Jones M., Fleming S.A. Organic Chemistry

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15.4 Infrared Spectroscopy (IR) 709

a disk with KBr and placed in the beam of IR radiation; alternatively, it can be dis-

solved in Nujol®, which is a petroleum oil.

Spectrometers are routinely attached to computers that can search for matches

between the spectrum of an unknown and a library of known spectra. As with mass

spectrometry, gas chromatographs can be attached to IR spectrometers and spectra

can be determined as the individual components of a mixture elute from a column.

As noted in Section 15.2, this technique is called gas chromatography/infrared spec-

troscopy, or GC/IR.

15.4b Characteristic Infrared Absorptions From the foregoing discussion,

one might expect IR spectroscopy to be a simple matter of noting what characteris-

tic absorptions appear in the spectrum.An IR spectrum might be a very simple phe-

nomenon, composed only of signals for each kind of bond present in the sample

molecule. There certainly are characteristic regions of absorption for certain bonds,

but the practical situation is not quite so simple because connected vibrating springs

interact.Two springs attached to the same object (or atom) do not vibrate independ-

ently—try it. If you set one spring in motion, the other moves as well. In fact, this

phenomenon occurs even if the two springs (bonds) are attached to different atoms

within the same molecule. The closer in energy the two vibrations are, the more

strongly they affect each other. A molecule is like a collection of interacting springs,

and it is often not possible to make a simple assignment of all the bands in an IR

spectrum, although computational methods are becoming more proficient at this

task. Most IR spectra are complicated series of bands from which we can extract

information, but which we cannot completely rationalize in more than a general way.

Figure 15.17 shows the IR spectrum of a typical organic molecule and includes a

few band assignments.

100

Transmittance (%)

Wavenumber (cm

–1

)

4006001200 100014001600180020003400 2800 220040004600 800

Microns (μ)

16 18 20 22 251514131211109875 5.5 63 3.5 4 4.52.4 2.6 2.82.2

0.0

0.4

0.5

0.6

0.8

1.0

2.0

0.3

0.05

0.2

0.1

Absorbance

Cyclohexanone

Overtone of C O

at 1710 cm

–1

C C

C O

C H

FIGURE 15.17 A typical IR spectrum.

Although this complexity may seem confusing at first,it is usually possible to gain

a lot of information about the functional groups present in a molecule, even if we

cannot assign all the bands, or draw a complete structure of the sample molecule. It

is a great help in structure assignment to know which types of bonds are present in

a molecule. There is also an important side benefit to these complicated IR spectra.

Their very complexity means that the IR spectrum of every molecule is different.

Each spectrum serves as a fingerprint of the molecule. If two IR spectra are identical

710 CHAPTER 15 Analytical Chemistry: Spectroscopy

(not similar—MJ’s research advisor insisted on there being no difference greater than

the width of the pen line drawn by the recorder) the compounds must be the same.

Table 15.3 gives the general positions of absorptions of a variety of functional

groups.

TABLE 15.3 Typical Infrared Absorptions of Functional Groups

Functional Group Position (cm

–1

) Intensity

2980–2850 m–s ( stretch)

1480–1420

( bend)

Alkanes

3350–3300

2260–2100

H s

( stretch)

m–w ( stretch)

Alkynes

free 3650–3580

3550–3300

1260–1000

1150–1050

hydrogen bonded

( stretches)

( stretch)

br, s ( stretch)

Alcohols

RNH

3500–3100 (two bands)

NH 3500–3100 (one band)

RNH

, R

NH, or R

N ~1200

br, m ( stretches)

br, m ( stretch)

( stretch)

Amines

3150–3000

1680–1620

1630–1600(conj)

995–985

915–905

980–960

730–665

895–885

840–790

( stretch)

m–w ( stretch)

( out-of-plane bends)

( out-of-plane bend)

(br, variable)

Alkenes

(conjugation generally lowers C“ C double-bond stretching vibrations by about 20 cm

–1

)

15.4 Infrared Spectroscopy (IR) 711

TABLE 15.3 Typical Infrared Absorptions of Functional Groups

(continued)

Functional Group Position (cm

–1

) Intensity

3080–3020

1600–1580

Monosubstituted 770–730

710–690

Ortho disubstituted 770–735

Meta disubstituted 900–860

810–750

725–680

Para disubstituted 860–800

( out-of-plane bends)

( out-of-plane bend)

Aromatic Compounds

m–w ( stretch)

~2250

Nitriles

s ( stretch)

CAUTION! ere certainly is some subjectivity in this table, and the values represent average positions for “normal” compounds.

Medium = m, strong = s, weak = w, broad = br.

1680–1640

( stretch)

Imines

RCHO

2900–2700 (two bands)

1730–1700 (higher in strained cyclic molecules)

1750–1735

1300–1000

1730–1700

3200–2800

1820–1770

1820–1750 (two bands)

1150–1000

Carbonyl Compounds

s ( stretch)

br, s ( stretch)

( stretch)

(subtract 20–30 cm

–1

for a conjugated carbonyl)

ORR

OHR

ClR

O O

RR

Once again, we come to the question of memorization. Should one learn this

chart by heart? We think not. It is important to know that this kind of

general correlation exists, and you should have a rough idea of where some impor-

tant functional groups absorb. If you come to use IR often, you will automati-

cally learn the relevant details of the chart, as you work out what the signals in

your IR spectra tell you.

712 CHAPTER 15 Analytical Chemistry: Spectroscopy

ANSWER (a) In the stretch region (3350–2850 cm

1

) almost all the bands

are below 3000 cm

1

, indicating that there are only alkane carbon–hydrogen bonds.

No hydrogens are attached to double or triple bonds, because such carbon–hydrogen

100

Transmittance (%)

Wavenumber (cm

–1

)

4006001200 100014001600180020003400 2800 220040004600 800

Microns (μ)

16 18 20 22 251514131211109875 5.5 63 3.5 4 4.52.4 2.6 2.82.2

0.0

0.4

0.5

0.6

0.8

1.0

2.0

0.3

0.05

0.2

0.1

Absorbance

100

Transmittance (%)

Wavenumber (cm

–1

)

4006001200 100014001600180020003400 2800 220040004600 800

Microns (μ)

16 18 20 22 251514131211109875 5.5 63 3.5 4 4.52.4 2.6 2.82.2

0.0

0.4

0.5

0.6

0.8

1.0

2.0

0.3

0.05

0.2

0.1

Absorbance

100

*(a)

(b)

(c)

Transmittance (%)

Wavenumber (cm

–1

)

4006001200 100014001600180020003400 2800 220040004600 800

Microns (μ)

16 18 20 22 251514131211109875 5.5 63 3.5 4 4.52.4 2.6 2.82.2

0.0

0.4

0.5

0.6

0.8

1.0

2.0

0.3

0.05

0.2

0.1

Absorbance

WORKED PROBLEM 15.7 Use the data in Table 15.3 to identify the functional

groups in the IR spectra below. Perhaps, given the molecular formulas, you can

make some guesses as to structure.

(continued)

15.5

H Nuclear Magnetic Resonance Spectroscopy (NMR) 713

bonds would absorb above 3000 cm

1

.There is a strong absorption in the carbonyl

region at about 1740 cm

1

. A 1740 cm

1

band is too high to be a ketone or an

acid. The compound must be an ester, or perhaps an aldehyde. Finally, there is a

strong band at 1200 cm

1

, appropriate for the stretch for an ester. Given

the formula of C

, one can make two guesses at the structure.

OCH

Methyl propanoate

OCH

Ethyl acetate

This problem tries to show that without further information there is no reli-

able way to choose between these two possibilities. Given the information so far,

either answer is possible. We could either compare the IR spectrum with those of

known samples, or obtain further spectra. An NMR spectrum would distinguish

the two possibilities easily, as you will soon see. In fact, this compound is methyl

propanoate. For (b) and (c), see the Study Guide.

Summary

The IR spectrum of a compound provides information on the

kinds of functional groups in the molecule. In addition, if

standard spectra of known compounds are available, compar-

ison of the highly detailed IR spectra can be used as a proof

of identity—two identical IR spectra can only come from

identical compounds.

15.5

H Nuclear Magnetic Resonance

Spectroscopy (NMR)

In contrast to IR spectroscopy, which gives an overall view of the

functional groups in a molecule, nuclear magnetic resonance

(NMR) spectra can give a wealth of detail. It is the general feel-

ing among organic chemists that given good NMR spectra of a

molecule, the structure must follow. Today, the frontier of struc-

ture determination by NMR lies in immense biomolecules, and

even these structures are yielding to an increasing variety of

sophisticated NMR techniques.

Nuclear magnetic resonance was introduced very early on in

Section 2.14 (p. 88). There we only developed a bare-bones

outline—just enough to let you work out the numbers of differ-

ent carbon or hydrogen atoms in a molecule and to have you get

used to working with symmetry. Here, we go on to many of the

gory details of this incredibly powerful spectroscopic technique.

We will start with

H NMR (hydrogen nuclear magnetic reso-

nance), and then pick up

C NMR (carbon nuclear magnetic

resonance) again.

An increasingly useful and powerful tool in the medical

field is NMR spectroscopy, called Magnetic Resonance

Imaging (MRI) in this setting.

714 CHAPTER 15 Analytical Chemistry: Spectroscopy

As you can see from Figure 15.16, NMR spectroscopy involves energies much

smaller than those in IR, UV, or visible spectroscopy. What happens when radia-

tion that has a wavelength of 1–100 m ( 10

6

kcal/mol) is absorbed? What change

can such a tiny amount of energy induce? Molecular vibrations require much

higher energies (1–10 kcal/mol). Molecular rotations demand much less energy than

vibrations ( 10

4

kcal/mol), but even these motions require about 100 times more

energy than that of the radio waves used for NMR.To see what happens when radio

waves are absorbed by molecules, we must first go back to the beginning of our

discussion of atomic structure, and learn a bit more about the structure of the

atomic nucleus.

Like the electron,the nucleus has spin.For some nuclei (

Si),

the value of the nuclear spin (I ) is

. For others, the spin can take different val-

ues, such as . A nonzero spin is

a requirement for the NMR phenomenon. It is convenient that some very com-

mon isotopes have zero spin (

O) and therefore are NMR inactive. However,

the most common nucleus in organic chemistry, hydrogen, is one of the nuclei

with a spin of

and is NMR active. Keep in mind that hydrogens in molecules

are sometimes referred to as protons. The NMR effect arises as follows: Any

spinning charged particle generates a magnetic field, and therefore we can think

of the proton as a bar magnet. In the absence of a magnetic field, these bar mag-

nets will be oriented randomly (Fig. 15.18a). However, when we apply a magnetic

field (B

), the proton can align either with the field or against it (Fig. 15.18b).

(

O), 1 (

N),

(

Cl), or

(

In the absence of an applied

magnetic field

In the presence of an applied

magnetic field B

–

Energy

Aligned against

the field: higher

energy  hydrogens

Aligned with the

field: lower energy

α hydrogens

Applied magnetic field

+ + +

–

–––

–Pole

+Pole

(b)(a)

FIGURE 15.18 (a) In the absence of an applied magnetic field, nuclear spins will

be randomly oriented. (b) When an external magnetic field is applied, alignment

with the field will be slightly favored energetically over alignment against the

field; there will be a small excess of molecules aligned with the field.

Alignment with the field (I 

, α hydrogen) will be slightly more favor-

able energetically than alignment against the field (I 

, β hydrogen), and

there will be an excess of nuclei in the lower energy, more favorable orienta-

tion. Because the energy difference between the two orientations is very small,

there will only be a slight excess, but it will exist, and there is the possibility of

inducing transitions between the two orientations if the proper amount of energy

is supplied. The function of the radio waves is to supply the energy necessary

to change the orientation of the nuclear spin (often called “flipping” the spin).

15.5

H Nuclear Magnetic Resonance Spectroscopy (NMR) 715

Energy

–Pole

+Pole

+ + +

–

–––

Aligned against

the field: higher

energy β hydrogens

Aligned with the

field: lower energy

α hydrogens

β β

αααα

+ + +

–

–––

β β

αααα

+ + +

–

–––

β β

–

ααα

Radiowave

energy absorbed

Energy

released

FIGURE 15.19 Absorption of energy can “flip” a nuclear spin, moving the low energy proton into the higher

energy level.

The disturbed equilibrium is reestablished when this absorbed energy is returned

to the environment as heat (Fig. 15.19).

The magnitude of the separation of the two states—alignment with and against

the applied magnetic field—depends on the strength of that field, B

, as shown in

Eq. (15.4) and Figure 15.20. The higher the field strength, the greater the energy

difference between hydrogens aligned with and opposed to the applied mag-

netic field. For values of B

in the range commonly used, the frequencies ν for

typically studied isotopes are in the range of 60–750 MHz.The gyromagnetic ratio,

γ, is characteristic of a given nucleus, which means there is a different γ for every

NMR active nucleus.

(15.4)ν = γB

/2π

Energy

ΔE is a function of B

In the absence of

an applied magnetic

field (B

= 0), the

nuclear spin states

are of equal energy

FIGURE 15.20 The energy difference

between the two spin states of

hydrogen depends on the strength

of the applied magnetic field.

If we can build an instrument capable of delivering the required radio waves and

of detecting the absorption of tiny amounts of energy, we should be able to find a

signal from any hydrogen-containing material. But won’t there be problems from

716 CHAPTER 15 Analytical Chemistry: Spectroscopy

WORKED PROBLEM 15.8 The magnetic field strength is measured in units of

Tesla (T). Calculate the energy involved in an NMR transition at a field

strength of 4.7 T. For H, γ is 2.7  10

1

, and Planck’s constant is

9.5  10

14

(kcal/mol)(s).

ANSWER Energy equals Planck’s constant (h)  frequency (ν)or E  hν. The

frequency is given by Equation (15.4), ν  γB

/2π, so the useful relationship for

this problem is

1.9  10

5

whats? Figure out the units using the same equation.

E  (kcal/mol)(s)(T

1

)(T)  kcal/mol

The answer is 1.9  10

5

kcal/mol or 0.019 cal/mol. Notice that this last value

is calories per mole! These NMR transitions involve tiny energies.

E = (9.5 * 10

-14

)(2.7 * 10

)(4.7)/2(3.14) = 1.9 * 10

-5

γB

/2π E =

the other nuclei? Many common nuclei are NMR active, and there might be too

many overlapping signals for us to make sense of the spectrum. We are saved from

this fate by the gyromagnetic ratios γ, which are sufficiently different so that for the

usual field strengths, there is a wide separation of signals from different nuclei. For

example, signals from

C or

F nuclei appear far from those for

In the early NMR spectrometers, the magnetic field was varied, while the

frequency of the radio wave was held constant.The magnetic field was swept over

the appropriate range for the nucleus in question, most often hydrogen or carbon.

As the chemically different nuclei absorbed energy (came into resonance) they

would be detected one by one. It took several minutes to acquire the spectrum and

rather large samples—several milligrams—were necessary. Today a different and

more sensitive spectrometer is used. These days, superconducting magnets supply

far higher magnetic fields than did the old permanent magnets. The magnetic

field is not varied; instead, the sample is irradiated with a very short pulse of

radio frequency that extends over the entire range for the nucleus in question.

All hydrogen atoms or all carbon atoms absorb energy at once. They return to

equilibrium over time, and a computer transforms the intensity versus time data

into intensity versus resonance frequency data through what is called a Fourier

transform (FT). An FT spectrum can be obtained in a few seconds on very small

samples, and the results from many pulses are averaged to increase the signal-to-

noise ratio.

Resonance frequencies are always referenced to the resonance frequency of a

standard signal, and given in parts per million (ppm) of the magnetic field. The

standard chosen for

H NMR is tetramethylsilane (TMS), which shows a sharp

signal at a frequency somewhat removed from resonance frequencies of hydrogens

in typical organic molecules.The position of TMS is defined as the zero point of

the ppm scale. Why is the somewhat odd ppm scale used? Let’s assume that we

are using a 60-MHz spectrometer and that the resonance frequency of the sam-

ple molecule appears at 120 Hz from TMS. If we measured the NMR spectrum

of the same sample on one of the higher field instruments common today, say,

15.6 NMR Measurements 717

300 MHz, the resonance frequency would be different because the gap between

the two nuclear spin states depends on the strength of the applied magnetic field

(Fig. 15.20). At 300 MHz, our sample hydrogen signal would appear at 600 Hz

from TMS.Although the measured resonance position is different for every applied

magnetic field, the position is always the same in ppm. In the example above, it

appears at 2.0 ppm each time: 120 Hz/(60  10

Hz), or 600 Hz/(300  10

Hz)

(Fig. 15.21). The ppm scale was devised so that chemists could report the same

values for resonance positions regardless of the magnetic field strength of their

spectrometer.

Spectrum taken at 60 MHz

Intensity

109876543210

(ppm)

Chemical shift (δ)

Signal

120 Hz from TMS or

2.0 ppm from TMS

Spectrum taken at 300 MHz

Intensity

109876543210

Chemical shift (δ)

(ppm)

Signal

600 Hz from TMS but

still 2.0 ppm from TMS

FIGURE 15.21 The ppm scale ensures

that peak positions will be the same

regardless of external field strength.

15.6 NMR Measurements

Up to this point, NMR spectroscopy might seem to be merely an expensive way of

seeing if a compound contained hydrogen. We pay some hundreds of thousands of

dollars for a 300-MHz machine, put a sample of an organic compound in it, and if

we see absorption of radio waves at an appropriate frequency, we know the sample

contains hydrogen. That’s usually not very surprising if we are dealing with organic

molecules. So what’s the big deal? It turns out that we don’t see just one big signal

from a hydrogen-containing molecule. Instead, there is a signal at a different posi-

tion for every different hydrogen in the molecule.The resonance frequency for each

hydrogen in the molecule is critically dependent on its precise local environment—

any difference in environment will product a different resonance frequency.

The NMR spectrum is obtained by dissolving approximately 10 mg of the

compound to be analyzed in about 1 mL of a deuterated solvent,usually deuteriochlo-

roform (CDCl

), with 0.05% TMS added.The spectrum is then taken and the sig-

nals analyzed.

718 CHAPTER 15 Analytical Chemistry: Spectroscopy

WORKED PROBLEM 15.9 The higher the external magnetic field of the NMR

spectrometer, the greater the separation between the different signals, and

therefore the greater the resolving power of the instrument. This resolution

is one of the reasons that chemists are willing to spend extra thousands of

dollars to get extra megahertz. For example, two signals appearing at 120

and 126 Hz in a spectrum taken on a 60-MHz machine will be much better

separated if the spectrum is obtained at 300 MHz. Show that this last statement

is true.

ANSWER At 60 MHz, the signals at 120 and 126 Hz are at 2.0 and 2.1 ppm,

respectively, [120 Hz/(60  10

Hz) and 126 Hz/(60  10

Hz)]. At 300 MHz,

these signals are at 600 and 630 Hz, respectively. So the two signals are separated

by 6 Hz at 60 MHz, but by 30 Hz at 300 MHz. That’s quite an improvement in

separation.

There are three pieces of data associated with every NMR signal. The first is

called the integral. One can determine the relative number of hydrogens giving rise

to the signal by measuring the integral, or the area under the peak.The second piece

of data is the chemical shift, ␦.The chemical shift is the location of the NMR sig-

nal along the x-axis (see Fig.15.21). It is reported in ppm; for example,as δ  4ppm.

The chemical shift of most organic molecules lies between 0.8 and 10 ppm. The

third piece of data is the coupling, or multiplicity—the number of lines in each sig-

nal. Coupling changes the signal from a single line into multiple lines.The coupled

signals are called “doublets”(d, two lines),“triplets”(t, three lines),“quartets”(q, four

lines), and so on. The distance between the lines of the multiplet (the coupling

constant, J ) can be measured in Hz. A typical coupling constant J is about 7 Hz,

but the range is 1 to 16 Hz.

15.6a The Integral The

H NMR spectrum of 4,4-dimethyl-2-pentanone is

shown in Figure 15.22. Three signals appear, and the integration shows that they

are in the ratio 2:3:9. These days, integrations are calculated electronically and

printed out as numbers below or next to the signal. In older spectra, one sees a line

traced by a pen,as in Figure 15.22.The integral is determined by measuring the ver-

tical (y-axis) distance the pen traverses as it moves across the signal along the x-axis.

10 9 8 7 6 5 4 3 2 1 0 (ppm)

Chemical shift (δ)

4,4-Dimethyl-2-pentanone

FIGURE 15.22 The

H NMR spectrum of 4,4-dimethyl-2-pentanone.