# Organic Chemistry I & II

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### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

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#### $301 Hardcover print text only ### Always up-to-date content, constantly revised by community of professors Content meets standard for Introduction to Anatomy & Physiology course, and is updated with the latest content ### In-Book Interactivity Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost ### Customizable Ability to revise, adjust and adapt content to meet needs of course and instructor ### All-in-one Platform Access to additional questions, test banks, and slides available within one platform ## Pricing Average price of textbook across most common format ### Top Hat Steven Forsey, “Organic Chemistry”, Only one edition needed #### Up to40-60%more affordable Lifetime access on any device ### McGraw-Hill Carey & Giuliano, “Organic Chemistry”, 10th Edition ####$219

Hardcover print text only

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

#### $301 Hardcover print text only ### Wiley David R. Klein, “Organic Chemistry”, 3rd Edition ####$301

Hardcover print text only

## Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## All-in-one Platform

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

#### Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

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# Chapter 16: Infra-red Spectroscopy and Mass Spectrometry

## Learning Objectives

• Develop a basic understanding of the chemistry and physics involved in the measurement of infrared spectra.
• Analyze infrared spectra to identify functional groups present in a molecule.
• Use infrared spectroscopy to determine structures of unknown molecules.
• Develop a basic understanding of the principles of mass spectrometry.
• Use mass spectrometry to determine molecular formulas.
• Analyze mass spectra to determine structures of unknown molecules.
• Use both infrared spectroscopy and mass spectrometry to determine molecular structure.

## 16.1 Spectroscopy

Electromagnetic radiation consists of mutually perpendicular, oscillating electric and magnetic fields. The measurement of the interaction of electromagnetic radiation with matter is called spectroscopy.

This is generally done by measuring the intensity of radiation before and after passing through a material of interest. The electromagnetic spectrum encompasses a continuum of radiation with an enormous range of energies, from extremely high energy cosmic rays to the very low energy cosmic background echo of the Big Bang, as shown in Figure 16.2.

The energy of a given frequency of electromagnetic radiation is given by Equation 16.1, where h is Planck's constant and ν is the frequency (number of waves per second) of the radiation. The frequency is equal to c, the speed of light (~3 x 108 m/s), divided by the wavelength (distance between peaks) of the radiation (c = λν). The reciprocal of the wavelength is called the wavenumber (ν̃), which is the number of waves in a given distance, usually 1 cm. The frequencies, wavelengths, and energies of common types of electromagnetic radiation are given in Table 16.1. An important thing to remember is that a higher frequency or wave number or a shorter wavelength of radiation gives higher energy, while a lower frequency or wave number or a longer wavelength gives lower energy.

Q 16.1 - Level 1

The near infrared region of the electromagnetic spectrum extends from the end of the visible region at a wavelength of 700 nm to the mid-infrared at 2500 nm. What is the frequency of near infrared light with a wavelength of 1200 nm? (1nm = 1x10$^{-9}$m)?

A

2.5x10$^{14}$s$^{-1}$

B

3.6x10$^{2}$s$^{-1}$

C

2.5x10$^{5}$s$^{-1}$

The interaction of radiation with matter only takes place when there is a match of the energy with a fundamental resonance of the substance which results in a transition. Thus, the energy levels in matter are known to be quantized. γ-Rays have enough energy to excite the nucleus of an atom to a higher level and may also cause the ejection of electrons and formation of ions. As a result of the high energy and high toxicity to cells, γ-rays, most commonly from decay of 60Co, are used to treat inoperable cancers.

In the case of X-rays, the energies match the energy required to completely eject one of the inner shell non-bonding electrons from an atom, also forming ions. Because of their lower energy, X-rays are not as dangerous to cells as γ-rays, so they have been used to visualize inside our bodies.

Ultraviolet (UV) radiation promotes σ or π bonding electrons to the corresponding σ* or π* antibonding orbitals. The energy of a single photon of γ-rays, x-rays, or UV radiation is comparable to or greater than most bond energies. Thus, absorption of these forms of radiation can result in broken bonds in molecules. It is the UV radiation in sunlight which gives us a tan and a sunburn.

Visible radiation, with wavelengths between 400 to 700 nm, excites electrons in highly conjugated π-systems to low-lying π* orbitals in organic compounds or excites d-electrons in transition metals. These absorptions result in the colors of objects that we perceive.

Microwave radiation initiates rotational motions in molecules. The radiation in a microwave oven excites the water molecules in food, resulting in rapid heating.

As we will see in Chapter 17, FM radio frequency radiation promotes nuclear spin flips in a magnetic field, although we may be more familiar with using it to listen to music.

The energy of infrared (IR) radiation matches with the energy needed to initiate molecular vibrations. IR radiation is what we feel as the heat radiating from an open fire.

The IR region of the electromagnetic spectrum is often divided into three parts.

First is near-IR, with wavelengths ranging from about 700 nm to 2.5 µm. Near-IR starts just beyond the visible region and it is dominated by absorptions due to overtones (multiples) and linear combinations of fundamental molecular vibrations. The frequency of an overtone can be two or three times that of the fundamental frequency. Even though water is transparent in the visible region of the spectrum, it is a strong absorber in most of the near-IR region. Although the near-IR does not provide us with any useful information for organic structure determination, it is nonetheless useful in quantitative analysis of materials — for example, to determine the water content of oils.

Second is mid-IR, from 2.5 µm to about 50 µm. Mid-IR is the region where most fundamental vibrations in molecules are observed and is thus the region of interest for determination of molecular structure.

Third is far-IR, from 50 to 1000 µm. The far-IR contains some low-frequency rotational and vibrational modes of molecules, as well as metal-ligand vibrations, but it is not generally useful for organic structure determination.

## 16.2 Measurement of IR Spectra

IR spectroscopy is the measurement of vibrational motions in molecules. In order to obtain an IR spectrum, one first needs a source of IR radiation, which covers the spectrum. A convenient source of broadband IR radiation is a glowing rod or coil of wire which is electrically heated to 1000-2000 ºC. In earlier IR instruments, because glass is not transparent to IR, the emitted IR radiation was passed through a salt prism or grating to disperse it. Then it was passed through a slit to select a specific frequency. The radiation was then split into two parallel beams, one of which passes through the sample material and other through a reference material. The two beams then passed through a chopper before reaching the detector, a heat-sensitive material that changes voltage with temperature. The chopper alternated the reference and sample beam striking the detector, so that it could be balanced in the absence of absorbance by the sample. The spectrum was then scanned by slowly rotating the prism or grating to change the frequency passing through the slit, so the collection of an IR spectrum generally took 5-10 minutes.

In more modern instruments, the IR spectrum is often collected much more rapidly using a physical principle called interferometry. The IR beam is split into two perpendicular beams, one of which is reflected by a fixed mirror and the other by a moving mirror (Figure 16.3).

The recombined beam then passes through the sample and hits the detector. As the moving mirror moves over a short distance (~2 mm), the frequency of IR radiation passing through the sample changes due to constructive and destructive interference of the two beams when they recombine. The beam intensity passing through the sample is recorded as a function of the position of the moving mirror. A Fourier transform (FT) of the resulting plot, called an interferogram, gives the IR spectrum. Thus, the instrument is commonly called an FT-IR.

The compound of interest can be spread on salt plates, if it is a liquid, or mixed with a salt, usually KBr, and pressed into a pellet, if it is a solid. The plate or pellet is then placed in the IR beam inside the cell compartment to measure the spectrum. The spectrum shows the percent transmittance versus the wavenumber of the IR radiation. This is therefore classified as a transmission IR experiment.

Newer instruments use a technique called Attenuated Total Reflectance (ATR), which just requires pressing the compound on the surface of a cell through which the IR beam passes. The wave nature of the IR radiation allows it to interact with materials within a distance of one wavelength from the surface of the cell.

Q16.2 - Level 1

What is the major advantage of FT-IR over dispersive IR instruments?

A

Greater sensitivity

B

Faster data collection

C

Easier sample preparation

## 16.3 Molecular Vibrations

All bonds vibrate back and forth about an average bond length at a specific frequency which can be approximated mathematically as a simple harmonic oscillator, with two weights connected by a spring (Figure 16.5).

However, as a bond reaches a high energy level, the bond deviates from this model since it can dissociate into the atoms (Figure 16.5). This system obeys Hooke's Law (Equation 16.2) for the lower energy levels, where k is the force constant and Q is the distance away from the equilibrium distance. The vibrational frequency, ν, of the oscillator is related to the force constant by Equation 16.3, where μ is the reduced mass of the system, given by Equation 16.4, in which m1 and m2 are the masses of the atoms in the bond.

Thus, the frequency of a particular molecular vibration can be affected both by the bond strength, equivalent to the oscillator force constant, as well as the masses of the atoms forming the bond. This makes IR spectroscopy particularly valuable for organic structure determination, since all the bonds in a molecule will in principle exhibit unique absorptions at different frequencies in the IR spectrum. Vibrational energies will be different between different atoms because of bond strengths and atomic masses. Light atoms or stiffer bonds vibrate at a higher frequency. Triple bonds are stiffer and vibrate at higher frequencies than double bonds, and double bonds are stiffer and vibrate at a higher frequency than single bonds. We can see some of these effects below. Notice that stretching frequencies of groups involving a light atom like hydrogen, such as C-H, N-H and O-H, all occur at relatively high frequencies.

The maximum possible number of vibrational modes in a molecule is equal to 3n-6, where n is the number of atoms in the molecule. Even a small molecule like ethanol, C2H6O, can have up to 21 vibrational modes, yet not all of them may be observed. In addition to simple stretches of two atoms, the types of vibrational modes include symmetric and asymmetric stretches, bending, twisting, scissoring and wagging motions involving three atoms.

Please watch Video 16.1 below on methane and propane vibrations.

​Note: This video does not feature audio​

As you can see from Figure 16.6, the mid-IR spectrum can be conveniently subdivided into several regions (Figure 16.8). The IR spectrum is most commonly scaled in wave numbers (cm-1) rather than wavelength. In the regions from about 4000 cm-1 (2.5 µM) to 2500 cm-1 (4 µM), the stretching of C-H, N-H, and O-H single bonds are observed. This is the X-H region. Between 2500 cm-1 and 2000 cm-1 (5 µM), stretching vibrations of C≡C, C≡N, and N≡N triple bonds are seen. This is the triple-bond region. Between 2000 cm-1 and 1500 cm-1 (6.7 µM), the stretching vibrations of C=O and C=C double bonds are observed. This is the double bond region. Finally, in the region between 1500 cm-1 and 200 cm-1 (50 µm), the various stretching, bending, wagging and twisting modes of single bonds between C, O, N, H and halogens are seen.

This last region is complex, with many overlapping bands, and is difficult to interpret, with a few exceptions—for example, C-O and C-X stretches. However, the spectrum in this region is characteristic and unique for every compound, so it is known as the fingerprint region. Comparison of the fingerprint region of an unknown compound with that of a reference sample can be used as a method of identification. Databases with libraries of reference IR spectra are available to help in molecule identification. Outside of the fingerprint region, IR spectroscopy is especially valuable for the determination of the presence of functional groups in a molecule, especially those with OH, double, or triple bonds. Conversely, if the IR spectrum lacks the absorption of a functional group, we can conclude that such a functional group is not present in the molecule.

### 16.3.1 Characteristic Alcohols and Amine C-H Stretching Vibrations

In general, in the absence of intermolecular interactions, the width of IR absorption bands is relatively narrow, about 10-20 cm-1 at half height. However, if there are strong intermolecular interactions between the molecules, then each molecule will have a slightly different environment and thus a different IR absorption spectrum. The individual spectra overlap, resulting in broadened absorption bands, with widths up to 100-200 cm-1. The most common intermolecular interaction leading to broadened absorptions is hydrogen bonding, which is found in structures containing -OH or -NH functional groups. Thus, the -OH and -NH stretching bands of alcohols, carboxylic acids, amines, and amides are usually broader than other vibrational bands. The broad OH absorption can be seen at about 3360 cm-1 in 2-butanol and the N-H stretching bands at 3362 and 3286 for butan-2-amine given below.

Primary, secondary and tertiary amines can be distinguished by looking in the region between 3300 and 3500 cm-1. Primary amines show a doublet at about 3300 cm-1, which is due to symmetric and asymmetric stretching of the H-N-H bonds (Figure 16.7). Since secondary amines have one N-H bond, only one stretching absorption peak is observed, and as expected a tertiary amine does not show absorption at approximately 3300 cm-1, as seen in Figure 16.10.

Please watch Video 16.2 below on primary and secondary amines.

Q16.3 - Level 1

The IR of an unknown compound is given. Which compound is it?

A

Primary amine

B

Secondary amine

C

Tertiary amine

### 16.3.2 Characteristic C-H Stretching Vibrations

The most characteristic C-H absorption signals are above 2700 cm-1 and are influenced significantly by the hybridization state of the carbon—sp, sp2 or sp3. As previously seen in Figure 16.6, the shorter and stronger the bond, the higher the absorption frequency. Thus the shorter, stronger acetylene sp C-H bond will absorb at the highest energy, followed by the sp2 C-H bond, and then the sp3 C-H bond as seen in Table 16.2.

Another very diagnostic C-H stretching vibration occurs with aldehydes. For most aldehydes, two moderately intense bands are observed between 2830-2695 cm-1. These are very diagnostic absorption bands because they are only observed for aldehydes and not for other compounds that contain a carbonyl group, such as ketones, carboxylic acids, amides, and anhydrides. The characteristic C-H stretching vibrations are shown in Figure 16.11. Note that the C-H absorption frequencies of sp3 alkanes stop at 3000 cm-1. Therefore, a good way to determine if you have an alkene is to draw a line at 3000 cm-1. Any sharp peak just to the left of 3000 cm-1 will be the sp2 C-H absorption of an alkene. A sharp band at approximately 3300 will be the sp C-H absorption of a terminal alkyne. If the peaks are broad above 3000 cm-1, the absorption may be caused by O-H or N-H vibrations.

Please watch Video 16.3 below on sp, sp2 and sp3 C-H stretching vibrations.

Q16.4 - Level 2

Match the IR spectrum to the functional group.

Premise
Response
1

1)

A

Alkyne

2

2)

B

Alcohol

3

3)

C

Alkane

4

4)

D

Amine

E

Alkene

F

Aldehyde

Q16.5 - Level 2

Would you expect to see absorption peaks greater than 3000 cm$^{-1}$ for each of these compounds (yes or no)?

Premise
Response
1

1)

A

Yes

2

2)

B

Yes

3

3)

C

No

4

4)

D

No

5

5)

E

Yes

6

6)

F

No

### 16.3.3 Characteristic C=O Carbonyl Stretching Vibration

The other characteristic feature of IR spectra is the peak intensity. Absorption of a photon of IR radiation results in the excitation of a vibration to the next higher vibrational level as shown in Figure 16.5. The likelihood of the absorption is greater if there is a change in the dipole moment of the bond in the transition. Therefore, IR absorption is strongly favorable for bonds between atoms with a large difference in electronegativity—for example, a C=O—and thus a strong dipole moment. The associated vibrational bands show a high intensity in the IR spectrum as seen in the strong absorption at 1715 cm-1 for pentan-2-one. The location of the strong C=O stretching absorption bands will vary between 1870-1540 cm-1, depending what it is bonded to.

The frequency of the molecular vibration is affected by the structure of the molecule. Carbonyl groups are particularly sensitive to resonance effects, since the carbonyl group is a resonance hybrid of single and double bonded resonance structures (Figure 16.13). The dipolar resonance structure has a formal single bond, so it has a smaller force constant than the double bonded resonance structure. The greater contribution that the dipolar structure makes to the resonance hybrid, the lower the observed vibrational frequency, because the single bond structure has lower bonding energy. Thus any group that stabilizes the partial positive charge on the carbonyl carbon decreases the C=O bond strength and lowers the absorption frequency. Conversely, any group that increases the C=O bond strength will increase the force constant and increase the frequency of absorption.

Heteroatoms such as halogens, oxygen and nitrogen are both electron donating and withdrawing. Since heteroatoms are generally more electronegative than carbon, they inductively increase the bond strength by pulling electron density towards themselves and strengthen the C=O bond. However, because they possess lone pairs of electrons, they are also electron donating through resonance and may decrease the bond strength (Figure 16.13). IR provides us a valuable tool in observing the ability of a heteroatom to donate electron density to the π bond. This affects the electrophilicity of the carbonyl carbon and will be examined in more detail in Chapter 25: Carboxylic Acid Derivatives.

As seen in Figure 16.14, acid chlorides and esters absorb at the highest C=O stretching frequencies. Thus, for acid chlorides and esters, electron withdrawing effects are stronger than electron donation. This is due to the electronegativity of the oxygen and chlorine atoms. But oxygen is more electronegative than chlorine. Why do acid chlorides absorb at higher frequencies?

The answer is in the orbital overlap. The p orbital overlap required for a C=Cl bond is not as effective as compared to oxygen because of the orbital size difference. Therefore, the Cl group donates much less electron density through resonance than it withdraws inductively. The orbital resonance or the ability to donate the lone pair of the heteroatom to the empty π orbital is illustrated in Figure 16.15.

Thus, for acid chlorides and esters, electron withdrawing effects are stronger than electron donation. This is due to the electronegativity of the oxygen and chlorine atoms. But oxygen is more electronegative than chlorine. Why do acid chlorides absorb at higher frequencies? The answer is in the orbital overlap. The p orbital overlap required for a C=Cl bond is not as effective as compared to oxygen because of the orbital size difference. Therefore, the Cl group donates much less electron density through resonance than it withdraws inductively. The orbital resonance or the ability to donate the lone pair of the heteroatom to the empty π orbital is illustrated in Figure 16.15.

Nitrogen is the least electronegative and shows a carbonyl stretching at lower frequencies, ~1650 cm-1. This is due to the strong resonance from the nitrogen’s lone pair. In carboxylic acid salts, the oxygen’s electrons are equally shared between the two oxygens. Therefore, conjugation with the carbonyl group is more efficient than with amides. Hyperconjugation also can influence carbonyl frequencies, so aldehydes generally show higher frequencies than ketones (1720-1740 cm-1 vs 1715 cm-1).

Please watch Video 16.4 below on carbonyl stretching of butanal below.

Other factors that affect the vibration frequencies of the C=O bond are hydrogen bonding, conjugation with double bonds, and ring strain:

Hydrogen bonding: Monomeric carboxylic acids can be observed in very dilute solutions, and due to the electron withdrawing nature of the adjacent oxygen atom, absorption occurs at approximately 1760 cm-1. However, carboxylic acids form hydrogen bonds in solution and will dimerize at higher concentration, in aprotic solvents, and in the solid state. For the majority of IR spectra of carboxylic acids, extensive hydrogen bonding is taking place. This weakens the C=O bond and lowers the frequency of saturated acids to 1730-1700 cm-1. Hydrogen bonding to the O by the solvent is another factor that can influence C=O frequencies, causing a shift to lower frequency, since it stabilizes the dipolar resonance structure, and broadening the absorption (Figure 16.16).

α, β Conjugation: Conjugated carbonyls also show lower frequency absorptions by 25-45 cm-1. The adjacent double bond weakens the C=O bond and increases the single bond character through delocalization. This causes the absorption bands of acid chlorides, esters, aldehydes and ketones to appear at frequencies of lower energy. The effect is generally not observed for amides. Further conjugation results in a small change of approximately 15 cm-1 to lower frequency. A characteristic consequence of conjugation is the increased peak intensity of the C=C bond because of the polarization of the bond. Resonance also decreases the bond strength of the C=C bond and absorption moves from ~1650cm-1 to a lower frequency of 1640-1620 cm-1 (Figure 16.16).

Ring Size: Ring strain can also influence carbonyl vibrational frequencies. Cyclobutanone exhibits the C=O stretch at ~1880 cm-1, while unstrained cyclohexanone has a “normal” ketone absorption at 1715 cm-1.

Figure 16.17 shows the shifting of the C=O stretching vibration to smaller wavenumbers and the increase in intensity of the C=C absorption band due to increased polarization.

Q16.6 - Level 2

Are the following statements true or false concerning the following pairs of molecules?

Premise
Response
1

The C=O stretching vibration of 1) will occur at a lower wave number than 2)

A

False

2

The C=O stretching vibration of 3) will occur at a lower wave number than 4)

B

True

3

The C=O stretching vibration of 5) will occur at a lower wave number than 6)

C

True

D

False

### 16.3.4 Characteristic C-C Bonds and Nitriles C≡N

The C-C stretching of alkanes occurs at very low frequencies below 500 cm-1 and does not appear in an IR spectrum; however, C=C and C≡C are observed. The stretching C=C absorption band for unconjugated alkenes is moderate to weak at 1670-1640 cm-1, and for many symmetric trans disubstituted or tetrasubstituted alkenes, an absorption band is not observed at all. Why wouldn’t a symmetrical alkene absorb IR radiation? As mentioned with C=O bond stretching, the larger the dipole between atoms, the more intense the absorption. Thus, many symmetrical trans disubstituted or tetrasubstituted alkenes do not have a dipole, and the absorption is very weak or not observed as seen in Figure 16.18a. The absorption will increase if a larger dipole is created within the double bond, as seen in the increased intensity of the C=C stretching of 1-chloro-2-methylprop-1-ene (Figure 16.18c).

Similarly, C≡C bonds in highly symmetrical molecules have very small dipole moments and correspondingly very weak, often unobservable, IR absorptions. The C≡C  stretching absorption for alkynes occurs in the region of 2260-2100 cm-1. The C≡N  bond has a significantly stronger dipole moment and thus a strong absorption is observed in the region 2260-2240 cm-1 (Figure 16.19).

Please watch Video 16.5 below on C-C stretching.

Q16.7 - Level 1

Which compound would produce the strongest signal for a carbon-carbon double bond?

A

1)

B

2)

C

3)

Bonds that have a small dipole moment can be observed using another technique known as Raman spectroscopy that involves changes in the frequency of scattered visible light. The scattered light can either lose or gain energy corresponding to the frequency of the molecular vibration, so the difference between the frequency of the incident light and scattered light gives the vibrational frequency. Since Raman scattering is an unlikely phenomenon that only occurs once in about 106 collisions between a photon and a molecule, it is necessary to use laser excitation to observe a spectrum. The relative intensity of Raman bands is related to differences in polarizability between ground and excited vibrational states rather than differences in dipole moment.

Characteristic aromatic C=C stretching vibrations occur between 1450-1600 cm-1. The peaks observed in this area can be used to determine the substitution pattern on the benzene ring, but that is beyond the scope of this textbook. Generally, three signals of medium intensity are observed around 1450, 1500 and 1600 cm-1. Characteristic weaker bands may be observed between 1700-2000 cm-1. These bands are called overtone bands and are created by higher energy absorptions.

Before going onto the next section watch Video 16.6. The video animates the characteristic vibrations of (E)-pent-2-enenitrile and shows their location on the IR spectrum.

Please watch Video 16.6 below on (E)-Pent-2-enenitrile.

## 16.4. Analyzing the IR Spectrum

The first step in analyzing the IR spectrum is to identify the position of the strongest bands. These can often be assigned to common functional groups by comparison with a table of approximate vibrational frequencies, as given in Table 16.3.

For example, the IR spectrum of formic acid (Figure 16.21b) shows strong absorption peaks at 3150, 1727, and 1180 cm-1, which can be confidently assigned to the O-H, C=O, and C-O stretching vibrations, respectively. If we compare the spectrum of formic acid with the spectrum of methanol (Figure 16.21a), we see strong peaks at 3340, 2945, and 1030 cm-1, which correspond O-H, C-H, and C-O stretches, respectively, but the C=O stretch at 1727 cm-1 is missing.

Let us work through another example where we need to identify which of the structures is most consistent with the IR spectrum data. Most of the diagnostic peaks are higher than 1500 cm-1, above the fingerprint region.

First, identify the peaks above 1500 cm-1 by answering the following question.

Q16.8 - Level 2

Match the absorption band wave numbers to the type of bond.

Premise
Response
1

The peak at 1640 represents

A

C=C double bond

2

The peak at 1720 represents

B

sp$^2$ C-H bond

3

The peak at 2983 represents

C

carbonyl C=O double bond

4

The peaks at 3031 and 3075 represent

D

sp$^3$ C-H bond

Answering the above question (Question 16.8) tells you that spectrum cannot be of 1 or 3. It cannot be 1 because the compound does not have a double bond. It cannot be 3 because the compound does not have a carbonyl functional group. Thus the spectrum must belong to either 2 or 4. Look at the difference between the two structures; 3 is conjugate to a double bond and 4 is not. The normal carbonyl stretching frequency for an ester is 1750-1735 cm-1 and the stretching frequency given in the spectrum is 1712cm-1. What would cause the normal absorption band to be shifted to a lower wave number?

Q16.9 - Level 2

Identify the structure below that is most consistent with the IR spectrum.

A

1)

B

2)

C

3)

D

4)

Q16.10 - Level 3

Identify the structure below that is most consistent with the IR spectrum.

A

1)

B

2)

C

3)

D

4)

E

5)

Q16.11 - Level 3

Identify the structure below that is most consistent with the IR spectrum.

A

1)

B

2)

C

3)

D

4)

E

5)

Q16.12 - Level 2

Look at this IR spectrum of methyl cyanoacetate. Click on the spectrum to locate the C-H stretch.

Q16.13 - Level 2

Look at this IR spectrum of methyl cyanoacetate. Click on the spectrum to locate the C≡N stretch.

Q16.14 - Level 2

Look at this IR spectrum of methyl cyanoacetate. Click on the spectrum to locate the C-O stretch.

Q16.15 - Level 2

Look at this IR spectrum of methyl cyanoacetate. Click on the spectrum to locate the C=O stretch.

Q16.16 - Level 3

Which of the structures of a compound with molecular formula $C_3H_6O$ is consistent with the IR spectrum shown?

A

Oxetane

B

Methyl vinyl ether

C

Acetone

D

Propylene oxide

In the following questions (16.17-16.20), all of the compounds below have the molecular formula C6H10O. Identify the spectrum with the compound.

Q16.17 - Level 3

Identify the structure below that is most consistent with the IR spectrum.

A

1)

B

2)

C

3)

D

4)

Q16.18 - Level 3

Identify the structure below that is most consistent with the IR spectrum.

A

1)

B

2)

C

3)

D

4)