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.

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.

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


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


Solution - 12.7 Interpreting Infrared Spectra
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 |

TABLE 12.1 Characteristic IR Absorptions of Some Functional Groups


Solution - 12.7 Interpreting Infrared Spectra
TABLE 12.1 Characteristic IR Absorptions of Some Functional Groups

| 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)

(b)

(c)

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.


Solution - 12.7 Interpreting Infrared Spectra
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.


Solution - 12.7 Interpreting Infrared Spectra
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.


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 $1660 \mathrm{~cm}^{-1}$.

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 $3100 \mathrm{~cm}^{-1}$

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.


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 $990 \mathrm{~cm}^{-1}$ |  |
| $\mathrm{R}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | $890 \mathrm{~cm}^{-1}$ |  |

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 -
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.

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

Cyclohexanol (FIGURE 12.25) is a good example.


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.

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.

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 -
$1730 \mathrm{~cm}^{-1}$
$1705 \mathrm{~cm}^{-1}$
$1705 \mathrm{~cm}^{-1}$
The $\mathrm{C}-\mathrm{H}$ group attached to the carbonyl is responsible for the characteristic IR absorbance for aldehydes at 2750 and $2850 \mathrm{~cm}^{-1}$ (FIGURE 12.27). Although these are not very intense, the absorbance at $2750 \mathrm{~cm}^{-1}$ is helpful when trying to distinguish between an aldehyde and a ketone.

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.


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 -
$1735 \mathrm{~cm}^{-1}$
$1715 \mathrm{~cm}^{-1}$

$1715 \mathrm{~cm}^{-1}$


Predicting IR Absorptions of Compounds -
Where might the following compounds have IR absorptions?


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.


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?

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.

Phenylacetaldehyde

PROBLEM The IR spectrum of phenylacetylene is shown in Figure 12.29. What absorption bands can you 12-9 identify?

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

PROBLEM Where might the following compounds have IR absorptions?
12-10 (a)

(b)

(c)

PROBLEM Where might the following compound have IR absorptions?


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?

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.

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.

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.


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.)

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, $\widetilde{v}$


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.

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.

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.


Visualizing Chemistry -
PROBLEM Where in the IR spectrum would you expect each of the following molecules to absorb?

PROBLEM Show the structures of the fragments you would expect in the mass spectra of the following 12-13 molecules:
(a)

(b)

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) $\mathrm{M}^{+}=123.0320$

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?

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.

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 .)

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, $\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}$

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 $m / z=59$

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?

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?

PROBLEM What fragments might you expect in the mass spectra of the following compounds?
12-26 (a)

(b)

(c)


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?

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.

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 $1500 \mathrm{~cm}^{-1}$

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)

(b)


Infrared Spectroscopy -
(b)

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

12-33 (a)

(b)

(c)

(d)

(e)

PROBLEM How would you use infrared spectroscopy to distinguish between the following pairs of 12-34 constitutional isomers?
(a)

and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$
(b)

and

(c)
and
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
PROBLEM At what approximate positions might the following compounds show IR absorptions?
12-35 (a)

(b)

(c)

(d)

(e)

(f)

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?

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.

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?

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)

(b)

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)

(b)

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 $2215 \mathrm{~cm}^{-1}$

PROBLEM 4-Methyl-2-pentanone and 3-methylpentanal are isomers. Explain how you could tell them apart,
12-44 both by mass spectrometry and by infrared spectroscopy.

4-Methyl-2-pentanone

3-Methylpentanal


General Problems -
4-Methyl-2-pentanone

3-Methylpentanal

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 $\mathrm{cm}^{-1}$ ?

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?

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 $\mathrm{M}^{+}=74$ ?

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.


General Problems -
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.

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.


CHAPTER 13 -
Structure Determination: Nuclear Magnetic Resonance Spectroscopy

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

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.

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.

| 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)

(b)

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.


CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy
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).

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 $v=500 \mathrm{MHz}, \Delta E=2.0 \times 10^{-4} \mathrm{~kJ} / \mathrm{mol}$.


CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy
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).


CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy
| 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?


CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions
From the description thus far, you might expect all ${ }^{1} \mathrm{H}$ nuclei in a molecule to absorb energy at the same frequency and all ${ }^{13} \mathrm{C}$ nuclei to absorb at the same frequency. If so, we would observe only a single NMR absorption signal in the ${ }^{1} \mathrm{H}$ or ${ }^{13} \mathrm{C}$ spectrum of a molecule, a situation that would be of little use. In fact, the absorption frequency is not the same for all ${ }^{1} \mathrm{H}$ or all ${ }^{13} \mathrm{C}$ nuclei.

All nuclei in molecules are surrounded by electrons. When an external magnetic field is applied to a molecule, the electrons moving around nuclei set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field so that the effective field actually felt by the nucleus is a bit weaker than the applied field.

$$
B_{\text {effective }}=B_{\text {applied }}=B_{\text {local }}
$$

In describing this effect of local fields, we say that nuclei experience shielding from the full effect of the applied field by the surrounding electrons. Because each chemically distinct nucleus in a molecule is in a slightly different electronic environment, each nucleus is shielded to a slightly different extent and the effective magnetic field felt by each is slightly different. These tiny differences in the effective magnetic fields experienced by different nuclei can be detected, and we thus see a distinct NMR signal for each chemically distinct ${ }^{13} \mathrm{C}$ or ${ }^{1} \mathrm{H}$ nucleus in a molecule. As a result, an NMR spectrum effectively maps the carbon-hydrogen framework of an organic molecule. With practice, it's possible to read this map and derive structural information.


CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions
FIGURE 13.4 shows both the ${ }^{1} \mathrm{H}$ and the ${ }^{13} \mathrm{C}$ NMR spectra of methyl acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}$. The horizontal axis shows the effective field strength felt by the nuclei, and the vertical axis indicates the intensity of absorption of rf energy. Each peak in the NMR spectrum corresponds to a chemically distinct ${ }^{1} \mathrm{H}$ or ${ }^{13} \mathrm{C}$ nucleus in the molecule. Note that NMR spectra are formatted with the zero absorption line at the bottom, whereas IR spectra are formatted with the zero absorption line at the top; Section 12.5 . Note also that ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ spectra can't be observed simultaneously on the same spectrometer because different amounts of energy are required to spinflip the different kinds of nuclei. The two spectra must be recorded separately.


CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions
FIGURE 13.4 (a) The ${ }^{\mathbf{1}} \mathrm{H}$ NMR spectrum and (b) the proton-decoupled ${ }^{13} \mathrm{C}$ NMR spectrum of methyl acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}$. The small peak
labeled "TMS" at the far right of each spectrum is a calibration peak, as explained in the next section.
The ${ }^{13}$ C NMR spectrum of methyl acetate in FIGURE 13.4b shows three peaks, one for each of the three chemically distinct carbon atoms in the molecule. The ${ }^{1} \mathrm{H}$ NMR spectrum in FIGURE 13.4a shows only two peaks, however, even though methyl acetate has six hydrogens. One peak is due to the $\mathrm{CH}_{3} \mathrm{C}=\mathrm{O}$ hydrogens, and the other to the $-\mathrm{OCH}_{3}$ hydrogens. Because the three hydrogens in each methyl group have the same electronic environment, they are shielded to the same extent and are said to be equivalent. Chemically equivalent nuclei always show the same absorption. The two methyl groups themselves, however, are not equivalent, so the two sets of hydrogens absorb at different positions.

The operation of a basic NMR spectrometer is illustrated in FIGURE 13.5. An organic sample is dissolved in a suitable solvent (usually deuteriochloroform, $\mathrm{CDCl}_{3}$, which has no hydrogens) and placed in a thin glass tube between the poles of a magnet. The strong magnetic field causes the ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ nuclei in the molecule to align in one of the two possible orientations, and the sample is irradiated with rf energy. If the frequency of the rf irradiation is held constant and the strength of the applied magnetic field is varied, each nucleus comes into resonance at a slightly different field strength. A sensitive detector monitors the absorption of rf energy, and its electronic signal is then amplified and displayed as a peak.


CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions
FIGURE 13.5 Schematic operation of a basic NMR spectrometer. A thin glass tube containing the sample solution is placed between the poles of a strong magnet and irradiated with rf energy.

NMR spectroscopy differs from IR spectroscopy (Section 12.6-Section 12.8) in that the timescales of the two techniques are quite different. The absorption of infrared energy by a molecule giving rise to a change in vibrational amplitude is an essentially instantaneous process (about $10^{-13} \mathrm{~s}$ ), but the NMR process is much slower (about $10^{-3} \mathrm{~s}$ ). This difference in timescales between IR and NMR spectroscopy is analogous to the difference between cameras operating at very fast and very slow shutter speeds. The fast camera (IR) takes an instantaneous picture and freezes the action. If two rapidly interconverting species are present, IR spectroscopy records the spectrum of both. The slow camera (NMR), however, takes a blurred, time-averaged picture. If two species interconverting faster than $10^{3}$ times per second are present in a sample, NMR records only a single, averaged spectrum, rather than separate spectra of the two discrete species.

Because of this blurring effect, NMR spectroscopy can be used to measure the rates and activation energies of very fast chemical processes. In cyclohexane, for example, a ring-flip (Section 4.6) occurs so rapidly at room temperature that axial and equatorial hydrogens can't be distinguished by NMR; only a single, averaged ${ }^{1} \mathrm{H}$ NMR absorption is seen for cyclohexane at $25^{\circ} \mathrm{C}$. At $-90^{\circ} \mathrm{C}$, however, the ring-flip is slowed down enough that two absorption peaks are visible, one for the six axial hydrogens and one for the six equatorial hydrogens. Knowing the temperature and the rate at which signal blurring begins to occur, it's possible to calculate that the activation energy for the cyclohexane ring-flip is $45 \mathrm{~kJ} / \mathrm{mol}(10.8 \mathrm{kcal} / \mathrm{mol})$.


CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions
PROBLEM 2-Chloropropene shows signals for three kinds of protons in its ${ }^{1}$ H NMR spectrum. Explain. 13-3


CHAPTER CONTENTS - 13.3 Chemical Shifts
NMR spectra are displayed on charts that show the applied field strength increasing from left to right (FIGURE 13.6). Thus, the left part of the chart is the low-field, or downfield, side, and the right part is the high-field, or upfield, side. Nuclei that absorb on the downfield side of the chart require a lower field strength for resonance, implying that they have less shielding. Nuclei that absorb on the upfield side require a higher field strength for resonance, implying that they have more shielding.

FIGURE 13.6 The NMR chart. The downfield, deshielded side is on the left, and the upfield, shielded side is on the right. The tetramethylsilane (TMS) absorption is used as reference point.

To define the position of an absorption, the NMR chart is calibrated and a reference point is used. In practice, a small amount of tetramethylsilane $\left[\mathrm{TMS} ;\left(\mathrm{CH}_{3}\right)_{4} \mathrm{Si}\right]$ is added to the sample so that a reference absorption peak is produced when the spectrum is run. TMS is used as reference for both ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ measurements because in both cases it produces a single peak that occurs upfield of other absorptions normally found in organic compounds. The ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ spectra of methyl acetate in FIGURE 13.4 have the TMS reference peak indicated.


CHAPTER CONTENTS - 13.3 Chemical Shifts
The position on the chart at which a nucleus absorbs is called its chemical shift. The chemical shift of TMS is set as the zero point, and other absorptions normally occur downfield, to the left on the chart. NMR charts are calibrated using an arbitrary scale called the delta ( $\boldsymbol{\delta}$ ) scale, where $1 \delta$ equals 1 part-per-million ( 1 ppm ) of the spectrometer operating frequency. For example, if we were measuring the ${ }^{1} \mathrm{H}$ NMR spectrum of a sample using an instrument operating at $200 \mathrm{MHz}, 1 \delta$ would be 1 part per million of $200,000,000 \mathrm{~Hz}$, or 200 Hz . If we were measuring the spectrum using a 500 MHz instrument, $1 \delta=500 \mathrm{~Hz}$. The following equation can be used for any absorption:

$$
\delta=\frac{\text { Observed chemical shift (number of Hz away from TMS) }}{\text { Spectrometer frequency in } \mathrm{MHz}}
$$

Although this method of calibrating NMR charts may seem complex, there's a good reason for it. As we saw earlier, the rf frequency required to bring a given nucleus into resonance depends on the spectrometer's magnetic field strength. But because there are many different kinds of spectrometers with many different magnetic field strengths available, chemical shifts given in frequency units ( Hz ) vary from one instrument to another. Thus, a resonance that occurs at 120 Hz downfield from TMS on one spectrometer might occur at 600 Hz downfield from TMS on another spectrometer with a more powerful magnet.

By using a system of measurement in which NMR absorptions are expressed in relative terms (parts per million relative to spectrometer frequency) rather than absolute terms (Hz), it's possible to compare spectra obtained
on different instruments. The chemical shift of an NMR absorption in $\delta$ units is constant, regardless of the operating frequency of the spectrometer. A ${ }^{1} \mathrm{H}$ nucleus that absorbs at $2.0 \delta$ on a 200 MHz instrument also absorbs at $2.0 \delta$ on a 500 MHz instrument.


CHAPTER CONTENTS - 13.3 Chemical Shifts
The range in which most NMR absorptions occur is quite narrow. Almost all ${ }^{1} \mathrm{H}$ NMR absorptions occur from 0 to $10 \delta$ downfield from the proton absorption of TMS, and almost all ${ }^{13} \mathrm{C}$ absorptions occur from 1 to $220 \delta$ downfield from the carbon absorption of TMS. Thus, there is a likelihood that accidental overlap of nonequivalent signals will occur. The advantage of using an instrument with higher field strength (say, 500 $\mathrm{MHz})$ rather than lower field strength $(200 \mathrm{MHz})$ is that different NMR absorptions are more widely separated at the higher field strength. The chances that two signals will accidentally overlap are therefore lessened, and interpretation of spectra becomes easier. For example, two signals that are only 20 Hz apart at $200 \mathrm{MHz}(0.1$ ppm) are 50 Hz apart at 500 MHz (still 0.1 ppm ).
PROBLEM The following ${ }^{1} \mathrm{H}$ NMR peaks were recorded on a spectrometer operating at 200 MHz . Convert each 13-4 into $\delta$ units.
(a) $\mathrm{CHCl}_{3} ; 1454 \mathrm{~Hz}$
(b) $\mathrm{CH}_{3} \mathrm{Cl} ; 610 \mathrm{~Hz}$
(c) $\mathrm{CH}_{3} \mathrm{OH} ; 693 \mathrm{~Hz}$
(d) $\mathrm{CH}_{2} \mathrm{Cl}_{2} ; 1060 \mathrm{~Hz}$


CHAPTER CONTENTS - 13.3 Chemical Shifts
PROBLEM When the ${ }^{1} \mathrm{H}$ NMR spectrum of acetone, $\mathrm{CH}_{3} \mathrm{COCH}_{3}$, is recorded on an instrument operating at 200 13-5 MHz, a single sharp resonance at $2.1 \delta$ is seen.
(a) How many hertz downfield from TMS does the acetone resonance correspond to?
(b) If the ${ }^{1} \mathrm{H}$ NMR spectrum of acetone is recorded at 500 MHz , what would the position of the absorption be in $\delta$ units?
(c) How many hertz downfield from TMS does this 500 MHz resonance correspond to?


CHAPTER CONTENTS - 13.4 Chemical Shifts in ${ }^{1} \mathrm{H}$ NMR Spectroscopy
As mentioned previously, differences in chemical shifts are caused by the small local magnetic field of electrons surrounding different nuclei. Nuclei that are more strongly shielded by electrons require a higher applied field to bring them into resonance so they absorb on the right side of the NMR chart. Nuclei that are less strongly shielded need a lower applied field for resonance so they absorb on the left of the NMR chart.

Most ${ }^{1} \mathrm{H}$ chemical shifts fall within the range 0 to $10 \delta$, which can be divided into the five regions shown in TABLE 13.2. By remembering the positions of these regions, it's often possible to tell at a glance what kinds of protons a molecule contains.

TABLE 13.3 shows the correlation of ${ }^{1} \mathrm{H}$ chemical shift with electronic environment in more detail. In general, protons bonded to saturated, $s p^{3}$-hybridized carbons absorb at higher fields, whereas protons bonded to $s p^{2}$-hybridized carbons absorb at lower fields. Protons on carbons that are bonded to electronegative atoms, such as N, O, or halogen, also absorb at lower fields.


CHAPTER CONTENTS - 13.4 Chemical Shifts in ${ }^{1} \mathrm{H}$ NMR Spectroscopy
| Type of hydrogen |  | Chemical shift ( $\mathbf{8}$ ) |
| :---: | :---: | :---: |
| Reference | $\mathrm{Si}\left(\mathrm{CH}_{3}\right)_{4}$ | 0 |
| Alkyl (primary) | $-\mathrm{CH}_{3}$ | 0.7-1.3 |
| Alkyl (secondary) | $-\mathrm{CH}_{2}-$ | 1.2-1.6 |
| Alkyl (tertiary) | <smiles>CC(C)C</smiles> | 1.4-1.8 |
| Allylic | <smiles>C=CC(C)C</smiles> | 1.6-2.2 |
| Methyl ketone | <smiles>CC(C)=O</smiles> | 2.0-2.4 |
| Aromatic methyl | $\mathrm{Ar}-\mathrm{CH}_{3}$ | 2.4-2.7 |
| Alkynyl | $-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$ | 2.5-3.0 |
| Alkyl halide | <smiles>CCC(C)C</smiles> | 2.5-4.0 |
| Alcohol | <smiles>CC(C)(C)O</smiles> | 2.5-5.0 |
| Alcohol, ether | <smiles>COC(C)C</smiles> | 3.3-4.5 |
| Vinylic | <smiles>CC=C(C)C</smiles> | 4.5-6.5 |
| Aryl | Ar-H | 6.5-8.0 |
| Aldehyde | <smiles>CC=O</smiles> | $9.7-10.0$ |
| Carboxylic acid | <smiles>CC(=O)O</smiles> | 11.0-12.0 |


Predicting Chemical Shifts in ${ }^{1}$ H NMR Spectra -
Methyl 2,2-dimethylpropanoate $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CCO}_{2} \mathrm{CH}_{3}$ has two peaks in its ${ }^{1} \mathrm{H}$ NMR spectrum. What are their approximate chemical shifts?


Strategy -
Identify the types of hydrogens in the molecule, and note whether each is alkyl, vinylic, or next to an electronegative atom. Then predict where each absorbs, using TABLE 13.3 if necessary.


Solution -
The $-\mathrm{OCH}_{3}$ protons absorb around 3.5 to $4.0 \delta$ because they are on carbon bonded to oxygen. The $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-$ protons absorb near $1.0 \delta$ because they are typical alkane-like protons.

PROBLEM Each of the following compounds has a single ${ }^{1} \mathrm{H}$ NMR peak. Approximately where would you 13-6 expect each compound to absorb?
(a)

(b)

(c)

(d)

(e)

PROBLEM Identify the different types of protons in the following molecule, and tell where you would expect
13-7 each to absorb:


Solution - 13.5 Integration of ${ }^{1} \mathrm{H}$ NMR Absorptions: Proton Counting
Look at the ${ }^{1} \mathrm{H}$ NMR spectrum of methyl 2,2-dimethylpropanoate in FIGURE 13.7. There are two peaks, corresponding to the two kinds of protons, but the peaks aren't the same size. The peak at $1.2 \delta$, due to the $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}$ - protons, is larger than the peak at $3.7 \mathrm{\delta}$, due to the $-\mathrm{OCH}_{3}$ protons.

FIGURE 13.7 The ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of methyl 2,2-dimethylpropanoate. Integrating the peaks in a stair-step manner shows that they have a $1: 3$ ratio, corresponding to the ratio of the numbers of protons $(3: 9)$ responsible for each peak. Modern instruments give a direct digital readout of relative peak areas.

The area under each peak is proportional to the number of protons causing that peak. By electronically measuring, or integrating, the area under each peak, it's possible to measure the relative numbers of the different kinds of protons in a molecule.

Modern NMR instruments provide a digital readout of relative peak areas, but an older, more visual method displays the integrated peak areas as a stair-step line, with the height of each step proportional to the area under the peak, and therefore proportional to the relative number of protons causing the peak. For example, the two steps for the peaks in methyl 2,2-dimethylpropanoate have a $1: 3$ (or $3: 9$ ) height ratio when integrated-exactly what we expect because the three $-\mathrm{OCH}_{3}$ protons are equivalent and the nine $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}$ - protons are equivalent.


Solution - 13.5 Integration of ${ }^{1} \mathrm{H}$ NMR Absorptions: Proton Counting
PROBLEM How many peaks would you expect in the ${ }^{1} \mathrm{H}$ NMR spectrum of 1,4-dimethylbenzene (para-xylene,
13-8 or $p$-xylene)? What ratio of peak areas would you expect on integration of the spectrum? Refer to Table 13.3 for approximate chemical shifts, and sketch what the spectrum would look like. (Remember from Section 2.4 that aromatic rings have two resonance forms.)

$p$-Xylene


Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra
In the ${ }^{1} H$ NMR spectra we've seen thus far, each different kind of proton in a molecule has given rise to a single peak. It often happens, though, that the absorption of a proton splits into multiple peaks, called a multiplet. For example, in the ${ }^{1} \mathrm{H}$ NMR spectrum of bromoethane shown in FIGURE 13.8, the $-\mathrm{CH}_{2} \mathrm{Br}$ protons appear as four peaks (a quartet) centered at $3.42 \delta$ and the $-\mathrm{CH}_{3}$ protons appear as three peaks (a triplet) centered at $1.68 \delta$.

FIGURE 13.8 The ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of bromoethane, $\mathbf{C H}_{\mathbf{3}} \mathbf{C H}_{\mathbf{2}} \mathbf{B r}$. The $-\mathrm{CH}_{2} \mathrm{Br}$ protons appear as a quartet at $3.42 \delta$, and the $-\mathrm{CH}_{3}$ protons appear as a triplet at $1.68 \delta$.

Called spin-spin splitting, multiple absorptions of a nucleus are caused by the interaction, or coupling, of the spins of nearby nuclei. In other words, the tiny magnetic field produced by one nucleus affects the magnetic field felt by a neighboring nucleus. Look at the $-\mathrm{CH}_{3}$ protons in bromoethane, for example. The three equivalent $-\mathrm{CH}_{3}$ protons are neighbored by two other magnetic nuclei-the two protons on the adjacent $-\mathrm{CH}_{2} \mathrm{Br}$ group. Each of the neighboring $-\mathrm{CH}_{2} \mathrm{Br}$ protons has its own nuclear spin, which can align either with or against the applied field, producing a tiny effect that is felt by the $-\mathrm{CH}_{3}$ protons.


Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra
There are three ways in which the spins of the two $-\mathrm{CH}_{2} \mathrm{Br}$ protons can align, as shown in FIGURE 13.9. If both proton spins align with the applied field, the total effective field felt by the neighboring $-\mathrm{CH}_{3}$ protons is slightly larger than it would be otherwise. Consequently, the applied field necessary to cause resonance is slightly reduced. Alternatively, if one of the $-\mathrm{CH}_{2} \mathrm{Br}$ proton spins aligns with the field and one aligns against the field, there is no effect on the neighboring $-\mathrm{CH}_{3}$ protons. (This arrangement can occur in two ways, depending on which of the two proton spins aligns which way.) Finally, if both $-\mathrm{CH}_{2} \mathrm{Br}$ proton spins align against the applied field, the effective field felt by the $-\mathrm{CH}_{3}$ protons is slightly smaller than it would be otherwise, and the applied field needed for resonance is slightly increased.

FIGURE 13.9 The origin of spin-spin splitting in bromoethane. The nuclear spins of neighboring protons, indicated by horizontal arrows, align either with or against the applied field, causing the splitting of absorptions into multiplets.

Any given molecule has only one of the three possible alignments of $-\mathrm{CH}_{2} \mathrm{Br}$ spins, but in a large collection of molecules, all three spin states are represented in a 1:2:1 statistical ratio. We therefore find that the neighboring $-\mathrm{CH}_{3}$ protons come into resonance at three slightly different values of the applied field, and we see a 1:2:1 triplet in the NMR spectrum. One resonance is a little above where it would be without coupling, one is at the same place it would be without coupling, and the third resonance is a little below where it would be without coupling.


Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra
In the same way that the $-\mathrm{CH}_{3}$ absorption of bromoethane is split into a triplet, the $-\mathrm{CH}_{2} \mathrm{Br}$ absorption is split into a quartet. The three spins of the neighboring $-\mathrm{CH}_{3}$ protons can align in four possible combinations: all three with the applied field, two with and one against (three ways), one with and two against (three ways), or all three against. Thus, four peaks are produced for the $-\mathrm{CH}_{2} \mathrm{Br}$ protons in a 1:3:3:1 ratio.

As a general rule, called the $\boldsymbol{n}+1$ rule, protons that have $n$ equivalent neighboring protons show $n+1$ peaks in their NMR spectrum. For example, the spectrum of 2-bromopropane in FIGURE 13.10 shows a doublet at 1.71 $\delta$ and a seven-line multiplet, or septet, at $4.28 \delta$. The septet is caused by splitting of the $-\mathrm{CHBr}-$ proton signal by six equivalent neighboring protons on the two methyl groups ( $n=6$ leads to $6+1=7$ peaks). The doublet is due to signal splitting of the six equivalent methyl protons by the single $-\mathrm{CHBr}-$ proton ( $n=1$ leads to 2 peaks). Integration confirms the expected $6: 1$ ratio.

FIGURE 13.10 The ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of 2-bromopropane. The $-\mathrm{CH}_{3}$ proton signal at $1.71 \delta$ is split into a doublet, and the $-\mathrm{CHBr}-\mathrm{proton}$ signal at $4.28 \delta$ is split into a septet. Note that the distance between peaks is the same in both multiplets. Note also that the two outer peaks of the septet are small enough to be nearly missed.


Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra
The distance between peaks in a multiplet is called the coupling constant and is denoted J. Coupling constants are measured in hertz and generally fall in the range 0 to 18 Hz . The exact value of the coupling constant between two neighboring protons depends on the geometry of the molecule, but a typical value for an openchain alkane is $J=6$ to 8 Hz . The same coupling constant is shared by both groups of hydrogens whose spins are coupled and is independent of spectrometer field strength. In bromoethane, for instance, the $-\mathrm{CH}_{2} \mathrm{Br}$ protons
are coupled to the $-\mathrm{CH}_{3}$ protons and appear as a quartet with $J=7 \mathrm{~Hz}$. The $-\mathrm{CH}_{3}$ protons appear as a triplet with the same $J=7 \mathrm{~Hz}$ coupling constant.

Because coupling is a reciprocal interaction between two adjacent groups of protons, it's sometimes possible to tell which multiplets in a complex NMR spectrum are related to each other. If two multiplets have the same coupling constant, they are probably related, and the protons causing those multiplets are therefore adjacent in the molecule.

The most commonly observed coupling patterns and the relative intensities of lines in their multiplets are listed in TABLE 13.4. Note that it's not possible for a given proton to have five equivalent neighboring protons. (Why not?) A six-line multiplet, or sextet, is therefore found only when a proton has five nonequivalent neighboring protons that coincidentally happen to be coupled with an identical coupling constant $J$.

TABLE 13.4 Some Common Spin Multiplicities


Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra
TABLE 13.4 Some Common Spin Multiplicities

| Number of equivalent adjacent protons | Multiplet | Ratio of intensities |
| :---: | :--- | :---: |
| 0 | Singlet | 1 |
| 1 | Doublet | $1: 1$ |
| 2 | Triplet | $1: 2: 1$ |
| 3 | Quartet | $1: 3: 3: 1$ |
| 4 | Quintet | $1: 4: 6: 4: 1$ |
| 6 | Septet | $1: 6: 15: 20: 15: 6: 1$ |

Spin-spin splitting in ${ }^{1} \mathrm{H}$ NMR can be summarized by three rules.


RULE 1 -
Chemically equivalent protons don't show spin-spin splitting. Equivalent protons may be on the same carbon or on different carbons, but their signals don't split.

Three C-H protons are chemically equivalent; no splitting occurs.

Four C-H protons are chemically equivalent; no splitting occurs.

RULE 2
The signal of a proton with $n$ equivalent neighboring protons is split into a multiplet of $n+1$ peaks with coupling constant $\boldsymbol{J}$. Protons that are farther than two carbon atoms apart don't usually couple, although they sometimes show weak coupling when they are separated by a $\pi$ bond.

Splitting observed

Splitting not usually observed

RULE 3
Two groups of protons coupled to each other have the same coupling constant, $J$.
The spectrum of para-methoxypropiophenone in FIGURE 13.11 further illustrates these three rules. The downfield absorptions at 6.91 and $7.93 \delta$ are due to the four aromatic-ring protons. There are two kinds of aromatic protons, each of which gives a signal that is split into a doublet by its neighbor. The $-\mathrm{OCH}_{3}$ signal is
unsplit and appears as a sharp singlet at $3.84 \delta$. The $-\mathrm{CH}_{2}$ - protons next to the carbonyl group appear at $2.93 \delta$ in the region expected for protons on carbon next to an unsaturated center, and their signal is split into a quartet by coupling with the protons of the neighboring methyl group. The methyl protons appear as a triplet at $1.20 \delta$ in the usual upfield region.

FIGURE 13.11 The ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of para-methoxypropiophenone.


Assigning a Chemical Structure from a ${ }^{1} \mathrm{H}$ NMR Spectrum -
Propose a structure for a compound, $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$, that fits the following ${ }^{1} \mathrm{H}$ NMR data: $0.92 \delta(3 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz}$ ), $1.20 \delta(6 \mathrm{H}$, singlet), $1.50 \delta(2 \mathrm{H}$, quartet, $J=7 \mathrm{~Hz}$, $1.64 \delta(1 \mathrm{H}$, broad singlet).


Strategy -
It's best to begin solving structural problems by calculating a molecule's degree of unsaturation (we'll see this again in Worked Example 13.4). In the present instance, a formula of $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$ corresponds to a saturated, openchain molecule, either an alcohol or an ether.

To interpret the NMR information, let's look at each absorption individually. The three-proton absorption at $0.92 \delta$ is due to a methyl group in an alkane-like environment, and the triplet-splitting pattern implies that the $\mathrm{CH}_{3}$ is next to a $\mathrm{CH}_{2}$. Thus, our molecule contains an ethyl group, $\mathrm{CH}_{3} \mathrm{CH}_{2}-$. The six-proton singlet at $1.20 \delta$ is due to two equivalent alkane-like methyl groups attached to a carbon with no hydrogens, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}$, and the twoproton quartet at $1.50 \delta$ is due to the $\mathrm{CH}_{2}$ of the ethyl group. All 5 carbons and 11 of the 12 hydrogens in the molecule are now accounted for. The remaining hydrogen, which appears as a broad one-proton singlet at 1.64 $\delta$, is probably due to an OH group, since there is no other way to account for it. Putting the pieces together gives the structure: 2-methyl-2-butanol.


Solution -
2-Methyl-2-butanol

PROBLEM Predict the splitting patterns you would expect for each proton in the following molecules:
13-9 (a) $\mathrm{CHBr}_{2} \mathrm{CH}_{3}$
(b)
(f)

$\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{Br}$
(c) $\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$
(d)

(e)

PROBLEM Draw structures for compounds that meet the following descriptions:
13-10
(a) $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$; one singlet
(b) $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{Cl}$; one doublet and one septet
(c) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Cl}_{2} \mathrm{O}$; two triplets
(d) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}$; one singlet, one triplet, and one quartet

PROBLEM The integrated ${ }^{1} \mathrm{H}$ NMR spectrum of a compound of formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ is shown in Figure 13.12.
13-11 Propose a structure.

FIGURE 13.12 An integrated ${ }^{1} \mathrm{H}$ NMR spectrum for Problem 11.


13.7 ${ }^{1} \mathrm{H}$ NMR Spectroscopy and Proton Equivalence -
Because each electronically distinct hydrogen in a molecule has its own unique absorption, one use of ${ }^{1} \mathrm{H}$ NMR is to find out how many kinds of electronically nonequivalent hydrogens are present. In the ${ }^{1} \mathrm{H}$ NMR spectrum of methyl acetate shown previously in FIGURE 13.4a, for instance, there are two signals, corresponding to the two kinds of nonequivalent protons present, $\mathrm{CH}_{3} \mathrm{C}=\mathrm{O}$ protons and $-\mathrm{OCH}_{3}$ protons.

For relatively small molecules, a quick look at the structure is often enough to decide how many kinds of protons are present and thus how many NMR absorptions might appear. If in doubt, though, the equivalence or nonequivalence of two protons can be determined by comparing the structures that would be formed if each hydrogen were replaced by an X group. There are four possibilities.

- One possibility is that the protons are chemically unrelated and thus nonequivalent. If so, the products formed on substitution of H by X would be different constitutional isomers. In butane, for instance, the $-\mathrm{CH}_{3}$ protons are different from the $-\mathrm{CH}_{2}$ - protons. They therefore give different products on substitution by X than the $-\mathrm{CH}_{2}$ protons and would likely show different NMR absorptions.

The $-\mathrm{CH}_{2}$ - and $-\mathrm{CH}_{3}$ hydrogens are unrelated and have different NMR absorptions.

The two substitution products are constitutional isomers.

- A second possibility is that the protons are chemically identical and thus electronically equivalent. If so, the same product would be formed regardless of which H is substituted by X . In butane, for instance, the six $-\mathrm{CH}_{3}$ hydrogens on C 1 and C 4 are identical, would give the identical structure on substitution by X , and would show an identical NMR absorption. Such protons are said to be homotopic.


13.7 ${ }^{1} \mathrm{H}$ NMR Spectroscopy and Proton Equivalence -
- The third possibility is a bit more subtle. Although they might at first seem homotopic, the two $-\mathrm{CH}_{2}-$ hydrogens on C 2 in butane (and the two $-\mathrm{CH}_{2}$ - hydrogens on C 3 ) are in fact not identical. Substitution by X of a hydrogen at C2 (or C3) would form a new chirality center, so different enantiomers (Section 5.1) would result, depending on whether the pro- $R$ or pro- $S$ hydrogen had been substituted (Section 5.11). Such hydrogens, whose substitution by X would lead to different enantiomers, are said to be enantiotopic. Enantiotopic hydrogens, even though not identical, are nevertheless electronically equivalent and thus have the same NMR absorption.

or

The two hydrogens on C2 (and
The two possible substitution
products are enantiomers.
on C3) are enantiotopic and have the same NMR absorption.

- The fourth possibility arises in chiral molecules, such as ( $R$ )-2-butanol. The two $-\mathrm{CH}_{2}$ - hydrogens at C 3 are neither homotopic nor enantiotopic. Because substitution of a hydrogen at C3 would form a second chirality center, different diastereomers (Section 5.6) would result, depending on whether the pro- $R$ or pro- $S$ hydrogen had been substituted. Such hydrogens, whose substitution by X leads to different diastereomers, are said to be diastereotopic. Diastereotopic hydrogens are neither chemically nor electronically equivalent. They are completely different and would likely show different NMR absorptions.

The two hydrogens on C3 are diastereotopic and have different NMR absorptions.

The two possible substitution products are diastereomers.

PROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:
13-12 (a)

(b)

(c)

(d)

(e)

(f)


13.7 ${ }^{1} \mathrm{H}$ NMR Spectroscopy and Proton Equivalence -
PROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:
13-12 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM How many kinds of electronically nonequivalent protons are present in each of the following
13-13 compounds, and thus how many NMR absorptions might you expect in each?
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$
(b) $\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NO}_{2}$
(d) Methylbenzene
(e) 2-Methyl-1-butene (f) cis-3-Hexene

PROBLEM How many absorptions would you expect ( $S$ )-malate, an intermediate in carbohydrate metabolism, 13-14 to have in its ${ }^{1} \mathrm{H}$ NMR spectrum? Explain.


(S)-Malate - 13.8 More Complex Spin-Spin Splitting Patterns
In the ${ }^{1} \mathrm{H}$ NMR spectra we've seen so far, the chemical shifts of different protons have been distinct and the spin-spin splitting patterns have been straightforward. It often happens, however, that different kinds of hydrogens in a molecule have accidentally overlapping signals. The spectrum of toluene (methylbenzene) in FIGURE 13.13, for example, shows that the five aromatic ring protons give a complex, overlapping pattern, even though they aren't all equivalent.

FIGURE 13.13 The ${ }^{1}$ H NMR spectrum of toluene, showing the accidental overlap of the five nonequivalent aromatic ring protons.
Yet another complication in ${ }^{1} \mathrm{H}$ NMR spectroscopy arises when a signal is split by two or more nonequivalent kinds of protons, as is the case with trans-cinnamaldehyde, isolated from oil of cinnamon (FIGURE 13.14). Although the $n+1$ rule predicts splitting caused by equivalent protons, splittings caused by nonequivalent protons are more complex.

FIGURE 13.14 The ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of trans-cinnamaldehyde. The signal of the proton at C 2 (blue) is split into four peaks-a doublet of doublets-by the two nonequivalent neighboring protons.

To understand the ${ }^{1} \mathrm{H}$ NMR spectrum of trans-cinnamaldehyde, we have to isolate the different parts and look at the signal of each proton individually.


(S)-Malate - 13.8 More Complex Spin-Spin Splitting Patterns
To understand the ${ }^{1} \mathrm{H}$ NMR spectrum of trans-cinnamaldehyde, we have to isolate the different parts and look at the signal of each proton individually.

- The five aromatic proton signals (black in FIGURE 13.14) overlap into a complex pattern with a large peak at $7.42 \delta$ and a broad absorption at $7.57 \delta$.
- The aldehyde proton signal at C1 (red) appears in the normal downfield position at $9.69 \delta$ and is split into a doublet with $J=6 \mathrm{~Hz}$ by the adjacent proton at C 2 .
- The vinylic proton at C3 (green) is next to the aromatic ring and is therefore shifted downfield from the normal vinylic region. This C3 proton signal appears as a doublet centered at $7.49 \delta$. Because it has one neighbor proton at C 2 , its signal is split into a doublet, with $J=12 \mathrm{~Hz}$.
- The C2 vinylic proton signal (blue) appears at $6.73 \delta$ and shows an interesting, four-line absorption pattern. It is coupled to the two nonequivalent protons at C 1 and C 3 with two different coupling constants: $J_{1-2}=6$ Hz and $J_{2-3}=12 \mathrm{~Hz}$.

A good way to understand the effect of multiple coupling, such as that occurring for the C2 proton of trans-cinnamaldehyde, is to draw a tree diagram, like that in FIGURE 13.15. The diagram shows the individual effect of each coupling constant on the overall pattern. Coupling with the C3 proton splits the signal of the C2 proton in trans-cinnamaldehyde into a doublet with $J=12 \mathrm{~Hz}$. Further coupling with the aldehyde proton then splits each peak of the doublet into new doublets with $J=6 \mathrm{~Hz}$, and we therefore observe a four-line spectrum for the C 2 proton.

$6.73 \delta$


(S)-Malate - 13.8 More Complex Spin-Spin Splitting Patterns
$6.73 \delta$

FIGURE 13.15 A tree diagram for the C2 proton of trans-cinnamaldehyde shows how it is coupled to the C1 and C3 protons with different coupling constants.

One further trait evident in the cinnamaldehyde spectrum is that the four peaks of the C 2 proton signal are not all the same size. The two left-hand peaks are somewhat larger than the two right-hand peaks. Such a size difference occurs whenever coupled nuclei have similar chemical shifts-in this case, $7.49 \delta$ for the C3 proton and $6.73 \delta$ for the C 2 proton. The peaks nearer the signal of the coupled partner are always larger, and the peaks farther from the signal of the coupled partner are always smaller. Thus, the left-hand peaks of the C2 proton multiplet at $6.73 \delta$ are closer to the C3 proton absorption at $7.49 \delta$ and are larger than the right-hand peaks. At the same time, the right-hand peak of the C3 proton doublet at $7.49 \delta$ is larger than the left-hand peak because it is closer to the C 2 proton multiplet at $6.73 \delta$. This skewing effect on multiplets can often be useful because it tells where to look in the spectrum to find the coupled partner: look in the direction of the larger peaks.

PROBLEM 3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which the vinylic proton at C2 is
13-15 coupled with both the C 1 vinylic proton $(J=16 \mathrm{~Hz}$ ) and the C 3 methylene protons ( $J=8 \mathrm{~Hz}$ ). Draw a tree diagram for the C 2 proton signal, and account for the fact that a five-line multiplet is observed.


3-Bromo-1-phenyl-1-propene - 13.9 Uses of ${ }^{1} \mathrm{H}$ NMR Spectroscopy
NMR is used to help identify the product of nearly every reaction run in the laboratory. For example, we said in Section 8.5 that hydroboration-oxidation of alkenes occurs with non-Markovnikov regiochemistry to yield the less highly substituted alcohol. With the help of NMR, we can now prove this statement.
Does hydroboration-oxidation of methylenecyclohexane yield cyclohexylmethanol or 1-methylcyclohexanol?

Methylenecyclohexane
Cyclohexylmethanol 1-Methylcyclohexanol
The ${ }^{1} \mathrm{H}$ NMR spectrum of the reaction product is shown in FIGURE 13.16a. The spectrum shows a two-proton peak at $3.40 \delta$, indicating that the product has a $-\mathrm{CH}_{2}$ - group bonded to an electronegative oxygen atom $\left(-\mathrm{CH}_{2} \mathrm{OH}\right)$. Furthermore, the spectrum shows no large three-proton singlet absorption near $1 \delta$, where we would expect the signal of a quaternary $-\mathrm{CH}_{3}$ group to appear. (FIGURE 13.16b) gives the spectrum of 1-methylcyclohexanol, the alternative product.) Thus, it's clear that cyclohexylmethanol is the reaction product.

FIGURE 13.16 (a) The ${ }^{\mathbf{1}} \mathrm{H}$ NMR spectrum of cyclohexylmethanol, the product from hydroboration-oxidation of methylenecyclohexane, and (b) the ${ }^{1} H$ NMR spectrum of 1-methylcyclohexanol, the possible alternative reaction product.

PROBLEM How could you use ${ }^{1} \mathrm{H}$ NMR to determine the regiochemistry of electrophilic addition to alkenes?
13-16 For example, does addition of HCl to 1-methylcyclohexene yield 1-chloro-1-methylcyclohexane or 1-chloro-2-methylcyclohexane?


$13.10{ }^{13} \mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR -
In some ways, it's surprising that carbon NMR is even possible. After all, ${ }^{12} \mathrm{C}$, the most abundant carbon isotope, has no nuclear spin and can't be seen by NMR. Carbon-13 is the only naturally occurring carbon isotope with a nuclear spin, but its natural abundance is only $1.1 \%$. Thus, only about 1 of every 100 carbon atoms in an organic sample can be observed by NMR. The problem of low abundance has been overcome, however, by the use of signal averaging and Fourier-transform NMR (FT-NMR). Signal averaging increases instrument sensitivity, and FT-NMR increases instrument speed.

The low natural abundance of ${ }^{13} \mathrm{C}$ means that any individual NMR spectrum is extremely "noisy." That is, the signals are so weak that they are cluttered with random background electronic noise, as shown in FIGURE 13.17a. If, however, hundreds or thousands of individual runs are added together by a computer and then averaged, a greatly improved spectrum results (FIGURE 13.17b). Background noise, because of its random nature, increases very slowly as the runs are added, while the nonzero signals stand out clearly. Unfortunately,
the value of signal averaging is limited when using the method of NMR spectrometer operation described in Section 13.2, because it takes about 5 to 10 minutes to obtain a single spectrum. Thus, a faster way to obtain spectra is needed if signal averaging is to be used.

FIGURE 13.17 Carbon-13 NMR spectra of 1-pentanol, $\mathbf{C H}_{\mathbf{3}} \mathbf{C H}_{\mathbf{2}} \mathbf{C H}_{\mathbf{2}} \mathbf{C H}_{\mathbf{2}} \mathbf{C H}_{\mathbf{2}} \mathbf{O H}$. Spectrum (a) is a single run, showing the large amount of background noise. Spectrum (b) is an average of 200 runs.


$13.10{ }^{13} \mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR -
In the method of NMR spectrometer operation described in Section 13.2, the rf frequency is held constant while the strength of the magnetic field is varied so that all signals in the spectrum are recorded sequentially. In the FT-NMR technique used by modern spectrometers, however, all the signals are recorded simultaneously. A sample is placed in a magnetic field of constant strength and is irradiated with a short pulse of rf energy that covers the entire range of useful frequencies. All ${ }^{1} \mathrm{H}$ or ${ }^{13} \mathrm{C}$ nuclei in the sample resonate at once, giving a complex, composite signal that is mathematically manipulated using so-called Fourier transforms and then displayed in the usual way. Because all resonance signals are collected at once, it takes only a few seconds rather than a few minutes to record an entire spectrum.

Combining the speed of FT-NMR with the sensitivity enhancement of signal averaging is what gives modern NMR spectrometers their power. Literally thousands of spectra can be taken and averaged in a few hours, resulting in sensitivity so high that a ${ }^{13} \mathrm{C}$ NMR spectrum can be obtained from less than 0.1 mg of sample and a ${ }^{1} \mathrm{H}$ spectrum can be recorded from only a few micrograms.

One further question needs to be answered before moving forward with our discussion of ${ }^{13} \mathrm{C}$ NMR. Why is spin-spin splitting seen only for ${ }^{1} \mathrm{H}$ NMR? Why is there no splitting of carbon signals into multiplets in ${ }^{13} \mathrm{C}$ NMR? After all, you might expect that the spin of a given ${ }^{13} \mathrm{C}$ nucleus would couple with the spin of an adjacent magnetic nucleus, either ${ }^{13} \mathrm{C}$ or ${ }^{1} \mathrm{H}$.


$13.10{ }^{13} \mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR -
No coupling of a ${ }^{13} \mathrm{C}$ nucleus with nearby carbons is seen because their low natural abundance makes it unlikely that two ${ }^{13} \mathrm{C}$ nuclei will be adjacent. No coupling of a ${ }^{13} \mathrm{C}$ nucleus with nearby hydrogens is seen because ${ }^{13} \mathrm{C}$ spectra are normally recorded using broadband decoupling. At the same time that the sample is irradiated with a pulse of rf energy to cover the carbon resonance frequencies, it is also irradiated by a second band of rf energy covering all the hydrogen resonance frequencies. This second irradiation makes the hydrogens spin-flip so rapidly that their local magnetic fields average to zero and no coupling with carbon spins occurs.


$13.10{ }^{13} \mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - 13.11 Characteristics of ${ }^{13}$ C NMR Spectroscopy
At its simplest, ${ }^{13} \mathrm{C}$ NMR makes it possible to count the number of different carbon atoms in a molecule. Look at the ${ }^{13} \mathrm{C}$ NMR spectra of methyl acetate and 1-pentanol shown previously in FIGURE 13.4b and FIGURE 13.17b. In each case, a single sharp resonance line is observed for each different carbon atom.
Most ${ }^{13} \mathrm{C}$ resonances are between 0 and 220 ppm downfield from the TMS reference line, with the exact chemical shift of each ${ }^{13} \mathrm{C}$ resonance dependent on that carbon's electronic environment within the molecule. FIGURE 13.18 shows the correlation of chemical shift with environment.

The factors that determine chemical shifts are complex, but it's possible to make some generalizations from the data in FIGURE 13.18. One trend is that a carbon atom's chemical shift is affected by the electronegativity of nearby atoms. Carbons bonded to oxygen, nitrogen, or halogen absorb downfield (to the left) of typical alkane carbons. Because electronegative atoms attract electrons, they pull electrons away from neighboring carbon atoms, causing those carbons to be deshielded and to come into resonance at a lower field.

Another trend is that $s p^{3}$-hybridized carbons generally absorb from 0 to $90 \delta$, while $s p^{2}$ carbons absorb from 110 to 220 . Carbonyl carbons ( $\mathrm{C}=0$ ) are particularly distinct in ${ }^{13} \mathrm{C}$ NMR and are always found at the lowfield end of the spectrum, from 160 to $220 \delta$. FIGURE 13.19 shows the ${ }^{13} \mathrm{C}$ NMR spectra of 2 -butanone and para-bromoacetophenone and indicates the peak assignments. Note that the $\mathrm{C}=\mathrm{O}$ carbons are at the left edge of the spectrum in each case.


$13.10{ }^{13} \mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - 13.11 Characteristics of ${ }^{13}$ C NMR Spectroscopy
FIGURE 13.19 Carbon-13 NMR spectra of (a) 2-butanone and (b) para-bromoacetophenone.
The ${ }^{13} \mathrm{C}$ NMR spectrum of para-bromoacetophenone is interesting in several ways. Note particularly that only six carbon absorptions are observed, even though the molecule contains eight carbons. para-bromoacetophenone has a symmetry plane that makes ring carbons 4 and $4^{\prime}$, and ring carbons 5 and $5^{\prime}$ equivalent. (Remember from Section 2.4 that aromatic rings have two resonance forms.) Thus, the six ring carbons show only four absorptions in the range 128 to $137 \delta$.

para-Bromoacetophenone
A second interesting point about both spectra in FIGURE 13.19 is that the peaks aren't uniform in size. Some peaks are larger than others even though they are one-carbon resonances (except for the two 2-carbon peaks of para-bromoacetophenone). This difference in peak size is a general feature of broadband-decoupled ${ }^{13} \mathrm{C}$ NMR spectra, and explains why we can't integrate ${ }^{13} \mathrm{C}$ NMR spectra in the same way we integrate the resonances in a ${ }^{1} \mathrm{H}$ NMR spectrum. The local environment of each carbon atom determines not only its chemical shift but also the time it takes for the nuclei to return to their equilibrium state after receiving a pulse of rf radiation and flipping their spins. Quaternary carbons, regardless of their hybridization state or substituents, typically give smaller resonances than primary, secondary, or tertiary carbons.


Predicting Chemical Shifts in ${ }^{13} \mathrm{C}$ NMR Spectra -
At what approximate positions would you expect ethyl acrylate, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$, to show ${ }^{13} \mathrm{C}$ NMR absorptions?


Strategy -
Identify the distinct carbons in the molecule, and note whether each is alkyl, vinylic, aromatic, or in a carbonyl group. Then predict where each absorbs, using FIGURE 13.18 as necessary.


Solution -
Ethyl acrylate has five chemically distinct carbons: two different $\mathrm{C}=\mathrm{C}$, one $\mathrm{C}=\mathrm{O}$, one $\mathrm{O}-\mathrm{C}$, and one alkyl C . From FIGURE 13.18, the likely absorptions are

The actual absorptions are at $14.1,60.5,128.5,130.3$, and $166.0 \delta$.

PROBLEM Predict the number of carbon resonance lines you would expect in the ${ }^{13}$ C NMR spectra of the 13-17 following compounds:
(a) Methylcyclopentane (b) 1-Methylcyclohexene (c) 1,2-Dimethylbenzene
(d) 2-Methyl-2-butene
(e)

PROBLEM Propose structures for compounds that fit the following descriptions:
13-18 (a) A hydrocarbon with seven lines in its ${ }^{13} \mathrm{C}$ NMR spectrum
(b) A six-carbon compound with only five lines in its ${ }^{13} \mathrm{C}$ NMR spectrum
(c) A four-carbon compound with three lines in its ${ }^{13} \mathrm{C}$ NMR spectrum

PROBLEM Classify the resonances in the ${ }^{13} \mathrm{C}$ NMR spectrum of methyl propanoate, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}$ (Figure 13-19 13.20).

FIGURE $13.20{ }^{13} \mathrm{C}$ NMR spectrum of methyl propanoate, Problem 19.


Solution - 13.12 DEPT ${ }^{13} \mathrm{C}$ NMR Spectroscopy
Numerous techniques developed in recent years have made it possible to obtain enormous amounts of information from ${ }^{13} \mathrm{C}$ NMR spectra. Among these techniques is one called DEPT-NMR, for distortionless enhancement by polarization transfer, which makes it possible to distinguish between signals due to $\mathrm{CH}_{3}, \mathrm{CH}_{2}$, CH , and quaternary carbons. That is, the number of hydrogens attached to each carbon in a molecule can be determined.

A DEPT experiment is usually done in three stages, as shown in FIGURE 13.21 for 6-methyl-5-hepten-2-ol. The first stage is to run an ordinary spectrum (called a broadband-decoupled spectrum) to locate the chemical shifts
of all carbons. Next, a second spectrum called a DEPT-90 is run, using special conditions under which only signals due to CH carbons appear. Signals due to $\mathrm{CH}_{3}, \mathrm{CH}_{2}$, and quaternary carbons are absent. Finally, a third spectrum called a DEPT-135 is run, using conditions under which $\mathrm{CH}_{3}$ and CH resonances appear as positive signals, $\mathrm{CH}_{2}$ resonances appear as negative signals-that is, as peaks below the baseline-and quaternary carbons are again absent.

FIGURE 13.21 DEPT-NMR spectra for 6-methyl-5-hepten-2-ol. Part (a) is an ordinary broadband-decoupled spectrum, which shows signals for all eight carbons. Part (b) is a DEPT-90 spectrum, which shows only signals for the two CH carbons. Part (c) is a DEPT-135 spectrum, which shows positive signals for the two CH and three $\mathrm{CH}_{3}$ carbons and negative signals for the two $\mathrm{CH}_{2}$ carbons.


Solution - 13.12 DEPT ${ }^{13} \mathrm{C}$ NMR Spectroscopy
Putting together the information from all three spectra makes it possible to tell the number of hydrogens attached to each carbon. The CH carbons are identified in the DEPT- 90 spectrum, the $\mathrm{CH}_{2}$ carbons are identified as negative peaks in the DEPT- 135 spectrum, the $\mathrm{CH}_{3}$ carbons are identified by subtracting the CH peaks from the positive peaks in the DEPT-135 spectrum, and quaternary carbons are identified by subtracting all peaks in the DEPT-135 spectrum from the peaks in the broadband-decoupled spectrum.

c Subtract DEPT-135 from broadband-decoupled spectrum
CH DEPT-90
$\mathrm{CH}_{2}$ Negative DEPT-135
$\mathrm{CH}_{3}$ Subtract DEPT-90 from positive DEPT-135


Assigning a Chemical Structure from a ${ }^{13}$ C NMR Spectrum -
Propose a structure for an alcohol, $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$, that has the following ${ }^{13} \mathrm{C}$ NMR spectral data:
Broadband decoupled ${ }^{13}$ C NMR: 19.0, 31.7, 69.5 ס;
DEPT-90: 31.7 ס;
DEPT-135: positive peak at $19.0 \delta$, negative peak at $69.5 \delta$.


Strategy -
As noted in Section 7.2, it usually helps with compounds of known formula but unknown structure to calculate the compound's degree of unsaturation. In the present instance, a formula of $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ corresponds to a saturated, open-chain molecule.

To gain information from the ${ }^{13} \mathrm{C}$ data, let's begin by noting that the unknown alcohol has four carbon atoms, yet has only three NMR absorptions, which implies that two of the carbons must be equivalent. Looking at chemical shifts, two of the absorptions are in the typical alkane region (19.0 and $31.7 \delta$ ), while one is in the region of a carbon bonded to an electronegative atom ( $69.5 \delta$ )-oxygen in this instance. The DEPT-90 spectrum tells us that the alkyl carbon at $31.7 \delta$ is tertiary (CH); the DEPT- 135 spectrum tells us that the alkyl carbon at $19.0 \delta$ is a methyl $\left(\mathrm{CH}_{3}\right)$ and that the carbon bonded to oxygen ( $69.5 \delta$ ) is secondary $\left(\mathrm{CH}_{2}\right)$. The two equivalent carbons are probably both methyls bonded to the same tertiary carbon, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-$. We can now put the pieces together to propose a structure: 2-methyl-1-propanol.


Solution -
PROBLEM Assign a chemical shift to each carbon in 6-methyl-5-hepten-2-ol (Figure 13.21).


13-20 -
PROBLEM Estimate the chemical shift of each carbon in the following molecule. Predict which carbons will 13-21 appear in the DEPT-90 spectrum, which will give positive peaks in the DEPT-135 spectrum, and which will give negative peaks in the DEPT-135 spectrum.

PROBLEM Propose a structure for an aromatic hydrocarbon, $\mathrm{C}_{11} \mathrm{H}_{16}$, that has the following ${ }^{13} \mathrm{C}$ NMR spectral 13-22 data:

Broadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, $139.8 \delta$
DEPT-90: 125.5, 127.5, $130.3 \delta$
DEPT-135: positive peaks at $29.5,125.5,127.5,130.3 \delta$; negative peak at $50.2 \delta$


13-20 - 13.13 Uses of ${ }^{13} \mathrm{C}$ NMR Spectroscopy
The information derived from ${ }^{13} \mathrm{C}$ NMR spectroscopy is extraordinarily useful for structure determination. Not only can we count the number of nonequivalent carbon atoms in a molecule, we can also get information about the electronic environment of each carbon and find how many protons are attached to each. As a result, we can address many structural questions that go unanswered by IR spectroscopy or mass spectrometry.

Here's an example: how do we know that the E2 reaction of an alkyl halide follows Zaitsev's rule (Section 11.7)? Does treatment of 1-chloro-1-methylcyclohexane with a strong base give predominantly the trisubstituted alkene 1-methylcyclohexene or the disubstituted alkene methylenecyclohexane?

1-Methylcyclohexene will have five $s p^{3}$-carbon resonances in the 20 to $50 \delta$ range and two $s p^{2}$-carbon resonances in the 100 to $150 \delta$ range. Methylenecyclohexane, however, because of its symmetry, will have only three $s p^{3}$-carbon resonance peaks and two $s p^{2}$-carbon peaks. The spectrum of the actual reaction product, shown in FIGURE 13.22, clearly identifies 1-methylcyclohexene as the product of this E2 reaction.
1-chloro-1-methylcyclohexane with base.

PROBLEM We saw in Section 9.3 that addition of HBr to a terminal alkyne leads to the Markovnikov addition
13-23 product, with the Br bonding to the more highly substituted carbon. How could you use ${ }^{13} \mathrm{C}$ NMR to identify the product of the addition of 1 equivalent of HBr to 1-hexyne?