# Organic Chemistry I & II

Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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## Comparison of Organic Chemistry Textbooks

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### Top Hat

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

### McGraw-Hill

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

### Wiley

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

### Wiley

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

### Pricing

Average price of textbook across most common format

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Hardcover print text only

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

Hardcover print text only

### Wiley

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

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

Hardcover print text only

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

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

### Top Hat

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

### McGraw-Hill

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

### Wiley

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

### Wiley

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

## In-book Interactivity

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

### Top Hat

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

### McGraw-Hill

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

### Wiley

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

### Wiley

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

## Customizable

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

### Top Hat

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

### McGraw-Hill

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

### Wiley

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

### Wiley

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

## All-in-one Platform

### Top Hat

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

### McGraw-Hill

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

### Wiley

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

### Wiley

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

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

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

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# Chapter 19: Conjugated Systems, Orbital Symmetry, and UV-Vis Spectroscopy

## Learning Objectives

• Define what is meant by conjugation on electronic structure.
• Understand the effect of conjugation on reaction mechanisms.
• Define what is meant by conjugation.
• Determine the major product of an addition reaction for a conjugated system.
• Determine the major product of a Diels-Alder reaction.

## 19.1 Introduction

In this chapter, we’ll be exploring a specific class of hydrocarbons called conjugated π systems, which involve an orderly arrangement of alternating  double bonds and single bonds. First, we’ll examine how and why they are thermodynamically more stable compared to non-conjugated hydrocarbons of similar size and shape. Molecular orbital theory will be used to account for this enhanced stability. Second, reactions and mechanisms of conjugated systems will be explored. An emphasis will be placed on how conjugation can influence the product of the reaction. Third, an application of conjugated systems in electronic spectroscopy will be investigated. We will introduce electronic spectroscopy and correlate a conjugated system’s structure with its color, or lack thereof.

### 19.1.1 Classification of Dienes

In chapter 2.8 you were introduced to the nomenclature of conjugated dienes and in chapter 16.5 the chirality of allenes or cumulated dienes was discussed. Shown in Figure 19.1 are the three classes of dienes: conjugated, cumulated, and isolated:

• Conjugated dienes: π bonds are separated by only one single bond.
• Cumulated dienes: adjacent π bonds that share the same vinylic carbon.
• Isolated dienes: π bonds are separated by more than one single bond. Isolated dienes are also called unconjugated dienes.

3D Molecule*: buta-1,3-diene

3D Molecule*: penta-1,4-diene

The proximity of the π bonds to each other plays a significant role in the reactivity and thermodynamic stability of molecules. In conjugated dienes, the $\pi$ bonds are separated by a σ bond, which creates a continuous system of overlapping p orbitals. This is not possible for cumulated or isolated dienes. In cumulated dienes, $\pi$ overlap is impossible because the π bonds are at 90o to each other and for isolated dienes, the double bonds are separated by sp3 carbons as shown in Figure 19.1.

In general, conjugated dienes are more stable than cumulated and isolated dienes. Cumulated dienes are the least stable of the three dienes. On the other hand, because the double bonds of conjugated dienes interact with each other, conjugated dienes are the most interesting of the three dienes. The interaction of the double bonds in conjugated dienes results in unexpected properties and reactions.

Q19.1 - Level 1

Identify each of the following compounds as either cumulated, conjugated, or isolated dienes.

Premise
Response
1

A)

A

Isolated

2

B)

B

Cumulated

3

C)

C

Conjugated

4

D)

D

Conjugated

5

E)

E

Cumulated

6

F)

F

Isolated

7

G)

G

Isolated

H

Isolated

I

Conjugated

J

Cumulated

### 19.1.2 Preparation of Conjugated Dienes

Conjugated dienes can be prepared through β-eliminations. For example, an allylic halide can react with a strong, nonnucleophilic base such as tert-butoxide to produce a conjugated diene. Also, two successive elimination reactions can be used on dihalides to generate dienes.

### 19.1.3 Bond Lengths

What effect does conjugation have on bond length? As seen below notice that the σ bond length between two sp2 carbons is shorter than two sp3 carbons or an spand an sp3 carbon.

To explain why atomic hybridization plays a role in bond length, let’s do a brief review from general chemistry.

Q19.2 - Level 1

Match the percent s-character with the atomic orbital

Premise
Response
1

s orbital

A

100

2

sp orbital

B

33

3

sp$^2$ orbital

C

50

4

sp$^3$ orbital

D

25

The higher the percent s-character, the closer the electron density to the nucleus. Building from the previous question, try to answer the following problem.

Q19.3 - Level 1

Rank the orbitals with respect to the closeness of the orbital electron density to the nucleus. Use “1” for closest and “4” for furthest.

Premise
Response
1

s

A

4

2

sp

B

3

3

sp$^2$

C

2

4

sp$^3$

D

1

The C-C single bond of a conjugated diene is formed from the overlap of two carbon sp2 hybrid orbitals. That of a saturated alkane is formed from the overlap of two carbon sp3 hybrid orbitals. Because sp2 hybrid orbitals have electron density closer to the nucleus, a bond between them will be shorter than between two sp3 orbitals.

Q19.4 - Level 1

Rank the indicated C-C bond lengths in the following molecule from longest to shortest (longest at the top).

A

1)

B

2)

C

4)

D

3)

### 19.2 Stabilities of Conjugated Dienes

We’ll first look at some thermodynamic data which supports the observation that conjugated systems have enhanced stability. Then, we’ll use MO theory to explain why.

### 19.2.1 Heats of Hydrogenation of Isolated Versus Conjugated Dienes

The following figure shows the heats of hydrogenation (ΔH°) of related hydrocarbons: two equivalents of 1-butene and one equivalent of 1,3-butadiene.

One would expect that the hydrogenation of two equivalents of 1-butene would release the same amount of energy as the hydrogenation on one equivalent of 1,3-butadiene since both situations deal with a pair of π bonds. However, the 1-butene reaction releases 254 kJ of energy while 239 kJ is released for the 1,3-butadiene reaction.

In other words, 1,3-butadiene is surprisingly more stable by roughly 15 kJ.

Because the two conjugated π bonds are not isolated from each other and there is some overlap between adjacent π electrons 1,3-butadiene is relatively more stable than the isolated diene (although this overlap is small). We can also examine this phenomenon with resonance structures. Figure 19.5 gives the possible resonance structures of butadiene. The best resonance form shows alternating double and single bonds, but some double bond character is possible between carbons 2 and 3.

As stated in section 19.1.3 the bond length between the C2-C3, σ(sp2-sp2), of butadiene is shorter than a bond between two sp3 carbons, σ(sp3-sp3). This was explained by the higher percentage of s character of the sp2 hybridized orbital. However, as seen above the shorter σ(sp2-sp2) is also caused by π overlap which gives this σ bond partial double bond characteristics.

Q19.5 - Level 1

Which molecule should have the larger heat of hydrogenation: 1,3- or 1,4-pentadiene?

A

B

### 19.2.2 Conformation of Dienes

At room temperature, a dynamic equilibrium occurs for 1,3-butadiene as shown below.

Conjugation favors a planar structure and because of the weak π interaction, there’s a barrier to rotation from one conformation to the other. The s-cis conformation is almost 12kJ mol-1 less stable than the s-trans conformation due to steric interaction.

## 19.3 Molecular Orbital Picture of a Conjugated System

Let’s first start off with a review from general chemistry about MO theory. When atoms form a molecule, each atom’s orbitals combine with each other. This is known as the symmetry-adapted linear combination of atomic orbitals (SALCAO). Remember that an atomic orbital is really a wave function. When two wave functions combine linearly, they can do so constructively or destructively, resulting in bonding and antibonding molecular orbitals. The former is lower in energy than the original atomic orbitals, while the latter is higher in energy than the original atomic orbitals. Shown below is the MO diagram for H2.

The highest occupied MO is called the HOMO, while the lowest unoccupied MO is called the LUMO. Simply put, a molecule should be expected to form and be stable if sufficient bonding MOs (lower energy) are occupied relative to antibonding MOs (higher energy).

The z axis is conventionally defined as the internuclear axis. Two 2pz orbitals can combine linearly to give a bonding and an antibonding MO. Because the linear combination results from direct overlap, the MO that is formed is σ bond in character. Hence, the overlap of 2pz orbitals produces one σ2p and one σ*2p.

One 2px orbital can combine with another 2px orbital. However, the orientation of the lobes is parallel to each other, rather than being head-to-head as in the case for the 2pz interaction. Because the overlap is above and below the σ bond , the resulting MOs are π bonding in character. The sideways overlap of two 2py orbitals is identical to the overlap between two 2px orbitals. Since both are perpendicular to the internuclear z-axis, π2p and π*2p MOs are formed.

Let’s now look at the MO diagram for ethene and the MOs that are formed from the sideways overlap of each carbon’s 2p unhybridized atomic orbital (with each other).

There are two important points to highlight. First, because the 2p orbitals are parallel to each other, the MOs that are formed are π in character (electron density above and below the internuclear axis). Second, for every atomic orbital the same number of molecular orbitals must be created.

Thus, the sideways overlap of two p orbitals will create two new MOs. They are created by mathematically combining them in phase and out of phase with each other to form π2p and π*2p. The shading in the above diagram represents the phase of each lobe. When identical colored lobes align with each other, the orbitals are in phase with each other and constructive interference occurs. This results in a bonding interaction between the two adjacent atoms.

When lobes of different colors align, the orbitals are out of phase with each other and destructive interference occurs. This creates a node between the two adjacent atoms. Because of this node, the electron density is on each atom but never in between the atoms, resulting in an antibonding interaction between the adjacent atoms.

In ethene, π2p is more stable than π*2p because of the bonding interaction between the two adjacent atoms that delocalizes the electrons over both carbon atoms above and below the plane of the σ bond.

### 19.3.1 MO Diagram of 1,3-butadiene

In a linear π system such as this, you will see a pattern. The lowest energy level has no nodes, the second 1 node, the third 2 nodes, etc. A good way to think of this linear system is to picture yourself holding a skipping rope with your friend. The lowest energy wave that you could create between you and your partner is a single wave with no nodes. Now if you put energy into the system by moving your hand at higher frequency you would create a node an equal distance between you and your partner. Increasing the energy and frequency of your hand more would create 3 nodes each equal distance from each other. Thus, the higher the energy the greater the number of nodes.

The π MO diagram of 1,3-butadiene can be constructed using each of the 4 carbon atom’s 2p orbitals The sideways linear combination of four atomic p orbitals will produce 4 π molecular orbitals. Each carbon atom contributes 1 electron in its 2p orbital (due to sp2 hybridization) to the π system.

The MO diagram for butadiene is given below. Notice that the linear combination of the four atomic p orbitals creates four molecular orbitals (MOs) of increasing energy. There is a clear correlation between energy and number of nodes of the resulting MO.

Q19.6 - Level 1

Match each MO from Figure 19.11 with the correct number of nodes, excluding the internuclear axis.

Premise
Response
1

Ψ$_1$

A

1

2

Ψ$_2$

B

4

3

Ψ$_3$

C

0

4

Ψ$_4$

D

3

E

2

Each atomic p orbital contributes one electron in the π conjugated system. Thus, these four electrons are added to the MO diagram starting at the lowest energy level. The lowest MO Ψ1is fully occupied. This MO is completely bonding in character since all lobes are in phase with each other.

The second MO Ψis of higher energy because it possesses one node. We can assess the energies of the MOs by looking at the types of bonding interactions between carbons. In Ψ2 there are bonding interactions between carbons C1-C2 and C3- C4 but because there is a node between C2-C3 this is an antibonding interaction. Thus overall this MO has net bonding interaction; two bonding and one antibonding,

The next two higher energy levels overall have net antibonding interaction;Ψ3* has one bonding interaction (C2-C3) and two anti bonding interactions (C1-C2 and C3-C4) and thus is an antibonding orbital overall.Ψ4* has 3 antibonding interactions(C1-C2, C2-C3, C3-C4) and is the highest in energy of all four MOs.

The MO diagram provides a way to explain the stability of conjugated systems. The 4 electrons occupy the lower energy bonding MOs Ψ1 and Ψ2. In the Ψ1 molecular orbital, there is no node between C2-C3, and thus there is bonding between C2-C3. However, in Ψ2 there is a node between C2 and C3. The highest occupied molecular orbital (HOMO) affects the ability of the π system to donate its electrons to an electrophile and influences the molecular orbital characteristics of a molecule. The HOMO thus shows that most of the double bond character for butadiene is between C1-C2 and C3-C4, just like a line diagram would predict, however, there is some double bond character between C2-C3 as seen in Ψ1. Hence, MO theory allows us explain the contracted bond length that exists between a pair of sp2 hybridized carbon atoms and the greater stability in a conjugated system.

The HOMO and LUMO are called the frontier molecular orbitals since they are at the frontier of reactivity and are the most likely orbitals involved in bond formation and breaking during a reaction.

### 19.3.2 MO Diagram of 1,3,5-hexatriene

Let’s now look at the MO diagram for the next largest conjugated system after 1,3-butadiene. There are now 6 carbon 2p orbitals involved in MO formation. Just as before, we’ll form an equal number of bonding MO’s and an equal number of antibonding MO’s, so in this case there are 3 of each type.

An important feature about the molecular orbitals presented in Figure 19.12 is their orbital symmetry.  Notice, that for  Ψ1 , Ψ3, and Ψ5 the phases on the terminal carbons are the same. These orbitals are said to be symmetric.  For Ψ2 , Ψ4 and Ψ6 the terminal carbons have the opposite phase from each other and are said to be antisymmetric. For polyenes, the lowest energy π molecular orbital will always have no nodes and will be symmetrical. The molecular orbitals will then alternate between symmetric and antisymmetric as the energy increases and the number of nodes increases.

Q19.7 - Level 1

Which MOs are bonding and which are antibonding in Figure 19.12?

Premise
Response
1

Ψ1

A

Bonding

2

Ψ2

B

Antibonding

3

Ψ3

C

Bonding

4

Ψ4

D

Bonding

5

Ψ5

E

Antibonding

6

Ψ6

F

Antibonding

Once again, as the number of nodes increases, the energy of the MO also goes up. All six 2p electrons occupy low energy bonding MOs Ψ1, Ψ2, and Ψ3. There is partial double bond character between C2-C3 and C4-C5 as seen in Ψ2, and Ψ3. The lower energy overlapping molecular orbitals increases the stability of 1,3,5-hexatriene, thus the conjugated π system of 1,3,5-hexatriene will be more stable than a molecule that has 3 isolated π bonds.

## 19.4 Allylic Cations

In this section, you will be introduced to a specific type of conjugated system and the reactions that it undergoes.

### 19.4.1 Definition

An allylic cation has the following structure.

Let’s examine the hybridization of each carbon. The two vinylic positions are sp2 hybridized. But what about the allylic carbocation? It’s sp2 hybridized also! Because we have three sp2 atoms adjacent to each other, we have three 2p orbitals adjacent to each other. This is another example of a conjugated system. The double bond is conjugated with an empty p orbital as discussed in Chapter 1.

### 19.4.2 Stability Relative to Other Carbocations

Just how stable is an allylic carbocation? The following image lists various carbocations by their relative stabilities.

Q19.8 - Level 1

Select the correct ranking of stability for the carbocations A-D, from lowest to highest.

A

Carbocation A

B

Carbocation C

C

Carbocation B

D

Carbocation D

Q19.9 - Level 1

Protonation of isoprene can occur in two different locations to form two different resonance stabilized allylic cations. Which is more stable?

A

1) Because the positive charge is on primary and secondary allylic carbocations etc.

B

1) Because the positive charge is on a primary and a tertiary carbon

C

2) Because the positive charge is on a primary and a secondary carbon

D

2) Because the positive charge is on a secondary and a tertiary carbon

Because allylic carbocations are exceptionally stable, they can form in the course of a reaction and serve as intermediates. We’ll explore this in the following section.

### 19.5 1,2- and 1,4-Addition to Conjugated Dienes

Previously, you learned about electrophilic addition to an alkene. We’re going to take this reaction and apply it to conjugated dienes. The allylic cation intermediate can play a major role and affect the regiochemistry of the product.

### 19.5.1 The General Reaction

Conjugated dienes most commonly undergo hydrohalogenation and halogenation reactions. In each case, two possible addition products are formed, as seen below.

The conditions which affect the relative percent yields for 1,2- and 1,4-addition will be discussed next.

### 19.5.2 The Mechanism

Just as with a regular alkene, a conjugated diene will react with a strong acid H-X, where X = Cl, Br, or I. The first step is Markovnikov addition of hydrogen to one of the two double bonds. This results in the formation of a resonance stabilized allylic carbocation.

The nucleophile can attack at either positive centers to give either the 1,2-addition or the 1,4-addition product.

It is interesting to note that below -80oC the 1,2-addition product is formed in an 80% yield, and if the temperature is increased to 40oC the 1,4-addition product is produced in an 85% yield.

The mechanism for the halogenation of butadiene is given below and again temperature plays a big role in which the product is produced. At OoC the 1,2-addition product is produced as the major product and if heated the 1,4-addition product is major.

The question is why. Let us explore this by answering a few questions.

Q19.10 - Level 1

The two resonance structures of an allylic carbocation intermediate is given below. If you consider these two resonance structures as isolated compounds, which structure would be more stable?

A

1) Because the charge is on a secondary carbon

B

2) Because the charge is on a primary carbon

Q19.11 - Level 1

Consider the two products formed. Which product is more stable?

A

1) Because it is a disubstituted alkene

B

1) Because it is a monosubstituted alkene

C

2) Because it is a disubstituted alkene

D

2) Because it is a monosubstituted alkene

These two questions create a third question. What has a greater effect on the product that is formed: the stability of the carbocation intermediate or the stability of the product? To answer this we must consider both kinetics and thermodynamics.

Q19.12 - Level 2

The following reactions show the 1,2 and 1,4 addition products of hydrohalogenation to 1,3-butadiene. Determine (match) if each of the following statements is true or false.

Premise
Response
1

The transition state leading to the 1¸2 addition product is expected to be lower in energy because the secondary allylic intermediate is lower in energy

A

False

2

The 1¸4-addition product is more thermodynamically stable and should form at a faster rate

B

True

3

The rate of reaction will be faster for the formation of the 1¸2-addition product because its activation energy is lower

C

True

From these questions, we see that the less stable 1,2-addition product will be formed at a faster rate than the more stable 1,4-addition product, because the 1,2-addition product's transition state is lower in energy. Can we control this with temperature? The explanation is given in the next section.

### 19.5.3 Kinetic vs Thermodynamic Control in the Addition of HBr

Temperature determines if the 1,2- or the 1,4-product is the major addition product formed. At or near 0oC, the 1,2-product is major, and is said to be under kinetic control. The less stable product is formed at a faster rate. At (or above) room temperature, or if the reaction vessel is warmed up from below 0oC to room temperature, the 1,4-product is formed in a much higher yield. This reaction is said to be under thermodynamic control. The more stable product is formed in greater yield.

When a conjugated diene is treated with HX (X = Cl, Br), 1,2-addition and 1,4-addition products can both form.

The pathway to produce the 1,2-product is kinetically favored because Ea2< Ea1. The pathway to produce the 1,4-product is thermodynamically favored since the product is more stable and at a lower potential energy.

### Kinetic Control

At lower temperatures more molecules will have enough kinetic energy to surmount Ea2 than Ea1 at a faster rate. These conditions will produce more of the less stable 1,2-addition product. This reaction is said to be kinetically controlled.

### Thermodynamic Control

At higher temperatures the 1,2-addition products will form faster but more molecules will have enough kinetic energy to surmount Ea1, thus some 1,4-addition product is formed initially. The activation for the reverse reaction for 1,4-addition is very large.Thus once the 1,4-addition product is formed the reaction will not likely go back to reform the carbocation intermediate. The activation energy for the reverse reaction to form the 1,2-addition product is not as large and the reverse reaction will readily occur at higher temperatures. Thus over time and at the higher temperature equilibrium concentrations will be achieved, favoring the formation of 1,4- addition product. This reaction is said to be thermodynamically controlled.

Q19.13 - Level 2

Choose the correct structure of the major product for the following reaction.

A

a

B

b

C

c

D

d

Watch the following video on ionic hydrohalogenation of conjugated dienes.

Q19.14 - Level 2

Choose the correct structure of the major product for the following reaction.

A

a

B

b

C

c

D

d

## 19.6 Pericyclic Reactions

Pericyclic reactions, which do not involve ionic or radical intermediates, are a unique type of reaction involving π electrons.

There are 3 major classes of pericyclic reactions: cycloaddition reactions, sigmatropic rearrangements, and electrocyclic reactions.

All pericyclic reactions have the following traits in common:

•  The reaction is concerted which means that all bond-making and bond-breaking occurs in one step.
• Electrons tend to be redistributed during the reaction in a cyclic pattern.
• The proposed transition state involves a cyclic structure.
• There are no intermediates involved.

Q19.15 - Level 1

Label the following reaction as either a cycloaddition, sigmatropic rearrangement, or an electrocyclic reaction.

A

B

Sigmatropic rearrangement

C

Electrocyclic reaction

## 19.7 The Diels-Alder Reaction

The Diels-Alder reaction is a very powerful synthetic tool that combines two organic fragments to form 6 member rings. The Diels-Alder reaction is an example of a cycloaddition reaction and was discovered and developed by Otto Diels (1876-1954) and Kurt Alder (1902-1958). They shared the 1950 Nobel Prize in Chemistry for their discovery.

### 19.7.1 The General Reaction and Mechanism

The simplest Diels-Alder reaction is shown in Figure 19.21. Buta-1,3-diene reacting with ethene  to form two new C-C bonds and one C-C double bound, producing cyclohexene. The alkene in a Diels-Alder reaction is called a dienophile, which means “diene loving” and the cyclic product is commonly referred to as the Diels-Alder adduct.

The Diels-Alder reaction is also known as a [4+2] cycloaddition, where the 4 and 2 indicate the number of π electrons contributed by the reagents.

• The diene contributes 4 π  electrons from two conjugated π bonds
• The dienophile contributes 2 π  electrons from a single π bond

The Diels-Alder reaction occurs in one concerted step, with all bond-making and bond-breaking occurring simultaneously.   The curved arrows for this mechanism are given in Figure 19.22 and depict the electron movement of the 3 pairs of π electrons involved in this reaction. The 3 pairs of π eletrons all move at the same time and in one step.

An easy way to predict the product is as follows:

• C1 and C4 of the diene forms 2 new sigma bonds with the vinylic carbons of the dienophile.
• There is only one π bond in the product, and it is always opposite the original dienophile C=C double bond.
• The result is a cyclohexene derivative.

Q19.16 - Level 1

Choose the correct structure of the major product for the following reaction.

A

a

B

b

C

c

D

d