sna spectroscopy interpreting ir spectra

What Does an IR Spectrum Look Like?

• A spectrum is a graph in which the amount of light absorbed is plotted on the y-axis and frequency is plotted on the x-axis. An example is shown below. You can run your finger along the graph and see whether any light of a particular frequency is absorbed; if so, you will see a "peak" at that frequency. If not, you will see "the baseline" at that frequency.

Figure IR1. IR spectrum of benzene. The x-axis labels are, from

right to left, 500, 1000, 1500, 2000, 3000 and 4000 cm-1.

In IR spectra (spectra = plural of spectrum):

• the y axis is usually labeled "transmittance". Transmittance is the amount of light that passes through the sample.

• the unit of transmittance is percent (%).• the x axis is labeled "wavenumbers". Wavenumbers are

proportional to frequency, so the higher the frequency, the higher the wavenumber.

• the symbol for wavenumber is reciprocal centimeters (cm-

1).• the x-axis is usually displayed with high wavenumber on

the left and lower wavenumber on the right.

• As you run your finger from left to right across an IR spectrum, you can see whether or not light is absorbed at particular frequencies. When the curve dips down, less light is transmitted. That means light is absorbed. The dip in the graph is called a peak. Different bonds absorb different frequencies of light, so the peaks tell you what kinds of bonds are present.

IR2. Hydrocarbon Spectra

• All organic and biological compounds contain carbon and hydrogen, usually with various other elements as well. Hydrocarbons are compounds containing only carbon and hydrogen, but no other types of atoms. Since all organic compounds contain carbon and hydrogen, looking at hydrocarbon spectra will tell us what peaks are due to the basic C&H part of these molecules. It is sometimes useful to think of the C&H part of a molecule as the basic skeleton or scaffolding used to construct the molecule. The other atoms often form more interesting and active features, like the doors, windows and lights on a building.

• The simplest hydrocarbons contain only single bonds between their carbons, and no double or triple bonds. These hydrocarbons are variously referred to as saturated hydrocarbons, paraffins or alkanes. Examples of alkanes include hexane and nonane. (You can take a look at the Glossary to see what these names tell you about the structure.)

Look at the IR spectrum of hexane. You should see:

• a set of peaks dipping down from the baseline at about 2900 cm-1.

• another set of peaks dipping down from the baseline at about 1400-1500 cm-1.

Figure IR2.IR spectrum of hexane.

• If you look at an IR spectrum of any other alkane, you will also see peaks at about 2900 and 1500 cm-1. The IR spectra of many organic compounds will show these peaks because the compound may contain paraffinic parts in addition to parts with other elements in them.

• These two kinds of peaks tell you that C-H bonds are present.• Specifically, the bonds involve sp3 or tetrahedral carbons.• Stretching C-H bonds in alkanes absorb light at around 2900

cm-1.• Bending H-C-H angles in alkanes absorb light at around 1500

cm-1.

IR3. Subtle Points of IR Spectroscopy

• Alkanes show two sets of peaks in the IR spectrum. Alkanes contain two kinds of bonds: C-C bonds and C-H bonds. However, these two facts are not related. The reasons are explained through bond polarity and molecular vibrations.

Bond polarity can play a role in IR spectroscopy

• nature rules that only bonds that contain dipoles can absorb infrared light.

• C-C bonds are usually nonpolar and usually do not show up as peaks in the IR spectrum.

• C-H bonds are not very polar and do not give rise to strong peaks in the IR spectrum.

• a whole lot of small C-H peaks can add up together to look like one big peak. This would happen if a molecule contained many C-H bonds (a common situation).

Molecular vibrations play a major role in IR spectroscopy.

• IR light interacts with vibrating bonds. When light is absorbed, the bond has a little more energy and vibrates at a higher frequency.

• a bond does not have an exact, fixed length; it can stretch and compress. This is called a bond stretching vibration.

• Stretching C-H bonds in alkanes absorb light at around 2900 cm-1.

• bond angles can also bend; for instance, the H-C-H bond angle can compress and stretch. This is called a bending vibration.

• Bending H-C-H angles in alkanes absorb light at around 1500 cm-1.

• The factors that govern what bonds (and what vibrations) show up at what frequencies are easily handled by computational chemistry software. In fact, prediction of absorption frequencies in IR spectra can be done using 17th century classical mechanics, specifically Hooke's Law (devised to explain the vibrational frequencies of springs). Computation is not the focus of this chapter but it may help you keep track of what kinds of vibrations absorb at what frequencies.

• Hooke's Law states:• the vibrational frequency is proportional to

the strength of the spring; the stronger the spring, the higher the frequency.

• the vibrational frequency is inversely proportional to the masses at the ends of the spring; the lighter the weights, the higher the frequency.

• Remember, there are two factors here, so you won't be able to make predictions knowing only one factor. Some strong bonds may not absorb at high frequency because they are between heavy atoms. The information is presented mostly to help you organize what bonds absorb at what general frequencies after you have learned about them.

• The reasons explaining why C-H bending vibrations are at lower frequency than C-H stretching vibrations are also related to Hooke's Law. An H-C-H bending vibration involves three atoms, not just two, so the mass involved is greater than in a C-H stretch. That means lower frequency. Also, it turns out that the "stiffness" of a bond angle (analogous to the strength of a spring) is less than the "stiffness" of a bond length; the angle has a little more latitude to change than does the length. Both factors lead to a lower bending frequency.

Problem IR.1

For each of the following pairs, identify which bond would show up at a higher wavenumber in the IR spectrum:• a) C-H or C-O b) C-O or C=O c) C=N or C=N• d) N-H or N-O e) a covalent O-H bond or a

hydrogen bond in water

IR4. Carbon Carbon Multiple Bonds

• Unsaturated hydrocarbons contain only carbon and hydrogen, but also have some multiple bonds between carbons. One type of unsaturated hydrocarbon is an olefin, also known as an alkene. Alkenes contain double bonds between carbons. One example of an alkene is 1-heptene. It looks similar to hexane, except for the double bond from the first carbon to the second.

Look at the IR spectrum of 1-heptene. You should see:

• a set of peaks dipping down from the baseline at about 2900 cm-1.

• another set of peaks dipping down from the baseline at about 1500 cm-1.

Figure IR3. IR spectrum of 1-heptene.

• So far, these peaks are the same as the ones seen for hexane. We can assign them as the C-H stretching and bending frequencies, respectively.

• Looking further, you will also see:• a small peak around 3100 cm-1.• a small peak near 1650 cm-1.• medium peaks near 800 and 1000 cm-1.

• The peak at 3100 cm-1 hardly seems different from the C-H stretch seen before. It is also a C-H stretch, but from a different type of carbon. This stretch involves the sp2 or trigonal planar carbon of the double bond, whereas the peak at 2900 involves an sp3 or tetrahedral carbon.

• The peak at 1650 cm-1 can be identified via computational methods as arising from a carbon-carbon double bond stretch. It is a weak stretch because this bond is not very polar. Sometimes it is obscured by other, larger peaks.

• The larger peaks near 800 and 1000 cm-1 are bending vibrations. They are due to a C=C-H bond angle that bends out of the plane of the double bond (remember that the carbons on either end of the double bond are trigonal planar). They are called oop bends. Oop bends are often prominent in alkenes and are easier to spot than an sp2 C-H stretching mode or a C=C stretching mode.

Problem IR.2.

• Given this information about the infrared spectra of alkenes, which bond do you think is stronger, an sp2 or an sp3 C-H bond?

Problem IR.3.

• What do you think would happen to the peak due to carbon-carbon double bond stretching if an electronegative atom were nearby in the molecule?

Problem IR.4.

• Oop bends can be diagnostic of the position and geometry of double bonds.

• Compare the oop bending modes or peaks seen in 1-heptene to those in Z-2-octene, aka cis-2-octene (in Z-2-octene, the double bond adopts a curled shape with alkyl substituents coming from the same side of the double bond).

Figure IR4. IR spectrum of Z-2-octene.

• 2. Also compare it to E-2-octene, aka trans-2-octene (in which the double bond has a zig-zag shape, with alkyl substituents coming from opposite sides).

Figure IR5. IR spectrum of E-2octene.

• c) Make predictions about the oop bending modes in 1-octene, Z-2-hexene and E-2-hexene.

Problem IR.5.

• Why do you think an sp3 CH2 bending mode occurs around 1500 cm-1 but a C=CH oop bending mode occurs around 800-1000 cm-1 ?

Problem IR.6.

• In the IR spectrum of 1-octyne, new peaks appear at 3300 and 2100 cm-1.

Figure IR6. IR spectrum of octyne.

• By comparison with the other hydrocarbons, can you identify the peak at 3300 cm-1?

• The peak at 2100 cm-1 is due to a carbon-carbon bond. Which one? Compare this peak to the one seen from a carbon-carbon bond in 1-hexene and explain the differences