Infrared Spectroscopy used to analyze the presence of functional groups (bond types) in organic molecules How IR spectroscopy works:
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1 Infrared Spectroscopy used to analyze the presence of functional groups (bond types) in organic molecules It is the study of the interaction of infrared energy with organic molecules; the process analyzes molecules for molecular bond vibrations, measuring how fast bonds vibrate. How IR spectroscopy works: 1. A sample is placed in the infrared instrument ( IR ) and allowed to interact with infrared energy. When the molecule has a bond vibration that matches that of the infrared energy, the molecule absorbs that energy, preventing all of the energy from just passing through the molecule. 2. When the energy is absorbed, the instrument records that energy as not being entirely transmitted through a molecule. The energy that is absorbed has a specific frequency, thus identifying the (same) frequency of vibration of the bond in the molecule. This process works only because every type of bond we commonly look at in organic molecules has a different vibrational frequency. So the FREQUENCY VALUE tells you what TYPE F BND is vibrating. 3. When the analysis is complete, every frequency (and therefore bond type) has been noted on the spectra. 4. With every frequency identified, every bond type is identified, and functional groups can then also be identified.
2 When the sample is placed in the instrument and the instrument scans a range of frequencies, looking to match a frequency with a frequency of vibration within the molecule, two things could happen: When nothing matches, 100% of the energy just passes through the molecule (100% photon energy is transmitted) When the frequency of the energy from the instrument matches that of the frequency of the vibration of the bond, the molecule absorbs the energy (less than 100% is transmitted). The IR shows the range of frequencies of energy (X-axis) and how much energy is passing through (Y-axis). The peaks are those shown at frequencies when less than 100% energy is being transmitted (upside down peaks!). 2
3 Y Axis: Percent Transmittance: Amount of energy that passes through the molecule If no bonds are vibrating at frequencies of energy, 100% energy photon transmittance occurs If bonds are vibrating at same frequency, energy is absorbed, less than 100% is transmitted. X Axis: Wavenumbers: the frequency of infrared energy, expressed as reciprocal wavelength in centimeters (1/cm or cm -1 ) There are TW Factors that affect the FREQUENCY of vibration (X axis): 1. Size of the atoms attached to the bond. Larger atoms vibrate slower, at a lower frequency. Smaller atoms vibrate faster, higher frequency. C-H versus C-Br 2. Bond length (the strength of the bond). Shorter stronger bonds vibrate faster, at a higher frequency. Longer, weaker bonds vibrate slower, lower frequency. versus C C C C sp C-H versus sp 2 C-H versus 3
4 The INTENSITY (Size) of the Peak (Y axis): 1. As far as molecule structure is concerned, the intensity of the peak (strong, medium, weak) is affected by the extent that the vibrational process changes the DIPLE of the molecule. When the molecule absorbs infrared energy that matches the vibrational frequency, the bond that is affected undergoes significant bond stretching and compressing to dissipate the excess energy. The greater the stretching effect has on the dipole moment while vibration occurs, the larger the peak will appear on the IR spectrum. Non-polar: C C When stretched: C C Polar: non-polar no change to dipole C When stretched: C polar Large change to dipole moment Polar covalent bonds will often be the largest (strongest) peaks on the IR spectrum. 2. The intensity of a peak is also affected by the number of any specific type of bond in a molecule. The more bonds of the same type will result in a larger number of data points acquired at a specific frequency, which will result in 4
5 the appearance of a larger peak than one would normally expect. This explains why bonds of alkyl groups can produce large peaks, even though they are non-polar. 3. Also remember that using too much or too little sample changes the amount of material available during acquisition of data on your molecule. Too much sample results in an overly large amount of data for EVERY bond, resulting in a spectrum where every peak looks strong (big). Too little sample will produce a spectrum where every peak appears weak (small) and nothing looks significant. Sample Infrared Spectrum: Analysis of an IR Spectrum: The Infrared Spectrum is divided into 6 Zones: Zone 1: cm -1 5
6 o Alcohol -H o Amine or Amide N-H o Terminal Alkyne sp C-H (CΞC-H) Zone 2: cm -1 o Aryl or vinyl sp 2 C-H (>3000 cm -1 ) o Alkyl (<3000 cm -1 ) o Aldehyde sp 2 C-H o ALS: Carboxylic Acid -H Zone 3: cm -1 o Alkyne CΞC o Nitrile CΞN Zone 4: cm -1 o Carbonyl (C=) functional groups Zone 5: cm -1 o Alkene C=C o Aromatic ring pseudo double bonds Zone 6: the Fingerprint Region cm -1 o Alcohol, Ester, Carboxylic Acid, Ether C- ( cm -1) o Amine, Amide C-N o Aromatic substitution patterns ( cm -1 ) 6
7 1. Alkanes do not have a functional group but they do contain ( cm -1 ) as do alkyl groups so these are often very prominent peaks on an IR spectrum. 3-methylpentane Each functional group in the infrared region has distinct characteristics with bonds that you need to recognize and identify Alcohols, Phenols, Amines all of these functional groups are capable of hydrogen bonding. Due to the intermolecular forces, the bond lengths for -H/N-H vary in length resulting in broad peaks. Zone 1 ( cm -1 ) H H N H NH 2 7
8 2. Alcohols: U shaped peak, Zone 1 (-H, cm -1 ) 4-methyl-2-pentanol H -H C- Look what happens to the U shaped peak when the alcohol is too sterically hindered to hydrogen bond to another molecule: H -H, NT hydrogen bonding 8
9 3. Amines: 1º W shaped peak, Zone 1 (2 N-H, cm -1 ) butylamine N H H NH 2 4. Amines: 2º V shaped peak, Zone 1 (N-H, cm -1 ) diethylamine N-H N H 9
10 5. Amines: 3º no visible peak in Zone 1 (no N-H, cm -1 ) N,N-dimethylethylamine N Note: Amides will also have W and V peaks for NH 2 and NH, respectively, as necessary. 6. Ethers: C- peak, Zone 6 ( cm -1 ) C- 10
11 7. Alkyl Halides - C-X in Zone 6 o C-Cl cm -1 o C-Br cm -1 2-chloro-2-methylpropane Cl C-Cl 8. Alkenes o C=C in Zone 5 ( cm -1 ) o Sp 2 C-H in Zone 2, unless tetrasubstituted ( cm -1 ) Each of the following contains an alkene, except for the aromatic ring, which has a different pi bond system. 11
12 2-methyl-1-pentene sp 2 C-H C=C Note how this disubstituted alkene shows a medium sized C=C peak, even though it is a non-polar bond. With two alkyl groups on the same carbon of the alkene, the electron cloud of this double bond is electronically lopsided, being electron-rich on one side (as drawn). Take a look at an alkene with a more electronically symmetrical double bond. The C=C peak is smaller between cm -1 due to more even distribution of electron density, caused by having two alkyl groups attached, one on each end of the alkene: 12
13 H H An even more electronically symmetrical alkene can t even see the peak for C=C between cm -1 : H H 9. Alkynes o CΞC in Zone 3 (visible in asymmetric alkynes) ( cm -1 ) Terminal Internal 13
14 o Sp C-H in Zone 2 (terminal alkyne only; 3300 cm -1 ) 1-hexyne C C spc-h spc-h Like the alkenes, the more asymmetrical the electron distribution of an alkyne, the more obvious the CΞC peak. This makes the terminal alkyne typically the easiest CΞC peak to view on a spectrum. With just one methyl group on one end of an alkyne, see the size difference in the alkyne peak for 6-methyl-2- heptyne (Tiny CΞC peak between cm -1 ): 14
15 Now see what having an ethyl group does: The peak for CΞC peak is pretty much non-existent between cm -1 : 15
16 10. Nitriles o CΞN in Zone 3 (more intense than CΞC) N C N hexanenitrile N C N Carbonyl-Containing Functional Groups Ketones, aldehydes, carboxylic acids, esters, acid anhydrides, acid halides o The carbonyl group, C=, is typically the most intense (largest, strongest) peak on an IR spectrum o Value shifts cm -1 LWER when next to aromatic ring due to resonance causing a lengthening of the C= bond 16
17 11. Ketones C= only ( cm -1 ) 4-methyl-2-pentanone C= 12. Aldehydes o Two major types of bonds: H H C= in Zone 3 ( cm -1 ) Sp 2 C-H fangs in Zone 2 ( and cm -1 ) 17
18 H sp 2 C-H C= 13. Carboxylic Acids o Three major types of bonds: Extremely broad H ( cm -1 ) C= in Zone 3 ( cm -1 ) C- in Zone 6 ( cm -1 ) H H Note that the spectrum below has VERLAPPING PEAKS the peak for the H and the peaks for the stretches are overlapped to form a very odd shaped peak. [The pink box contains the portion from the H bond.] 18
19 -H C- H C= It is not unusual to have overlapping peaks on an IR. Another example: The aldehyde peak between cm -1 is blending in with the, making it harder to distinguish this as containing an aldehyde. Always easiest if you focus on finding the peak between cm -1. sp 2 C-H (visible) sp 2 C-H (buried by peaks from ) H 19
20 In this example the alcohol peak is oddly pointed due to the overlap of the H peak with the sp C-H of the terminal alkyne: -H H spc-h 14. Esters o C= in Zone 3 ( cm -1 ) o C- in Zone 6 ( cm -1 ) 20
21 C= C- 15. Acid Anhydrides o Two C= in Zone 3 ( ; cm -1 ) o C- in Zone 6 ( cm -1 ) propionic anhydride 2 C= C- 21
22 16. Amides o C= in Zone 3 ( cm -1 ) o NH peak(s) in Zone 1 ( V or W shaped) ( cm -1 ) NH 2 N H N pentanamide NH 2 2N-H C= 17. Acid Halides o C= in Zone 3 ( cm -1 ) o C-X in Zone 6 Cl Br 22
23 acetyl chloride Cl C= C-Cl 18. Aromatic Rings Monosubstituted 1,2-disubstituted 1,3-disubstituted 1,4-disubstituted ortho meta para o Pseudo double bond peaks one must appear ~1600 cm -1 and 1-2 more ~ cm -1 (Zone 5) o Sp 2 C-H peak (Zone 2, cm -1 ) o Substitution patterns for monosubstituted and disubstituted aromatic rings (Zone 6) Mono Substituted (Zone 6: , cm -1 ) 23
24 sp 2 C-H pseudo C=C monosubstituted rtho Substituted (Zone 6: cm -1 ) sp 2 C-H pseudo C=C rtho 24
25 Meta Substituted (Zone 6: , cm -1 ) sp 2 C-H pseudo C=C Meta Para Substituted (Zone 6: cm -1 ) sp 2 C-H pseudo C=C Para 19. Nitro (-N 2 ) Spans Zones 5 and 6 o N=/N- ( and cm -1 ) 25
26 N N N 2 N=/N- N=/N- So How do you SLVE an IR? You need to break the IR up into the basic Zones, moving from left to right and only delving into the Fingerprint Zone if need be (i.e. you think you have a functional group that applies). 26
27 Example 1: Zone 1: ( cm -1 ) Alcohol H present ( U ) (note the C- in Zone 6) Amine or Amide absent no V or W No sharp stretch at 3300 cm -1 Zone 2: ( cm -1 ) No Aryl/vinyl sp 2 C-H (>3000 cm -1 ) Sp 3 C-H present (<3000 cm -1 ) No aldehyde fangs between No broad H of carboxylic acid (no C= in Zone 4) Zone 3: ( cm -1 ) No CΞC or CΞN Zone 4: ( cm -1 ) No C= peak (must be longest, strongest peak) Zone 5: ( cm -1 ) No peak at 1600, etc no aromatic ring No peak no alkene Functional group? Alcohol 27
28 Example 2: Zone 1: ( cm -1 ) Alcohol H absent ( U ) Amine or Amide absent no V or W No spc-h - sharp stretch at 3300 cm -1 Zone 2: ( cm -1 ) No Aryl/vinyl sp 2 C-H (>3000 cm -1 ) Sp 3 C-H present (<3000 cm -1 ) No aldehyde fangs between No broad H of carboxylic acid (no C= in Zone 4) Zone 3: ( cm -1 ) No CΞC or CΞN Zone 4: ( cm -1 ) C= present (note the C- in Zone 6) Zone 5: ( cm -1 ) No peak at 1600, etc no aromatic ring No peak no alkene Functional group? Ester 28
29 Example 3: Zone 1: ( cm -1 ) Alcohol H absent ( U ) Amine or Amide absent no V or W No spc-h - sharp stretch at 3300 cm -1 Zone 2: ( cm -1 ) Aryl/vinyl sp 2 C-H present (>3000 cm -1 ) Sp 3 C-H present (<3000 cm -1 ) No aldehyde fangs between No broad H of carboxylic acid (no C= in Zone 4) Zone 3: ( cm -1 ) No CΞC or CΞN Zone 4: ( cm -1 ) No C= present Zone 5: ( cm -1 ) Peak at 1600, etc Aromatic Ring No peak no alkene Zone 6: ( cm -1 ) Peaks ~690, 750 monosubstituted aromatic ring Functional group? Monosubstituted aromatic ring 29
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