Magnetic Resonance Imaging (MRI) -
As practiced by organic chemists, NMR spectroscopy is a powerful method of structure determination. A small amount of sample, typically a few milligrams or less, is dissolved in a small amount of solvent, the solution is placed in a thin glass tube, and the tube is placed into the narrow ( $1-2 \mathrm{~cm}$ ) gap between the poles of a strong magnet. Imagine, though, that a much larger NMR instrument were available. Instead of a few milligrams, the sample size could be tens of kilograms; instead of a narrow gap between magnet poles, the gap could be large enough for a whole person to climb into so that an NMR spectrum of body parts could be obtained. That large instrument is exactly what's used for magnetic resonance imaging (MRI), a diagnostic technique of enormous value to the medical community.

Like NMR spectroscopy, MRI takes advantage of the magnetic properties of certain nuclei, typically hydrogen, and of the signals emitted when those nuclei are stimulated by radiofrequency energy. Unlike what happens in NMR spectroscopy, though, MRI instruments use data manipulation techniques to look at the threedimensional location of magnetic nuclei in the body rather than at the chemical nature of the nuclei. As noted, most MRI instruments currently look at hydrogen, present in abundance wherever there is water or fat in the body.

The signals detected by MRI vary with the density of hydrogen atoms and with the nature of their surroundings, allowing identification of different types of tissue and even allowing the visualization of motion. For example, the volume of blood leaving the heart in a single stroke can be measured, and heart motion can be observed. Soft tissues that don't show up well on X-ray images can be seen clearly, allowing diagnosis of brain tumors, strokes, and other conditions. This technique is also valuable in diagnosing damage to knees or other joints and is a noninvasive alternative to surgical explorations.

Several types of atoms in addition to hydrogen can be detected by MRI, and the applications of images based on
${ }^{31} \mathrm{P}$ atoms are being explored. This approach holds great promise for studies of metabolism.


Magnetic Resonance Imaging (MRI) -
Several types of atoms in addition to hydrogen can be detected by MRI, and the applications of images based on
${ }^{31} \mathrm{P}$ atoms are being explored. This approach holds great promise for studies of metabolism.

FIGURE 13.23 If you're a runner, you really don't want this to happen to you. The MRI of this left knee shows bone formation in the tissue in a patient with osteochondroma. (credit: "Osteochondroma MRI" by Michael R Carmont, Sian Davies, Daniel Gey van Pittius and Robin Rees/ Wikimedia Commons, CC BY 2.0)


Key Terms -
```
• chemical shift • homotopic
- coupling
- integration
- coupling constant (J)
- multiplet
- delta (\delta) scale
- n+1 rule
- DEPT-NMR
- nuclear magnetic resonance (NMR) spectroscopy
- diastereotopic
- shielding
- downfield
- spin-spin splitting
- enantiotopic
- upfield
- FT-NMR
```


Summary -
Nuclear magnetic resonance spectroscopy, or NMR, is the most valuable of the numerous spectroscopic techniques used for structure determination. Although we focused in this chapter on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding.

When magnetic nuclei, such as ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$, are placed in a strong magnetic field, their spins orient either with or against the field. On irradiation with radiofrequency (rf) waves, energy is absorbed and the nuclei "spin-flip" from the lower energy state to the higher energy state. This absorption of rf energy is detected, amplified, and displayed as an NMR spectrum.

Each electronically distinct ${ }^{1} \mathrm{H}$ or ${ }^{13} \mathrm{C}$ nucleus in a molecule comes into resonance at a slightly different value of the applied field, thereby producing a unique absorption signal. The exact position of each peak is called the chemical shift. Chemical shifts are caused by electrons setting up tiny local magnetic fields that shield a nearby nucleus from the applied field.

The NMR chart is calibrated in delta units ( $\boldsymbol{\delta}$ ), where $1 \delta=1 \mathrm{ppm}$ of spectrometer frequency. Tetramethylsilane (TMS) is used as a reference point because it shows both ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ absorptions at unusually high values of applied magnetic field. The TMS absorption occurs on the right-hand (upfield) side of the chart and is arbitrarily assigned a value of $0 \delta$.
${ }^{13} \mathrm{C}$ spectra are run on Fourier-transform NMR (FT-NMR) spectrometers using broadband decoupling of proton spins so that each chemically distinct carbon shows a single unsplit resonance line. As with ${ }^{1} \mathrm{H}$ NMR, the chemical shift of each ${ }^{13} \mathrm{C}$ signal provides information about a carbon's chemical environment in the sample. In addition, the number of protons attached to each carbon can be determined using the DEPT-NMR technique.


Summary -
In ${ }^{1} H$ NMR spectra, the area under each absorption peak can be electronically integrated to determine the relative number of hydrogens responsible for each peak. In addition, neighboring nuclear spins can couple, causing the spin-spin splitting of NMR peaks into multiplets. The NMR signal of a hydrogen neighbored by $n$ equivalent adjacent hydrogens splits into $n+1$ peaks (the $\boldsymbol{n}+\mathbf{1}$ rule) with coupling constant $\boldsymbol{J}$.


Visualizing Chemistry -
PROBLEM Into how many peaks would you expect the ${ }^{1} \mathrm{H}$ NMR signals of the indicated protons to be split? 13-24 $($ Green $=\mathrm{Cl}$.)
(a)

(b)

PROBLEM How many absorptions would you expect the following compound to have in its ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR 13-25 spectra?

PROBLEM Sketch what you might expect the ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR spectra of the following compound to look like 13-26 (green $=\mathrm{Cl}$ ):

PROBLEM How many electronically nonequivalent kinds of protons and how many kinds of carbons are 13-27 present in the following compound? Don't forget that cyclohexane rings can ring-flip.

PROBLEM Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or 13-28 diastereotopic.
(a)

(b)

Cysteine


Chemical Shifts and NMR Spectroscopy -
PROBLEM The following ${ }^{1} \mathrm{H}$ NMR absorptions were obtained on a spectrometer operating at 200 MHz and are 13-29 given in hertz downfield from the TMS standard. Convert the absorptions to $\delta$ units.
(a) 436 Hz
(b) 956 Hz
(c) 1504 Hz

PROBLEM The following ${ }^{1} \mathrm{H}$ NMR absorptions were obtained on a spectrometer operating at 300 MHz . Convert 13-30 the chemical shifts from $\delta$ units to hertz downfield from TMS.
(a) $2.1 \mathrm{\delta}$
(b) $3.45 \delta$
(c) $6.30 \delta$
(d) $7.70 \delta$

PROBLEM When measured on a spectrometer operating at 200 MHz , chloroform $\left(\mathrm{CHCl}_{3}\right)$ shows a single sharp 13-31 absorption at $7.3 \delta$.
(a) How many parts per million downfield from TMS does chloroform absorb?
(b) How many hertz downfield from TMS would chloroform absorb if the measurement were carried out on a spectrometer operating at 360 MHz ?
(c) What would be the position of the chloroform absorption in $\delta$ units when measured on a 360 MHz spectrometer?
PROBLEM Why do you suppose accidental overlap of signals is much more common in ${ }^{1} \mathrm{H}$ NMR than in ${ }^{13} \mathrm{C}$ 13-32 NMR?

PROBLEM Is a nucleus that absorbs at $6.50 \delta$ more shielded or less shielded than a nucleus that absorbs at 13-33 3.20 $\delta$ ? Does the nucleus that absorbs at $6.50 \delta$ require a stronger applied field or a weaker applied field to come into resonance than the one that absorbs at $3.20 \delta$ ?


${ }^{\mathbf{1}} \mathrm{H}$ NMR Spectroscopy -
PROBLEM How many types of nonequivalent protons are present in each of the following molecules?
13-34 (a)

(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}$
(c)

Naphthalene
(d)

Styrene
(e)


Ethyl acrylate -
PROBLEM The following compounds all show a single line in their ${ }^{1} \mathrm{H}$ NMR spectra. List them in order of 13-35 expected increasing chemical shift:
$\mathrm{CH}_{4}, \mathrm{CH}_{2} \mathrm{Cl}_{2}$, cyclohexane, $\mathrm{CH}_{3} \mathrm{COCH}_{3}, \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$, benzene
PROBLEM How many signals would you expect each of the following molecules to have in its ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ 13-36 spectra?
(a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Propose structures for compounds with the following formulas that show only one peak in their ${ }^{1} \mathrm{H}$
13-37 NMR spectra:
(a) $\mathrm{C}_{5} \mathrm{H}_{12}$
(b) $\mathrm{C}_{5} \mathrm{H}_{10}$
(c) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}$

PROBLEM Predict the splitting pattern for each kind of hydrogen in the following molecules:
13-38
(a) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CH}$ (b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}$
(c) trans-2-Butene

PROBLEM Predict the splitting pattern for each kind of hydrogen in isopropyl propanoate,
13-39 $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$.
PROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:

13-40 (a)

(b)

(c)

PROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:


13-41 (a) -
(b)

(c)

PROBLEM The acid-catalyzed dehydration of 1-methylcyclohexanol yields a mixture of two alkenes. How
13-42 could you use ${ }^{1}$ H NMR to help you decide which was which?

PROBLEM How could you use ${ }^{1}$ H NMR to distinguish between the following pairs of isomers?
13-43 (a) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{3}$ and

(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}$
and $\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
(c)
and

(d)
and

PROBLEM Propose structures for compounds that fit the following ${ }^{1} \mathrm{H}$ NMR data:
13-44
(a) $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$
(b) $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br}$
$0.95 \delta(6 \mathrm{H}$, doublet, $J=7 \mathrm{~Hz})$
2.32 ठ (3 H, singlet)
$2.10 \delta(3 \mathrm{H}$, singlet)
2.32 ( 3 H , singlet)
$2.43 \delta(1 \mathrm{H}$, multiplet)
2.32 (3 H, singlet)

PROBLEM Propose structures for the two compounds whose ${ }^{1} \mathrm{H}$ NMR spectra are shown.
13-45
(a) $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}$

(b) $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Cl}_{2}$


${ }^{13}$ C NMR Spectroscopy -
PROBLEM How many ${ }^{13} \mathrm{C}$ NMR absorptions would you expect for cis-1,3-dimethylcyclohexane? For 13-46 trans-1,3-dimethylcyclohexane? Explain.

PROBLEM How many absorptions would you expect to observe in the ${ }^{13} \mathrm{C}$ NMR spectra of the following 13-47 compounds?
(a) 1,1-Dimethylcyclohexane
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{3}$
(c) tert-Butylcyclohexane
(d) 3-Methyl-1-pentyne
(e) cis-1,2-Dimethylcyclohexane
(f) Cyclohexanone

PROBLEM Suppose you ran a DEPT-135 spectrum for each substance in Problem 13.47. Which carbon atoms 13-48 in each molecule would show positive peaks, and which would show negative peaks?

PROBLEM How could you use ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR to help distinguish the following isomeric compounds of the 13-49 formula $\mathrm{C}_{4} \mathrm{H}_{8}$ ?

PROBLEM How could you use ${ }^{1} \mathrm{H}$ NMR, ${ }^{13} \mathrm{C}$ NMR, and IR spectroscopy to help you distinguish between the 13-50 following structures?

3-Methyl-2-cyclohexenone


3-Cyclopentenyl methyl ketone -
PROBLEM Assign as many resonances as you can to specific carbon atoms in the ${ }^{13}$ C NMR spectrum of ethyl 13-51
benzoate.


General Problems -
PROBLEM Assume that you have a compound with the formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$.
13-52 (a) How many double bonds and/or rings does your compound contain?
(b) Propose as many structures as you can that fit the molecular formula.
(c) If your compound shows an infrared absorption peak at $1715 \mathrm{~cm}^{-1}$, what functional group does it have?
(d) If your compound shows a single ${ }^{1} \mathrm{H}$ NMR absorption peak at $2.1 \delta$, what is its structure?

PROBLEM The compound whose ${ }^{1} \mathrm{H}$ NMR spectrum is shown has the molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}$. Propose a
13-53 structure.

PROBLEM The compound whose ${ }^{1} \mathrm{H}$ NMR spectrum is shown has the molecular formula $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{Cl}$ and has an
13-54 infrared absorption peak at $1740 \mathrm{~cm}^{-1}$. Propose a structure.


General Problems -
PROBLEM Propose structures for compounds that fit the following ${ }^{1} \mathrm{H}$ NMR data:
13-55
(a) $\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{Cl}_{2}$
$2.18 \delta(3 \mathrm{H}$, singlet)
$4.16 \delta(2 \mathrm{H}$, doublet, $J=7 \mathrm{~Hz}$ )
$5.71 \delta(1 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz}$ )
(d) $\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{Br}$
$2.15 \delta(2 \mathrm{H}$, quintet, $J=7 \mathrm{~Hz})$
$2.75 \delta(2 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz})$
$3.38 \delta(2 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz})$
7.22 ( 5 H , singlet)
(b) $\mathrm{C}_{10} \mathrm{H}_{14}$
$1.30 \delta(9 \mathrm{H}$, singlet)
$7.30 \delta(5 \mathrm{H}$, singlet)
(c) $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{BrO}$
$2.11 \delta(3 \mathrm{H}$, singlet)
$3.52 \delta(2 \mathrm{H}$, triplet, $J=6 \mathrm{~Hz}$ )
$4.40 \delta(2 \mathrm{H}$, triplet, $J=6 \mathrm{~Hz})$

PROBLEM Long-range coupling between protons more than two carbon atoms apart is sometimes observed 13-56 when $\pi$ bonds intervene. An example is found in 1-methoxy-1-buten-3-yne. Not only does the acetylenic proton, $\mathrm{H}_{a}$, couple with the vinylic proton $\mathrm{H}_{b}$, it also couples with the vinylic proton $\mathrm{H}_{\mathcal{C}}$, four carbon atoms away. The data are:


General Problems -
$$
\begin{array}{lll}
\mathrm{H}_{a}(3.08 \delta) & \mathrm{H}_{b}(4.52 \delta) & \mathrm{H}_{c}(6.35 \delta) \\
J_{a-b}=3 \mathrm{~Hz} & J_{a-c}=1 \mathrm{~Hz} & J_{b-c}=7 \mathrm{~Hz}
\end{array}
$$


1-Methoxy-1-buten-3-yne -
Construct tree diagrams that account for the observed splitting patterns of $H_{a}, H_{b}$, and $H_{c}$.
PROBLEM The ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR spectra of compound $\mathbf{A}, \mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$, are shown. Propose a structure for $\mathbf{A}$, and 13-57 assign peaks in the spectra to your structure.

PROBLEM Propose structures for the three compounds whose ${ }^{1} \mathrm{H}$ NMR spectra are shown.
13-58
(a) $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$

(b) $\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{Br}$

(c) $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$

PROBLEM The mass spectrum and ${ }^{13} \mathrm{C}$ NMR spectrum of a hydrocarbon are shown. Propose a structure for 13-59 this hydrocarbon, and explain the spectral data.

PROBLEM Compound A, a hydrocarbon with $\mathrm{M}^{+}=96$ in its mass spectrum, has the following ${ }^{13} \mathrm{C}$ spectral
13-60 data. On reaction with $\mathrm{BH}_{3}$, followed by treatment with basic $\mathrm{H}_{2} \mathrm{O}_{2}$, $\mathbf{A}$ is converted into $\mathbf{B}$, whose ${ }^{13} \mathbf{C}$ spectral data are also given. Propose structures for A and B.


Compound A -
Broadband-decoupled ${ }^{13}$ C NMR: 26.8, 28.7, 35.7, 106.9, $149.7 \delta$
DEPT-90: no peaks
DEPT-135: no positive peaks; negative peaks at 26.8, $28.7,35.7,106.9 \delta$


Compound B -
Broadband-decoupled ${ }^{13}$ C NMR: 26.1, 26.9, 29.9, 40.5, $68.2 \delta$
DEPT-90: 40.5 ס
DEPT-135: positive peak at $40.5 \delta$; negative peaks at $26.1,26.9,29.9,68.2 \delta$

PROBLEM Propose a structure for compound $\mathbf{C}$, which has $\mathrm{M}^{+}=86$ in its mass spectrum, an IR absorption at
13-61 $3400 \mathrm{~cm}^{-1}$, and the following ${ }^{13} \mathrm{C}$ NMR spectral data:


Compound C -
Broadband-decoupled ${ }^{13}$ C NMR: 30.2, 31.9, 61.8, 114.7, $138.4 \delta$
DEPT-90: $138.4 \delta$
DEPT-135: positive peak at $138.4 \delta$; negative peaks at $30.2,31.9,61.8,114.7 \delta$
PROBLEM Compound $\mathbf{D}$ is isomeric with compound $\mathbf{C}$ (Problem 13.61) and has the following ${ }^{13}$ C NMR spectral 13-62 data. Propose a structure.


Compound D -
Broadband-decoupled ${ }^{13}$ C NMR: 9.7, 29.9, 74.4, 114.4, $141.4 \delta$
DEPT-90: 74.4, $141.4 \delta$
DEPT-135: positive peaks at $9.7,74.4,141.4 \delta$; negative peaks at $29.9,114.4 \delta$

PROBLEM Propose a structure for compound $\mathbf{E}, \mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{2}$, which has the following ${ }^{13} \mathrm{C}$ NMR spectral data:
13-63


Compound E -
Broadband-decoupled ${ }^{13}$ C NMR: 19.1, 28.0, 70.5, 129.0, 129.8, $165.8 \delta$
DEPT-90: 28.0, $129.8 \delta$
DEPT-135: positive peaks at 19.1, 28.0, $129.8 \delta$; negative peaks at $70.5,129.0 \delta$

PROBLEM Compound $\mathbf{F}$, a hydrocarbon with $\mathrm{M}^{+}=96$ in its mass spectrum, undergoes reaction with HBr to 13-64 yield compound $\mathbf{G}$. Propose structures for $\mathbf{F}$ and $\mathbf{G}$, whose ${ }^{13} \mathrm{C}$ NMR spectral data are given below.


Compound $\mathbf{F}$ -
Broadband-decoupled ${ }^{13}$ C NMR: 27.6, 29.3, 32.2, $132.4 \delta$
DEPT-90: 132.4 ס
DEPT-135: positive peak at $132.4 \delta$; negative peaks at $27.6,29.3,32.2 \delta$


Compound G -
Broadband-decoupled ${ }^{13}$ C NMR: 25.1, 27.7, 39.9, $56.0 \delta$
DEPT-90: 56.0 ס
DEPT-135: positive peak at $56.0 \delta$; negative peaks at $25.1,27.7,39.9 \delta$
PROBLEM 3-Methyl-2-butanol has five signals in its ${ }^{13}$ C NMR spectrum at 17.90, 18.15, 20.00, 35.05, and
13-65 72.75 $\delta$. Why are the two methyl groups attached to C3 nonequivalent? Making a molecular model should be helpful.


3-Methyl-2-butanol -
PROBLEM A ${ }^{13} \mathrm{C}$ NMR spectrum of commercially available 2,4-pentanediol, shows five peaks at 23.3, 23.9,
13-66 46.5, 64.8, and 68.1 $\delta$. Explain.


2,4-Pentanediol -
PROBLEM Carboxylic acids $\left(\mathrm{RCO}_{2} \mathrm{H}\right)$ react with alcohols $\left(\mathrm{R}^{\prime} \mathrm{OH}\right)$ in the presence of an acid catalyst. The reaction
13-67 product of propanoic acid with methanol has the following spectroscopic properties. Propose a structure.

Propanoic acid
MS: $\mathrm{M}^{+}=88$
IR: $1735 \mathrm{~cm}^{-1}$
${ }^{1} \mathrm{H}$ NMR: $1.11 \delta(3 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz}) ; 2.32 \delta(2 \mathrm{H}$, quartet, $J=7 \mathrm{~Hz})$;
$3.65 \delta$ (3 H, singlet)
${ }^{13}$ C NMR: 9.3, 27.6, 51.4, $174.6 \delta$
PROBLEM Nitriles $(R C \equiv N)$ react with Grignard reagents ( $\mathrm{R}^{\prime} \mathrm{MgBr}$ ). The reaction product from
13-68 2-methylpropanenitrile with methylmagnesium bromide has the following spectroscopic properties. Propose a structure. MS: M ${ }^{+}=86$ IR: $1715 \mathrm{~cm}^{-1}{ }^{1} \mathrm{H}$ NMR: $1.05 \delta(6 \mathrm{H}$, doublet, J = 7 Hz ); 2.12 ( 3 H , singlet); $2.67 \delta\left(1 \mathrm{H}\right.$, septet, $\mathrm{J}=7 \mathrm{~Hz}$ ) ${ }^{13} \mathrm{C}$ NMR: 18.2, 27.2, 41.6, $211.2 \delta$


2,4-Pentanediol -
PROBLEM The proton NMR spectrum is shown for a compound with the formula $\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{NO}_{4}$. The infrared 13-69 spectrum displays strong bands at 1750 and $1562 \mathrm{~cm}^{-1}$ and a medium-intensity band at $1320 \mathrm{~cm}^{-1}$. The normal carbon-13 and the DEPT experimental results are tabulated. Draw the structure of this compound.

| Normal Carbon | DEPT-135 | DEPT-90 |
| :---: | :---: | :---: |
| 14 ppm | Positive | No peak |
| 16 | Positive | No peak |
| 63 | Negative | No peak |
| 83 | Positive | Positive |
| 165 | No peak | No peak |

PROBLEM The proton NMR spectrum of a compound with the formula $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$ is shown. The normal carbon-13
13-70 and the DEPT experimental results are tabulated. The infrared spectrum shows a broad peak at about $3340 \mathrm{~cm}^{-1}$ and a medium-sized peak at about $1651 \mathrm{~cm}^{-1}$. Draw the structure of this compound.

| Normal Carbon | DEPT-135 | DEPT-90 |
| :---: | :---: | :---: |
| 22.2 ppm | Positive | No peak |
| 40.9 | Negative | No peak |
| 60.2 | Negative | No peak |
| 112.5 | Negative | No peak |
| 142.3 | No peak | No peak |

PROBLEM The proton NMR spectrum of a compound with the formula $\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{2}$ is shown. The infrared
13-71 spectrum displays a strong band at $1738 \mathrm{~cm}^{-1}$ and a weak band at $1689 \mathrm{~cm}^{-1}$. The normal carbon-13 and the DEPT experimental results are tabulated. Draw the structure of this compound.


2,4-Pentanediol -
| Normal Carbon | DEPT-135 | DEPT-90 |
| :---: | :---: | :---: |
| 18 ppm | Positive | No peak |
| 21 | Positive | No peak |
| 26 | Positive | No peak |
| 61 | Negative | No peak |
| 119 | Positive | Positive |
| 139 | No peak | No peak |
| 171 | No peak | No peak |


CHAPTER 14 -
Conjugated Compounds and Ultraviolet Spectroscopy

FIGURE 14.1 Saffron, derived from stigmas of the saffron crocus, is the world's most expensive spice. Its color is caused by the presence of alternating single and double bonds. (credit: modification of work "Welcome to My World! ~ Huron River Watershed" by j van cise photos/ Flickr, CC BY 2.0)


CHAPTER CONTENTS -
14.1 Stability of Conjugated Dienes: Molecular Orbital Theory
14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations
14.3 Kinetic versus Thermodynamic Control of Reactions
14.4 The Diels-Alder Cycloaddition Reaction
14.5 Characteristics of the Diels-Alder Reaction
14.6 Diene Polymers: Natural and Synthetic Rubbers
14.7 Ultraviolet Spectroscopy
14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation
14.9 Conjugation, Color, and the Chemistry of Vision

WHY THIS CHAPTER? Conjugated compounds of many different sorts are common in nature. Many of the pigments responsible for the brilliant colors of fruits, flowers, and even animals have numerous alternating single and double bonds. $\beta$-Carotene, for instance, the orange pigment responsible for the color of carrots and an important source of vitamin A , is a conjugated polyene with 11 double bonds. Conjugated enones (alkene + ketone) are common structural features of many biologically important molecules such as progesterone, the hormone that prepares the uterus for implantation of a fertilized ovum. Cyclic conjugated molecules such as benzene are a major field of study in themselves. In this chapter, we'll look at some of the distinctive properties
of conjugated molecules and at the reasons for those properties.

Most of the unsaturated compounds we looked at in the chapters on Alkenes: Structure and Reactivity and Alkenes: Reactions and Synthesis had only one double bond, but many compounds have numerous sites of unsaturation. If the different unsaturations are well separated in a molecule, they react independently, but if they're close together, they may interact. In particular, compounds that have alternating single and double bonds-so-called conjugated compounds-have some distinctive characteristics. The conjugated diene 1,3-butadiene, for instance, has some properties quite different from those of the nonconjugated 1,4-pentadiene.

1,3-Butadiene (conjugated; alternating double and single bonds)

1,4-Pentadiene (nonconjugated; nonalternating double and single bonds)


CHAPTER CONTENTS - 14.1 Stability of Conjugated Dienes: Molecular Orbital Theory
Conjugated dienes can be prepared by some of the methods previously discussed for preparing alkenes (Section 11.7-Section 11.11). The base-induced elimination of HX from an allylic halide is one such reaction.


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) -
Simple conjugated dienes used in polymer synthesis include 1,3-butadiene, chloroprene (2-chloro-1,3-butadiene), and isoprene (2-methyl-1,3-butadiene). Isoprene has been prepared industrially by several methods, including the acid-catalyzed double dehydration of 3-methyl-1,3-butanediol.

3-Methyl-1,3-butanediol

Isoprene
2-Methyl-1,3-butadiene

One of the properties that distinguishes conjugated from nonconjugated dienes is that the central single bond is shorter than might be expected. The C2-C3 single bond in 1,3-butadiene, for instance, has a length of 147 pm , some 6 pm shorter than the C2-C3 bond in butane ( 153 pm ).

1,3-Butadiene

Butane

Another distinctive property of conjugated dienes is their unusual stability, as evidenced by their heats of hydrogenation (TABLE 14.1). We saw in Section 7.6 that monosubstituted alkenes, such as 1-butene, have $\Delta H^{\circ}{ }_{\text {hydrog }}$ near $-126 \mathrm{~kJ} / \mathrm{mol}(-30.1 \mathrm{kcal} / \mathrm{mol})$, whereas disubstituted alkenes, such as 2-methylpropene, have $\Delta H^{\circ}{ }_{\text {hydrog }}$ near $-119 \mathrm{~kJ} / \mathrm{mol}(-28.4 \mathrm{kcal} / \mathrm{mol})$, which is approximately $7 \mathrm{~kJ} / \mathrm{mol}$ less negative. We concluded from these data that more highly substituted alkenes are more stable than less substituted ones. That is, more highly substituted alkenes release less heat on hydrogenation because they contain less energy to start with. A similar conclusion can be drawn for conjugated dienes.

TABLE 14.1 Heats of Hydrogenation for Some Alkenes and Dienes


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) -
TABLE 14.1 Heats of Hydrogenation for Some Alkenes and Dienes

| Alkene or diene | Product | $\Delta H^{\circ} \mathrm{hydrog}$ |  |
| :---: | :---: | :---: | :---: |
|  |  | (kJ/mol) | ( $\mathrm{kcal} / \mathrm{mol}$ ) |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$ | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | -126 | -30.1 |
| <smiles>C=C(C)C</smiles> | <smiles>CC(C)C</smiles> | -119 | -28.4 |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$ | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | -253 | -60.5 |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$ | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | -236 | -56.4 |
| <smiles>C=CC(=C)C</smiles> | <smiles>CCC(C)C</smiles> | -229 | -54.7 |


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) -
Because a monosubstituted alkene has a $\Delta H^{\circ}{ }_{\text {hydrog }}$ of approximately $-126 \mathrm{~kJ} / \mathrm{mol}$, we might expect that a compound with two monosubstituted double bonds would have a $\Delta H^{\circ}{ }_{\text {hydrog }}$ approximately twice that value, or $-252 \mathrm{~kJ} / \mathrm{mol}$. Nonconjugated dienes, such as 1,4 -pentadiene ( $\Delta H^{\circ}{ }_{\text {hydrog }}=-253 \mathrm{~kJ} / \mathrm{mol}$ ), meet this expectation, but the conjugated diene 1,3-butadiene ( $\Delta H^{\circ}{ }_{\text {hydrog }}=-236 \mathrm{~kJ} / \mathrm{mol}$ ) does not. 1,3-Butadiene is approximately $16 \mathrm{~kJ} / \mathrm{mol}(3.8 \mathrm{kcal} / \mathrm{mol})$ more stable than expected.

|  | $\Delta \boldsymbol{H}^{\circ}$ hydrog $(\mathbf{k J} / \mathbf{m o l})$ |  |
| :---: | :--- | :--- |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$ | $-126+(-126)=-252$ | Expected
 1,4-Pentadiene |
|  | -253 | Observed
 Difference |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}=\mathrm{CH}_{2}$ | $-126+(-126)=-252$ | Expected |
| 1,3-Butadiene | -236 | Observed |
|  | -16 | Difference |


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) -
What accounts for the stability of conjugated dienes? According to valence bond theory (Section 1.5 and Section 1.8), their stability is due to orbital hybridization. Typical $\mathrm{C}-\mathrm{C}$ single bonds, like those in alkanes, result from $\sigma$ overlap of $s p^{3}$ orbitals on both carbons, but in a conjugated diene, the central $\mathrm{C}-\mathrm{C}$ single bond results from $\sigma$ overlap of $s p^{2}$ orbitals on both carbons. Because $s p^{2}$ orbitals have more $s$ character ( $33 \% s$ ) than $s p^{3}$ orbitals $(25 \% s)$, the electrons in $s p^{2}$ orbitals are closer to the nucleus and the bonds they form are somewhat shorter
and stronger. Thus, the "extra" stability of a conjugated diene results in part from the greater amount of $s$ character in the orbitals forming the $\mathrm{C}-\mathrm{C}$ single bond.

Bond formed by overlap of $\boldsymbol{s p} \boldsymbol{p}^{2}$ orbitals
According to molecular orbital theory (Section 1.11), the stability of conjugated dienes arises because of an interaction between the $\pi$ orbitals of the two double bonds. To review briefly, when two $p$ atomic orbitals combine to form a $\pi$ bond, two $\pi$ molecular orbitals (MOs) result. One is lower in energy than the starting $p$ orbitals and is therefore bonding; the other is higher in energy, has a node between nuclei, and is antibonding. The two $\pi$ electrons occupy the low-energy, bonding orbital, resulting in formation of a stable bond between atoms (FIGURE 14.2).

FIGURE 14.2 Two $\boldsymbol{p}$ orbitals combine to form two $\boldsymbol{\pi}$ molecular orbitals. Both electrons occupy the low-energy, bonding orbital, leading to a net lowering of energy and formation of a stable bond. The asterisk on $\psi_{2} *$ indicates an antibonding orbital.


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) -
Now let's combine four adjacent $p$ atomic orbitals, as occurs in a conjugated diene. In so doing, we generate a set of four $\pi$ molecular orbitals, two of which are bonding and two of which are antibonding (FIGURE 14.3). The four $\pi$ electrons occupy the two bonding orbitals, leaving the antibonding orbitals vacant.

FIGURE 14.3 Four $\boldsymbol{\pi}$ molecular orbitals in 1,3-butadiene. Note that the number of nodes between nuclei increases as the energy level of the orbital increases.

The lowest-energy $\pi$ molecular orbital (denoted $\psi_{1}$, Greek psi) has no nodes between the nuclei and is therefore bonding. The $\pi$ MO of next-lowest energy, $\psi_{2}$, has one node between nuclei and is also bonding. Above $\psi_{1}$ and $\psi_{2}$ in energy are the two antibonding $\pi \operatorname{MOs}, \psi_{3} *$ and $\psi_{4}{ }^{*}$. (The asterisks indicate antibonding orbitals.) Note that the number of nodes between nuclei increases as the energy level of the orbital increases. The $\psi_{3} *$ orbital has two nodes between nuclei, and $\psi_{4}{ }^{*}$, the highest-energy MO, has three nodes between nuclei.

Comparing the $\pi$ molecular orbitals of 1,3-butadiene (two conjugated double bonds) with those of 1,4-pentadiene (two isolated double bonds) shows why the conjugated diene is more stable. In a conjugated diene, the lowest-energy $\pi \mathrm{MO}\left(\psi_{1}\right)$ has a favorable bonding interaction between C 2 and C 3 that is absent in a nonconjugated diene. As a result, there is a certain amount of double-bond character to the $\mathrm{C} 2-\mathrm{C} 3$ single bond, making that bond both stronger and shorter than a typical single bond. Electrostatic potential maps show clearly the additional electron density in the central single bond (FIGURE 14.4).

1,3-Butadiene (conjugated)


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) -
1,3-Butadiene (conjugated)

1,4-Pentadiene
(nonconjugated)

FIGURE 14.4 Electrostatic potential maps of 1,3-butadiene (conjugated) and 1,4-pentadiene (nonconjugated). Additional electron density is present in the central $\mathrm{C}-\mathrm{C}$ bond of 1,3-butadiene, corresponding to partial double-bond character.

In describing 1,3-butadiene, we say that the $\pi$ electrons are spread out, or delocalized, over the entire $\pi$ framework, rather than localized between two specific nuclei. Delocalization allows the bonding electrons to be closer to more nuclei, thus leading to lower energy and greater stability.

PROBLEM Allene, $\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}$, has a heat of hydrogenation of $-298 \mathrm{~kJ} / \mathrm{mol}(-71.3 \mathrm{kcal} / \mathrm{mol})$. Rank a 14-1 conjugated diene, a nonconjugated diene, and an allene in order of stability.


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) - 14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations
One of the most striking differences between conjugated dienes and typical alkenes is their behavior in electrophilic addition reactions. To review briefly, the addition of an electrophile to a carbon-carbon double bond is a general reaction of alkenes (Section 7.7). Markovnikov regiochemistry is observed because the more stable carbocation is formed as an intermediate. Thus, addition of HCl to 2-methylpropene yields 2-chloro-2-methylpropane rather than 1-chloro-2-methylpropane, and addition of 2 equivalents of HCl to the nonconjugated diene 1,4-pentadiene yields 2,4-dichloropentane.

Conjugated dienes also undergo electrophilic addition reactions readily, but mixtures of products are invariably obtained. Addition of HBr to 1,3 -butadiene, for instance, yields a mixture of two products (not counting cis-trans isomers). 3-Bromo-1-butene is the typical Markovnikov product of $\mathbf{1 , 2}$-addition to a double bond, but 1-bromo-2-butene seems unusual. The double bond in this product has moved to a position between carbons 2 and 3 , and HBr has added to carbons 1 and 4, a result described as $\mathbf{1 , 4}$-addition.

Many other electrophiles besides HBr add to conjugated dienes, and mixtures of products are usually formed. For example, $\mathrm{Br}_{2}$ adds to 1,3-butadiene to give a mixture of 3,4-dibromo-1-butene and 1,4-dibromo-2-butene.


Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\%) - 14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations
How can we account for the formation of 1,4-addition products? The answer is that allylic carbocations are involved as intermediates (recall that the word allylic means "next to a double bond"). When 1,3-butadiene reacts with an electrophile such as $\mathrm{H}^{+}$, two carbocation intermediates are possible-a primary nonallylic carbocation and a secondary allylic cation. Because an allylic cation is stabilized by resonance between two forms (Section 11.5), it is more stable and forms faster than a nonallylic carbocation.

When the allylic cation reacts with $\mathrm{Br}^{-}$to complete the electrophilic addition, the reaction can occur either at C 1 or at C3 because both carbons share the positive charge (FIGURE 14.5). Thus, a mixture of 1,2-and 1,4-addition products results. You might recall that a similar product mixture was seen for NBS bromination of alkenes in Section 10.3, a reaction that proceeds through an allylic radical.

FIGURE 14.5 An electrostatic potential map of the allylic carbocation produced by protonation of 1,3-butadiene shows that the positive charge is shared by carbons 1 and 3. Reaction of $\mathrm{Br}^{-}$with the more positive carbon (C3) predominantly yields the 1,2-addition product.


Predicting the Product of an Electrophilic Addition Reaction of a Conjugated Diene -
Give the structures of the likely products from reaction of 1 equivalent of HCl with 2 -methyl-1,3-cyclohexadiene. Show both 1,2 and 1,4 adducts.


Strategy -
Electrophilic addition of HCl to a conjugated diene involves the formation of allylic carbocation intermediate. Thus, the first step is to protonate the two ends of the diene and draw the resonance forms of the two allylic carbocations that result. Then, allow each resonance form to react with $\mathrm{Cl}^{-}$, generating a maximum of four possible products.

In the present instance, protonation of the $\mathrm{C} 1-\mathrm{C} 2$ double bond gives a carbocation that can react further to give the 1,2 adduct 3 -chloro-3-methylcyclohexene and the 1,4 adduct 3-chloro-1-methylcyclohexene. Protonation of the C3-C4 double bond gives a symmetrical carbocation, whose two resonance forms are equivalent. Thus, the 1,2 adduct and the 1,4 adduct have the same structure: 6 -chloro-1-methylcyclohexene. Of the two possible modes of protonation, the first is more likely because it yields a more stable, tertiary allylic cation rather than a less-stable, secondary allylic cation.


Solution -
PROBLEM Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HCl with 14-2 1,3-pentadiene.

PROBLEM Look at the possible carbocation intermediates produced during addition of HCl to 1,3-pentadiene 14-3 (Problem 14-2), and predict which 1,2 adduct predominates. Which 1,4 adduct predominates?

PROBLEM Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HBr with
14-4 the following compound:


Solution - 14.3 Kinetic versus Thermodynamic Control of Reactions
Electrophilic addition to a conjugated diene at or below room temperature normally leads to a mixture of products in which the 1,2 adduct predominates over the 1,4 adduct. When the same reaction is carried out at higher temperatures, however, the product ratio often changes and the 1,4 adduct predominates. For example, addition of HBr to 1,3 -butadiene at $0^{\circ} \mathrm{C}$ yields a $71: 29$ mixture of 1,2 and 1,4 adducts, but the same reaction carried out at $40^{\circ} \mathrm{C}$ yields a $15: 85$ mixture. Furthermore, when the product mixture formed at $0{ }^{\circ} \mathrm{C}$ is heated to $40^{\circ} \mathrm{C}$ in the presence of HBr , the ratio of adducts slowly changes from $71: 29$ to $15: 85$. Why?

To understand the effect of temperature on product distribution, let's briefly review what we said in Section 6.7 about rates and equilibria. Imagine a reaction that can give either or both of two products, B and C.

Let's assume that $\mathbf{B}$ forms faster than $\mathbf{C}$ (in other words, $\Delta G^{\ddagger} \mathrm{B}^{<\Delta G^{\ddagger}} \mathrm{C}$ ) but that $\mathbf{C}$ is more stable than $\mathbf{B}$ (in other words, $\Delta G^{\circ}{ }_{\mathrm{C}}>\Delta G^{\circ}{ }_{\mathrm{B}}$ ). An energy diagram for the two processes might look like that shown in FIGURE 14.6.

FIGURE 14.6 An energy diagram for two competing reactions in which the less stable product $B$ forms faster than the more stable product C .


Solution - 14.3 Kinetic versus Thermodynamic Control of Reactions
FIGURE 14.6 An energy diagram for two competing reactions in which the less stable product $B$ forms faster than the more stable product C .

Let's first carry out the reaction at a lower temperature so that both processes are irreversible and no equilibrium is reached. Because $\mathbf{B}$ forms faster than $\mathbf{C}, \mathbf{B}$ is the major product. It doesn't matter that $\mathbf{C}$ is more stable than B, because the two are not in equilibrium. The product of an irreversible reaction depends only on relative rates, not on stability. Such reactions are said to be under kinetic control.

Now let's carry out the same reaction at some higher temperature so that both processes are readily reversible and an equilibrium is reached. Because $\mathbf{C}$ is more stable than $\mathbf{B}, \mathbf{C}$ is the major product obtained. It doesn't matter that $\mathbf{C}$ forms more slowly than $\mathbf{B}$, because the two are in equilibrium. The product of a readily reversible reaction depends only on stability, not on relative rates. Such reactions are said to be under equilibrium control, or thermodynamic control.

We can now explain the effect of temperature on the electrophilic addition reactions of conjugated dienes. At low temperature $\left(0^{\circ} \mathrm{C}\right)$, HBr adds to 1,3-butadiene under kinetic control to give a $71: 29$ mixture of products, with the more rapidly formed 1,2 adduct predominating. Because these low-temperature conditions don't allow the reaction to reach equilibrium, the product that forms faster predominates. At higher temperature $\left(40^{\circ} \mathrm{C}\right)$, however, the reaction occurs under thermodynamic control to give a $15: 85$ mixture of products, with the more stable 1,4 adduct predominating. The higher temperature allows the addition process to become reversible, so an equilibrium mixture of products results. FIGURE 14.7 shows this situation in an energy diagram.


Solution - 14.3 Kinetic versus Thermodynamic Control of Reactions
FIGURE 14.7 Energy diagram for the electrophilic addition of $\mathbf{H B r}$ to 1,3-butadiene. The 1,2 adduct is the kinetic product because it forms faster, but the 1,4 adduct is the thermodynamic product because it is more stable.

The electrophilic addition of HBr to 1,3-butadiene is a good example of how a change in experimental conditions can change the product of a reaction. The concept of thermodynamic control versus kinetic control is a useful one that we can sometimes take advantage of in the laboratory.

PROBLEM The 1,2 adduct and the 1,4 adduct formed by reaction of HBr with 1,3-butadiene are in equilibrium 14-5 at $40^{\circ} \mathrm{C}$. Propose a mechanism by which the interconversion of products takes place.

PROBLEM Why do you suppose 1,4 adducts of 1,3-butadiene are generally more stable than 1,2 adducts?
14-6


Solution - 14.4 The Diels-Alder Cycloaddition Reaction
Perhaps the most striking difference between conjugated and nonconjugated dienes is that conjugated dienes undergo an addition reaction with alkenes to yield substituted cyclohexene products. For example, 1,3-butadiene and 3-buten-2-one give 3-cyclohexenyl methyl ketone.

This process, named the Diels-Alder cycloaddition reaction after its discoverers, is extremely useful in the laboratory because it forms two carbon-carbon bonds in a single step and is one of the few general methods available for making cyclic molecules. (As the name implies, a cycloaddition reaction is one in which two reactants add together to give a cyclic product.) The 1950 Nobel Prize in Chemistry was awarded to Otto Diels and Kurt Alder in recognition of the importance of their discovery.

The mechanism of Diels-Alder cycloaddition is different from that of other reactions we've studied because it is neither polar nor radical. Rather, the Diels-Alder reaction is a pericyclic process. Pericyclic reactions, which we'll discuss in more detail in Chapter 30, take place in a single step by a cyclic redistribution of bonding electrons. The two reactants simply join together through a cyclic transition state in which the two new $\mathrm{C}-\mathrm{C}$ bonds form at the same time.

We can picture a Diels-Alder addition as occurring by head-on ( $\sigma$ ) overlap of the two alkene $p$ orbitals with the two $p$ orbitals on carbons 1 and 4 of the diene (FIGURE 14.8). This is, of course, a cyclic orientation of the reactants.

FIGURE 14.8 Mechanism of the Diels-Alder cycloaddition reaction. The reaction occurs in a single step through a cyclic transition state in which the two new $\mathrm{C}-\mathrm{C}$ bonds form simultaneously.


Solution - 14.4 The Diels-Alder Cycloaddition Reaction
In the Diels-Alder transition state, the two alkene carbons and carbons 1 and 4 of the diene rehybridize from $s p^{2}$ to $s p^{3}$ to form two new single bonds, while carbons 2 and 3 of the diene remain $s p^{2}$-hybridized to form the new double bond in the cyclohexene product. We'll study this mechanism in more detail in Section $\mathbf{3 0 . 5}$ but will concentrate for the present on learning about the characteristics and uses of the Diels-Alder reaction.


The Dienophile -
The Diels-Alder cycloaddition reaction occurs most rapidly if the alkene component, called the dienophile ("diene lover"), has an electron-withdrawing substituent group. Thus, ethylene itself reacts sluggishly, but propenal, ethyl propenoate, maleic anhydride, benzoquinone, propenenitrile, and similar compounds are highly reactive. Note also that alkynes, such as methyl propynoate, can act as Diels-Alder dienophiles.

In all cases, the double or triple bond of the dienophile is adjacent to the positively polarized carbon of an electron-withdrawing substituent. As a result, the double-bond carbons in these substances are substantially less electron-rich than the carbons in ethylene, as indicated by the electrostatic potential maps in FIGURE 14.9.

FIGURE 14.9 Electrostatic potential maps of ethylene, propenal, and propenenitrile show that electron-withdrawing groups make the double-bond carbons less electron-rich.

One of the most useful features of the Diels-Alder reaction is that it is stereospecific, meaning that a single product stereoisomer is formed. Furthermore, the stereochemistry of the dienophile is retained. If we carry out cycloaddition with methyl cis-2-butenoate, only the cis-substituted cyclohexene product is formed. With methyl trans-2-butenoate, only the trans-substituted cyclohexene product is formed.

1,3-Butadiene Methyl(Z)-2-butenoate
Cis product

Another stereochemical feature of the Diels-Alder reaction is that the diene and dienophile partners orient so that the endo product, rather than the alternative exo product, is formed. The words endo and exo are used to indicate relative stereochemistry when referring to bicyclic structures like substituted norbornanes (Section 4.9). A substituent on one bridge is said to be endo if it is syn (cis) to the larger of the other two bridges and is said to be exo if it is anti (trans) to the larger of the other two.


The Dienophile -
Endo products result from Diels-Alder reactions because the amount of orbital overlap between diene and dienophile is greater when the reactants lie directly on top of one another, so that the electron-withdrawing substituent on the dienophile is underneath the diene double bonds. In the reaction of 1,3-cyclopentadiene with maleic anhydride, for instance, the following result is obtained:

Maleic anhydride


Predicting the Product of a Diels-Alder Reaction -
Predict the product of the following Diels-Alder reaction:


Strategy -
Draw the diene so that the ends of its two double bonds are near the dienophile double bond. Then form two single bonds between the partners, convert the three double bonds into single bonds, and convert the former single bond of the diene into a double bond. Because the dienophile double bond is cis to begin with, the two attached hydrogens must remain cis in the product.


Solution -
PROBLEM Predict the product of the following Diels-Alder reaction:
14-7


The Diene -
Just as the dienophile component has certain constraints that affect its reactivity, so too does the conjugated diene component. The diene must adopt what is called an $\boldsymbol{s}$-cis conformation, meaning "cis-like" about the single bond, to undergo a Diels-Alder reaction. Only in the $s$-cis conformation are carbons 1 and 4 of the diene close enough to react through a cyclic transition state.


$s$-Cis conformation -
$\boldsymbol{s}$-Trans conformation
In the alternative $s$-trans conformation, the ends of the diene partner are too far apart to overlap with the dienophile $p$ orbitals.

Successful reaction

No reaction (ends too far apart)

Two examples of dienes that can't adopt an $s$-cis conformation, and thus don't undergo Diels-Alder reactions, are shown in FIGURE 14.10. In the bicyclic diene, the double bonds are rigidly fixed in an $s$-trans arrangement by geometric constraints of the rings. In $(2 Z, 4 Z)-2,4$-hexadiene, steric strain between the two methyl groups prevents the molecule from adopting $s$-cis geometry.

A bicyclic diene (rigid $s$-trans diene)

(2Z,4Z)-2,4-Hexadiene (s-trans, more stable)

FIGURE 14.10 Two dienes that can't achieve an $\boldsymbol{s}$-cis conformation and thus can't undergo Diels-Alder reactions.
In contrast to these unreactive dienes that can't achieve an $s$-cis conformation, other dienes are fixed only in the correct $s$-cis geometry and are therefore highly reactive in Diels-Alder cycloaddition. 1,3-Cyclopentadiene, for example, is so reactive that it reacts with itself. At room temperature, 1,3-cyclopentadiene dimerizes. One molecule acts as diene and a second molecule acts as dienophile in a self-Diels-Alder reaction.

Biological Diels-Alder reactions are also known but are uncommon. One example occurs in the biosynthesis of the cholesterol-lowering drug lovastatin (trade name Mevacor) isolated from the bacterium Aspergillus terreus. The key step is the intramolecular Diels-Alder reaction of a triene, in which the diene and dienophile components are within the same molecule.

PROBLEM Which of the following alkenes would you expect to be good Diels-Alder dienophiles?
14-8
(a)

(b)

(c)

(d)

(e)


$s$-Cis conformation -
PROBLEM Which of the following alkenes would you expect to be good Diels-Alder dienophiles?
14-8
(a)

(b)

(c)

(d)

(e)

PROBLEM Which of the following dienes have an $s$-cis conformation, and which have an $s$-trans conformation?
14-9 Of the $s$-trans dienes, which can readily rotate to $s$-cis?
(a)

(b)

(c)

PROBLEM Predict the product of the following Diels-Alder reaction:
14-10


$s$-Cis conformation - 14.6 Diene Polymers: Natural and Synthetic Rubbers
Conjugated dienes can be polymerized just as simple alkenes can (Section 8.10). Diene polymers are structurally more complex than simple alkene polymers, however, because double bonds occur every four carbon atoms along the chain, leading to the possibility of cis-trans isomers. The initiator (In) for the reaction can be either a radical, as occurs in ethylene polymerization, or an acid. Note that the polymerization is a 1,4 addition of the growing chain to a conjugated diene monomer.

trans-Polybutadiene
Rubber is a naturally occurring diene polymer of isoprene (2-methyl-1,3-butadiene) and is produced by more than 400 different plants. The major source is the so-called rubber tree, Hevea brasiliensis, from which the crude material, called latex, is harvested as it drips from a slice made through the bark. The double bonds of rubber have $Z$ stereochemistry, but gutta-percha, the $E$ isomer of rubber, also occurs naturally. Harder and more brittle than rubber, gutta-percha has a variety of applications, including use in dental endodontics and as the covering on some golf balls.

A number of different synthetic rubbers are produced commercially by diene polymerization. Both cis- and trans-polyisoprene can be made, and the synthetic rubber thus produced is similar to the natural material. Chloroprene ( 2 -chloro-1,3-butadiene) is polymerized to yield neoprene, an excellent, although expensive, synthetic rubber with good weather resistance. Neoprene is used in the production of industrial hoses and gloves, among other things.

Both natural and synthetic rubbers are too soft and tacky to be useful until they are hardened by heating with elemental sulfur, a process called vulcanization. Vulcanization cross-links the rubber chains by forming carbon-sulfur bonds between them, thereby hardening and stiffening the polymer. The exact degree of hardening can be varied, yielding material soft enough for automobile tires or hard enough for bowling balls (ebonite).


$s$-Cis conformation - 14.6 Diene Polymers: Natural and Synthetic Rubbers
The unusual ability of rubber to stretch and then contract to its original shape is due to the irregular structure of the polymer chains caused by the double bonds. These double bonds introduce bends and kinks into the polymer chains, thereby preventing neighboring chains from nestling together. When stretched, the randomly coiled chains straighten out and orient along the direction of the pull but are kept from sliding over one another by the cross-links. When the tension is released, the polymer reverts to its original random state.

PROBLEM Draw a segment of the polymer that might be prepared from 2-phenyl-1,3-butadiene. 14-11

PROBLEM Show the mechanism of the acid-catalyzed polymerization of 1,3-butadiene. 14-12


$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy
As we've seen, mass spectrometry, infrared spectroscopy, and nuclear magnetic resonance spectroscopy are techniques of structure determination applicable to all organic molecules. In addition to these three generally useful methods, there is a fourth-ultraviolet (UV) spectroscopy-that is applicable only to conjugated compounds. UV is less commonly used than the other three spectroscopic techniques because of the specialized information it gives, so we'll only discuss it briefly.

| Mass spectrometry | Molecular size and formula |
| :--- | :--- |
| IR spectroscopy | Functional groups present |
| NMR spectroscopy | Carbon-hydrogen framework |
| UV spectroscopy | Conjugated $\boldsymbol{\pi}$ electron systems |

The ultraviolet region of the electromagnetic spectrum extends from the short-wavelength end of the visible region $\left(4 \times 10^{-7} \mathrm{~m}\right)$ to the long-wavelength end of the X-ray region $\left(10^{-8} \mathrm{~m}\right)$, but the narrow range from $2 \times$ $10^{-7} \mathrm{~m}$ to $4 \times 10^{-7} \mathrm{~m}$ is the part of greatest interest to organic chemists. Absorptions in this region are usually measured in nanometers ( nm ), where $1 \mathrm{~nm}=10^{-9} \mathrm{~m}$. Thus, the ultraviolet range of interest is from 200 to 400 nm (FIGURE 14.11).

FIGURE 14.11 The ultraviolet (UV) and neighboring regions of the electromagnetic spectrum.
We saw in Section 12.5 that when an organic molecule is irradiated with electromagnetic energy, the radiation either passes through the sample or is absorbed, depending on its energy. With IR irradiation, the energy absorbed corresponds to the amount needed to increase molecular vibrations. With UV radiation, the energy absorbed corresponds to the amount needed to promote an electron from a lower-energy orbital to a higherenergy one in a conjugated molecule. The conjugated diene 1,3-butadiene, for instance, has four $\pi$ molecular orbitals, as shown previously in FIGURE 14.3. The two lower-energy, bonding MOs are occupied in the ground state, and the two higher-energy, antibonding MOs are unoccupied.


$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy
On irradiation with ultraviolet light (hv), 1,3-butadiene absorbs energy and a $\pi$ electron is promoted from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). Because the electron is promoted from a bonding $\pi$ molecular orbital to an antibonding $\pi^{*}$ molecular orbital, we call this a $\pi \rightarrow \pi^{*}$ excitation (read as "pi to pi star"). The energy gap between the HOMO and the LUMO of 1,3-butadiene is such that UV light of 217 nm wavelength is required to effect the $\pi \rightarrow \pi^{*}$ electronic transition (FIGURE 14.12).

FIGURE 14.12 Ultraviolet excitation of 1,3-butadiene results in the promotion of an electron from $\Psi_{2}$, the highest occupied molecular orbital (HOMO), to $\Psi_{3}{ }^{*}$, the lowest unoccupied molecular orbital (LUMO).

An ultraviolet spectrum is recorded by irradiating a sample with UV light of continuously changing wavelength. When the wavelength corresponds to the energy level required to excite an electron to a higher level, energy is absorbed. This absorption is detected and displayed on a chart that plots wavelength versus absorbance (A), defined as

$$
A=\log \frac{I_{0}}{I}
$$

where $I_{0}$ is the intensity of the incident light and $I$ is the intensity of the light transmitted through the sample.
Note that UV spectra differ from IR spectra in how they are presented. For historical reasons, IR spectra are usually displayed so that the baseline corresponding to zero absorption runs across the top of the chart and a valley indicates an absorption, whereas UV spectra are displayed with the baseline at the bottom of the chart so that a peak indicates an absorption (FIGURE 14.13).

The amount of UV light absorbed is expressed as the sample's molar absorptivity ( $\boldsymbol{\epsilon}$ ), defined by the equation

$$
\varepsilon=\frac{A}{c \times l}
$$


$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy
The amount of UV light absorbed is expressed as the sample's molar absorptivity ( $\boldsymbol{\epsilon}$ ), defined by the equation

$$
\varepsilon=\frac{A}{c \times l}
$$

where
$A=$ Absorbance
$c=$ Concentration in $\mathrm{mol} / \mathrm{L}$
$l=$ Sample pathlength in cm
Molar absorptivity is a physical constant, characteristic of the particular substance being observed and thus characteristic of the particular $\pi$ electron system in the molecule. Typical values for conjugated dienes are in
the range $\varepsilon=10,000$ to 25,000 . The units for molar absorptivity, $\mathrm{L} /(\mathrm{mol} \cdot \mathrm{cm})$, are usually dropped.
A particularly important use of this equation comes from rearranging it to the form $c=A /(\varepsilon \cdot I)$, which lets us measure the concentration of a sample in solution when $A, \varepsilon$, and $l$ are known. As an example, $\beta$-carotene, the pigment responsible for the orange color of carrots, has $\varepsilon=138,000 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{cm})$. If a sample of $\beta$-carotene is placed in a cell with a pathlength of 1.0 cm and the UV absorbance reads 0.37 , then the concentration of $\beta$-carotene in the sample is

$$
\begin{aligned}
c & =\frac{A}{\varepsilon l}=\frac{0.37}{\left(1.38 \times 10^{5} \frac{\mathrm{~L}}{\mathrm{~mol} \cdot \mathrm{~cm}}\right)(1.00 \mathrm{~cm})} \\
& =2.7 \times 10^{-6} \mathrm{~mol} / \mathrm{L}
\end{aligned}
$$


$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy
Unlike IR and NMR spectra, which show many absorptions for a given molecule, UV spectra are usually quite simple-often only a single peak. The peak is usually broad, and we identify its position by noting the wavelength at the top of the peak- $\lambda_{\max }$, read as "lambda max."

PROBLEM Calculate the energy range of electromagnetic radiation in the UV region of the spectrum from 200 14-13 to 400 nm (see Section 12.5). How does this value compare with the values calculated previously for IR and NMR spectroscopy?

PROBLEM If pure vitamin A has $\lambda_{\max }=325(\varepsilon=50,100)$, what is the vitamin A concentration in a sample whose 14-14 absorbance at 325 nm is $A=0.735$ in a cell with a pathlength of 1.00 cm ?


$s$-Cis conformation - 14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation
The wavelength necessary to effect the $\pi \rightarrow \pi^{*}$ transition in a conjugated molecule depends on the energy gap between HOMO and LUMO, which in turn depends on the nature of the conjugated system. Thus, by measuring the UV spectrum of an unknown, we can derive structural information about the nature of any conjugated $\pi$ electron system present in a molecule.

One of the most important factors affecting the wavelength of UV absorption by a molecule is the extent of conjugation. Molecular orbital calculations show that the energy difference between HOMO and LUMO decreases as the extent of conjugation increases. Thus, 1,3-butadiene absorbs at $\lambda_{\max }=217 \mathrm{~nm}$, 1,3,5-hexatriene absorbs at $\lambda_{\max }=258 \mathrm{~nm}$, and 1,3,5,7-octatetraene absorbs at $\lambda_{\max }=290 \mathrm{~nm}$. (Remember: longer wavelength means lower energy.)

Other kinds of conjugated systems, such as conjugated enones and aromatic rings, also have characteristic UV absorptions that are useful in structure determination. The UV absorption maxima of some representative conjugated molecules are given in TABLE 14.2.

| Name | Structure | $\lambda_{\text {max }}(\mathrm{nm})$ |
| :---: | :---: | :---: |
| 2-Methyl-1,3-butadiene | <smiles>C=CC(=C)C</smiles> | 220 |
| 1,3-Cyclohexadiene | <smiles>C1=CCCC=C1</smiles> | 256 |
| 1,3,5-Hexatriene | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$ | 258 |
| 1,3,5,7-Octatetraene | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$ | 290 |


$s$-Cis conformation - 14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation
TABLE 14.2 Ultraviolet Absorptions of Some Conjugated Molecules

| Name | Structure | $\lambda_{\text {max }}(\mathrm{nm})$ |
| :---: | :---: | :---: |
| 3-Buten-2-one | <smiles>C=CC(C)=O</smiles> | 219 |
| Benzene | $8$ | 203 |

PROBLEM Which of the following compounds would you expect to show ultraviolet absorptions in the 200 to
14-15 400 nm range?
(a)

(b)

(c)

(d)

(e)

Aspirin


Indole - 14.9 Conjugation, Color, and the Chemistry of Vision
Why are some organic compounds colored while others are not? $\beta$-Carotene, the pigment in carrots, is yelloworange, for instance, while cholesterol is colorless. The answer involves both the chemical structures of colored molecules and the way we perceive light.

The visible region of the electromagnetic spectrum is adjacent to the ultraviolet region, extending from approximately 400 to 800 nm . Colored compounds have such elaborate systems of conjugation that their "UV" absorptions extend into the visible region. $\beta$-Carotene, for instance, has 11 double bonds in conjugation, and its absorption occurs at $\lambda_{\max }=455 \mathrm{~nm}$ (FIGURE 14.14).

FIGURE 14.14 Ultraviolet spectrum of $\boldsymbol{\beta}$-carotene, a conjugated molecule with 11 double bonds. The absorption occurs in the visible region.
"White" light from the sun or from a lamp consists of all wavelengths in the visible region. When white light strikes $\beta$-carotene, the wavelengths from 400 to 500 nm (blue) are absorbed while all other wavelengths are transmitted and can reach our eyes. We therefore see the white light with the blue removed, which we perceive as a yellow-orange color for $\beta$-carotene.

Conjugation is crucial not only for the colors we see in organic molecules but also for the light-sensitive molecules on which our visual system is based. The key substance for vision is dietary $\beta$-carotene, which is converted to vitamin A by enzymes in the liver, oxidized to an aldehyde called 11-trans-retinal, and then isomerized by a change in geometry of the C11-C12 double bond to produce 11-cis-retinal.


Indole - 14.9 Conjugation, Color, and the Chemistry of Vision
There are two main types of light-sensitive receptor cells in the retina of the human eye: rod cells and cone cells. The 3 million or so rod cells are primarily responsible for seeing dim light and shades of gray, whereas the 100 million cone cells are responsible for seeing bright light and colors. In the rod cells of the eye, 11-cis-retinal is converted into rhodopsin, a light-sensitive substance formed from the protein opsin and 11-cis-retinal. When light strikes the rod cells, isomerization of the C11-C12 double bond occurs and trans-rhodopsin, called metarhodopsin II, is produced. In the absence of light, this cis-trans isomerization takes approximately 1100 years, but in the presence of light, it occurs within $2 \times 10^{-13}$ seconds! Isomerization of rhodopsin is accompanied by a change in molecular geometry, which in turn causes a nerve impulse to be sent through the optic nerve to the brain, where it is perceived as vision.

Rhodopsin
Metarhodopsin II
Metarhodopsin II is then recycled back into rhodopsin by a multistep sequence involving cleavage to all-trans-retinal and cis-trans isomerization back to 11-cis-retinal.


Photolithography -
Fifty years ago, someone interested in owning a computer would have paid approximately $\$ 150,000$ for 16 megabytes of random-access memory that would have occupied a volume the size of a small desk. Today, anyone can buy 60 times as much computer memory for $\$ 10$ and fit the small chip into their shirt pocket. The difference between then and now is due to improvements in photolithography, the process by which integratedcircuit chips are made.

Photolithography begins by coating a layer of $\mathrm{SiO}_{2}$ onto a silicon wafer and further coating with a thin (0.5-1.0 $\mu \mathrm{m})$ film of a light-sensitive organic polymer called a resist. A mask is then used to cover those parts of the chip that will become a circuit, and the wafer is irradiated with UV light. The nonmasked sections of the polymer undergo a chemical change when irradiated, which makes them more soluble than the masked, unirradiated sections. On washing the irradiated chip with solvent, solubilized polymer is selectively removed from the irradiated areas, exposing the $\mathrm{SiO}_{2}$ underneath. This $\mathrm{SiO}_{2}$ is then chemically etched away by reaction with hydrofluoric acid, leaving behind a pattern of polymer-coated $\mathrm{SiO}_{2}$. Further washing removes the remaining polymer, leaving a positive image of the mask in the form of exposed ridges of $\mathrm{SiO}_{2}$ (FIGURE 14.16). Additional cycles of coating, masking, and etching then produce the completed chips.

FIGURE 14.15 Manufacturing the ultrathin circuitry on this computer chip depends on the organic chemical reactions of special polymers. (credit: "Integrated circuit on microchip" by Jon Sullivan/Wikimedia Commons, Public Domain)


Photolithography -
FIGURE 14.15 Manufacturing the ultrathin circuitry on this computer chip depends on the organic chemical reactions of special polymers. (credit: "Integrated circuit on microchip" by Jon Sullivan/Wikimedia Commons, Public Domain)

The polymer resist currently used in chip manufacturing is based on the two-component diazoquinone-novolac system. Novolac resin is a soft, relatively low-molecular-weight polymer made from methylphenol and formaldehyde, while the diazoquinone is a bicyclic (two-ring) molecule containing a diazo
group $=\mathrm{N}=\mathrm{N}$ ) adjacent to a ketone carbonyl ( $\mathrm{C}=\mathrm{O}$ ). The diazoquinone-novolac mix is relatively insoluble when fresh, but on exposure to ultraviolet light and water vapor, the diazoquinone component undergoes reaction to yield $\mathrm{N}_{2}$ and a carboxylic acid, which can be washed away with dilute base. Novolac-diazoquinone technology is capable of producing features as small as $0.5 \mu \mathrm{~m}\left(5 \times 10^{-7} \mathrm{~m}\right)$, but further improvements in miniaturization are still being developed.

Diazonaphthoquinone

Novolac resin


Key Terms -
- 1,2-addition
- 1,4-addition
- conjugated
- Diels-Alder cycloaddition reaction
- dienophile
- endo
- exo
- highest occupied molecular orbital (HOMO)
- kinetic control
- lowest unoccupied molecular orbital (LUMO)
- molar absorptivity ( $\varepsilon$ )
- $s$-cis conformation
- stereospecific
- thermodynamic control
- ultraviolet (UV) spectroscopy


Summary -
The unsaturated compounds we've looked at previously have had only one double bond, but many compounds have numerous sites of unsaturation, which gives them some distinctive properties. Many such compounds are common in nature, including pigments and hormones.

A conjugated diene or other compound is one that contains alternating double and single bonds. One characteristic of conjugated dienes is that they are more stable than their nonconjugated counterparts. This stability can be explained by a molecular orbital description in which four $p$ atomic orbitals combine to form four $\pi$ molecular orbitals. Only the two bonding orbitals are occupied; the two antibonding orbitals are unoccupied. A $\pi$ bonding interaction in the lowest-energy MO introduces some partial double-bond character between carbons 2 and 3, thereby strengthening the $\mathrm{C} 2-\mathrm{C} 3$ bond and stabilizing the molecule.

Conjugated dienes undergo several reactions not observed for nonconjugated dienes. One is the 1,4-addition of electrophiles. When a conjugated diene is treated with an electrophile such as $\mathrm{HCl}, \mathbf{1 , 2 -}$ and $\mathbf{1 , 4}$-addition products are formed. Both result from the same resonance-stabilized allylic carbocation intermediate and are produced in varying amounts depending on the reaction conditions. The 1,2 adduct is usually formed faster and is said to be the product of kinetic control. The 1,4 adduct is usually more stable and is said to be the product of thermodynamic control.

Another reaction unique to conjugated dienes is Diels-Alder cycloaddition. Conjugated dienes react with electron-poor alkenes (dienophiles) in a single step through a cyclic transition state to yield a cyclohexene product. The reaction is stereospecific, meaning that only a single product stereoisomer is formed, and can occur only if the diene is able to adopt an $s$-cis conformation.


Summary -
Ultraviolet (UV) spectroscopy is a method of structure determination applicable specifically to conjugated $\pi$-electron systems. When a conjugated molecule is irradiated with ultraviolet light, energy absorption occurs and a $\pi$ electron is promoted from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). For 1,3-butadiene, radiation of $\lambda_{\max }=217 \mathrm{~nm}$ is required. The greater the extent of conjugation, the less the energy needed and the longer the wavelength of radiation required.


Summary of Reactions -
1. Electrophilic addition reactions (Section 14.2 and Section 14.3)

2. Diels-Alder cycloaddition reaction (Section 14.4 and Section 14.5)


Visualizing Chemistry -
PROBLEM Show the structures of all possible adducts of the following diene with 1 equivalent of HCl :


14-16 -
PROBLEM Show the product of the Diels-Alder reaction of the following diene with 3-buten-2-one,
14-17 $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCOCH}_{3}$. Make sure you show the full stereochemistry of the reaction product.

PROBLEM The following diene does not undergo Diels-Alder reactions. Explain.
14-18

PROBLEM The following model is that of an allylic carbocation intermediate formed by protonation of a
14-19 conjugated diene with HBr . Show the structure of the diene and the structures of the final reaction products.


Mechanism Problems -
PROBLEM Predict the major product(s) from the addition of 1 equivalent of HX and show the mechanism for 14-20 each of the following reactions.
(a)

(b)

$\xrightarrow[40^{\circ} \mathrm{C}]{\mathrm{HBr}}$
(c)

PROBLEM We've seen that the Diels-Alder cycloaddition reaction is a one-step, pericyclic process that occurs
14-21 through a cyclic transition state. Propose a mechanism for the following reaction:

PROBLEM In light of your answer to Problem 14-21 propose mechanisms for the following reactions.
14-22 (a)

(b)

PROBLEM Luminol, which is used by forensic scientists to find blood, fluoresces as a result of Diels-Alder-like
14-23 process. The dianion of luminol reacts with $\mathrm{O}_{2}$ to form an unstable peroxide intermediate that then loses nitrogen to form a dicarboxylate and emit light. The process is similar to that in Problems 14-21 and 14-22. Propose a mechanism for this process.

Luminol dianion
PROBLEM A useful diene in the synthesis of many naturally occurring substances is known as Danishefsky's
14-24 diene. It's useful because after the Diels-Alder reaction it can be converted into a product that can't be accessed by a typical Diels-Alder reaction. Show the Diels-Alder adduct and propose a mechanism that accounts for the final products.


Conjugated Dienes -
PROBLEM Name the following compounds:
14-25 (a)

(b)
$\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}=\mathrm{CHCH}=\mathrm{CHCH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}=\mathrm{CHCH}=\mathrm{CHCH}_{3}$
(d)

PROBLEM Draw and name the six possible diene isomers of formula $\mathrm{C}_{5} \mathrm{H}_{8}$. Which of the six are conjugated
14-26 dienes?
PROBLEM What product(s) would you expect to obtain from reaction of 1,3-cyclohexadiene with each of the
14-27 following?
(a) $1 \mathrm{~mol} \mathrm{Br}_{2}$ in $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(b) $\mathrm{O}_{3}$ followed by Zn
(c) 1 mol HCl in ether (d) 1 mol DCl in ether
(e) 3-Buten-2-one $\left(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCOCH}_{3}\right)$ (f) Excess $\mathrm{OsO}_{4}$, followed by $\mathrm{NaHSO}_{3}$

PROBLEM Electrophilic addition of $\mathrm{Br}_{2}$ to isoprene (2-methyl-1,3-butadiene) yields the following product 14-28 mixture:

Of the 1,2 -addition products, explain why 3,4-dibromo-3-methyl-1-butene (21\%) predominates over 3,4-dibromo-2-methyl-1-butene (3\%).

PROBLEM Propose a structure for a conjugated diene that gives the same product from both 1,2 and
14-29 1,4-addition of HBr .
PROBLEM Draw the possible products resulting from addition of 1 equivalent of HCl to
14-30 1-phenyl-1,3-butadiene. Which would you expect to predominate, and why?


Conjugated Dienes -
Diels-Alder Reactions
PROBLEM Predict the products of the following Diels-Alder reactions:
14-31 (a)

(b)
?

PROBLEM 2,3-Di-tert-butyl-1,3-butadiene does not undergo Diels-Alder reactions. Explain.
14-32


2,3-Di-tert-butyl-1,3-butadiene -
PROBLEM Show the structure, including stereochemistry, of the product from the following Diels-Alder 14-33 reaction:

PROBLEM How can you account for the fact that cis-1,3-pentadiene is much less reactive than 14-34 trans-1,3-pentadiene in the Diels-Alder reaction?

PROBLEM Would you expect a conjugated diyne such as 1,3-butadiyne to undergo Diels-Alder reaction with a
14-35 dienophile? Explain.
PROBLEM Reaction of isoprene (2-methyl-1,3-butadiene) with ethyl propenoate gives a mixture of two 14-36 Diels-Alder adducts. Show the structure of both, and explain why a mixture is formed.

PROBLEM Rank the following dienophiles in order of their expected reactivity in the Diels-Alder reaction.
14-37

PROBLEM 1,3-Cyclopentadiene is very reactive in Diels-Alder cycloaddition reactions, but
14-38 1,3-cyclohexadiene is less reactive and 1,3-cycloheptadiene is nearly inert. Explain. (Molecular models are helpful.)

PROBLEM 1,3-Pentadiene is much more reactive in Diels-Alder reactions than 2,4-pentadienal. Why might 14-39 this be?


1,3-Pentadiene -
2,4-Pentadienal

PROBLEM How could you use Diels-Alder reactions to prepare the following products? Show the starting diene
14-40 and dienophile in each case.
(a)

(b)

(c)

(d)

PROBLEM Show the product of the following reaction.
14-41


Diene Polymers -
PROBLEM Tires whose sidewalls are made of natural rubber tend to crack and weather rapidly in areas around
14-42 cities where high levels of ozone and other industrial pollutants are found. Explain.
PROBLEM 1,3-Cyclopentadiene polymerizes slowly at room temperature to yield a polymer that has no double
14-43 bonds except on the ends. On heating, the polymer breaks down to regenerate 1,3-cyclopentadiene. Propose a structure for the product.


UV Spectroscopy -
PROBLEM Arrange the molecules in each of the following sets according to where you would expect to find
14-44 their wavelength of maximum absorption in UV spectroscopy, from shortest to longest wavelength.
(a)

(b)

(c)

PROBLEM Which of the following compounds would you expect to have a $\pi \rightarrow \pi^{*}$ UV absorption in the 200 to 14-45 400 nm range?
(a)

(b)

(c) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}=\mathrm{O}$
A ketene

Pyridine
PROBLEM Would you expect allene, $\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}$, to show a UV absorption in the 200 to 400 nm range?
14-46 Explain.
PROBLEM The following ultraviolet absorption maxima have been measured:
14-47

| 1,3-Butadiene | 217 nm |
| :--- | :--- |
| 2-Methyl-1,3-butadiene | 220 nm |
| 1,3-Pentadiene | 223 nm |
| 2,3-Dimethyl-1,3-butadiene | 226 nm |
| 2,4-Hexadiene | 227 nm |
| 2,4-Dimethyl-1,3-pentadiene | 232 nm |
| 2,5-Dimethyl-2,4-hexadiene | 240 nm |

What conclusion can you draw about the effect of alkyl substitution on UV absorption maxima? Approximately what effect does each added alkyl group have?


UV Spectroscopy -
What conclusion can you draw about the effect of alkyl substitution on UV absorption maxima? Approximately what effect does each added alkyl group have?

PROBLEM 1,3,5-Hexatriene has $\lambda_{\max }=258 \mathrm{~nm}$. In light of your answer to Problem 14-47, approximately
14-48 where would you expect 2,3-dimethyl-1,3,5-hexatriene to absorb?
PROBLEM $\beta$-Ocimene is a pleasant-smelling hydrocarbon found in the leaves of certain herbs. It has the
14-49 molecular formula $\mathrm{C}_{10} \mathrm{H}_{16}$ and a UV absorption maximum at 232 nm . On hydrogenation with a palladium catalyst, 2,6-dimethyloctane is obtained. Ozonolysis of $\beta$-ocimene, followed by treatment with zinc and acetic acid, produces the following four fragments:

Acetone

Formaldehyde

Pyruvaldehyde

(a) How many double bonds does $\beta$-ocimene have?
(b) Is $\beta$-ocimene conjugated or nonconjugated? (c) Propose a structure for $\beta$-ocimene.
(d) Write the reactions, showing starting material and products.


General Problems -
PROBLEM Draw the resonance forms that result when the following dienes are protonated. If the resonance 14-50 forms differ in energy, identify the most stable one.
(a)

(b)

(c)

PROBLEM Answer the following questions for 1,3,5-cycloheptatriene.
14-51 (a) How many $p$ atomic orbitals are in the conjugated system?
(b) How many molecular orbitals describe the conjugated system?
(c) How many molecular orbitals are bonding molecular orbitals?
(d) How many molecular orbitals are anti-bonding molecular orbitals?
(e) Which molecular orbitals are filled with electrons?
(f) If this molecule were to absorb a photon of UV light an electron would move between which two molecular orbitals (be specific)?

PROBLEM Treatment of 3,4-dibromohexane with strong base leads to loss of 2 equivalents of HBr and
14-52 formation of a product with formula $\mathrm{C}_{6} \mathrm{H}_{10}$. Three products are possible. Name each of the three, and tell how you would use ${ }^{1} \mathrm{H}$ and ${ }^{13} \mathrm{C}$ NMR spectroscopy to help identify them. How would you use UV spectroscopy?

PROBLEM Addition of HCl to 1-methoxycyclohexene yields 1-chloro-1-methoxycyclohexane as a sole product.
14-53 Use resonance structures to explain why none of the other regioisomer is formed.

PROBLEM Aldrin, a chlorinated insecticide now banned from use in most countries since 1990, can be made
14-54 by Diels-Alder reaction of hexachloro-1,3-cyclopentadiene with norbornadiene. What is the structure of aldrin?


Norbornadiene -
PROBLEM Norbornadiene (Problem 14-54) can be prepared by reaction of chloroethylene with
14-55 1,3-cyclopentadiene, followed by treatment of the product with sodium ethoxide. Write the overall scheme, and identify the two kinds of reactions.

PROBLEM The triene shown here reacts with 2 equivalents of maleic anhydride to yield a product with the
14-56 formula $\mathrm{C}_{17} \mathrm{H}_{16} \mathrm{O}_{6}$. Predict a structure for the product.

PROBLEM Myrcene, $\mathrm{C}_{10} \mathrm{H}_{16}$, is found in oil of bay leaves and is isomeric with $\beta$-ocimene (Problem 14-49).
14-57 It has an ultraviolet absorption at 226 nm and can be hydrogenated to yield 2,6-dimethyloctane. On ozonolysis followed by zinc/acetic acid treatment, myrcene yields formaldehyde, acetone, and 2-oxopentanedial:

2-Oxopentanedial
Propose a structure for myrcene, and write the reactions, showing starting material and products.

PROBLEM Hydrocarbon $\mathbf{A}, \mathrm{C}_{10} \mathrm{H}_{14}$, has a UV absorption at $\lambda_{\max }=236 \mathrm{~nm}$ and gives hydrocarbon $\mathbf{B}, \mathrm{C}_{10} \mathrm{H}_{18}$, on
14-58 hydrogenation. Ozonolysis of $\mathbf{A}$, followed by zinc/acetic acid treatment, yields the following diketo dialdehyde:

An illustration shows the structure of diketo dialdehyde. It shows two carbonyl groups single bonded to each other. Each carbonyl group is bonded to a chain of three methylene groups and then to an aldehyde group.
(a) Propose two possible structures for $\mathbf{A}$.
(b) Hydrocarbon $\mathbf{A}$ reacts with maleic anhydride to yield a Diels-Alder adduct. Which of your structures for A is correct?
(c) Write the reactions, showing the starting material and products.


Norbornadiene -
PROBLEM Adiponitrile, a starting material used in the manufacture of nylon, can be prepared in three steps
14-59 from 1,3-butadiene. How would you carry out this synthesis?
$\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}=\mathrm{CH}_{2} \quad \xrightarrow{3 \text { steps }} \mathrm{N} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$
Adiponitrile
PROBLEM Ergosterol, a precursor of vitamin D, has $\lambda_{\max }=282 \mathrm{~nm}$ and molar absorptivity $\epsilon=11,900$. What is
14-60 the concentration of ergosterol in a solution whose absorbance $A=0.065$ with a sample pathlength $l=1.00 \mathrm{~cm}$ ?

PROBLEM Dimethyl butynedioate undergoes a Diels-Alder reaction with ( $2 E, 4 E$ )-2,4-hexadiene. Show the
14-61 structure and stereochemistry of the product.

Dimethyl butynedioate
PROBLEM Dimethyl butynedioate also undergoes a Diels-Alder reaction with (2E,4Z)-2,4-hexadiene, but the
14-62 stereochemistry of the product is different from that of the ( $2 E, 4 E$ ) isomer (Problem 14-61). Explain.

PROBLEM How would you carry out the following synthesis (more than one step is required)? What
14-63 stereochemical relationship between the $-\mathrm{CO}_{2} \mathrm{CH}_{3}$ group attached to the cyclohexane ring and the -CHO groups would your synthesis produce?

PROBLEM The double bond of an enamine (alkene + amine) is much more nucleophilic than a typical alkene
14-64 double bond. Assuming that the nitrogen atom in an enamine is $s p^{2}$-hybridized, draw an orbital picture of an enamine, and explain why the double bond is electron-rich.


An enamine -
PROBLEM Benzene has an ultraviolet absorption at $\lambda_{\max }=204 \mathrm{~nm}$, and para-toluidine has $\lambda_{\max }=235 \mathrm{~nm}$.
14-65 How do you account for this difference?

Benzene
( $\lambda_{\text {max }}=204 \mathrm{~nm}$ )

$p$-Toluidine
( $\lambda_{\text {max }}=\mathbf{2 3 5} \mathbf{n m}$ )


CHAPTER 15 -
Benzene and Aromaticity

FIGURE 15.1 A fennel plant is an aromatic herb used in cooking. A phenyl group-pronounced exactly the same way-is the characteristic structural unit of "aromatic" organic compounds. (credit: modificatio of work "Fennel" by Ruth Hartnup/Flickr, CC BY 2.0)


CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds
15.2 Structure and Stability of Benzene
15.3 Aromaticity and the Hückel $4 n+2$ Rule
15.4 Aromatic Ions
15.5 Aromatic Heterocycles: Pyridine and Pyrrole
15.6 Polycyclic Aromatic Compounds
15.7 Spectroscopy of Aromatic Compounds

WHY THIS CHAPTER? The reactivity of substituted aromatic compounds, more than that of any other class of substances, is intimately tied to their exact structure. As a result, aromatic compounds provide an extraordinarily sensitive probe for studying the relationship between structure and reactivity. We'll examine this relationship here and in the following chapter, and we'll find that the lessons learned are applicable to all other organic compounds, including important substances such as the nucleic acids that control our genetic makeup.

In the early days of organic chemistry, the word aromatic was used to describe such fragrant substances as benzaldehyde (from cherries, peaches, and almonds), toluene (from Tolu balsam), and benzene (from coal distillate). It was soon realized, however, that substances grouped as aromatic differed from most other organic compounds in their chemical behavior.

Benzene

Benzaldehyde

Toluene

Today, the association of aromaticity with fragrance has long been lost, and we now use the word aromatic to refer to the class of compounds that contain six-membered rings with three alternating double bonds. Many naturally occurring compounds are aromatic in part, including steroids such as estrone and well-known pharmaceuticals such as the cholesterol-lowering drug atorvastatin, marketed as Lipitor. Benzene itself causes a depressed white blood cell count (leukopenia) on prolonged exposure and should not be used as a laboratory solvent.

Estrone

Atorvastatin (Lipitor)


CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds
Simple aromatic hydrocarbons come from two main sources: coal and petroleum. Coal is an enormously complex mixture consisting primarily of large arrays of benzene-like rings joined together. Thermal breakdown of coal occurs when heated to $1000^{\circ} \mathrm{C}$ in the absence of air, and a mixture of volatile products called coal tar boils off. Fractional distillation of coal tar yields benzene, toluene, xylene (dimethylbenzene), naphthalene, and a host of other aromatic compounds (FIGURE 15.2).

Benzene (bp $80^{\circ} \mathrm{C}$ )

Toluene (bp $111{ }^{\circ} \mathrm{C}$ )

Xylene (bp: ortho, $144{ }^{\circ} \mathrm{C}$; meta, $139{ }^{\circ} \mathrm{C}$; para, $138^{\circ} \mathrm{C}$ )

Indene (bp $182{ }^{\circ} \mathrm{C}$ )

Phenanthrene (mp $101{ }^{\circ} \mathrm{C}$ )
FIGURE 15.2 Some aromatic hydrocarbons found in coal tar.

Unlike coal, petroleum contains few aromatic compounds and consists largely of alkanes (see Chapter 3 Chemistry Matters). During petroleum refining, however, aromatic molecules are formed when alkanes are passed over a catalyst at about $500^{\circ} \mathrm{C}$ under high pressure.

Aromatic substances, more than any other class of organic compounds, have acquired a large number of nonsystematic names. IUPAC rules discourage the use of most such names but do allow some of the more widely used ones to be retained (TABLE 15.1). Thus, methylbenzene is known commonly as toluene; hydroxybenzene as phenol; aminobenzene as aniline; and so on.


CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds
| TABLE 15.1 Common Aromatic Compounds | Names of Some |
| :---: | :---: |
| Structure | Name |
| <smiles>Cc1ccccc1</smiles> | Toluene (bp $111^{\circ} \mathrm{C}$ ) |
| <smiles>Oc1ccccc1</smiles> | Phenol $\left(\mathrm{mp} 43^{\circ} \mathrm{C}\right)$ |
| <smiles>Nc1ccccc1</smiles> | Aniline $\left(\mathrm{bp} 184^{\circ} \mathrm{C}\right)$ |
| <smiles>CC(=O)c1ccccc1</smiles> | Acetophenone (mp $21^{\circ} \mathrm{C}$ ) |
| <smiles>O=Cc1ccccc1</smiles> | Benzaldehyde (bp $178{ }^{\circ} \mathrm{C}$ ) |
| <smiles>O=C(O)c1ccccc1</smiles> | Benzoic acid (mp $122^{\circ} \mathrm{C}$ ) |
| <smiles>Cc1ccccc1C</smiles> | ortho-Xylene (bp $144{ }^{\circ} \mathrm{C}$ ) |
| <smiles>C=Cc1ccccc1</smiles> | Styrene <br> (bp $145^{\circ} \mathrm{C}$ ) |

Monosubstituted benzenes are named systematically in the same manner as other hydrocarbons, with -benzene as the parent name. Thus, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}$ is bromobenzene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$ is nitrobenzene, and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ is propylbenzene.

Bromobenzene

Nitrobenzene

Propylbenzene


CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds
Bromobenzene

Nitrobenzene

Propylbenzene

Alkyl-substituted benzenes are sometimes referred to as arenes and are named in different ways depending on the size of the alkyl group. If the alkyl substituent is smaller than the ring (six or fewer carbons), the arene is referred to as an alkyl-substituted benzene. If the alkyl substituent is larger than the ring (seven or more carbons), the compound is referred to as a phenyl-substituted alkane. The name phenyl, pronounced fen-nil and sometimes abbreviated as Ph or $\Phi$ (Greek phi), is used for the $-\mathrm{C}_{6} \mathrm{H}_{5}$ unit when the benzene ring
is considered a substituent. The word is derived from the Greek pheno ("I bear light"), commemorating the discovery of benzene by Michael Faraday in 1825 from the oily residue left by the illuminating gas used in London street lamps. In addition, the name benzyl is used for the $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}$ - group.

Disubstituted benzenes are named using the prefixes ortho (o), meta (m), or para (p). An ortho-disubstituted benzene has its two substituents in a 1,2 relationship on the ring, a meta-disubstituted benzene has its two substituents in a 1,3 relationship, and a para-disubstituted benzene has its substituents in a 1,4 relationship.

ortho-Dichlorobenzene 1,2 disubstituted

meta-Dimethylbenzene (meta-xylene) 1,3 disubstituted

para-Chlorobenzaldehyde
1,4 disubstituted

The ortho, meta, para system of nomenclature is also useful when discussing reactions. For example, we might describe the reaction of bromine with toluene by saying, "Reaction occurs at the para position," meaning at the position para to the methyl group already present on the ring.


CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds
Toluene $\quad \boldsymbol{p}$-Bromotoluene
As with cycloalkanes (Section 4.1), benzenes with more than two substituents are named by choosing a point of attachment as carbon 1 and numbering the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found. The substituents are listed alphabetically when writing the name.

Note in the second and third examples shown that -phenol and -toluene are used as the parent names rather than -benzene. Any of the monosubstituted aromatic compounds shown in TABLE 15.1 can serve as a parent name, with the principal substituent ( -OH in phenol or $-\mathrm{CH}_{3}$ in toluene) attached to C 1 on the ring.

PROBLEM Tell whether the following compounds are ortho-, meta-, or para-disubstituted:
15-1 (a)

(b)

(c)

PROBLEM Give IUPAC names for the following compounds:
15-2 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Draw structures corresponding to the following IUPAC names:
15-3 (a) $p$-Bromochlorobenzene (b) $p$-Bromotoluene (c) $m$-Chloroaniline
(d) 1-Chloro-3,5-dimethylbenzene


CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene
Benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$ has six fewer hydrogens than the six-carbon cycloalkane cyclohexane $\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)$ and is clearly unsaturated, usually being represented as a six-membered ring with alternating double and single bonds. Yet it has been known since the mid-1800s that benzene is much less reactive than typical alkenes and fails to undergo typical alkene addition reactions. Cyclohexene, for instance, reacts rapidly with $\mathrm{Br}_{2}$ and gives the addition product 1,2-dibromocyclohexane, but benzene reacts slowly with $\mathrm{Br}_{2}$ and gives the substitution product $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}$ rather than the addition product.

We can get a quantitative idea of benzene's stability by measuring heats of hydrogenation (Section 7.6). Cyclohexene, an isolated alkene, has $\Delta H_{\text {hydrog }}^{\circ}=-118 \mathrm{~kJ} / \mathrm{mol}(-28.2 \mathrm{kcal} / \mathrm{mol})$, and 1,3-cyclohexadiene, a conjugated diene, has $\Delta H^{\circ}{ }_{\text {hydrog }}=-230 \mathrm{~kJ} / \mathrm{mol}(-55.0 \mathrm{kcal} / \mathrm{mol})$. As noted in Section 14.1, this value for 1,3-cyclohexadiene is a bit less than twice that for cyclohexene because conjugated dienes are more stable than isolated dienes.


CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene
Carrying the process one step further, we might expect $\Delta H^{\circ}{ }_{\text {hydrog }}$ for "cyclohexatriene" (benzene) to be a bit less than $3 \times 118=354 \mathrm{~kJ} / \mathrm{mol}$, or three times the cyclohexene value. The actual value, however, is $-206 \mathrm{~kJ} /$ mol , some $148 \mathrm{~kJ} / \mathrm{mol}(35.4 \mathrm{kcal} / \mathrm{mol})$ less than expected. Because of this difference in actual and expected energy released during hydrogenation, benzene must have $148 \mathrm{~kJ} / \mathrm{mol}$ less energy to begin with. In other words, benzene is more stable than expected by $148 \mathrm{~kJ} / \mathrm{mol}$ (FIGURE 15.3).

FIGURE 15.3 A comparison of the heats of hydrogenation for cyclohexene, 1,3-cyclohexadiene, and benzene. Benzene is $148 \mathrm{~kJ} / \mathrm{mol}$ $(35.4 \mathrm{kcal} / \mathrm{mol})$ more stable than might be expected for "cyclohexatriene."

Further evidence for the unusual nature of benzene is that all its carbon-carbon bonds have the same length -139 pm , which is intermediate between typical single ( 154 pm ) and double ( 134 pm ) bonds. In addition, an electrostatic potential map shows that the electron density in all six $\mathrm{C}-\mathrm{C}$ bonds is identical. Thus, benzene is a planar molecule with the shape of a regular hexagon. All $\mathrm{C}-\mathrm{C}-\mathrm{C}$ bond angles are $120^{\circ}$, all six carbon atoms are $s p^{2}$-hybridized, and each carbon has a $p$ orbital perpendicular to the plane of the six-membered ring.


CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene
Because all six carbon atoms and all six $p$ orbitals in benzene are equivalent, it's impossible to define three localized $\pi$ bonds in which a given $p$ orbital overlaps only one neighboring $p$ orbital. Rather, each $p$ orbital overlaps equally well with both neighboring $p$ orbitals, leading to a picture of benzene in which all six $\pi$ electrons are free to move about the entire ring (FIGURE 15.4b). In resonance terms (Section 2.4 and Section 2.5), benzene is a hybrid of two equivalent forms. Neither form is correct by itself; the true structure of benzene is somewhere in between the two resonance forms but is impossible to draw with our usual conventions. Because of this resonance, benzene is more stable and less reactive than a typical alkene.

(b)

FIGURE 15.4 (a) An electrostatic potential map of benzene and (b) an orbital picture. Each of the six carbon atoms has a $p$ orbital that can overlap equally well with neighboring $p$ orbitals on both sides. As a result, all $\mathrm{C}-\mathrm{C}$ bonds are equivalent and benzene must be represented as a hybrid of two resonance forms.

Chemists sometimes represent the two benzene resonance forms by using a circle to indicate the equivalence of the carbon-carbon bonds, but this representation has to be used carefully because it doesn't indicate the number of $\pi$ electrons in the ring. (How many electrons does a circle represent?) In this book, benzene and
other aromatic compounds will be represented by single line-bond structures. We'll be able to keep count of $\pi$ electrons this way but must be aware of the limitations of the drawings.

Having just seen a resonance description of benzene, let's now look at the alternative molecular orbital description. We can construct $\pi$ molecular orbitals for benzene just as we did for 1,3-butadiene in Section 14.1. If six $p$ atomic orbitals combine in a cyclic manner, six benzene molecular orbitals result, as shown in FIGURE 15.5. The three low-energy molecular orbitals, denoted $\psi_{1}, \psi_{2}$, and $\psi_{3}$, are bonding combinations, and the three high-energy orbitals are antibonding.


CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene
Six benzene molecular orbitals
FIGURE 15.5 The six benzene $\boldsymbol{\pi}$ molecular orbitals. The bonding orbitals $\psi_{2}$ and $\psi_{3}$ have the same energy and are said to be degenerate, as are the antibonding orbitals $\psi_{4}{ }^{*}$ and $\psi_{5}{ }^{*}$. The orbitals $\psi_{3}$ and $\psi_{4}{ }^{*}$ have no $\pi$ electron density on two carbons because of a node passing through these atoms.

Note that the two bonding orbitals $\psi_{2}$ and $\psi_{3}$ have the same energy, as do the two antibonding orbitals $\psi_{4}$ * and $\psi_{5}{ }^{*}$. Such orbitals with the same energy are said to be degenerate. Note also that the two orbitals $\psi_{3}$ and $\psi_{4} *$ have nodes passing through ring carbon atoms, thereby leaving no $\pi$ electron density on these carbons. The six $p$ electrons of benzene occupy the three bonding molecular orbitals and are delocalized over the entire conjugated system, leading to the observed $150 \mathrm{~kJ} / \mathrm{mol}$ stabilization of benzene.

PROBLEM Pyridine is a flat, hexagonal molecule with bond angles of $120^{\circ}$. It undergoes substitution rather
15-4 than addition and generally behaves like benzene. Draw a picture of the $\pi$ orbitals of pyridine to explain its properties. Check your answer by looking ahead to Section 15.5.

Pyridine


CHAPTER CONTENTS - 15.3 Aromaticity and the Hückel $4 n+2$ Rule
Let's list what we've said thus far about benzene and, by extension, about other benzene-like aromatic molecules.

- Benzene is cyclic and conjugated.
- Benzene is unusually stable, having a heat of hydrogenation $150 \mathrm{~kJ} / \mathrm{mol}$ less negative than we might expect for a conjugated cyclic triene.
- Benzene is planar and has the shape of a regular hexagon. All bond angles are $120^{\circ}$, all carbon atoms are $s p^{2}$-hybridized, and all carbon-carbon bond lengths are 139 pm .
- Benzene undergoes substitution reactions that retain the cyclic conjugation rather than electrophilic addition reactions that would destroy it.
- Benzene can be described as a resonance hybrid whose structure is intermediate between two line-bond structures.

This list might seem to be a good description of benzene and other aromatic molecules, but it isn't enough. Something else, called the Hückel $4 \boldsymbol{n}+2$ rule, is needed to complete a description of aromaticity. According to a theory devised in 1931 by the German physicist Erich Hückel, a molecule is aromatic only if it has a planar, monocyclic system of conjugation and contains a total of $4 n+2 \pi$ electrons, where $n$ is an integer ( $n=0,1$, 2,3 ,). In other words, only molecules with $2,6,10,14,18, \pi$ electrons can be aromatic. Molecules with $4 n \pi$ electrons ( $4,8,12,16$,) can't be aromatic, even though they may be cyclic, planar, and apparently conjugated. In fact, planar, conjugated molecules with $4 n \pi$ electrons are said to be antiaromatic because delocalization of their $\pi$ electrons would lead to their destabilization.

Let's look at several examples to see how the Hückel $4 n+2$ rule works.

- Cyclobutadiene has four $\pi$ electrons and is antiaromatic. As indicated by the electrostatic potential map, the $\pi$ electrons are localized in two double bonds rather than delocalized around the ring.


CHAPTER CONTENTS - 15.3 Aromaticity and the Hückel $4 n+2$ Rule
- Cyclobutadiene has four $\pi$ electrons and is antiaromatic. As indicated by the electrostatic potential map, the $\pi$ electrons are localized in two double bonds rather than delocalized around the ring.

Cyclobutadiene is highly reactive and shows none of the properties associated with aromaticity. In fact, it was not even prepared until 1965, when Rowland Pettit at the University of Texas was able to make it at low temperature. Even at $-78^{\circ} \mathrm{C}$, however, cyclobutadiene is so reactive that it dimerizes by a Diels-Alder reaction. One molecule behaves as a diene and the other as a dienophile.

- Benzene has six $\pi$ electrons $(4 n+2=6$ when $n=1)$ and is aromatic.