Infrared Spectroscopy: How to use the 5 zone approach to identify functional groups

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1 Infrared Spectroscopy: How to use the 5 zone approach to identify functional groups Definition: Infrared Spectroscopy is the study of the Infrared Spectrum. An Infrared Spectrum is the plot of photon energy (x axis) versus the amount of photons (y axis) X axis: the stretching frequency (wavelength) Y axis: the amount of photons absorbed (% Transmittance) An example of an Infrared Spectrum: The Infrared Spectrum is divided into 5 Zones Zone 1 o cm -1 o Alcohol O-H o Amide/Amine N-H o Terminal Alkyne C-H Zone 2 o cm -1 o Alkyl C-H (peak < 3000 cm -1 ) o Aryl or vinyl C-H (peak > 3000 cm -1 ) o Aldehyde C-H o Carboxylic Acid O-H Zone 3 o cm -1 o Alkyne C C o Nitrile C N

2 Zone 4 o cm -1 o Carbonyl functional groups Zone 5 o cm -1 o Alkene C=C o Benzene Each functional group located in the infrared spectrum has distinct characteristics Alcohol and Amine o Because of hydrogen bonding the O-H and the N-H peaks are broader Carboxylic Acid o There needs to be two peaks One O-H peak located in zone 2 A C=O stretch located in zone 4 Aldehyde o There needs to be two sp2 C-H peaks that are between approximately 2900 and 2700 cm in zone 2 o A C=O peak in zone 4 Alkyne o This peak is found in molecules when there are different atoms on either side of the triple bond (molecule is unsymmetrical.) Carbonyl o It is usually the most intense peak in the spectrum Benzene ring o There must be a peak at around 1600 cm-1 and another at 1500 cm-1 o An sp2 C-H peak should be located in zone 2 because many benzene rings have at east one hydrogen atom attached to it. *** When analyzing the infrared spectrum we must always list the functional group and bond***

3 Example #1 Ethanol Step 1: Calculate DBE DBE = C (H/2) + (N/2) + 1 = 2 (6/2) + (0/2) + 1 = = 0 No pi bonds or rings Step 2: We now need to analyze the spectrum. The spectrum above is already broken into color coated according to the different zones. Zone 1: ( cm -1 ) Alcohol O-H: Present Amine or Amide: Absent no N in formula Terminal alkyne C-H: Absent not enough DBE, no peak ~2200 cm -1 Zone 2: ( cm -1 ) Aryl/vinyl sp 2 : Absent not enough DBE no peak > 3000 cm -1 Alkyl sp 3 : Present (peak < 3000 cm -1 ) Aldehyde C-H: Absent no 2700 cm -1 Carboxylic acid O-H: Absent not broad enough (There needs to be two peaks a broad O-H peak in zone 2 and a C = O peak in zone 4)

4 Zone 3 ( cm -1 ) Alkyne C C: Absent no peaks not enough DBE Nitrile C N: Absent no peaks not enough DBE no N in formula Zone 4 ( cm -1 ) C=O: Absent no peaks Zone 5 ( cm -1 ) Benzene ring: Absent no peak ~1600 cm -1 not enough DBE Alkene C=C: Absent no peaks ~1600 cm -1 not enough DBE (C=C plus C=O) Example #2 Ethyl Ethanoate (C 4 H 8 O 2 ) Step 1: Calculate the DBE DBE = C (H/2) + (N/2) + 1 = 4 (8/2) + (0/2) + 1 = = 1 1 pi bond or one ring Step 1:

5 Step 2: We now need to analyze the spectrum. The spectrum above is already broken into color coated according to the different zones. Zone 1: ( cm -1 ) A lcohol O-H: Absent Amine or amide N-H: Absent, there is no Nitrogen in the formula Terminal alkyne C-H: Absent, there is not enough DBE and no peak at ~ 2200 cm -1 Zone 2: ( cm -1 ) Aryl/Vinyl sp 2 C-H: Absent not enough DBE no peaks > 3000 cm -1 Alkyl sp 3 C-H: Present peaks < 3000 cm -1 Aldehyde C-H: Absent no peak at 2700 cm -1 (even though there is a peak at 2900 cm -1, it needs to have both in order for there to be an aldehyde) Carboxylic Acid O-H: Absent not broad enough (There needs to be two peaks a broad O-H peak in zone 2 and a C = O peak in zone 4) Zone 3 ( cm -1 ) Alkyne C C: Absent no peaks not enough DBE (requires 2 DBE s and we only have 1 DBE in this example) Nitrile C N: Absent no peaks not enough DBE (requires 2 DBE s and we only have 1 DBE in this example (same reasoning as the Alkyne C C)) Zone 4 ( cm -1 ) C=O: Present -- peak at 1700 cm-1 The possible carbonyl groups that it can be are Ketone: No, because we have 2 oxygens in our formula and a ketone only has 1 oxygen. Ester: Possible because we have 2 oxygens in our formula, and there is a peak at about 1740 cm -1 and we have a DBE of 1. Aldehyde: Absent because there is no peak at 2700 cm -1 Carboxylic acid: Absent because there is no broad peak in zone 2 Amide: Absent because there are no N in our formula Zone 5 ( cm -1 ) Benzene ring: Absent no peak at ~1600 cm -1 not enough DBE Alkene C=C: Absent no peak at ~1600 cm -1 not enough DBE

6 Example #3 1-aminobutane (CH 3 (CH2) 3 NH 2 ) Step 1: Calculate DBE DBE = C (H/2) + (N/2) + 1 = 4 (11/2) + (1/2) + 1 = = 1 1 ring or one pi bond Step 2: We now need to analyze the spectrum. The spectrum above is already broken into color coated according to the different zones. Zone 1: ( cm -1 ) Alcohol O-H: Absent not broad enough Amine or Amide: Present Terminal alkyne C-H: Absent not enough DBE, no peak ~2200 cm -1 Zone 2: ( cm -1 ) Aryl/vinyl sp 2 : Absent not enough DBE no peak > 3000 cm -1 Alkyl sp 3 : Present (peak < 3000 cm -1 ) Aldehyde C-H: Absent no 2700 cm -1 Carboxylic acid O-H: Absent not broad enough (There needs to be two peaks a broad O-H peak in zone 2 and a C = O peak in zone 4)

7 Zone 3 ( cm -1 ) Alkyne C C: Absent no peaks not enough DBE Nitrile C N: Absent no peaks not enough DBE Zone 4 ( cm -1 ) C=O: Absent no peaks Zone 5 ( cm -1 ) Benzene ring: Absent no peak ~1600 cm -1 not enough DBE Alkene C=C: Absent no peaks ~1600 cm -1 not enough DBE (C=C plus C=O) Works Cited: Dr. Hardinger s Lecture Supplement and Think book m

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