The Infrared One-Two

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Rafe Bailey
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1 The Infrared One-Two Introduction: What is IR Spec and Why use it? Infrared spectroscopy is built on the principle that absorption of photons causes changes in molecular vibrations. Molecular vibrations are influenced by bond dipoles, as higher bond lengths and higher electronegativity differences yields a higher bond dipole, which causes an increase in absorption. But why do we need to know all of this technical jargon? The fact of the matter is that infrared spectroscopy allows us to determine what functional groups are present in a molecule. Thus, by analyzing the spectrum produced by infrared spectroscopy, we can go beyond a known formula and provide ourselves with more knowledge about the actual connectivity of functional groups in an atom. Combined with Proton NMR, IR spectroscopy allows us to put together the structure of a molecule that we are presented with or that can be deciphered from mass spectrometry techniques. However, our focus in this tutorial is to breakdown the different zones present in the IR spectrum to make it easier to read these graphs. With practice, it will become second nature in recognizing what is going on in this funky looking graph. Learning the Zones & What to Look For In infrared spectroscopy, the spectrum is organized with stretching frequency on the x- axis and transmittance on the y-axis. Stretching frequency, defined as the energy of photons absorbed to cause molecular vibrations, range from a frequency of 0 cm -1 to 4000 cm -1. It may seem weird that frequency is in centimeters rather than Hz, but that's just how it works. Note that transmittance %, or how strongly light was absorbed has stronger peaks near 0 on the y axis as opposed to 100. This is done because 100% transmittance actually means that no absorption occurs at that frequency. Do not get caught up in the uncanny convention; just understand that this is how the IR spectrum is to be read.

2 The spectrum itself is divided into 5 zones and an area known as the fingerprint region. Each of these zones have been outlined below and their characteristic stretching frequencies are very telling in what functional groups may be present in that zone. While the stretching frequencies and peaks can be memorized, it is best to commit them to memory through practice problems and not pure remembering. This way, it just becomes second nature when given an IR spectrum to analyze.

3 Zone 1 ( cm -1 ) Alcohol O-H Strong & Broad Broad b/c of H bonding Alkyne C H Strong & Sharp also must have peak in zone 3 Amine or amide N- H Medium/Broad Broad b/c of H bonding Zone 2 ( cm -1 ) Aryl or vinyl sp 2 C Varies n/a H Alkyl sp 3 C-H Varies n/a Aldehyde C-H ~2900, ~2700 Medium; two peaks Two sets of stretching in zone 2, 1 in zone 4 Carboxylic acid O- H Strong, broad Must have 2 bands: broad O-H in zone 2, C=O in zone 4 *Some people find it helpful to draw a vertical line at the 3000 cm -1 mark to identify the peaks; >3000 cm -1 shows the presence of an aryl or vinyl, while <3000 cm -1 is an alkyl Zone 3 ( cm -1 ) Alkyne C C Variable, sharp When symmetric, peak not observed Nitrile C N Variable, sharp n/a Zone 4 ( cm -1 ) Stretching Intensity/Shape Special Notes Ketone C=O Strong ^ Ester C=O Strong ^ Aldehyde C=O Strong ^ Carboxylic Acid Strong ^ C=O Amide C=O Strong ^ *Carbonyl C=O frequencies cm -1 lower when conjugated to a pi bond

4 ^almost always the most intense peak in the entire spectrum Zone 5 ( cm -1 ) Alkene C=C ~1600 Varies n/a Benzene C=C ~ Varies; often 2 peaks , 1500; often C-H peak in zone 2 Fingerprint Region (below 1450 cm -1 ) Often contains many absorptions Ignored because it is often too difficult to make precise assignments Hard to differentiate if a functional group is absent or present However, same fingerprint for same compound Therefore useful for identifying an unknown compound by matching fingerprint to known compound IR Two Step Procedure With the basic functional groups outlined, we will now follow a set procedure to analyze the IR spectrum in an organized manner. This systematic way will allow us to perform the steps without giving it too much thought. The two steps are: 1. Calculate the DBE from the given formula DBE = C - H/2 + N/2 + 1 This will tell us how many pi bonds and/or rings are present in the molecule, which gives us a way to check if we came up with the right functional groups May have to obtain formula from mass spec procedure first 2. Analyze the IR Spectrum by going through the zones. Eliminate any functional groups that contain an element not present in the molecular formula Do not use more DBE than you have If it helps, draw lines to separate the zones Always list the characteristic stretching frequencies of the five zones and their structure next to it. Don't just simply say "There is nothing in zone 1";

5 Example list the functional group, and say that is absent/why (no peak, no oxygen in formula, not enough DBE, etc). Step One: Calculate DBE DBE = C - H/2 + N/2 + 1 = = 1 pi bond or ring Step Two: Analyze the IR Spectrum by going through the zones: Zone 1 ( cm -1 ) Alcohol O-H?...Absent; No broad peak ~3500 cm -1 Alkyne C H?...Absent; No peak ~3300 cm -1 ; not enough DBE Amide/Amine N-H?...Absent, No N in formula, no peak Zone 2 ( cm -1 ) Aryl or vinyl C-H...Absent; no peak >3000 cm -1 ; not enough DBE Alkyl C-H...Present Aldehyde C-H...Absent; no peak ~2700 cm -1 (both 2900/2700 peaks required) Carboxylic Acid O-H...Absent; peak not broad enough Zone 3 ( cm -1 )

6 Alkyne C C...Absent; no peak, not enough DBE Nitrile C N...Absent; no peak, not enough DBE, no N in formula Zone 4 ( cm -1 ) Carbonyl C=O...PRESENT at 1718 cm -1 Ketone?...Present Ester?...Not in stretching frequency Aldehyde?...No; 2700 cm -1 peak in zone 2 ABSENT Carboxylic Acid?...No; peak in zone 2 ABSENT Amide?...No; no N in given formula; out of stretching frequency Zone 5 ( cm -1 ) Benzene Ring C=C...Absent; no peak, not enough DBE Alkene C=C...Absent; no peak, not enough DBE After all of this is done, we are left with sp 3 alkyl C-H from zone 2 and a carbonyl (probably a ketone) C=O from zone 4. Our IR spectrum has given us this insight that will help us put together the structure. For this one, we can predict that it is 2-heptanone because it is a relatively simple molecule (C 7 H 14 O). However, for other molecules, we will have to rely on proton NMR to give us more insight on equivalency to come up with a reliable structure. In summary, IR spectroscopy is as hard as you make it. The only way to learn this type of material is to PRACTICE, PRACTICE, PRACTICE. For more examples, go through the problems listed here, and make sure you do the OWLS and the Practice Problems. If something is still hazy, ask for clarification on the forums. Happy studying!

7 Works Cited 1. Dr. Hardinger's Chemistry 14C Lecture Supplement & Thinkbook (9th Ed.) logical_emphasis/chapter 4%3A_Structure_Determination_I/Section_4.2%3A Infrared_spectroscopy gif

Symmetric Stretch: allows molecule to move through space

Symmetric Stretch: allows molecule to move through space BACKGROUND INFORMATION Infrared Spectroscopy Before introducing the subject of IR spectroscopy, we must first review some aspects of the electromagnetic spectrum. The electromagnetic spectrum is composed