Infrared Spectral Interpretation - chromacademy.com

1 I Wherever you see this symbol, it is important to access the on-line course as there is interactive material that cannot be fully shown in this reference manual. Infrared Spectral Interpretation 1 Contents Page 1 Infrared Spectral Quality 2-3 2 Orbital Hybridization sp, sp2, and sp3 Carbon 4-6 sp3 Hybridization 4-5 sp2 Hybridization 5-6 sp Hybridization 6 3 General Infrared Interpretation Concepts 7-8 4 IR Frequencies 9 5 Alkanes 10-11 6 Alkenes 12-13 7 Alkynes 14 8 Aromatic Compounds 15-17 9 Alcohols 18-19 10 The OH Group and Hydrogen Bonding 20 11 Ethers 21-22 12 Amines 23-24 13 Carbonyl Frequencies 25-26 14 Aldehydes 27 15 Ketones 28-29 16 Carboxylic Acids 30-31 17 Esters 32-33 18 Amides 34-35 19 Anhydrides 36 20 Acid Chlorides 37 21 Fermi Resonance 38-39 22 Stretching Frequencies 40-44 Carbonyls 40-41 Nitrogen 42 Sulfur 43 Phosphorus, Boron.

2 And Halogen 44 Inorganic 44 References 45 2 1. Infrared Spectral Quality Infrared Interpretation must be performed on a high quality spectrum; otherwise, misleading results can be obtained. More often than not, bad sample preparation is the main cause of poor quality IR spectra. A high quality Infrared spectrum must possess a flat (level) baseline (positioned near or around 100% transmittance and with a low level of noise). Shifting of the baseline position would indicate at least one of the following conditions: The background spectrum does not correspond to the sample spectrum ( different solvent). The aperture size (where appropriate) for the background and sample spectra do not correspond. The sample strongly absorbs. The sample solvent absorbs strongly when obtaining spectra in the solution phase.

3 The NaCl or KBr discs are cloudy. A sloping baseline usually indicates the electromagnetic radiation has been diffracted or scattered as it interacts with the sample. This may happen when the particles in the KBr pellet are not ground properly or if the sample surface of a thin film is not homogeneous (air bubbles etc.). Additional bands should be avoided, in particular, carbon dioxide and water which can absorb Infrared radiation, therefore, their presence should be minimized (usually below 2% transmittance, Figure 1). This can be achieved by ensuring samples are made with dry solvents (solution IR), dry KBr (solid state IR), and by taking a background spectrum to account for atmospheric water and carbon dioxide within the instrument. Figure 1: IR background spectrum.

4 Dispersive IR instruments are dual beam instruments with equivalent beams passing independently through both the sample and reference chambers. During analysis the sample and reference beams are alternately focused on the detector by an optical chopper, allowing for comparison and subtraction of the sample and reference spectra. Therefore, the reference chamber should be filled with the same matrix as the sample to account for any IR active components (solvent, water, CO2). FTIR instruments are single beam instruments, therefore, before a sample spectrum is obtained a spectrum of the sample matrix is acquired which can be subtracted from the sample spectrum to 3 account for any IR active solvents or dissolved gases within the sample matrix which may obscure analytically relevant peaks in the sample spectrum.

5 Remember: Water produces small, sharp absorption bands in the regions from 4000-3000 and 1800-1600 cm-1. Carbon dioxide produces a strong doublet near 2340 cm-1. 4 2. Orbital Hybridization sp, sp2, and sp3 Carbon Hybridization is used to explain molecular structures and describes the various orbital types which are involved in the bonding between atoms. In infra-red analysis, the nature of the bonding can make a big difference to the region of the spectrum in which the signal appears, and as such, this brief refresher is intended to remind you of the concept of Orbital Hybridization which can be used to help interpret Infrared spectra. Using carbon as an example it is known that for a tetrahedrally coordinated carbon ( methane CH4) the carbon atom will have four orbitals with the correct symmetry to bond to the four hydrogen atoms.

6 With respect to IR spectroscopy, the energy of the Infrared light absorbed by a C-H bond is dependent on the type of hybridization of the bonding orbitals. The C-H bond strengths are in the order sp3>sp2>sp, due to the increased s character of the hybrid orbital which results in better overlap with the hydrogen s-orbital. This results in the different IR stretching frequencies that are observed for sp3, sp2, and sp carbons. The orbitals that are used by carbon to form hybrid bonding orbitals are s- and p-orbitals (Figure 2). Figure 2: s- and p-orbitals. sp3 Hybridization The ground state configuration of carbon is 1s2 2s2 2px1 2py1. The p orbitals are equal in energy and said to be degenerate. The two singly occupied p orbitals can be utilized for bonding to give methylene CH2, an unstable free radical (Figure 3).

7 Excitation of an electron from the doubly occupied 2s orbital to the empty 2p orbital results in four singly occupied orbitals. Excitation of an electron from the 2s to the 2p orbital requires an input of energy; this is offset by the release of energy that is obtained by the formation of the two additional bonds, making this an energetically favored process. Quantum mechanics states that the lowest energy will be obtained if the four bonds are equivalent which requires that they are formed from equivalent orbitals on the carbon. Therefore, a set of four equivalent orbitals can be obtained via hybridization of the valence-shell (core orbitals that are not involved in bonding) s- and p-orbitals to form four sp3 orbitals, each consisting of 25% s character and 75% p character (Figure 3).

8 The hybrid orbitals are orientated at a bond angle of from each other, giving tetrahedral geometry as seen in methane (Figure 4). In methane the four sp3 hybrid orbitals will overlap with the hydrogen 1s orbital resulting in four covalent bonds ( bonds) which will be of equal length and have equal bond strengths. 5 Figure 3: Formation of sp3 hybrid orbitals. Figure 4: Molecular geometry arising from sp3 hybridization, for example in methane. sp2 Hybridization sp2 hybridization results in the formation of molecules with trigonal planar structures ( aluminium trihydride). The three sp2 hybridized orbitals are formed as follows (Figure 5): Figure 5: Formation of sp2 hybrid orbitals. Each of the three sp2 orbitals has 33% s character and 67% p character.

9 The orbitals are orientated to minimize electron repulsion giving bond angles of 120 . The p-orbital that is not used to form the hybrid orbitals remains unchanged and sits perpendicular to the plane of the three sp2 orbitals (Figure 6). In compounds such as alkenes there is a double bond between the carbon atoms, ethene C2H2. In the case of ethene, two of the sp2 hybridized orbitals (on each of the individual carbon atoms) are used to form bonds with the 1s orbitals on hydrogen. The remaining sp2 orbitals on each of the carbon atoms overlap to form a C-C -bond. The remaining two p-orbitals contain a single electron; overlap of these orbitals forms a -bond, creating a double bond between the two carbon atoms. The double bond results in ethene having linear geometry with the carbon atoms being trigonal planar (Figure 6).

10 6 Figure 6: Geometry and bonding resulting from sp2 hybridization. sp Hybridization The linear geometry of molecules such as alkynes can be explained by sp hybridization. The 2s orbital and one 2p orbital hybridize to form two sp hybrid orbitals, which will each have 50% s and 50% p character (Figure 7). These orbitals are aligned to minimize electron repulsion and give a bond angle of 180 , as seen in linear molecules (Figure 8). There are two remaining p-orbitals that contain a single electron that can be utilized by the molecule. As in sp2 hybridization these p-orbitals are orientated perpendicular to the two sp orbitals. Figure 7: Formation of sp hybrid orbitals. The singly occupied p-orbitals can be used in molecules such as ethyne to form two additional -bonds resulting in a triple bond.