Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

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

Up to 40-60% more affordable

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

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

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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

Access to additional questions, test banks, and slides available within 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

About this textbook

Lead Authors

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.

Contributing Authors

Felix NgassaGrand Valley State University

Neil GargUCLA

Jennifer ChaytorSaginaw Valley State University

Greg DomskiAugustana College

Christian E. MaduCollin Community College

Christopher NicholsonUniversity of West Florida

Franklin OwEast Los Angeles College, UCLA

Robert S. PhillipsUniversity of Georgia

Grigoriy SeredaUniversity of South Dakota

Simon E. LopezUniversity of Florida

Brannon McCulloughNorthern Arizona University

Jason JonesKennesaw State University

José BoquinAugustana College

Stephanie BrouetSaginaw Valley State University

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Chapter 9: E1 Reactions to Form Alkenes

One of the uses of ethylene is in the artificial ripening of fruit. [1]​


Contents

Learning Objectives

  • Rank a series of alkenes from most to least stable.
  • Write the mechanism for an E1 reaction and explain the term unimolecular elimination.
  • Understand how a mutual carbocation intermediate produces both E1 and SN1 products.
  • Comprehend the reaction condition necessary for an SN1/E1 reaction to occur.
  • Understand Zaitsev’s rule and how it is used to determine the major E1 product.
  • Predict the major and minor products formed in E1/SN1 reactions.
  • Understand the reaction condition necessary for a dehydration reaction to occur and why the E1 reaction becomes the dominant mechanism.
  • Predict the major and minor alkene products produced in dehydration reactions.
  • Recognize when a carbocation rearrangement will occur and predict the products formed in the SN1/E1 or E1 reactions.

9.1 Definitions

Labeling carbons and hydrogens: The carbons bonded adjacent to a specific functional group are labeled outward, using Greek letters. The nomenclature can also apply to the hydrogens bonded to the labeled carbons.

Figure 9.1. Greek letter labelling of carbons and hydrogens in hexan-3-one and 3-bromo-3-methylpentane ​

For instance, in the examples given above, 3-bromo-3-methylpentane has 3 β carbons with 7 β hydrogens and 2 γ carbons with 6 γ hydrogens.

Elimination reaction: An elimination reaction of a neutral molecule involves the removal or elimination of substituents of the molecule to form an alkene in a one- or two-step mechanism. This type of elimination is often called beta-elimination because the elimination occurs between the alpha and beta carbons or adjacent atoms. When elimination happens in a one-step mechanism, it is called an E2 reaction while the two-step mechanism is referred to as an E1 reaction.

Figure 9.2. Generic elimination reaction​

Leaving group (LG): Leaving group must be able to leave as a relatively stable (i.e. non-reactive), weakly basic molecule or ion.

Unimolecular (E1): The rate of reaction is only dependent on the concentration of the substrate. The reaction rate is first-order (unimolecular). The rate of reaction = k[substrate], where k is equal to the rate constant.

Bimolecular (E2): The rate of reaction is dependent on the concentration of the base and the substrate. The reaction rate is second-order (bimolecular). The rate of reaction = k[base][substrate], where k is equal to the rate constant.

Stereoselective reaction: A reaction in which a single reactant can produce two or more stereoisomeric products and one of these products is preferred over another. For example, the reaction of either enantiomer of 2-bromobutane will stereoselectively produce (E)-but-2-ene as the major product. Notice that it does not matter which enantiomer the starting material is; the product ratio is the same.

Figure 9.3. Stereoselectivity in elimination reactions of enantiomers of 2-bromobutane ​

Stereospecific reaction: A reaction in which a single reactant can produce two or more stereoisomeric products and one of these products is exclusively formed over the other(s). For example; the elimination reaction of (1R,2R)-1-bromo-1,2-diphenylpropane with a strong base produces the (Z) stereoisomer whereas the elimination of the (1R,2S) diastereomer produces the (E) stereoisomer.

Figure 9.4. Stereospecificity in elimination reactions of diastereomers of 1-bromo-1,2-diphenylpropane ​

Similarly, an SN2 reaction with 2-bromobutane and sodium methanethiolate (CH3SNa) will produce either (S)-sec-butyl(methyl)sulfane or (R)-sec-butyl(methyl)sulfane depending on the enantiomer with which you start. This is another example of a stereospecific reaction.

Figure 9.5. Stereospecificity in SN2 reactions of enantiomers of 2-bromobutane ​

Regiochemistry: A reaction that can occur with different regions on a molecule to produce different products. A region is defined as a site on a molecule where a reaction can occur. Regiochemistry is the difference in the reactivity of the various sites.

Regioselective reaction: When there is more than one reactive site a regioselective reaction can occur. A regioselective reaction is one that produces one major product when many possible products are possible. For example, the molecule 2-bromo-2-methylbutane has eight hydrogens on β carbons that can undergo elimination reaction in the presence of a base such as sodium ethoxide. Removal of a Ha hydrogen produces the minor product 2-methyl-but-1-ene while removal of a Hb hydrogen produces the major product. An empirical rule that states; when two or more alkenes can be produced in an elimination reaction, the thermodynamically most stable alkene will predominate. The most thermodynamically stable alkene will be the alkene that has the most alkyl groups attached to the alkene carbons.

Figure 9.6. Regioselective outcomes of elimination of 2-bromo-2-methylbutan​e ​

Zaitsev’s Rule: An empirical rule that states; when two or more alkenes can be produced in an elimination reaction, the thermodynamically most stable alkene will predominate. The most thermodynamically stable alkene will be the alkene that has the most alkyl groups attached to the alkene carbons.

Figure 9.7. Elimination of 2-bromo-2-methylbutane with a strong base to produce major (Zaitsev) product and minor product


Hofmann Rule: This rule refers to a special β-elimination reaction in which the alkene having the smallest number of alkyl groups attached to the double bonded carbon atoms will be the predominant product (less stable alkene forms as the major product). The Hofmann rule is observed in elimination reactions with leaving groups like quaternary ammonium salts, tertiary sulfonium salts, and with bulky strong bases (NaOC(CH3)3) reacting with tertiary alkyl halides. 

Figure 9.8. Elimination of 2-bromo-2-methylbutane with a bulky, strong base to produce major (Hofmann) product and minor (Zaitsev) product ​


9.2 Stability of Alkenes

The relative stability of alkenes can be determined by comparing their heats of combustion. The amount of heat given off during combustion is related to the stability of the molecule. The larger the amount of heat given off, the less relatively stable the molecule. A good analogy is the stress in the rubber of a balloon. A balloon that is inflated so that the rubber is stretched and is under a lot of stress will produce a much bigger bang when popped than a balloon whose rubber is soft and is not under a lot of stress. Comparing the heats of combustion of six isomeric alkenes shows how alkyl substitution affects the stability of alkenes. (Data obtained from: Handbook of Chemistry and Physics, 94th Edition, 2013–2014.)

Figure 9.9. Alkyl substituent effects on stability by enthalpy of combustion ​

Trend: Double bond stability increases with increasing substitution (hyperconjugation) and the trans or (E) isomer is more stable than the cis or (Z) isomer (steric interaction).

Hyperconjugation: Alkyl groups are electron donating and contribute electron density to the empty π* molecular orbital of the π bond. The simplified diagram below shows the favorable molecular orbital interaction that helps stabilize the double bond. Thus, it can be said that the electrons from the sigma bonds are delocalized to some extent in the double bond through space. This is similar to resonance, although not as stabilizing as resonance. 

Alkenes are classified based on the number of carbon atoms that are bonded to the carbons bearing the double bond. A monosubstituted alkene has one carbon atom bonded to the carbons of the double bond, a disubstituted alkene will have two carbon atoms, and so forth. The figure below shows the stability trend for alkenes.

Figure 9.10. Stabilization of alkenes by hyperconjugation​

The stability of alkenes increases with increasing substitution (the addition of alkyl groups) and decreasing steric interaction between disubstituted alkenes.

Figure 9.11. Alkene stability with substitution from most to least stable ​


Q9.1 - Level 1

Order the given alkenes from most stable to least stable.

question description
Premise
Response
1

Most stable

A

2)

2

Second most stable

B

3)

3

Least stable

C

1)


Q9.2 - Level 1

Order the given alkenes from most stable to least stable.

question description
Premise
Response
1

Most stable

A

2)

2

Second most stable

B

3)

3

Least stable

C

1)


Q9.3 - Level 1

Order the given alkenes from most stable to least stable.

question description
Premise
Response
1

Most stable

A

3)

2

Second most stable

B

1)

3

Least stable

C

2)


9.3 E1 Reaction with Alkyl Halides

(Note: electrons are colored to help you visualize their flow and see where bonds are formed and broken during the mechanism.)

When 2-bromo-2-methyl propane is dissolved in methanol, a solvolysis (SN1) reaction occurs. However, an alkene is also produced as a minor product (20%). (Solvolysis reaction: the solvent is the nucleophile.) Notice how methanol is acting as a nucleophile and a base in this mechanism.

Figure 9.12. Reaction mechanisms of 2-bromo-2-methylpropane in methanol. a, top) SN1 solvolysis. b, bottom) E1. ​

Kinetic analysis shows that the rate of alkene formation is first-order and only depends upon the concentration of starting alkyl halide. Experimentally, this means that if the alkyl halide concentration is doubled, then the rate of reaction doubles. If the nucleophile/base (CH3OH) concentration is doubled, then the rate of the reaction does not change. The nucleophile/base is not involved in the rate-determining step.

This reaction is termed “E1” and has the same rate-determining step as the SN1 reaction: the endothermic formation of a carbocation. In this reaction, the solvent is acting as both a nucleophile and a base. If it reacts as a nucleophile, then an SN1 reaction occurs; however, if the solvent acts as a base, then a proton is abstracted from the carbon adjacent to the carbocation to form an alkene.

Click on the video to see a unimolecular E1 reaction in action with an audio explanation.


9.3.1 Which Reaction Mechanism Dominates with Alkyl Halides: the SN1 or E1 Reaction?

For alkyl halides, the SN1 reaction dominates because the activation energy for an E1 reaction is larger than for an SN1 reaction. The activation energy for an E1 reaction is larger because more bond changes are occurring in the second transition state. The base is forming a new σ bond with the hydrogen, the C–H bond is breaking and a new C–C π bond is forming, whereas for the SN1 reaction bond formation only occurs between the nucleophile and the carbocation (neglecting solvent interactions).

Figure 9.13​. a, top) Overall reactions of 2-bromo-2-methylpropane in methanol. b, middle left) Transition state for elimination mechanism. c, middle right) Transition state for SN1 mechanism. d, bottom) Potential energy diagram for SN1 and E1 reactions of 2-bromo-2-methylpropane in methanol. ​

Since the activation energy for the SN1 reaction is lower, the rate of formation of the SN1 product is greater and the substitution product will be produced in greater yield.

9.3.2 Effect of Temperature

When higher temperatures are used, the amount of the E1 product increases because of the proportion of molecules with enough kinetic energy to surmount the energy barrier increases. However, with alkyl halides, increasing the temperature still produces the substituted product as the major product and only changes the ratio of E1/SN1 products. SN1 dominates with alkyl halides and in very few cases elevated temperatures might favor an E1 reaction.

9.4 SN1/E1 Reactions, Regiochemistry, and Selectivity of E1 products (Zaitsev’s Rule)

In many SN1/E1 reactions, more than one product can form. For example, the reaction of 2-bromo-2-methylbutane produces one SN1 product (60%) and two E1 products (40%). Two E1 products are formed because elimination can occur in two different regions. Abstraction of any of the 6 β hydrogens on the methyl carbon or carbon1 will produce 2-methylbut-1-ene. Abstraction of either two β hydrogens on carbon 3 will produce 2-methylbut-2-ene.

Figure 9.14a. E1 mechanisms for the formation of 2-methylbut-1-ene from 2-bromo-2-methylbutane. ​


Figure 9.14b. E1 mechanisms for the formation of 2-methylbut-2-ene from 2-bromo-2-methylbutane.


Figure 9.14c. SN1 mechanisms for the formation of 2-methoxy-2-methylbutane from 2-bromo-2-methylbutane





We can answer this by looking at the transition state of both E1 reactions. After the formation of the carbocation, a proton can be removed at any of the β carbons. Since alkyl substitution increases the stability of an alkene through hyperconjugation (Sec. 2.0), it makes sense that the lower energy transition state would be the one in which the forming alkene is more substituted. Since the trisubstituted alkene-like transition state is more stable and has a lower activation energy than the disubstituted alkene-like transition state, the rate of formation of the more stable product should be greater than the rate of formation of the less stable product, thus producing the more stable alkene in greater yield.

Note: we can relate thermodynamic stability of the product to its kinetic rate of formation because we are associating the stability of the product to the assumed stability of the transition state, which, in most cases, is a correct assumption, but not always.

9.15.png
Figure 9.15. a, top) Overall reactions of 2-bromo-2-methylbutane in methanol. B, middle) Zaitsev product transition state. c, middle- right) Hofmann product transition state. d, bottom) Potential energy diagram for E1 reactions of 2-bromo-2-methylbutane in methanol​.


Zaitsev’s Rule: When more than one elimination product is possible, the dominant product will be the most substituted alkene. The more substituted alkene is referred to as the Zaitsev's product and occurs through a Zaitsev’s elimination. However, this is not always the case as a reaction that produces the less substituted product arises from what is called a Hofmann elimination. Hofmann eliminations will be investigated in E2 reactions.

Q9.4 - Level 2

Identify the major and minor E1 products formed in the following SN_N1 / E1 reaction.

question description
Premise
Response
1

1)

A

Minor

2

2)

B

Major


9.5 Dehydration of Alcohols: the E1 Reaction Becomes Dominant

Tertiary and secondary alcohols heated in concentrated acids, such as H2SO4 and H3PO4, undergo E1 reactions. The products formed exhibit regiochemistry preference for the Zaitsev product or, stated another way, the major product produced will be the most stable alkene.

Figure 9.16. Dehydration reactions of tertiary and secondary alcohols​

Because the E1 mechanism involves a carbocation intermediate, the ratio of products and the regiochemistry and stereochemistry cannot be controlled using different reagents. Note: the yields given in the above two reactions are only approximate because the ratio of products will depend on reaction conditions.

To understand the reaction, let us examine the mechanism of a tertiary alcohol reacting with sulfuric acid. We know that alcohols are poor leaving groups. Thus the first step is to make the alcohol a better leaving group. This is accomplished using a strong acid and protonating the alcohol to form an oxonium ion. The leaving group is now water.

Figure 9.17. Proton transfer and loss of leaving group in dehydration reaction of 2-methylbutan-2-ol

Notice that the intermediate formed is the same intermediate that was formed in the reaction of 2-bromo-2-methylbutane in methanol (Sec. 3.0), but in this case only E1 products are formed and no SN1 products (see below). Why?

Figure 9.18. Proton transfer as the final step in dehydration reaction of 2-methylbutan-2-ol.​

The reason for this is that the hydrogen sulfate ion (HSO4-) is a very poor nucleophile and has a greater affinity for protons than the carbon atom. Also, notice that all steps are reversible; thus, conditions must be used to favor the formation of the alkene. To produce the alkene as the major product, low concentrations of water and higher temperatures are required. This equilibrium is discussed in more detail in the alkenes Pages Chapter 12.4.

Chemical procedure to produce only E1 products starting with alcohols: use concentrated acids (low H2O content) whose conjugate bases are poor nucleophiles and heat the reaction mixture.


Q9.5 - Level 2

What would be the major product of the following dehydration reaction?

question description
A

1)

B

2)

C

3)


Q9.6 - Level 3

What would be the major product of the following reaction?

question description
A

1)

B

2)

C

3)

D

4)

Watch the following video to see the explanation to Q9.5 and Q9.6.


Q9.7 - Level 1

Identify the major and minor products formed in the following reaction.

question description
Premise
Response
1

1)

A

Major

2

2)

B

Minor


9.5.1 Dehydration of Alcohols and Carbocation Rearrangements

Review SN1 reactions with carbocation rearrangements and answer the following questions.

When 3,3-dimethylbutan-2-ol is heated in concentrated sulfuric acid, the following products are formed:

Figure 9.19. Dehydration reaction of 3,3-dimethylbutan-2-ol heated in sulfuric acid​.


Q9.8 - Level 3

Can you do the 4 step mechanism? Try writing out the mechanism then match the step with the intermediates and major product formed.

question description
Premise
Response
1

Intermediate formed in step 1

A

1)

2

Intermediate formed in step 2

B

3)

3

Intermediate formed in step 3

C

5)

4

Major product formed in step 4

D

2)

E

4)

F

6)



Q9.9 - Level 2

Now that you have the correct order to produce the major product, match each step with the mechanistic process.

question description
Premise
Response
1

Mechanistic process to form the intermediate in step 1

A

Proton transfers (acid/base) and loss of a leaving group

2

Mechanistic process to form the intermediate in step 2

B

Loss of a leaving group

3

Mechanistic process to form the intermediate in step 3

C

Rearrangement

4

Mechanistic process to form the product in step 4

D

Nucleophilic attack

E

Proton transfer (acid/base)

F

Proton transfer (acid/base)

Watch the following video to see the explanation to Q9.9.


The products produced from rearrangements almost always dominate. This is because intramolecular rearrangements occur at a faster rate than the intermolecular (between molecules) proton transfer reactions. Below is a Free Energy–Reaction Coordinate diagram describing the energetics of the reaction. Calculations have shown that the energy barrier for rearrangement reactions is very small and, when you think about the two different processes, it is reasonable to assume that a proton transfer would have a higher free energy of activation because more bond breaking and formation is occurring than in a carbocation rearrangement. Also, the proton transfer reaction requires two molecules to move through the solvent and collide with each other. This should take more energy than the intramolecular rearrangement of the molecules' molecular orbitals.

Figure 9.20​a. Dehydration reaction of 3,3-dimethylbutan-2-ol affected by heating in sulfuric acid.


Figure 9.20b. Potential energy diagram for dehydration of 3,3-dimethylbutan-2-ol in hot sulfuric acid.


Q9.10 - Level 3

What is the major product formed in the following dehydration reaction?

question description
A

1)

B

2)

C

3)

D

4)


Q9.11 - Level 2

Both dehydration reactions go through carbocation rearrangements. Does the mechanism in each reaction involve a hydride shift or a methyl shift?

question description
Premise
Response
1

1)

A

Methyl shift

2

2)

B

Methyl shift

C

Hydride shift

D

Hydride shift


9.5.2 E1 Reactions and Stereochemistry

Consider the following reaction; two stereoisomers are formed and, as expected, the more stable trans isomer is produced in greater yield.

9.21.png
Figure 9.21. Stereoisomeric products of dehydration reaction of pentan-3-ol, elimination of either β hydrogens will produce the same products.

Why do two stereoisomers form?

In the above reaction, elimination at either β hydrogens will produce the same product. Consider only one set of β hydrogens and label them as blue Ha and red Hb.

9.22.png
Figure 9.22. Removal of beta hydrogens in dehydration reaction of pentan-3-ol ​

When answering the following question, look at the alkyl groups with respect to each other and the hydrogen that is being abstracted in the transition state.


Q9.12 - Level 3

Match the transitions state to the alkene that it produces, (Z)-pent-2-ene or (E)-pent-2-ene.

question description
Premise
Response
1

Transition State 1

A

(EE)-pent-2-ene

2

Transition State 2

B

(ZZ)-pent-2-ene


Thus, the reason two stereoisomers are formed is that there are two β hydrogens and the product produced depends on which hydrogen atom is removed. Elimination of the β hydrogen (Hb) produces the (Z) product and abstraction of (Ha) produces the (E) product.

Figure 9.33


Q9.13 - Level 3

Why is (E)-pent-2-ene produced in a greater yield than (Z)-pent-2-ene? Select two correct answers.

A

The transition state of the second step for the formation of (Z)-pent-2-ene has a smaller Ea than the transition state of (EE)-pent-2-ene.

B

The transition state of the second step for the formation of (Z)-pent-2-ene has a larger Ea than the transition state of (EE)-pent-2-ene.

C

The Ea for the transitions state of (EE)-pent-2-ene is higher because of the steric interaction between the alkyl groups.

D

The Ea for the transitions state (ZZ)-pent-2-ene is higher because of the steric interaction between the alkyl groups.


Q9.14 - Level 3

Dehydration reactions are not stereospecific, however a major stereoisomer generally forms. Identify the major and minor products formed in the following reaction.

question description
Premise
Response
1

1)

A

Major

2

2)

B

Minor


Q9.15 - Level 2

Dehydration reactions are not stereospecific, however a major stereoisomer generally forms. Identify the major and minor products formed in the following reaction.

question description
Premise
Response
1

1)

A

Major

2

2)

B

Minor


For the SN1 reaction, it was shown that increasing the stability of the transition state and carbocation, the rate of reaction increases because the activation energy of the rate-determining step decreases. If necessary, then please review Chapter 8.3, and answer the following question.

Q9.16 - Level 1

Arrange the alcohols in order of reactivity in an acid catalyzed dehydration reaction (most reactive at the top).

question description
A

2)

B

3)

C

1)

Need help? Think about the stability of the carbocation intermediates formed in each reaction and the rate of their formation. What is more stable: 1°, 2°, or 3°?


Q9.17 - Level 1

Arrange the given alcohols in order of reactivity in an acid catalyzed dehydration reaction.

question description
Premise
Response
1

Fastest

A

1)

2

Second Fastest

B

2)

3

Slowest

C

3)

4

No reaction

D

4)


Need help? Think about the stability of the carbocation intermediates formed in each reaction and the rate of their formation. What is more stable: 1°,2°, 3°, benzylic, or aryl?

9.5.3 SN1/E1 with Carbocation Rearrangement

Q9.18 - Level 2

When the given alkyl halide was heated in ethanol a number of E1 products formed. Three E1 products are given. Did the mechanisms for these E1 products involve a hydride shift or a methyl shift before elimination occurred?

question description
Premise
Response
1

1)

A

Hydride shift

2

2)

B

Methyl shift

3

3)

C

Hydride shift

D

Hydride shift

E

Methyl shift

F

Methyl shift