 or CH3 groups) or O–H bending. However, O–H bending is normally shown as a broad peak whereas in the case the peak is very sharp. Thus, it is more likely to be C–H bending. The peak at 1458.15 cm-1 is likely to be CH2 scissor bend or CH3 asymmetric bend and that at 1375.92 cm-1 is CH3 symmetric bend. At the right end of spectrum, there are two small peaks with wavenumbers around 900 cm-1, which could be C–H rocking, N–H rocking, or experimental error (noise, contaminations eg. fingerprint, or sample is too old). Because from the possible functional groups found in this spectrum (CH2 or CH3), it is suggested that the sample might be either polyethylene or polypropylene, in which no C = X = Y functional group present.
This spectrum is compared with the data in the library using software OMNIC. The computer’s first choice is polypropylene, which has a repeating unit. Sample 3 Peaks near 3000 cm-1 are suggested to be C–H stretching, which is CH3 asymmetric stretching. That around 1700 cm-1 is C=O stretching, and those around 1400 cm-1 could be C=C stretching or N–H bending. The broad intense peaks at about 1200 cm-1 is C–O or C–N stretching. Peaks between 912.56 cm-1 and 1063.89 cm-1 are probably C–C or C–N stretching. Next set of peaks are again due to C–H or N–H rocking. The possible functional groups are C=O, C–C, also C–O and C–N. But C–O and C–N are less likely to be present at the same time. Therefore, a possible polymer could be PMMA. Sample 4 Start from the right end of spectrum in this case. Peaks between 695.94 cm-1 and 906.57 cm-1 are due to aromatic C–H bend according to the numerical listing. Peaks from about 1154.53 cm-1 to 1601.07 cm-1 might be caused by C–H, O–H, N–H bending, C = C or C = N stretching. The peak at 1371.85 cm-1 is likely to be CH3 symmetric bend and that at 1452.09 cm-1 is CH3 asymmetric bend or CH2 scissor bend. The small peaks on the left from around 1750 cm-1 to 1950 cm-1 are likely to be aromatic C–H overtones (weak). Again, peaks just below 3000 cm-1 are probably C–H stretching. The peak at 2849.47 cm-1 is likely to be CH2 symmetric stretch and that at 29** cm-1 (not shown in the figure) deduced to be CH2 asymmetric stretch. Peaks just above 3000 cm-1 are suggested to be O–H or N–H stretching. Nevertheless, considering the rest of possible functional groups present (CH2 CH3, aromatic ring), these peaks are unlikely to be either of these two functional groups. A suggestion is that these peaks are caused by aromatic ring. The rest of small wavy peaks are possibly due to noise or contaminations. Hence this sample is likely to be polystyrene with repeating unit. Sample 5 Analogous to samples above, peaks just below 3000 cm-1 are likely to be C–H stretching. The peak at 2950.47 cm-1 could be CH3 asymmetric stretch, the peak at 2919.18 cm-1 is likely to be CH2 asymmetric stretch and that at 2849.36 cm-1 is CH3 symmetric stretch. Peaks at around 1400 cm-1 are probably due to C–H bending or O–H bending (but less likely to be). The peak at 1462.53 cm-1 is likely to be CH2 scissor bend or CH3 asymmetric bend and that at 1376.03 cm-1 is CH3 symmetric bend. In addition, small peaks from 719.61 cm-1 to about 1000 cm-1 could
be C–H or N–H rocking. It is less likely to be noise because these peaks are more intensive than those in sample 2. A possible guess for this polymer, with functional groups CH2 or CH3, would be either polyethylene or polypropylene with repeating unit. The ruler sample Similarly to sample 4, peaks just above 3000 cm-1 are probably due to aromatic ring. Peaks just below 3000 cm-1 are caused by C–H stretching. The peak at 2923.93 cm-1 is likely to be CH2 asymmetric stretch and that at 2850.46 cm-1 is CH2 symmetric stretch. Small peaks in the range 2000 cm-1 and 1700 cm-1 are suggested to be aromatic C–H overtones (weak). Next set of peaks are possibly caused by O–H bending (again less likely to be) N–H bending, C = C, C = O or C = N stretching or C–H bending (more likely). The peak at 1452.35 cm-1 is likely to be CH3 asymmetric bend or CH2 scissor bend and that at 1374.01 cm-1 is CH3 symmetric bend. Peaks from 906.69 cm-1 to 1069.18 cm-1 could be aromatic C–H bend (strong). Peaks at the right end might be C–H or N–H rocking. The possible functional groups are CH2, CH3, phenyl group, which suggest the sample to be polystyrene.
Discussion When the electric part of infrared radiation interacts with matter it can be absorbed, causing the chemical bonds in the material to vibrate, The presence of chemical bonds in a material is a necessary condition for infrared absorbance to occur. The functional groups then to absorb infrared radiation in the wavenumber range regardless of the structure of the rest of the molecule that the functional group is in. For instance C=O stretch of a carbonyl group at ∼1700 cm⁻¹ in ketones, aldehydes, and carboxylic acids.❶ This means there is a correlation between the wavenumbers at which a molecule absorbs infrared radiation and its structure. This correlation allows the structure of unknown molecules to be identified from the molecule’s infrared spectrum. And this is the mechanism of the useful chemical analysis tool we used in this experiment.
As what explained above, when infrared light encounter with molecules, it causes vibration to the molecules’ bonds. The vibration of two atoms joined together by a chemical bond can be likened to the vibrations of two balls joined by a spring. See fig 6. When electromagnetic radiation passes through a polar bond, It will stretch or compress the bond( repel and attract). If the bond is non-polar, there will no repel or attract occurs. This is why IR spectra is not suitable for diatomic molecules like O=O. As infrared spectroscopy is an established analyzing tools, it has lots of advantages. It’s high sensitive and universal. Non destructive. The informations it could get are rich. It really fast and easy control. And compare with other analyzing tools like XRD, it’s relative inexpensive. Despite it got lots of advantages,some limitation should be mentioned. It cannot detect atoms or monatomic ions. As they contain no chemical bonds possess vibrational motion, hence do not absorb infrared radiation. The noble gases such as He, Ar can not be detected as they exist as individual molecules. Another class of substance that do not absorb infrared radiation is diatomic molecules, like what I mentioned above. If the substance we measured was too complex, likes some complex mixture. There is a limitation for infrared spectroscopy. Because these samples give rise to complex spectra, which has lots of bands. It’s hard to interpret which bands are from which molecules in a sample. Aqueous solution are also difficult to analyze using infrared spectroscopy. Because there are a lot of water molecules, the Van De Waals bonds between could absorb large amount of infrared radiation. Background spectrum is always produced before sample spectrum. The background spectrum obtained during the experiment fluctuates a lot. This might because of noise, or the infrared beam hit particles in the air and caused a loss in intensity. When the spectra of sample are compared with library data using the software, the overlay of spectra is never 100%. This is because there are always some experimental errors present, as mentioned before; these could be noise, contaminations, aging of sample and so on.
Another definition defined as overtone should be discussed here as it’s quite crucial for deducing sample form these figure above. An overtone transition is a state where a molecule is exited form fundamental vibrational energy level v=0 to higher level v=2 ,3,4 . The same molecule can rises to several bands in the spectra due to overtone( see fig 7). Normally , an overtone band is very weak in comparison with a fundamental , because these transitions are not allowed.
Conclusion According to the graphs and the deduction. Sample 5 is polyethylene or polypropylene with repeating unit. Sample 4 is polystyrene with repeating unit. Sample 3 is PMMA. Sample 2 is polyethylene or polypropylene. The material of the ruler deduced from the graph is polystyrene.
References ➊ Organic Spectroscopy by William Kemp. ➋ Identification and analysis of plastic by J Haslam And H A Willis ➌ Organic Structural Spectroscopy by Joseph B.Lambert ,Herbert F.Shurvell and David A. Lightner ➍Material science Engineering by Willam D. Callister ➎The science and engineering of materials by Donald R. Askeland