Some instruments, called double-focusing mass spectrometers, have two magnetic sectors in their mass analyzers, giving these spectrometers have such high resolution that they provide mass measurements accurate to 5 ppm , or about 0.0005 amu , making it possible to distinguish between two formulas with the same nominal mass. For example, both $\mathrm{C}_{5} \mathrm{H}_{12}$ and $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ have MW $=72$, but they differ slightly beyond the decimal point: $\mathrm{C}_{5} \mathrm{H}_{12}$ has an exact mass of 72.0939 amu , whereas $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ has an exact mass of 72.0575 amu . A highresolution instrument can easily distinguish between them. Note, however, that exact mass measurements refer to molecules with specific isotopic compositions. Thus, the sum of the exact atomic masses of the specific isotopes in a molecule is measured -1.00783 amu for ${ }^{1} \mathrm{H}, 12.00000 \mathrm{amu}$ for ${ }^{12} \mathrm{C}, 14.00307 \mathrm{amu}$ for ${ }^{14} \mathrm{~N}, 15.994$ 91 amu for ${ }^{16} \mathrm{O}$, and so on-rather than the sum of the average atomic masses of elements, as found on a periodic table.


CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra

Unfortunately, not every compound shows a molecular ion in its electron-impact mass spectrum. Although $\mathrm{M}^{+}$ is usually easy to identify if it's abundant, some compounds, such as 2,2-dimethylpropane, fragment so easily that no molecular ion is observed (FIGURE 12.5). In such cases, alternative "soft" ionization methods that don't use electron bombardment can prevent or minimize fragmentation (see Section 12.4).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0377.jpg?height=467&width=1351&top_left_y=1753&top_left_x=384)

FIGURE 12.5 Mass spectrum of 2,2-dimethylpropane ( $\mathbf{C}_{\mathbf{5}} \mathbf{H}_{\mathbf{1 2}}$; MW = 72). No molecular ion is observed when electron-impact ionization is used. What do you think is the formula and structure of the $M^{+}$peak at $m / z=57$ ?


Knowing the molecular weight makes it possible to narrow considerably the choices of molecular formula. For example, if the mass spectrum of an unknown compound shows a molecular ion at $m / z=110$, the molecular formula is likely to be $\mathrm{C}_{8} \mathrm{H}_{14}, \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{O}, \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$, or $\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{~N}_{2}$. There are always a number of molecular formulas possible for all but the lowest molecular weights, and a computer can easily generate a list of the choices.


A further point about mass spectrometry, noticeable in the spectra of both propane (FIGURE 12.4) and 2,2-dimethylpropane (FIGURE 12.5), is that the peak for the molecular ion is not at the highest $\mathrm{m} / \mathrm{z}$ value. There is also a small peak at $\mathrm{M}+1$ due to the presence of different isotopes in the molecules. Although ${ }^{12} \mathrm{C}$ is the most abundant carbon isotope, a small amount ( $1.10 \%$ natural abundance) of ${ }^{13} \mathrm{C}$ is also present. Thus, a certain percentage of the molecules analyzed in the mass spectrometer are likely to contain a ${ }^{13} \mathrm{C}$ atom, giving rise to the observed $\mathrm{M}+1$ peak. In addition, a small amount of ${ }^{2} \mathrm{H}$ (deuterium; $0.015 \%$ natural abundance) is present, making a further contribution to the $\mathrm{M}+1$ peak.


Mass spectrometry would be useful even if molecular weight and formula were the only information that could be obtained, but in fact it provides much more. For one thing, the mass spectrum of a compound serves as a kind of "molecular fingerprint." Every organic compound fragments in a unique way depending on its structure, and the likelihood of two compounds having identical mass spectra is small. Thus, it's sometimes possible to identify an unknown by computer-based matching of its mass spectrum to one of the more than 785,061 searchable spectra recorded in a database called the Registry of Mass Spectral Data.


It's also possible to derive structural information about a molecule by interpreting its fragmentation pattern. Fragmentation occurs when the high-energy cation radical flies apart by spontaneous cleavage of a chemical bond. One of the two fragments retains the positive charge and is a carbocation, while the other fragment is a neutral radical.


Not surprisingly, the positive charge often remains with the fragment that is best able to stabilize it. In other words, a relatively stable carbocation is often formed during fragmentation. For example, 2,2-dimethylpropane tends to fragment in such a way that the positive charge remains with the tert-butyl group. 2,2-Dimethylpropane therefore has a base peak at $m / z=57$, corresponding to $\mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}$(FIGURE 12.5).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0378.jpg?height=183&width=720&top_left_y=1204&top_left_x=697)
$m / z=57$
Because mass-spectral fragmentation patterns are usually complex, it's often difficult to assign structures to fragment ions. Most hydrocarbons fragment in many ways, as demonstrated by the mass spectrum of hexane in FIGURE 12.6. The hexane spectrum shows a moderately abundant molecular ion at $m / z=86$ and fragment ions at $m / z=71,57,43$, and 29 . Since all the carbon-carbon bonds of hexane are electronically similar, all break to a similar extent, giving rise to the observed mixture of ions.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0378.jpg?height=478&width=1367&top_left_y=1666&top_left_x=371)

FIGURE 12.6 Mass spectrum of hexane $\left(\mathbf{C}_{\mathbf{6}} \mathbf{H}_{\mathbf{1 4}} \mathbf{;} \mathbf{M W}=\mathbf{8 6}\right)$. The base peak is at $m / z=57$, and numerous other ions are present.
FIGURE 12.7 shows how the hexane fragments might arise. The loss of a methyl radical $\left(\mathrm{CH}_{3}, \mathrm{M}=15\right)$ from the hexane cation radical $\left(\mathrm{M}^{+}=86\right)$ gives rise to a fragment of mass $86-15=71$; the loss of an ethyl radical $\left(\mathrm{C}_{2} \mathrm{H}_{5}\right.$, $M=29$ ) accounts for a fragment of mass $86-29=57$; the loss of a propyl radical ( $\mathrm{C}_{3} \mathrm{H}_{7}, \mathrm{M}=43$ ) accounts for a fragment of mass $86-43=43$; and the loss of a butyl radical accounts for a fragment of mass 29 . With practice, it's sometimes possible to analyze the fragmentation pattern of an unknown compound and work backward to a structure that is compatible with the data.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0379.jpg?height=663&width=1182&top_left_y=189&top_left_x=466)

FIGURE 12.7 Fragmentation of hexane in a mass spectrometer.
We'll see in the next section and in later chapters that specific functional groups, such as alcohols, ketones, aldehydes, and amines, show specific kinds of mass spectral fragmentations that can be interpreted to provide structural information.


WORKED EXAMPLE 12.1 -


Using Mass Spectra to Identify Compounds -

Assume that you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell them apart? The mass spectra of both are shown in FIGURE 12.8.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0379.jpg?height=983&width=1186&top_left_y=1319&top_left_x=464)


FIGURE 12.8 Mass spectra of unlabeled samples A and B for Worked Example 12.1.


Strategy -

Look at the possible structures and decide on how they differ. Then think about how any of these differences in structure might give rise to differences in mass spectra. Methyl cyclohexane, for instance, has a $-\mathrm{CH}_{3}$ group,
and ethylcyclopentane has a $-\mathrm{CH}_{2} \mathrm{CH}_{3}$ group, which should affect the fragmentation patterns.


Solution -

Both mass spectra show molecular ions at $\mathrm{M}^{+}=98$, corresponding to $\mathrm{C}_{7} \mathrm{H}_{14}$, but they differ in their fragmentation patterns. Sample $A$ has its base peak at $m / z=69$, corresponding to the loss of a $\mathrm{CH}_{2} \mathrm{CH}_{3}$ group (29 mass units), but $\mathbf{B}$ has a rather small peak at $m / z=69$. Sample $\mathbf{B}$ shows a base peak at $m / z=83$, corresponding to the loss of a $\mathrm{CH}_{3}$ group ( 15 mass units), but sample $\mathbf{A}$ has only a small peak at $\mathrm{m} / \mathrm{z}=83$. We can therefore be reasonably certain that $\mathbf{A}$ is ethylcyclopentane and $\mathbf{B}$ is methylcyclohexane.


Solution -

PROBLEM The sex hormone testosterone contains only C, H, and O and has a mass of 288.2089 amu , as 12-1 determined by high-resolution mass spectrometry. What is the likely molecular formula of testosterone?


Solution -

PROBLEM Two mass spectra are shown in Figure 12.9. One spectrum is that of 2-methyl-2-pentene; the other 12-2 is of 2-hexene. Which is which? Explain.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0380.jpg?height=487&width=1356&top_left_y=857&top_left_x=479)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0380.jpg?height=451&width=1353&top_left_y=1363&top_left_x=478)
(b)


Solution -

FIGURE 12.9 Mass spectra for Problem 12-2.


Solution - 12.3 Mass Spectrometry of Some Common Functional Groups

As each functional group is discussed in future chapters, mass-spectral fragmentations characteristic of that group will be described. As a preview, though, we'll point out some distinguishing features of several common functional groups.


Alcohols -

Alcohols undergo fragmentation in a mass spectrometer by two pathways: alpha cleavage and dehydration. In the $\alpha$-cleavage pathway, a $\mathrm{C}-\mathrm{C}$ bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation. This type of fragmentation is seen in the spectrum of 2-pentanol in FIGURE 12.10.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0381.jpg?height=825&width=1215&top_left_y=195&top_left_x=452)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0381.jpg?height=150&width=1012&top_left_y=1031&top_left_x=554)


Alcohols -

In the dehydration pathway, water is eliminated, yielding an alkene radical cation with a mass 18 amu less than $\mathrm{M}^{+}$. For simplicity, we have drawn the dehydration below as an E2-type process. Often the hydrogen that is lost is not beta to the hydroxyl. Only a small peak from dehydration is observed in the spectrum of 2-pentanol (FIGURE 12.10).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0381.jpg?height=156&width=823&top_left_y=1375&top_left_x=651)


Amines -

The nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. The logic behind the rule comes from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. The presence of nitrogen in a molecule is often detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms.


Amines -

Aliphatic amines undergo a characteristic $\alpha$ cleavage in a mass spectrometer, similar to that observed for alcohols. A C-C bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0381.jpg?height=139&width=1039&top_left_y=2053&top_left_x=538)


Amines -

The mass spectrum of triethylamine has a base peak at $m / z=86$, which arises from an alpha cleavage resulting in the loss of a methyl group (FIGURE 12.11).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0382.jpg?height=812&width=1348&top_left_y=185&top_left_x=383)

FIGURE 12.11 Mass spectrum of triethylamine.


Halides -

The fact that some elements have two common isotopes gives their mass spectra a distinctive appearance. Chlorine, for example, exists as two isotopes, ${ }^{35} \mathrm{Cl}$ and ${ }^{37} \mathrm{Cl}$, in roughly a $3: 1$ ratio. In a sample of chloroethane, three out of four molecules contain a ${ }^{35} \mathrm{Cl}$ atom and one out of four has a ${ }^{37} \mathrm{Cl}$ atom. In the mass spectrum of chloroethane (FIGURE 12.12 we see the molecular ion (M) at $m / z=64$ for ions that contain a ${ }^{35} \mathrm{Cl}$ and another peak at $m / z=66$, called the $M+2$ peak, for ions containing a ${ }^{37} C l$. The ratio of the relative abundance of $M: M+$ 2 is about $3: 1$, a reflection of the isotopic abundances of chlorine.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0382.jpg?height=871&width=833&top_left_y=1402&top_left_x=646)

FIGURE 12.12 Mass spectrum of chloroethane.
In the case of bromine, the isotopic distribution is $50.7 \%{ }^{79} \mathrm{Br}$ and $49.3 \%{ }^{81} \mathrm{Br}$. In the mass spectrum of 1-bromohexane (FIGURE 12.13) the molecular ion appears at $\mathrm{m} / \mathrm{z}=164$ for ${ }^{79} \mathrm{Br}$-containing ions and the $\mathrm{M}+2$ peak is at $m / z=166$ for ${ }^{81} \mathrm{Br}$-containing ions. The ions at $\mathrm{m} / \mathrm{z}=135$ and 137 are informative as well. The two nearly equally large peaks tell us that the ions at those $m / z$ values still contain the bromine atom. The peak at $m / z=85$, on the other hand, does not contain bromine because there is not a large peak at $m / z=87$.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0383.jpg?height=847&width=1259&top_left_y=189&top_left_x=430)

FIGURE 12.13 Mass spectrum of 1-bromohexane.


Carbonyl Compounds -

Ketones and aldehydes that have a hydrogen on a carbon three atoms away from the carbonyl group undergo a characteristic mass-spectral cleavage called the McLafferty rearrangement. The hydrogen atom is transferred to the carbonyl oxygen, a $\mathrm{C}-\mathrm{C}$ bond between the alpha and beta carbons is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0383.jpg?height=226&width=983&top_left_y=1356&top_left_x=571)


Carbonyl Compounds -

In addition, ketones and aldehydes frequently undergo $\alpha$ cleavage of the bond between the carbonyl carbon and the neighboring carbon to yield a neutral radical and a resonance-stabilized acyl cation. Because the carbon neighboring the carbonyl carbon is called the alpha carbon, the reaction is called an alpha cleavage.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0383.jpg?height=178&width=812&top_left_y=1727&top_left_x=654)
(To be more general about neighboring positions in carbonyl compounds, Greek letters are used in alphabetical order: alpha, beta, gamma, delta, and so on.)
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Carbonyl Compounds -

The mass spectrum of butyrophenone illustrates both alpha cleavage and the McLafferty rearrangement (FIGURE 12.14). Alpha cleavage of the propyl substituent results in the loss of $\mathrm{C}_{3} \mathrm{H}_{7}=43$ mass units from the parent ion at $m / z=148$ to give the fragment ion at $m / z=105$. A McLafferty rearrangement of butyrophenone results in the loss of ethylene, $\mathrm{C}_{2} \mathrm{H}_{4}=28$ mass units, from the parent leaving the ion at $\mathrm{m} / \mathrm{z}=120$.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0384.jpg?height=798&width=1175&top_left_y=192&top_left_x=472)


Carbonyl Compounds -

FIGURE 12.14 Mass spectrum of butyrophenone.


WORKED EXAMPLE 12.2 -


Identifying Fragmentation Patterns in a Mass Spectrum -

The mass spectrum of 2-methyl-3-pentanol is shown in FIGURE 12.15. What fragments can you identify?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0384.jpg?height=478&width=1353&top_left_y=1263&top_left_x=386)

FIGURE 12.15 Mass spectrum of 2-methyl-3-pentanol, for Worked Example 12.2.


Strategy -

Calculate the mass of the molecular ion, and identify the functional groups in the molecule. Then write the fragmentation processes you might expect, and compare the masses of the resultant fragments with the peaks present in the spectrum.


Solution -

2-Methyl-3-pentanol, an open-chain alcohol, has $\mathrm{M}^{+}=102$ and might be expected to fragment by $\alpha$ cleavage and by dehydration. These processes would lead to fragment ions of $m / z=84,73$, and 59 . Of the three expected fragments, dehydration is not observed (no $m / z=84$ peak), but both $\alpha$ cleavages take place ( $m / z=73,59$ ).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0384.jpg?height=267&width=849&top_left_y=2214&top_left_x=635)


Solution -

PROBLEM What are the masses of the charged fragments produced in the following cleavage pathways?
12-3 (a) Alpha cleavage of 2-pentanone $\left(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$
(b) Dehydration of cyclohexanol (hydroxycyclohexane)
(c) McLafferty rearrangement of 4-methyl-2-pentanone $\left[\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\right]$
(d) Alpha cleavage of triethylamine $\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right]$


Solution -

PROBLEM List the masses of the parent ion and of several fragments you might expect to find in the mass
12-4 spectrum of the following molecule:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0385.jpg?height=310&width=396&top_left_y=544&top_left_x=428)


Solution - 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments

MS analyses of sensitive biological samples rarely use magnetic sector ionization. Instead, they typically use either electrospray ionization ( $E S I$ ) or matrix-assisted laser desorption ionization (MALDI), typically linked to a time-of-flight (TOF) mass analyzer. Both ESI and MALDI are soft ionization methods that produce charged molecules with little fragmentation, even with sensitive biological samples of very high molecular weight.


In an ESI source, as a sample solution exits the tube, it is subjected to a high voltage that causes the droplets to become charged. The sample molecules gain one or more protons from charged solvent molecules in the droplet. The volatile solvent quickly evaporates, giving variably protonated sample molecules $\left(M+H_{n}{ }^{n+}\right)$. In a MALDI source, the sample is adsorbed onto a suitable matrix compound, such as 2,5-dihydroxybenzoic acid, which is ionized by a short burst of laser light. The matrix compound then transfers the energy to the sample and protonates it, forming $\mathrm{M}+\mathrm{H}_{n}{ }^{n+}$ ions.


Following ion formation, the variably protonated sample molecules are electrically focused into a small packet with a narrow spatial distribution, and the packet is given a sudden kick of energy by an accelerator electrode. As each molecule in the packet is given the same energy, $E=m v^{2} / 2$, it begins moving with a velocity that depends on the square root of its mass, $v=\sqrt{2 E / m}$. Lighter molecules move faster, and heavier molecules move slower. The analyzer itself-the drift tube-is simply an electrically grounded metal tube inside which the different charged molecules become separated as they move at different velocities and take different amounts of time to complete their flight.


The Time of Flight technique is considerably more sensitive than the magnetic sector alternative, and protein samples of up to 100 kilodaltons ( $100,000 \mathrm{amu}$ ) can be separated with a mass accuracy of 3 ppm. FIGURE 12.16 shows a MALDI-TOF spectrum of chicken egg-white lysozyme, MW = 14,306.7578 daltons. Biochemists generally use the unit dalton, abbreviated Da, instead of amu, although the two are equivalent ( 1 dalton = 1 amu ).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0386.jpg?height=741&width=1386&top_left_y=193&top_left_x=364)


FIGURE 12.16 MALDI-TOF mass spectrum of chicken egg-white lysozyme. The peak at $14,306.7578$ daltons (amu) is due to the monoprotonated protein, $\mathrm{M}+\mathrm{H}^{+}$, and the peak at $28,614.2188$ daltons is due to an impurity formed by dimerization of the protein. Other peaks at lower $m / z$ values are various protonated species, $M+H_{n}{ }^{n+}$.


Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum

Infrared, ultraviolet, and nuclear magnetic resonance spectroscopies differ from mass spectrometry in that they are nondestructive and involve the interaction of molecules with electromagnetic energy rather than with an ionizing source. Before beginning a study of these techniques, however, let's briefly review the nature of radiant energy and the electromagnetic spectrum.


Visible light, X rays, microwaves, radio waves, and so forth are all different kinds of electromagnetic radiation. Collectively, they make up the electromagnetic spectrum, shown in FIGURE 12.17. The electromagnetic spectrum is arbitrarily divided into regions, with the familiar visible region accounting for only a small portion, from $3.8 \times 10^{-7} \mathrm{~m}$ to $7.8 \times 10^{-7} \mathrm{~m}$ in wavelength. The visible region is flanked by the infrared and ultraviolet regions.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0386.jpg?height=671&width=1243&top_left_y=1562&top_left_x=441)


FIGURE 12.17 The electromagnetic spectrum covers a continuous range of wavelengths and frequencies, from radio waves at the lowfrequency end to gamma ( $\gamma$ ) rays at the high-frequency end. The familiar visible region accounts for only a small portion near the middle of the spectrum.


Electromagnetic radiation is often said to have dual behavior. In some respects, it has the properties of a particle, called a photon, yet in other respects it behaves as an energy wave. Like all waves, electromagnetic radiation is characterized by a wavelength, a frequency, and an amplitude (FIGURE 12.18). The wavelength, $\boldsymbol{\lambda}$ (Greek lambda), is the distance from one wave maximum to the next. The frequency, $\boldsymbol{\nu}$ (Greek nu), is the
number of waves that pass by a fixed point per unit time, usually given in reciprocal seconds $\left(\mathrm{s}^{-1}\right)$, or hertz, $\mathbf{H z}$ ( $1 \mathrm{~Hz}=1 \mathrm{~s}^{-1}$ ). The amplitude is the height of a wave, measured from midpoint to peak. The intensity of radiant energy, whether a feeble glow or a blinding glare, is proportional to the square of the wave's amplitude.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0387.jpg?height=736&width=981&top_left_y=346&top_left_x=471)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0387.jpg?height=226&width=630&top_left_y=749&top_left_x=1016)


Infrared radiation


$$
\left(\nu=3.75 \times 10^{14} \mathrm{~s}^{-1}\right)
$$


FIGURE 12.18 Electromagnetic waves are characterized by a wavelength, a frequency, and an amplitude. (a) Wavelength ( $\lambda$ ) is the distance between two successive wave maxima. Amplitude is the height of the wave measured from the center. (b)-(c) What we perceive as different kinds of electromagnetic radiation are simply waves with different wavelengths and frequencies.
Multiplying the wavelength of a wave in meters ( m ) by its frequency in reciprocal seconds ( $\mathrm{s}^{-1}$ ) gives the speed of the wave in meters per second ( $\mathrm{m} / \mathrm{s}$ ). The rate of travel of all electromagnetic radiation in a vacuum is a constant value, commonly called the "speed of light" and abbreviated $c$. Its numerical value is defined as exactly $2.99792458 \times 10^{8} \mathrm{~m} / \mathrm{s}$, usually rounded off to $3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$.


$$
\begin{aligned}
& \text { Wavelength } \times \text { Frequency }=\text { Speed } \\
& \qquad \begin{array}{c}
\lambda(\mathrm{m}) \times v\left(\mathrm{~s}^{-1}\right)=c(\mathrm{~m} / \mathrm{s}) \\
\lambda=\frac{c}{v} \text { or } v=\frac{c}{\lambda}
\end{array}
\end{aligned}
$$


Just as matter comes only in discrete units called atoms, electromagnetic energy is transmitted only in discrete amounts called quanta. The amount of energy $\varepsilon$ corresponding to 1 quantum of energy ( 1 photon) of a given frequency $\nu$ is expressed by the Planck equation


$$
\varepsilon=h v=\frac{h c}{\lambda}
$$


where $h=$ Planck's constant $\left(6.62 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}=1.58 \times 10^{-34} \mathrm{cal} \cdot \mathrm{s}\right)$.
The Planck equation says that the energy of a given photon varies directly with its frequency $\nu$ but inversely with its wavelength $\lambda$. High frequencies and short wavelengths correspond to high-energy radiation such as gamma rays; low frequencies and long wavelengths correspond to low-energy radiation such as radio waves. Multiplying $\varepsilon$ by Avogadro's number $N_{\mathrm{A}}$ gives the same equation in more familiar units, where $E$ represents the energy of Avogadro's number (one "mole") of photons of wavelength $\lambda$ :


$$
E=\frac{N_{A} h c}{\lambda}=\frac{1.20 \times 10^{-4} \mathrm{~kJ} / \mathrm{mol}}{\lambda(\mathrm{~m})} \text { or } \frac{2.86 \times 10^{-5} \mathrm{kcal} / \mathrm{mol}}{\lambda(\mathrm{~m})}
$$


When an organic compound is exposed to a beam of electromagnetic radiation, it absorbs energy of some wavelengths but passes, or transmits, energy of other wavelengths. If we irradiate the sample with energy of many different wavelengths and determine which are absorbed and which are transmitted, we can measure the absorption spectrum of the compound.


An example of an absorption spectrum-that of ethanol exposed to infrared radiation-is shown in FIGURE
12.19. The horizontal axis records the wavelength, and the vertical axis records the intensity of the various energy absorptions in percent transmittance. The baseline corresponding to $0 \%$ absorption (or $100 \%$ transmittance) runs along the top of the chart, so a downward spike means that energy absorption has occurred at that wavelength.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0388.jpg?height=460&width=1397&top_left_y=388&top_left_x=364)

FIGURE 12.19 An infrared absorption spectrum for ethanol, $\mathbf{C H}_{\mathbf{3}} \mathbf{C H}_{\mathbf{2}} \mathbf{O H}$. A transmittance of $100 \%$ means that all the energy is passing through the sample, whereas a lower transmittance means that some energy is being absorbed. Thus, each downward spike corresponds to an energy absorption.


The energy a molecule gains when it absorbs radiation must be distributed over the molecule in some way. With infrared radiation, the absorbed energy causes bonds to stretch and bend more vigorously. With ultraviolet radiation, the energy causes an electron to jump from a lower-energy orbital to a higher-energy one. Different radiation frequencies affect molecules in different ways, but each provides structural information when the results are interpreted.


There are many kinds of spectroscopies, which differ according to the region of the electromagnetic spectrum used. We'll look at three: infrared spectroscopy, ultraviolet spectroscopy, and nuclear magnetic resonance spectroscopy. Let's begin by seeing what happens when an organic sample absorbs infrared energy.


WORKED EXAMPLE 12.3 -


Correlating Energy and Frequency of Radiation -

Which is higher in energy, FM radio waves with a frequency of $1.015 \times 10^{8} \mathrm{~Hz}(101.5 \mathrm{MHz})$ or visible green light with a frequency of $5 \times 10^{14} \mathrm{~Hz}$ ?


Strategy -

Remember the equations $\varepsilon=h \nu$ and $\varepsilon=h c / \lambda$, which say that energy increases as frequency increases and as wavelength decreases.


Solution -

Since visible light has a higher frequency than radio waves, it is higher in energy.


Solution -

PROBLEM Which has higher energy, infrared radiation with $\lambda=1.0 \times 10^{-6} \mathrm{~m}$ or an X ray with $\lambda=3.0 \times 10^{-9} \mathrm{~m}$ ? 12-5 Radiation with $\nu=4.0 \times 10^{9} \mathrm{~Hz}$ or with $\lambda=9.0 \times 10^{-6} \mathrm{~m}$ ?


Solution -

PROBLEM It's useful to develop a feeling for the amounts of energy that correspond to different parts of 12-6 the electromagnetic spectrum. Calculate the energies in $\mathrm{kJ} / \mathrm{mol}$ of each of the following kinds of radiation:
(a) A gamma ray with $\lambda=5.0 \times 10^{-11} \mathrm{~m}$ (b) An X ray with $\lambda=3.0 \times 10^{-9} \mathrm{~m}$
(c) Ultraviolet light with $\nu=6.0 \times 10^{15} \mathrm{~Hz}$ (d) Visible light with $\nu=7.0 \times 10^{14} \mathrm{~Hz}$
(e) Infrared radiation with $\lambda=2.0 \times 10^{-5} \mathrm{~m}$ (f) Microwave radiation with $\nu=1.0 \times 10^{11} \mathrm{~Hz}$


Solution - 12.6 Infrared Spectroscopy

In infrared (IR) spectroscopy, the IR region of the electromagnetic spectrum covers the range from just above the visible $\left(7.8 \times 10^{-7} \mathrm{~m}\right)$ to approximately $10^{-4} \mathrm{~m}$, but only the midportion from $2.5 \times 10^{-6} \mathrm{~m}$ to $2.5 \times 10^{-5} \mathrm{~m}$
is used by organic chemists (FIGURE 12.20. Wavelengths within the IR region are usually given in micrometers $\left(1 \mu \mathrm{~m}=10^{-6} \mathrm{~m}\right)$, and frequencies are given in wavenumbers rather than in hertz. The wavenumber $\widetilde{v}$ is the reciprocal of wavelength in centimeters and is therefore expressed in units of $\mathrm{cm}^{-1}$.


Solution - 12.6 Infrared Spectroscopy

$$
\text { Wavenumber: } \widetilde{v}\left(\mathrm{~cm}^{-1}\right)=\frac{1}{\lambda(\mathrm{~cm})}
$$


Solution - 12.6 Infrared Spectroscopy

Thus, the useful IR region is from 4000 to $400 \mathrm{~cm}^{-1}$, corresponding to energies of $48.0 \mathrm{~kJ} / \mathrm{mol}$ to $4.80 \mathrm{~kJ} / \mathrm{mol}$ (11.5-1.15 kcal/mol).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0389.jpg?height=440&width=1381&top_left_y=561&top_left_x=372)


Solution - 12.6 Infrared Spectroscopy

FIGURE 12.20 The infrared and adjacent regions of the electromagnetic spectrum.
Why does an organic molecule absorb some wavelengths of IR radiation but not others? All molecules have a certain amount of energy and are in constant motion. Their bonds stretch and contract, atoms wag back and forth, and other molecular vibrations occur. Some of the kinds of allowed vibrations are shown below:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0389.jpg?height=408&width=1449&top_left_y=1246&top_left_x=346)


Solution - 12.6 Infrared Spectroscopy

The amount of energy a molecule contains is not continuously variable but is quantized. That is, a molecule can stretch or bend only at specific frequencies corresponding to specific energy levels. Take bond stretching, for example. Although we usually speak of bond lengths as if they were fixed, the numbers given are really averages. In fact, a typical $\mathrm{C}-\mathrm{H}$ bond with an average bond length of 110 pm is actually vibrating at a specific frequency, alternately stretching and contracting as if there were a spring connecting the two atoms.


Solution - 12.6 Infrared Spectroscopy

When a molecule is irradiated with electromagnetic radiation, energy is absorbed if the frequency of the radiation matches the frequency of the vibration. The result of this energy absorption is an increased amplitude for the vibration; in other words, the "spring" connecting the two atoms stretches and compresses a bit further. Since each frequency absorbed by a molecule corresponds to a specific molecular motion, we can find what kinds of motions a molecule has by measuring its IR spectrum. By interpreting these motions, we can find out what kinds of bonds (functional groups) are present in the molecule.


Solution - 12.6 Infrared Spectroscopy

$$
\text { IR spectrum } \rightarrow \text { What molecular motions? } \rightarrow \text { What functional groups? }
$$


Solution - 12.7 Interpreting Infrared Spectra

The complete interpretation of an IR spectrum is difficult because most organic molecules have dozens of different bond stretching and bending motions, and thus have dozens of absorptions. On the one hand, this complexity is a problem because it generally limits the laboratory use of IR spectroscopy to pure samples of fairly small molecules-little can be learned from IR spectroscopy about large, complex biomolecules. On the other hand, this complexity is useful because an IR spectrum acts as a unique fingerprint of a compound. In fact,
the complex region of the IR spectrum, from $1500 \mathrm{~cm}^{-1}$ to around $400 \mathrm{~cm}^{-1}$, is called the fingerprint region. If two samples have identical IR spectra, they are almost certainly identical compounds.


Fortunately, we don't need to interpret an IR spectrum fully to get useful structural information. Most functional groups have characteristic IR absorption bands that don't change much from one compound to another. The $\mathrm{C}=\mathrm{O}$ absorption of a ketone is almost always in the range 1680 to $1750 \mathrm{~cm}^{-1}$; the $\mathrm{O}-\mathrm{H}$ absorption of an alcohol is almost always in the range 3400 to $3650 \mathrm{~cm}^{-1}$; the $\mathrm{C}=\mathrm{C}$ absorption of an alkene is almost always in the range 1640 to $1680 \mathrm{~cm}^{-1}$; and so forth. By learning where characteristic functional-group absorptions occur, it's possible to get structural information from IR spectra. TABLE 12.1 lists the characteristic IR bands of some common functional groups.


TABLE 12.1 Characteristic IR Absorptions of Some Functional Groups


| Functional Group |  | Absorption (cm ${ }^{-1}$ )$2850-2960$ | Intensity <br> Medium |
| :---: | :---: | :---: | :---: |
| Alkane | $\mathrm{C}-\mathrm{H}$ |  |  |
| Alkene | $=\mathrm{C}-\mathrm{H}$ | 3020-3100 | Medium |
|  | $C=C$ | 1640-1680 | Medium |
| Alkyne | $\equiv \mathrm{C}-\mathrm{H}$ | 3300 | Strong |
|  | $\mathrm{C} \equiv \mathrm{C}$ | 2100-2260 | Medium |
| Alkyl halide | $\mathrm{C}-\mathrm{Cl}$ | 600-800 | Strong |
|  | $\mathrm{C}-\mathrm{Br}$ | 500-600 | Strong |
| Alcohol | $\mathrm{O}-\mathrm{H}$ | 3400-3650 | Strong, broad |
|  | $\mathrm{C}-\mathrm{O}$ | 1050-1150 | Strong |
| Arene | C-H | 3030 | Weak |
| Aromatic ring |  | 1660-2000 | Weak |
|  |  | 1450-1600 | Medium |
| Amine | $\mathrm{N}-\mathrm{H}$ | 3300-3500 | Medium |
|  | $\mathrm{C}-\mathrm{N}$ | 1030-1230 | Medium |
| Carbonyl compound | $\mathrm{C}=\mathrm{O}$ | 1670-1780 | Strong |
|  | Aldehyde | 1730 | Strong |
|  | Ketone | 1715 | Strong |
|  | Ester | 1735 | Strong |
|  | Amide | 1690 | Strong |
|  | Carboxylic acid | 1710 | Strong |


| Functional Group | Absorption (cm ${ }^{\mathbf{1}}$ ) |  | Intensity |
| :--- | :--- | :---: | :--- |
| Carboxylic acid | $\mathrm{O}-\mathrm{H}$ | $2500-3100$ | Strong, broad |
| Nitrile | $\mathrm{C} \equiv \mathrm{N}$ | $2210-2260$ | Medium |
| Nitro | $\mathrm{NO}_{2}$ | 1540 | Strong |


Look at the IR spectra of hexane, 1-hexene, and 1-hexyne in FIGURE 12.21 to see an example of how IR spectroscopy can be used. Although all three IR spectra contain many peaks, there are characteristic absorptions of $\mathrm{C}=\mathrm{C}$ and $\mathrm{C} \equiv \mathrm{C}$ functional groups that allow the three compounds to be distinguished. Thus, 1-hexene shows a characteristic $\mathrm{C}=\mathrm{C}$ absorption at $1660 \mathrm{~cm}^{-1}$ and a vinylic $=\mathrm{C}-\mathrm{H}$ absorption at $3100 \mathrm{~cm}^{-1}$, whereas 1-hexyne has a $\mathrm{C} \equiv \mathrm{C}$ absorption at $2100 \mathrm{~cm}^{-1}$ and a terminal alkyne $\equiv \mathrm{C}-\mathrm{H}$ absorption at $3300 \mathrm{~cm}^{-1}$.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0391.jpg?height=464&width=1381&top_left_y=847&top_left_x=399)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0391.jpg?height=459&width=1378&top_left_y=1410&top_left_x=401)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0391.jpg?height=454&width=1380&top_left_y=2017&top_left_x=397)

FIGURE 12.21 IR spectra of (a) hexane, (b) 1-hexene, and (c) 1-hexyne. Spectra like these are easily obtained from sub-milligram amounts of material in a few minutes using commercially available instruments.


It helps in remembering the position of specific IR absorptions to divide the IR region from $4000 \mathrm{~cm}^{-1}$ to 400 $\mathrm{cm}^{-1}$ into four parts, as shown in FIGURE 12.22.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0392.jpg?height=459&width=1389&top_left_y=302&top_left_x=365)


FIGURE 12.22 The four regions of the infrared spectrum: single bonds to hydrogen, triple bonds, double bonds, and fingerprint.


- The region from 4000 to $2500 \mathrm{~cm}^{-1}$ corresponds to absorptions caused by $\mathrm{N}-\mathrm{H}, \mathrm{C}-\mathrm{H}$, and $\mathrm{O}-\mathrm{H}$ singlebond stretching motions. $\mathrm{N}-\mathrm{H}$ and $\mathrm{O}-\mathrm{H}$ bonds absorb in the 3300 to $3600 \mathrm{~cm}^{-1}$ range; $\mathrm{C}-\mathrm{H}$ bond stretching occurs near $3000 \mathrm{~cm}^{-1}$.
- The region from 2500 to $2000 \mathrm{~cm}^{-1}$ is where triple-bond stretching occurs. Both $\mathrm{C} \equiv \mathrm{N}$ and $\mathrm{C} \equiv \mathrm{C}$ bonds absorb here.
- The region from 2000 to $1500 \mathrm{~cm}^{-1}$ is where double bonds $(\mathrm{C}=\mathrm{O}, \mathrm{C}=\mathrm{N}$, and $\mathrm{C}=\mathrm{C})$ absorb. Carbonyl groups generally absorb in the range 1680 to $1750 \mathrm{~cm}^{-1}$, and alkene stretching normally occurs in the narrow range of 1640 to $1680 \mathrm{~cm}^{-1}$.
- The region below $1500 \mathrm{~cm}^{-1}$ is the fingerprint portion of the IR spectrum. A large number of absorptions due to a variety of $\mathrm{C}-\mathrm{C}, \mathrm{C}-\mathrm{O}, \mathrm{C}-\mathrm{N}$, and $\mathrm{C}-\mathrm{X}$ single-bond vibrations occur here.


Why do different functional groups absorb where they do? As noted previously, a good analogy is that of two weights (atoms) connected by a spring (a bond). Short, strong bonds vibrate at a higher energy and higher frequency than do long, weak bonds, just as a short, strong spring vibrates faster than a long, weak spring. Thus, triple bonds absorb at a higher frequency than double bonds, which in turn absorb at a higher frequency than single bonds. In addition, $\mathrm{C}-\mathrm{H}, \mathrm{O}-\mathrm{H}$, and $\mathrm{N}-\mathrm{H}$ bonds vibrate at a higher frequency than bonds between heavier $\mathrm{C}, \mathrm{O}$, and N atoms.


WORKED EXAMPLE 12.4 -


Distinguishing Isomeric Compounds by IR Spectroscopy -

Acetone $\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)$ and 2-propen-1-ol $\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{OH}\right)$ are isomers. How could you distinguish them by IR spectroscopy?


Strategy -

Identify the functional groups in each molecule, and refer to TABLE 12.1.


Solution -

Acetone has a strong $\mathrm{C}=\mathrm{O}$ absorption at $1715 \mathrm{~cm}^{-1}$, while 2-propen-1-ol has an -OH absorption at $3500 \mathrm{~cm}^{-1}$ and a $\mathrm{C}=\mathrm{C}$ absorption at $1660 \mathrm{~cm}^{-1}$.


Solution -

PROBLEM What functional groups might the following molecules contain?
12-7 (a) A compound with a strong absorption at $1710 \mathrm{~cm}^{-1}$
(b) A compound with a strong absorption at $1540 \mathrm{~cm}^{-1}$
(c) A compound with strong absorptions at $1720 \mathrm{~cm}^{-1}$ and 2500 to $3100 \mathrm{~cm}^{-1}$


Solution -

PROBLEM How might you use IR spectroscopy to distinguish between the following pairs of isomers?
12-8 (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ and $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ (b) Cyclohexane and 1-hexene
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ and $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$


Solution - 12.8 Infrared Spectra of Some Common Functional Groups

As each functional group is discussed in future chapters, the spectroscopic properties of that group will be described. For the present, we'll point out some distinguishing features of the hydrocarbon functional groups already studied and briefly preview some other common functional groups. We should also point out, however, that in addition to interpreting absorptions that are present in an IR spectrum, it's also possible to get structural information by noticing which absorptions are not present. If the spectrum of a compound has no absorptions at 3300 and $2150 \mathrm{~cm}^{-1}$, the compound is not a terminal alkyne; if the spectrum has no absorption near 3400 $\mathrm{cm}^{-1}$, the compound is not an alcohol; and so on.


Alkanes -

The IR spectrum of an alkane is fairly uninformative because no functional groups are present and all absorptions are due to $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{C}$ bonds. Alkane $\mathrm{C}-\mathrm{H}$ bonds show a strong absorption from 2850 to 2960 $\mathrm{cm}^{-1}$, and saturated $\mathrm{C}-\mathrm{C}$ bonds show a number of bands in the 800 to $1300 \mathrm{~cm}^{-1}$ range. Since most organic compounds contain saturated alkane-like portions, most organic compounds have these characteristic IR absorptions. The $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{C}$ bands are clearly visible in the three spectra shown previously in FIGURE 12.21.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0393.jpg?height=253&width=652&top_left_y=898&top_left_x=734)


Alkenes -

Alkenes show several characteristic stretching absorptions. Vinylic $=\mathrm{C}-\mathrm{H}$ bonds absorb from 3020 to 3100 $\mathrm{cm}^{-1}$, and alkene $\mathrm{C}=\mathrm{C}$ bonds usually absorb near $1650 \mathrm{~cm}^{-1}$, although in some cases their peaks can be rather small and difficult to see clearly when the alkene is symmetric, or nearly so. Both absorptions are visible in the 1-hexene spectrum in FIGURE 12.21b.

Alkenes have characteristic $=\mathrm{C}-\mathrm{H}$ out-of-plane bending absorptions in the 700 to $1000 \mathrm{~cm}^{-1}$ range, thereby allowing the substitution pattern on a double bond to be determined (FIGURE 12.23). For example, monosubstituted alkenes such as 1-hexene show strong characteristic bands at 910 and $990 \mathrm{~cm}^{-1}$, and 1,1-disubstituted alkenes $\left(\mathrm{R}_{2} \mathrm{C}=\mathrm{CH}_{2}\right)$ have an intense band at $890 \mathrm{~cm}^{-1}$.

| Alkenes | $=\mathrm{C}-\mathrm{H}$ | $3020-3100 \mathrm{~cm}^{-1}$ |
| :--- | :--- | :--- |
|  |  |  |
|  |  | $1640-1680 \mathrm{~cm}^{-1}$ |
| $\mathrm{RCH}=\mathrm{CH}_{2}$ | 910 and $990 \mathrm{~cm}^{-1}$ |  |
| $\mathrm{R}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | $890 \mathrm{~cm}^{-1}$ |  |


Alkenes -


Alkenes -

FIGURE 12.23 C-H out-of-plane bending vibrations for substituted alkenes.


Alkynes -

Alkynes show a $\mathrm{C} \equiv \mathrm{C}$ stretching absorption at 2100 to $2260 \mathrm{~cm}^{-1}$, an absorption that is much more intense for terminal alkynes than for internal alkynes. Terminal alkynes such as 1-hexyne also have a characteristic $\equiv \mathrm{C}-\mathrm{H}$ stretching absorption at $3300 \mathrm{~cm}^{-1}$ (FIGURE 12.21c). This band is diagnostic for terminal alkynes because it is fairly intense and quite sharp.

$$
\begin{array}{lll}
\text { Alkynes } & -\mathrm{C} \equiv \mathrm{C}- & 2100-2260 \mathrm{~cm}^{-1} \\
& \equiv \mathrm{C}-\mathrm{H} & 3300 \mathrm{~cm}^{-1}
\end{array}
$$


Aromatic Compounds -

Aromatic compounds, such as benzene, have a weak C-H stretching absorption at $3030 \mathrm{~cm}^{-1}$, just to the left of a typical saturated C-H band. In addition, they have a series of weak absorptions in the 1660 to $2000 \mathrm{~cm}^{-1}$ range and a series of medium-intensity absorptions in the 1450 to $1600 \mathrm{~cm}^{-1}$ region. These latter absorptions are due to complex molecular motions of the entire ring. The $\mathrm{C}-\mathrm{H}$ out-of-plane bending region for benzene derivatives, between 650 to $1000 \mathrm{~cm}^{-1}$, gives valuable information about the ring's substitution pattern, as it does for the substitution pattern of alkenes (FIGURE 12.24).


Aromatic compounds -


Aromatic compounds -

FIGURE 12.24 C-H out-of-plane bending vibrations for substituted benzenes.
The IR spectrum of phenylacetylene, shown in Figure 12.29 at the end of this section, gives an example, clearly showing the following absorbances: $\equiv \mathrm{C}-\mathrm{H}$ stretch at $3300 \mathrm{~cm}^{-1}, \mathrm{C}-\mathrm{H}$ stretches from the benzene ring at 3000 to $3100 \mathrm{~cm}^{-1}, \mathrm{C}=\mathrm{C}$ stretches of the benzene ring between 1450 and $1600 \mathrm{~cm}^{-1}$, and out-of-plane bending of the ring's $\mathrm{C}-\mathrm{H}$ groups, indicating monosubstitution at $750 \mathrm{~cm}^{-1}$.


Alcohols -

The $\mathrm{O}-\mathrm{H}$ functional group of alcohols is easy to spot. Alcohols have a characteristic band in the range 3400 to $3650 \mathrm{~cm}^{-1}$ that is usually broad and intense. Hydrogen bonding between $\mathrm{O}-\mathrm{H}$ groups is responsible for making the absorbance so broad. If an $\mathrm{O}-\mathrm{H}$ stretch is present, it's hard to miss this band or to confuse it with anything else.

$$
\text { Alcohols } \quad-\mathrm{O}-\mathrm{H} \quad 3400-3650 \mathrm{~cm}^{-1} \text { (broad, intense) }
$$


Alcohols -

Cyclohexanol (FIGURE 12.25) is a good example.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0395.jpg?height=502&width=1387&top_left_y=1614&top_left_x=369)


Amines -

The $\mathrm{N}-\mathrm{H}$ functional group of amines is also easy to spot in the IR, with a characteristic absorption in the 3300 to $3500 \mathrm{~cm}^{-1}$ range. Although alcohols absorb in the same range, an $\mathrm{N}-\mathrm{H}$ absorption band is much sharper and less intense than an $\mathrm{O}-\mathrm{H}$ band.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0395.jpg?height=91&width=936&top_left_y=2356&top_left_x=589)


Amines -

Primary amines ( $\mathrm{R}-\mathrm{NH}_{2}$ ) have two absorbances-one for the symmetric stretching mode and one for the asymmetric mode (FIGURE 12.26). Secondary amines ( $\mathrm{R}_{2} \mathrm{~N}-\mathrm{H}$ ) only have one $\mathrm{N}-\mathrm{H}$ stretching absorbance in this
region.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0396.jpg?height=464&width=1386&top_left_y=253&top_left_x=367)


Amines -

FIGURE 12.26 IR spectrum of cyclohexylamine.


Carbonyl Compounds -

Carbonyl functional groups are the easiest to identify of all IR absorptions because of their sharp, intense peak in the range 1670 to $1780 \mathrm{~cm}^{-1}$. Most important, the exact position of absorption within this range can often be used to identify the exact kind of carbonyl functional group-aldehyde, ketone, ester, and so forth.


ALDEHYDES -

Saturated aldehydes absorb at $1730 \mathrm{~cm}^{-1}$; aldehydes next to either a double bond or an aromatic ring absorb at $1705 \mathrm{~cm}^{-1}$.


Aldehydes -


Aldehydes -

FIGURE 12.27 The IR spectrum of benzaldehyde.


KETONES -

Saturated open-chain ketones and six-membered cyclic ketones absorb at $1715 \mathrm{~cm}^{-1}$. Ring strain stiffens the $\mathrm{C}=\mathrm{O}$ bond, making five-membered cyclic ketones absorb at $1750 \mathrm{~cm}^{-1}$ and four-membered cyclic ketones absorb at $1780 \mathrm{~cm}^{-1}$, about 20 to $30 \mathrm{~cm}^{-1}$ lower than the corresponding saturated ketone.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0396.jpg?height=266&width=1126&top_left_y=2271&top_left_x=494)


ESTERS -

Saturated esters have a $C=O$ absorbance at $1735 \mathrm{~cm}^{-1}$ and two strong absorbances in the 1300 to $1000 \mathrm{~cm}^{-1}$ range from the $\mathrm{C}-\mathrm{O}$ portion of the functional group. Like other carbonyl functional groups, esters next to either an aromatic ring or a double bond absorb at $1715 \mathrm{~cm}^{-1}$, about 20 to $30 \mathrm{~cm}^{-1}$ lower than a saturated ester.


Esters -


WORKED EXAMPLE 12.5 -


Predicting IR Absorptions of Compounds -


Strategy -

Identify the functional groups in each molecule, and then check TABLE 12.1 to see where those groups absorb.
Solution
(a) Absorptions: 3400 to $3650 \mathrm{~cm}^{-1}(\mathrm{O}-\mathrm{H}), 3020$ to $3100 \mathrm{~cm}^{-1}(=\mathrm{C}-\mathrm{H}), 1640$ to $1680 \mathrm{~cm}^{-1}(\mathrm{C}=\mathrm{C})$. This molecule has an alcohol $\mathrm{O}-\mathrm{H}$ group and an alkene double bond.
(b) Absorptions: $3300 \mathrm{~cm}^{-1}(\equiv \mathrm{C}-\mathrm{H}), 2100$ to $2260 \mathrm{~cm}^{-1}(\mathrm{C} \equiv \mathrm{C}), 1735 \mathrm{~cm}^{-1}(\mathrm{C}=\mathrm{O})$. This molecule has a terminal alkyne triple bond and a saturated ester carbonyl group.


WORKED EXAMPLE 12.6 -


Identifying Functional Groups from an IR Spectrum -

The IR spectrum of an unknown compound is shown in FIGURE 12.28. What functional groups does the compound contain?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0397.jpg?height=462&width=1389&top_left_y=1840&top_left_x=365)


FIGURE 12.28 IR spectrum for Worked Example 12.6.


Strategy -

All IR spectra have many absorptions, but those useful for identifying specific functional groups are usually found in the region from $1500 \mathrm{~cm}^{-1}$ to $3300 \mathrm{~cm}^{-1}$. Pay particular attention to the carbonyl region ( 1670 to 1780
$\mathrm{cm}^{-1}$ ), the aromatic region ( 1660 to $2000 \mathrm{~cm}^{-1}$ ), the triple-bond region ( 2000 to $2500 \mathrm{~cm}^{-1}$ ), and the $\mathrm{C}-\mathrm{H}$ region (2500 to $3500 \mathrm{~cm}^{-1}$ ).


Solution -

The spectrum shows an intense absorption at $1725 \mathrm{~cm}^{-1}$ due to a carbonyl group (perhaps an aldehyde, -CHO ), a series of weak absorptions from 1800 to $2000 \mathrm{~cm}^{-1}$ characteristic of aromatic compounds, and a $\mathrm{C}-\mathrm{H}$ absorption near $3030 \mathrm{~cm}^{-1}$, also characteristic of aromatic compounds. In fact, the compound is phenylacetaldehyde.
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Solution -

Phenylacetaldehyde


Solution -

PROBLEM The IR spectrum of phenylacetylene is shown in Figure 12.29. What absorption bands can you 12-9 identify?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0398.jpg?height=475&width=1389&top_left_y=890&top_left_x=455)

FIGURE 12.29 The IR spectrum of phenylacetylene, Problem 12-9.


Solution -


Solution -

PROBLEM Where might the following compound have IR absorptions?


12-11 -


(8.0) CHEMISTRY MATTERS -


X-Ray Crystallography -

The various spectroscopic techniques described in this and the next two chapters are enormously important in chemistry and have been fine-tuned to such a degree that the structure of almost any molecule can be found. Nevertheless, wouldn't it be nice if you could simply look at a molecule and "see" its structure with your eyes?


X-Ray Crystallography -

Determining the three-dimensional shape of an object around you is easy-you just look at it, let your eyes focus the light rays reflected from the object, and let your brain assemble the data into a recognizable image. If the object is small, you use a microscope and let the microscope lens focus the visible light. Unfortunately, there is a limit to what you can see, even with the best optical microscope. Called the diffraction limit, you can't see anything smaller than the wavelength of light you are using for the observation. Visible light has wavelengths of several hundred nanometers, but atoms in molecules have dimensions on the order of 0.1 nm . Thus, to see a molecule-whether a small one in the laboratory or a large, complex enzyme with a molecular weight in the tens of thousands-you need wavelengths in the 0.1 nm range, which corresponds to X rays.


X-Ray Crystallography -

Let's say that we want to determine the structure and shape of an enzyme or other biological molecule. The technique used is called X-ray crystallography. First, the molecule is crystallized (which often turns out to be the most difficult and time-consuming part of the entire process) and a small crystal of 0.4 to 0.5 mm on its longest axis is glued to the end of a glass fiber. The fiber and attached crystal are then mounted in an instrument called an X-ray diffractometer, which consists of a radiation source, a sample positioning and orienting device that can rotate the crystal in any direction, a detector, and a controlling computer.


X-Ray Crystallography -

Once mounted in the diffractometer, the crystal is irradiated with X rays, usually so-called $\mathrm{Cu} K_{\alpha}$ radiation with a wavelength of 0.154 nm . When the X rays strike the enzyme crystal, they interact with electrons in the molecule and are scattered into a diffraction pattern which, when detected and visualized, appears as a series of intense spots against a null background.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0399.jpg?height=923&width=1028&top_left_y=1441&top_left_x=543)


X-Ray Crystallography -

FIGURE 12.30 The structure of human muscle fructose-1,6-bisphosphate aldolase, as determined by X-ray crystallography. (credit: modification of work Protein Data Bank, 1ALD. PDB ID: 1ALD, Gamblin, S.J. Davies, G.J. Grimes, J.M. Jackson, R.M. Littlechild, J.A. Watson, H.C. (1991) J. Mol. Biol. 219: 573-576, CC BY 1.0.)


X-Ray Crystallography -

Manipulation of the diffraction pattern to extract three-dimensional molecular data is a complex process, but the final result is an electron-density map of the molecule. Because electrons are largely localized around atoms, any two centers of electron density located within bonding distance of each other are assumed to represent bonded atoms, leading to a recognizable chemical structure. So important is this structural information for biochemistry that an online database of approximately 145,000 biological substances has been created. Operated by Rutgers University and funded by the U.S. National Science Foundation, the Protein Data Bank ( PDB ) is a worldwide repository for processing and distributing three-dimensional structural data for biological macromolecules. We'll see how to access the PDB in the Chapter 26 Chemistry Matters.


Key Terms -

- absorption spectrum
- amplitude
- base peak
- cation radical
- electromagnetic spectrum
- frequency, $v$
- hertz, Hz,
- infrared (IR) spectroscopy
- MALDI
- mass spectrometry (MS)
- McLafferty Rearrangement
- parent peak
- photon
- quadrupole mass analyzer
- Time of Flight (TOF)
- wavelength, $\lambda$
- wavenumber, $\widetilde{v}$


Key Terms -


Summary -

Finding the structure of a new molecule, whether a small one synthesized in the laboratory or a large protein found in living organisms, is central to the progression of chemistry and biochemistry. The structure of an organic molecule is usually determined using spectroscopic methods, including mass spectrometry and infrared spectroscopy. Mass spectrometry (MS) tells the molecular weight and formula of a molecule; infrared (IR) spectroscopy identifies the functional groups present in the molecule.


Summary -

In small-molecule mass spectrometry, molecules are first ionized by collision with a high-energy electron beam. The ions then fragment into smaller pieces, which are magnetically sorted according to their mass-to-charge ratio $(\mathrm{m} / \mathrm{z})$. The ionized sample molecule is called the molecular ion, $M^{+}$, and measurement of its mass gives the molecular weight of the sample. Structural clues about unknown samples can be obtained by interpreting the fragmentation pattern of the molecular ion. Mass-spectral fragmentations are usually complex, however, and interpretation is often difficult. In biological mass spectrometry, molecules are protonated using either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI), and the protonated molecules are separated by time-of-flight (TOF) mass analysis.


Summary -

Infrared spectroscopy involves the interaction of a molecule with electromagnetic radiation. When an organic molecule is irradiated with infrared energy, certain frequencies are absorbed by the molecule. The frequencies absorbed correspond to the amounts of energy needed to increase the amplitude of specific molecular vibrations such as bond stretching and bending. Since every functional group has a characteristic combination of bonds, every functional group has a characteristic set of infrared absorptions. For example, the terminal alkyne $\equiv \mathrm{C}-\mathrm{H}$ bond absorbs IR radiation of $3300 \mathrm{~cm}^{-1}$, and the alkene $\mathrm{C}=\mathrm{C}$ bond absorbs in the range 1640 to $1680 \mathrm{~cm}^{-1}$. By observing which frequencies of infrared radiation are absorbed by a molecule and which are not, it's possible to determine the functional groups a molecule contains.


Additional Problems -


Visualizing Chemistry -

PROBLEM Where in the IR spectrum would you expect each of the following molecules to absorb?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0401.jpg?height=354&width=1386&top_left_y=189&top_left_x=407)


Visualizing Chemistry -

PROBLEM Show the structures of the fragments you would expect in the mass spectra of the following 12-13 molecules:
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0401.jpg?height=304&width=543&top_left_y=691&top_left_x=493)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0401.jpg?height=305&width=471&top_left_y=688&top_left_x=1125)

Mass Spectrometry
PROBLEM Propose structures for compounds that fit the following mass-spectral data:
12-14 (a) A hydrocarbon with $\mathrm{M}^{+}=132$ (b) A hydrocarbon with $\mathrm{M}^{+}=166$
(c) A hydrocarbon with $\mathrm{M}^{+}=84$

PROBLEM Write molecular formulas for compounds that show the following molecular ions in their high-
12-15 resolution mass spectra, assuming that $\mathrm{C}, \mathrm{H}, \mathrm{N}$, and O might be present. The exact atomic masses are: $1.00783\left({ }^{1} \mathrm{H}\right), 12.00000\left({ }^{12} \mathrm{C}\right), 14.00307\left({ }^{14} \mathrm{~N}\right), 15.99491\left({ }^{16} \mathrm{O}\right)$.
(a) $\mathrm{M}^{+}=98.0844$
(b) $\mathrm{M}^{+}=123.0320$


Visualizing Chemistry -

PROBLEM Camphor, a saturated monoketone from the Asian camphor tree, is used among other things as a
12-16 moth repellent and as a constituent of embalming fluid. If camphor has $\mathrm{M}^{+}=152.1201$ by highresolution mass spectrometry, what is its molecular formula? How many rings does camphor have?


Visualizing Chemistry -

PROBLEM The nitrogen rule of mass spectrometry says that a compound containing an odd number of
12-17 nitrogens has an odd-numbered molecular ion. Conversely, a compound containing an even number of nitrogens has an even-numbered $\mathrm{M}^{+}$peak. Explain.


Visualizing Chemistry -

PROBLEM In light of the nitrogen rule mentioned in Problem 12-17, what is the molecular formula of pyridine,
12-18 $\mathrm{M}^{+}=79$ ?
PROBLEM Nicotine is a diamino compound isolated from dried tobacco leaves. Nicotine has two rings and
12-19 $\mathrm{M}^{+}=162.1157$ by high-resolution mass spectrometry. Give a molecular formula for nicotine, and calculate the number of double bonds.


Visualizing Chemistry -

PROBLEM The hormone cortisone contains $\mathrm{C}, \mathrm{H}$, and O , and shows a molecular ion at $\mathrm{M}^{+}=360.1937$ by
$\mathbf{1 2 - 2 0}$ high-resolution mass spectrometry. What is the molecular formula of cortisone? (The degree of unsaturation for cortisone is 8 .)


Visualizing Chemistry -

PROBLEM Halogenated compounds are particularly easy to identify by their mass spectra because both
12-21 chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as ${ }^{35} \mathrm{Cl}$ ( $75.8 \%$ ) and ${ }^{37} \mathrm{Cl}(24.2 \%)$; and bromine occurs as ${ }^{79} \mathrm{Br}(50.7 \%)$ and ${ }^{81} \mathrm{Br}(49.3 \%)$. At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion?
(a) Bromomethane, $\mathrm{CH}_{3} \mathrm{Br}$
(b) 1-Chlorohexane, $\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}$


Visualizing Chemistry -

PROBLEM By knowing the natural abundances of minor isotopes, it's possible to calculate the relative heights
12-22 of $M^{+}$and $M+1$ peaks. If ${ }^{13} \mathrm{C}$ has a natural abundance of $1.10 \%$, what are the relative heights of the $M^{+}$and $M+1$ peaks in the mass spectrum of benzene, $C_{6} H_{6}$ ?


Visualizing Chemistry -

PROBLEM Propose structures for compounds that fit the following data:
12-23 (a) A ketone with $\mathrm{M}^{+}=86$ and fragments at $m / z=71$ and $m / z=43$
(b) An alcohol with $\mathrm{M}^{+}=88$ and fragments at $m / z=73, m / z=70$, and $m / z=59$


Visualizing Chemistry -

PROBLEM 2-Methylpentane $\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)$ has the mass spectrum shown. Which peak represents $\mathrm{M}^{+}$? Which is the
12-24 base peak? Propose structures for fragment ions of $m / z=71,57,43$, and 29 . Why does the base peak have the mass it does?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0402.jpg?height=476&width=1359&top_left_y=651&top_left_x=410)


Visualizing Chemistry -

PROBLEM Assume that you are in a laboratory carrying out the catalytic hydrogenation of cyclohexene to 12-25 cyclohexane. How could you use a mass spectrometer to determine when the reaction is finished?


Visualizing Chemistry -


Infrared Spectroscopy -

PROBLEM How might you use IR spectroscopy to distinguish among the three isomers 1-butyne, 12-27 1,3-butadiene, and 2-butyne?


Infrared Spectroscopy -

PROBLEM Would you expect two enantiomers such as ( $R$ )-2-bromobutane and ( $S$ )-2-bromobutane to have
12-28 identical or different IR spectra? Explain.
PROBLEM Would you expect two diastereomers such as meso-2,3-dibromobutane and ( $2 R, 3 R$ )-12-29 dibromobutane to have identical or different IR spectra? Explain.


Infrared Spectroscopy -

PROBLEM Propose structures for compounds that meet the following descriptions:
12-30 (a) $\mathrm{C}_{5} \mathrm{H}_{8}$, with IR absorptions at 3300 and $2150 \mathrm{~cm}^{-1}$
(b) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$, with a strong IR absorption at $3400 \mathrm{~cm}^{-1}$
(c) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$, with a strong IR absorption at $1715 \mathrm{~cm}^{-1}$
(d) $\mathrm{C}_{8} \mathrm{H}_{10}$, with IR absorptions at 1600 and $1500 \mathrm{~cm}^{-1}$


Infrared Spectroscopy -

PROBLEM How could you use infrared spectroscopy to distinguish between the following pairs of isomers?
12-31 (a) $\mathrm{HC} \equiv \mathrm{CCH}_{2} \mathrm{NH}_{2}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$ (b) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
PROBLEM Two infrared spectra are shown. One is the spectrum of cyclohexane, and the other is the spectrum
12-32 of cyclohexene. Identify them, and explain your answer.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0403.jpg?height=467&width=1380&top_left_y=208&top_left_x=481)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0403.jpg?height=467&width=1387&top_left_y=704&top_left_x=480)


Infrared Spectroscopy -

PROBLEM At what approximate positions might the following compounds show IR absorptions?


Infrared Spectroscopy -


Infrared Spectroscopy -

PROBLEM Assume that you are carrying out the dehydration of 1-methylcyclohexanol to yield
12-36 1-methylcyclohexene. How could you use infrared spectroscopy to determine when the reaction is complete?


Infrared Spectroscopy -

PROBLEM Assume that you are carrying out the base-induced dehydrobromination of 12-37 3-bromo-3-methylpentane (Section 11.7) to yield an alkene. How could you use IR spectroscopy to tell which of three possible elimination products is formed, if one includes $E / Z$ isomers?


General Problems -

PROBLEM Which is stronger, the $\mathrm{C}=\mathrm{O}$ bond in an ester ( $1735 \mathrm{~cm}^{-1}$ ) or the $\mathrm{C}=\mathrm{O}$ bond in a saturated ketone 12-38 ( $1715 \mathrm{~cm}^{-1}$ )? Explain.


General Problems -

PROBLEM Carvone is an unsaturated ketone responsible for the odor of spearmint. If carvone has $\mathrm{M}^{+}=150$ in 12-39 its mass spectrum and contains three double bonds and one ring, what is its molecular formula?
PROBLEM Carvone (Problem 12-39) has an intense infrared absorption at $1690 \mathrm{~cm}^{-1}$. What kind of ketone 12-40 does carvone contain?


General Problems -

PROBLEM The mass spectrum (a) and the infrared spectrum (b) of an unknown hydrocarbon are shown.
12-41 Propose as many structures as you can.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0404.jpg?height=483&width=1365&top_left_y=840&top_left_x=472)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0404.jpg?height=467&width=1384&top_left_y=1352&top_left_x=476)

PROBLEM The mass spectrum (a) and the infrared spectrum (b) of another unknown hydrocarbon are shown. 12-42 Propose as many structures as you can.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0404.jpg?height=488&width=1360&top_left_y=1938&top_left_x=480)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0405.jpg?height=473&width=1382&top_left_y=208&top_left_x=480)


General Problems -

PROBLEM Propose structures for compounds that meet the following descriptions:
12-43 (a) An optically active compound $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$ with an IR absorption at $1730 \mathrm{~cm}^{-1}$
(b) A non-optically active compound $\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{~N}$ with an IR absorption at $2215 \mathrm{~cm}^{-1}$


General Problems -


General Problems -

PROBLEM Grignard reagents (alkylmagnesium halides) undergo a general and very useful reaction with
12-45 ketones. Methylmagnesium bromide, for example, reacts with cyclohexanone to yield a product with the formula $\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}$. What is the structure of this product if it has an IR absorption at 3400 $\mathrm{cm}^{-1}$ ?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0405.jpg?height=131&width=431&top_left_y=1385&top_left_x=435)


General Problems -

Cyclohexanone
PROBLEM Ketones undergo a reduction when treated with sodium borohydride, $\mathrm{NaBH}_{4}$. What is the structure 12-46 of the compound produced by reaction of 2-butanone with $\mathrm{NaBH}_{4}$ if it has an IR absorption at 3400 $\mathrm{cm}^{-1}$ and $\mathrm{M}^{+}=74$ in the mass spectrum?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0405.jpg?height=113&width=392&top_left_y=1754&top_left_x=422)


General Problems -

2-Butanone
PROBLEM Nitriles, $\mathrm{R}-\mathrm{C} \equiv \mathrm{N}$, undergo a hydrolysis reaction when heated with aqueous acid. What is the
12-47 structure of the compound produced by hydrolysis of propanenitrile, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$, if it has IR absorptions from $2500-3100 \mathrm{~cm}^{-1}$ and at $1710 \mathrm{~cm}^{-1}$, and has $\mathrm{M}^{+}=74$ ?


General Problems -

PROBLEM The infrared spectrum of the compound with the following mass spectrum lacks any significant
12-48 absorption above $3000 \mathrm{~cm}^{-1}$. There is a prominent peak near $1740 \mathrm{~cm}^{-1}$ and another strong peak near $1200 \mathrm{~cm}^{-1}$. Propose a structure.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0406.jpg?height=871&width=1162&top_left_y=188&top_left_x=411)

PROBLEM The infrared spectrum of the compound with the following mass spectrum has a medium-intensity 12-49 peak at about $1650 \mathrm{~cm}^{-1}$. There is also a C-H out-of-plane bending peak near $880 \mathrm{~cm}^{-1}$. Propose a structure.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0406.jpg?height=798&width=1232&top_left_y=1233&top_left_x=414)

PROBLEM The infrared spectrum of the compound with the following mass spectrum has strong absorbances
$\mathbf{1 2 - 5 0}$ at 1584,1478 , and $1446 \mathrm{~cm}^{-1}$. Propose a structure.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0407.jpg?height=809&width=1370&top_left_y=189&top_left_x=413)


CHAPTER 13 -

Structure Determination: Nuclear Magnetic Resonance Spectroscopy
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0408.jpg?height=898&width=1885&top_left_y=451&top_left_x=239)


CHAPTER 13 -

FIGURE 13.1 NMR spectroscopy is an invaluable aid in carrying out the design and synthesis of new drugs. (credit: modification of work by Unknown/Pxhere, CCO 1.0)


CHAPTER CONTENTS -

13.1 Nuclear Magnetic Resonance Spectroscopy
13.2 The Nature of NMR Absorptions
13.3 Chemical Shifts
13.4 Chemical Shifts in ${ }^{1}$ H NMR Spectroscopy
13.5 Integration of ${ }^{1} \mathrm{H}$ NMR Absorptions: Proton Counting
13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra
$13.7^{1} \mathrm{H}$ NMR Spectroscopy and Proton Equivalence
13.8 More Complex Spin-Spin Splitting Patterns
13.9 Uses of ${ }^{1}$ H NMR Spectroscopy
$13.10{ }^{13} \mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR
13.11 Characteristics of ${ }^{13} \mathrm{C}$ NMR Spectroscopy
13.12 DEPT ${ }^{13}$ C NMR Spectroscopy
13.13 Uses of ${ }^{13} \mathrm{C}$ NMR Spectroscopy


CHAPTER CONTENTS -

WHY THIS CHAPTER? Nuclear magnetic resonance (NMR) spectroscopy has far-reaching applications in many scientific fields, particularly in chemical structure determination. Although we'll just give an overview of the subject in this chapter, focusing on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding.

Nuclear magnetic resonance (NMR) spectroscopy is the most valuable spectroscopic technique available to organic chemists. It's the method of structure determination that organic chemists usually turn to first.


CHAPTER CONTENTS -

We saw in the chapter on Structure Determination: Mass Spectrometry and Infrared Spectroscopy that mass spectrometry gives a molecule's formula and infrared spectroscopy identifies a molecule's functional groups. Nuclear magnetic resonance spectroscopy complements these other techniques by mapping a molecule's carbon-hydrogen framework. Taken together, MS, IR, and NMR make it possible to determine the structures of even very complex molecules.


CHAPTER CONTENTS -

| Mass spectrometry | Molecular size and formula |
| :--- | :--- |
| Infrared spectroscopy | Functional groups present |
| NMR spectroscopy | Map of carbon-hydrogen framework |


CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy

Many kinds of nuclei behave as if they were spinning about an axis, somewhat as the earth spins daily. Because they're positively charged, these spinning nuclei act like tiny magnets and can interact with an external magnetic field, denoted $\boldsymbol{B}_{0}$. Not all nuclei act this way, but fortunately for organic chemists, both the proton $\left({ }^{1} \mathrm{H}\right)$ and the ${ }^{13} \mathrm{C}$ nucleus do have spins. The more common ${ }^{12} \mathrm{C}$ isotope, however, does not have nuclear spin. (In speaking about NMR, the words proton and hydrogen are often used interchangeably, since a hydrogen nucleus is just a proton.) Let's see what the consequences of nuclear spin are and how we can use the results.


In the absence of an external magnetic field, the spins of magnetic nuclei are oriented randomly. When a sample containing these nuclei is placed between the poles of a strong magnet, however, the nuclei adopt specific orientations, much as a compass needle orients in the earth's magnetic field. A spinning ${ }^{1} \mathrm{H}$ or ${ }^{13} \mathrm{C}$ nucleus can orient so that its own tiny magnetic field is aligned either with (parallel to) or against (antiparallel to) the external field. The two orientations don't have the same energy, however, and aren't equally likely. The parallel orientation is slightly lower in energy by an amount that depends on the strength of the external field, making this spin state very slightly favored over the antiparallel orientation (FIGURE 13.2).
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0409.jpg?height=372&width=413&top_left_y=1321&top_left_x=585)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0409.jpg?height=373&width=478&top_left_y=1315&top_left_x=1100)

FIGURE 13.2 (a) Nuclear spins are oriented randomly in the absence of an external magnetic field but (b) have a specific orientation in the presence of an external field, $\boldsymbol{B}_{\mathbf{0}}$. Some of the spins (red) are aligned parallel to the external field while others (blue) are antiparallel. The parallel spin state is slightly lower in energy and therefore favored.


If the oriented nuclei are irradiated with electromagnetic radiation of the proper frequency, energy absorption occurs and the lower-energy spin state "flips" to the higher-energy state. When this spin-flip occurs, the magnetic nuclei are said to be in resonance with the applied radiation-hence the name nuclear magnetic resonance.


The exact frequency necessary for resonance depends both on the strength of the external magnetic field, the identity of the nucleus, and the electronic environment of the nucleus. If a very strong magnetic field is applied, the energy difference between the two spin states is larger and higher-frequency (higher-energy) radiation is required for a spin-flip. If a weaker magnetic field is applied, less energy is required to effect the transition between nuclear spin states (FIGURE 13.3).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0410.jpg?height=500&width=1381&top_left_y=192&top_left_x=369)

FIGURE 13.3 The energy difference $\Delta E$ between nuclear spin states depends on the strength of the applied magnetic field. Absorption of energy with frequency $v$ converts a nucleus from a lower spin state to a higher spin state. (a) Spin states have equal energies in the absence of an applied magnetic field but (b) have unequal energies in the presence of a magnetic field. At $v=200 \mathrm{MHz}, \Delta E=8.0 \times 10^{-5} \mathrm{~kJ} / \mathrm{mol}(1.9$ $\left.\times 10^{-5} \mathrm{kcal} / \mathrm{mol}\right)$. (c) The energy difference between spin states is greater at larger applied fields. At $v=500 \mathrm{MHz}, \Delta E=2.0 \times 10^{-4} \mathrm{~kJ} / \mathrm{mol}$.


In practice, superconducting magnets that produce enormously powerful fields up to 23.5 tesla (T) are sometimes used, but field strengths in the range of 4.7 to 7.0 T are more common. At a magnetic field strength of 4.7 T , so-called radiofrequency (rf) energy in the 200 MHz range ( $1 \mathrm{MHz}=10^{6} \mathrm{~Hz}$ ) brings a ${ }^{1} \mathrm{H}$ nucleus into resonance, and rf energy of 50 MHz brings a ${ }^{13} \mathrm{C}$ nucleus into resonance. At the highest field strength currently available in commercial instruments ( 23.5 T ), 1000 MHz energy is required for ${ }^{1} \mathrm{H}$ spectroscopy. These energies needed for NMR are much smaller than those required for IR spectroscopy; 200 MHz rf energy corresponds to only $8.0 \times 10^{-5} \mathrm{~kJ} / \mathrm{mol}$ versus the 4.8 to $48 \mathrm{~kJ} / \mathrm{mol}$ needed for IR spectroscopy.
${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ nuclei are not unique in their ability to exhibit the NMR phenomenon. All nuclei with an odd number of protons ( ${ }^{1} \mathrm{H},{ }^{2} \mathrm{H},{ }^{14} \mathrm{~N},{ }^{19} \mathrm{~F},{ }^{31} \mathrm{P}$, for example) and all nuclei with an odd number of neutrons $\left({ }^{13} \mathrm{C}\right.$, for example) show magnetic properties. Only nuclei with even numbers of both protons and neutrons $\left({ }^{12} \mathrm{C},{ }^{16} \mathrm{O},{ }^{32} \mathrm{~S}\right)$ do not give rise to magnetic phenomena (TABLE 13.1).


| TABLE 13.1 The NMR Behavior of Some Common |  |
| :---: | :---: |
| Nuclei |  |
| Magnetic nuclei | Nonmagnetic nuclei |
| ${ }^{1} \mathrm{H}$ | ${ }^{12} \mathrm{C}$ |
| ${ }^{13} \mathrm{C}$ | ${ }^{16} \mathrm{O}$ |
| ${ }^{2} \mathrm{H}$ | ${ }^{32} \mathrm{~S}$ |
| ${ }^{14} \mathrm{~N}$ |  |
| ${ }^{19} \mathrm{~F}$ |  |
| ${ }^{31} \mathrm{P}$ |  |


PROBLEM The amount of energy required to spin-flip a nucleus depends both on the strength of the external 13-1 magnetic field and on the nucleus. At a field strength of 4.7 T , rf energy of 200 MHz is required to bring a ${ }^{1} \mathrm{H}$ nucleus into resonance, but energy of only 187 MHz will bring a ${ }^{19} \mathrm{~F}$ nucleus into resonance. Calculate the amount of energy required to spin-flip a ${ }^{19} \mathrm{~F}$ nucleus. Is this amount greater or less than that required to spin-flip a ${ }^{1} \mathrm{H}$ nucleus?


PROBLEM Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at 300 13-2 MHz. Does increasing the spectrometer frequency from 200 to 300 MHz increase or decrease the amount of energy necessary for resonance?