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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McGraw-Hill

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

$219

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

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David R. Klein, “Organic Chemistry”, 3rd 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 10: E2 Reactions to Form Alkenes

Propene is used to form polypropylene, the main material used in ropes that come into frequent contact with water. Rope uses such as swimming lines, fishing nets, and golf course barriers are often made from polypropylene. [1]

 

Contents

Learning Objectives

  • Comprehend the reaction condition necessary for an E2 reaction to occur.
  • Write the mechanism for an E2 reaction and explain the term bimolecular elimination.
  • Understand the regioselectivity and stereoselectivity of an E2 reaction and predict the major and minor products produced in an E2 reaction.
  • Understand how the anti-coplanar transition state can produce E and Z stereoisomers.
  • Differentiate between strong unhindered bases and strong bulky bases.
  • Predict the products formed when a strong bulky base is used in an E2 reaction.
  • Recognize the reaction conditions and predict the products produced in a Hofmann Elimination.
  • Recognize the conditions and predict the products formed when a reaction is kinetically or thermodynamically controlled.

10.1 Definitions 

Note: Section 10.1 is the same as Section 9.1, review as needed.

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

Figure 10.1. Greek letter labelling of carbons and hydrogens in hexan-3-one and 3-bromo-3-methylpentane





In the example 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 α and β 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 10.2. 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). Rate of Reaction = k[base][substrate], where k is equal to the rate constant.

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

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 reflects 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 other 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.

Figure 10.4. Regioselective outcomes of elimination of 2-bromo-2-methylbutane​

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 10.5. 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 10.6. Stereospecificity in elimination reactions of diastereomers of 1-bromo-1,2-diphenylpropane​

Similarly, a 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 which enantiomer you start with. This is another example of a stereospecific reaction.

Figure 10.8. Stereospecificity in SN2 reactions of enantiomers of 2-bromobutane​

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


10.2 Stability of Alkenes

Note: Section 10.2 is the same as Section 9.2, review as needed.

The stability of alkenes can be determined relatively through values for 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 stable the molecule is. 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. The liquid heats of combustion of six isomeric alkenes in the diagram below show how alkyl substitution affects the stability of alkenes. (Data obtained from: Handbook of Chemistry and Physics, 94th Edition, 2013–2014.)

Figure 10.10. Alkyl substituent effects on stability by enthalpy of combustion​

Trend: Double bond stability increases with increasing substitution (hyperconjugation) and the trans isomer is more stable than the cis 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 into the double bond through space. This is similar to resonance, although not as stabilizing. 

Figure 10.11. Stabilization of alkenes by hyperconjugation​

Stability of alkenes increases with increasing substitution (the addition of alkyl groups) and decreasing steric interaction.

Alkenes are classified based on the number of carbon atoms bonded to the carbons that comprise 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 alkenes' stability trend.

Figure 10.12. Alkene stability with substitution from most to least stable​


10.3 E2 Reaction with Alkyl Halides

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

The following section will exclusively examine E2 reactions even though SN2 reactions may be possible. It is presented in this way so that you can thoroughly understand E2 reactions before adding the complication of the competing SN2 reaction. The competition between SN2 and E2 reactions is examined in the pages: Sub-Elim: Discerning the differences.

When 2-bromopropane is added to sodium ethoxide in ethanol, propene is produced as the major product.

Figure 10.13. Elimination reaction of 2-bromopropane with sodium ethoxide in ethanol​

The rate of this reaction is second-order or bimolecular (the same as an SN2 reaction).

Rate of Reaction = k[base][substrate], where k is equal to the rate constant.

E2 stands for Elimination Bimolecular. Thus, the reaction is dependent on the concentration of the base and the substrate and the transition state must involve the collision of the base with the substrate. E2 reactions are concerted reactions, which means they occur in one step and no intermediates are formed.

The following animation illustrates the collision between the base and substrate. The animation also illustrates the flow of electrons as described when using arrows to explain a mechanism and does not truly represent the molecular orbital interaction of an E2 reaction. 


For this reaction to occur, the base must physically contact the β proton (bimolecular) with enough kinetic energy to surmount the energy barrier and reach the transition state. In the transition state, the base is forming a σ bond with the proton, the electrons from the C–H σ bond are moving to form a new π bond, and the electrons from the C–Leaving Group σ bond are moving onto the leaving group. All of this occurs in one step in the transition state (a concerted reaction).

Figure 10.14​. E2 reaction mechanism for elimination of 2-bromopropane with sodium ethoxide in ethanol. Lone pairs are coloured and shown in sigma and pi bonds to help show the flow of electrons in the mechanism.


The following Potential Energy–Reaction Coordinate diagram depicts the energetics of the exothermic E2 reaction. Notice that in the reaction given below, the leaving group is a much weaker base (more stable) than the ethoxide anion and the products are more stable than the starting material. In this reaction, the reverse reaction would not occur because the bromide ion will not displace the more basic ethoxide anion. You can also think about the bond energies. The C–Br σ bond is weaker than the C–O σ bond; once the C–O bond has formed, the reverse reaction to form the weaker C–Br will not take place since it is thermodynamically less stable.

Figure 10.15. Potential energy diagram of E2 elimination of 2-bromopropane with sodium ethoxide in ethanol​




A common question by students: Why does elimination only occur between the α and β carbons? Why can’t the base remove a γ proton?

Let’s try and see what happens. This is how an organic chemist thinks. Chemists think about how a reaction might take place by pushing arrows. They think about what will attack and where. Then, they look at the possible products and contemplate if the reaction is possible, both thermodynamically and kinetically. The chemist will then try the reaction in the lab to see if it will take place as imagined.

The following questions allow you to use this approach.

Q10.1 - Level 2

Is the following reaction possible?

question description
A

No, because the carbanion is a stronger base than the ethoxide ion.

B

Yes, because the carbanion is a weaker base than the ethoxide ion.

C

No, because the ethoxide ion is a stronger base than the carbanion.

D

Yes, because the ethoxide ion is a weaker base than the carbanion.

E

No, because the ethoxide ion is a strong acid.


How about elimination between the γ and β carbons?

Q10.2 - Level 2

Is the following elimination reaction possible?

question description
A

Yes, because the hydride ion leaving group is a weaker base than the ethoxide ion.

B

No, because the hydride ion leaving group is a stronger base than the ethoxide ion.

C

No, because the ethoxide ion is a stronger base than the hydride ion leaving group.

D

Yes, because the ethoxide ion is a weaker base than the hydride ion leaving group.


How about elimination between the β and α carbons?

Q10.3 - Level 2

Is the following reaction possible?

question description
A

No, because the bromide ion leaving group is a stronger base than the ethoxide ion.

B

Yes, because the bromide ion leaving group is a weaker base than the ethoxide ion.

C

No, because the ethoxide ion is a stronger base than the bromide ion leaving group.

D

Yes, because the ethoxide ion is a weaker base than the bromide ion leaving group.

By answering the above questions you have answered the question as to where and why an E2 reaction takes place.

The E2 reaction takes place because more stable products are formed and the only acidic proton that the base can remove to accomplish this is the proton β to the leaving group. If the leaving group is a stronger base than the attacking base, then no reaction will take place.

Keeping this in mind, answer the following question.


Q10.4 - Level 3

Is the following reaction possible?

question description
A

No because the hydroxide ion leaving group is a stronger base than the ethanethiolate ion

B

Yes because the hydroxide ion leaving group is a weaker base than the ethanethiolate ion

C

No because the ethanethiolate ion is a stronger base than the hydroxide ion leaving group

D

Yes because the ethanethiolate ion is a weaker base than the hydroxide ion leaving group


10.3.1 E2: Zaitsev's Rule and Regiochemistry

Zaitsev’s rule: In elimination reactions, the most stable, most highly substituted alkene dominates(i.e., greatest number of alkyl groups attached to the double bonded carbon atoms).

Stability of alkenes increases with increasing substitution (the addition of alkyl groups) and decreasing steric interaction (sec. 2)

Figure 10.16. Changes in alkene stability as a function of substiution from most stable to least stable.


E2 reactions are regioselective, meaning that a reaction leads to a preferred product due to the different reactivities of the various regions of the molecule.

Knowing what you know about Zaitsev’s rule, answer the following question.


Q10.5 - Level 2

What is the major product produced in the following reaction?

question description
A

1)

B

2)

Intuitively, you knew the more thermodynamically stable alkene was the major product because of Zaitsev’s rule. This can also be explained kinetically and thermodynamically.

10.3.1.1 Kinetics and the Rate of Product Formation

For a bimolecular reaction: Rate of Reaction = k[base][substrate], where k is equal to the rate constant.

According to Arrhenius, the rate constant = Ae(−Ea/RT).

The term A incorporates the collision rate and the probability of the molecules having a favorable orientation for the reaction to occur. The activation energy Ea is the minimum kinetic energy that a molecule must have when colliding with another molecule for their reaction to occur (See Chapter 10.4). Thus, when comparing two reactions, the reaction that has a smaller Ea will occur faster.

In the following reaction (Figure 10.17), there are two regions where an E2 reaction can occur. They are depicted below with blue Ha and red Hb hydrogens. Elimination of any of the six red Hb hydrogens will produce the disubstituted alkene and either of the two blue Ha hydrogens will produce the trisubstituted alkene.

Figure 10.17. Regioseletivity of elimination of 2-bromo-2-methylbutane​

For this reaction, approximately 69% of 2-methylbut-2-ene and 31% of 2-methylbut-1-ene is produced. This means that the energy barrier to produce 2-methylbut-2-ene, the trisubstituted alkene, is smaller than 2-methylbut-1-ene because more of 2-methylbut-2-ene is produced (i.e., more molecules have enough kinetic energy to surmount the energy barrier to form the trisubstituted alkene in major proportions).

Why are energy barriers different in the above reaction? The answer is in the transition states.

Experimentally we know that alkyl groups attached to the sp2 carbons of an alkene help to stabilize the molecule. In the transition state of an E2 reaction a double bond is forming. Thus, it is expected that the activation energy of a more substituted alkene would be smaller than a less substituted alkene because the alkyl groups would assist in the double bond formation.

Figure 10.18a​. Reaction pathway with transition state for the formation of the disubstituted and trisubstitued E2 elimination products of 2-bromo-2-methylbutane.
Figure 10.18b. Potential energy diagram of the E2 elimination of 2-bromo-2-methylbutane.

Note: When thinking about the above reaction, we are relating thermodynamic stability of the product to its kinetic rate of formation because we are making the assumption that the stability of the product is similar to the stability of the transition state and, in most cases this is correct, but not always as we will discover in Sec. 3.5 Hofmann Eliminations and Bulky Bases and Sec. 4.0 Kinetics vs. Thermodynamics.

Q10.6 - Level 2

The major alkene obtained when the given alkyl halide is heated with potassium ethoxide is:

question description
A

1)

B

2)

C

3)

D

4)


10.3.2 E2: Anti-Coplanar Elimination and Stereospecific Reactions

Before discussing the stereochemistry of an E2 reaction, we must look closer at the transition states and the conformations the molecules must take before an E2 reaction can occur.

Review: Dihedral angle is the angle between two planes, where each plane is defined by three atoms.

Figure 10.19. Newman projections showing dihedral angle between carbon-hydrogen bond plane in blue and carbon-bromine bond plane in red​

Click on the video see the E2 mechanism shown as a Newman projection. Look carefully at the conformation of the substrate.

In the video, you can see that at the same time the proton is being removed, the bromine is leaving and the π bond is forming. The carbons are transforming from sp3 to sp2 and the sp2 orbitals between the C–H and C–Br σ bonds are transforming into p orbitals. Since a π bond is created from the sideways overlap of p orbitals, the C–H and C–Br sigma bonds must be in the same plane or be anti-coplanar. The dihedral angle between the hydrogen and bromine must be 180o for a π bond to form.

Figure 10.20. a) Elimination reaction of 2-bromobutane. b) Newman projection of transition state (including pi orbitals) of elimination reaction. Notice hydrogen is anti-coplanar to bromine. c) Side view of transition state (including pi orbitals)​

There is evidence to suggest that a strict dihedral angle of 180° is not necessary for an E2 mechanism and angles between 175° and 179° are sufficient. Many textbooks use the term anti-periplanar to describe the fact that the angle does not have to be exactly 180o.

The conformational preference for the anti-coplanar elimination can also be explained by examining the orbitals.

10.3.2.1 E2: Anti-Coplanar Elimination Orbital Explanation

The nucleophile donates its electrons into the C–Leaving Group empty antibonding orbital. For this to happen, the nucleophile must attack from the backside to have good overlap between the filled donor orbital of the nucleophile and the empty accepting orbital of the electrophile (Chapter 7.4).

Figure 10.21.​ a) Curved arrow mechanism of generic SN2 reaction. b) orbital view of reaction pathway of an SN2 reaction​

Similarly, in the elimination reaction, the filled donor orbital, the C–H σ bond that is breaking, must align itself with the empty σ antibonding orbital that is accepting the electrons to have a reaction. To have good overlap between the donor and acceptor orbitals, they must be in the same plane as shown below.

Figure 10.22. a) Curved arrow mechanism of generic E2 reaction. b) Orbital view of reaction pathway of an E2 reaction​

Question: A dihedral angle of zero also aligns the β hydrogen and the leaving group in the same plane. Is it possible for an E2 reaction to occur if the dihedral angle is zero or syn-coplanar?

Answer: Yes, but under special conditions. In the majority of cases, syn elimination will not occur because the eclipsed conformation is much less stable than the staggered conformation and has a larger activation energy. Studies have shown that for acyclic molecules, anti-coplanar elimination is favored over syn-coplanar elimination typically greater than 104.

Syn elimination occurs only under the following conditions:

  • When an anti-coplanar arrangement is not possible and the reaction occurs through a syn-coplanar arrangement.
  • When strong steric factors favor the syn elimination.

Generally, anti-elimination will occur in the majority of E2 reactions, meaning that a hydrogen that is not in the same plane as the leaving group cannot be removed by the base.

Figure 10.23. E2 reactions can only occur if the hydrogen and the leaving group are anti-coplanar or syn-coplanar​

Why does this matter? Answer the following questions to understand why this is important.

Q10.7 - Level 2

Which of the following alkenes is the most stable?

question description
A

1

B

2


Q10.8 - Level 3

In the following E2 reaction only one product is formed. Which alkene is produced?

question description
A

1

B

2

Question 10.8 is an example of a stereospecific reaction. The stereochemistry of the reactant determines the stereochemistry of the product. Only one stereospecific product could be formed because only one specific β hydrogen was available for the elimination reaction to occur.

Click on the video for the video explanation of Question 10.8.


Q10.9 - Level 3

Which Newman projection correctly depicts the conformation that would produce the given (Z)-alkene under strictly E2 conditions?

question description
A

1)

B

2)

C

3)

D

4)


Q10.10 - Level 3

(2S,3S)-2-Chloro-3-methylpentane is heated with sodium ethoxide in ethyl alcohol under strictly E2 conditions to give alkenes. The major product is:

question description
A

1)

B

2)

C

3)

D

4)


Q10.11 - Level 2

What is the major product produced in the following reaction?

question description
A

1)

B

2)

C

3)

D

4)


10.3.3 Cyclohexanes and E2, Diaxial Elimination

When (1R,2R,4S)-1-bromo-4-(tert-butyl)-2-methylcyclohexane is reacted with sodium ethoxide in ethanol only one product is formed. Why?

Figure 10.24. (1R,2R,4S)-1-bromo-4-(tert-butyl)-2-methylcyclohexane undergoes elimination reaction with sodium ethoxide in ethanol.

To answer this, we must look at the chair conformations.

Examine the two chair conformations of cyclohexane given below. Notice that only one chair conformation aligns the β hydrogens, with the leaving group in an anti-coplanar orientation. This occurs when the β hydrogens and leaving group are in the axial positions or diaxial. If the leaving group is in the equatorial position, then the β hydrogens are gauche, with a dihedral angle of 60o and, as discussed previously, elimination only occurs with dihedral angles of 180o or, under special cases, 0o.

Figure 10.25a​. Analysis of E2 elimination for bromocyclohexane conformations when the leaving group is axial.​


Figure 10.25b​. Analysis of E2 elimination for bromocyclohexane conformations when the leaving group is equatorial.​

Now consider the two chair conformations of (1R,2R,4S)-1-bromo-4-(tert-butyl)-2-methylcyclohexane given below. When the bromine is in the axial position, there is only one β hydrogen that is also in the axial position. The other β hydrogens occupy equatorial positions and the methyl group on carbon 2 occupies the adjacent axial position. Since there is only one hydrogen that is anti-coplanar to the axial bromine, elimination can only occur between carbons 1 and 6, which means that only one product can form.

Figure 10.26. a) E2 reaction for (1R,2R,4S)-1-bromo-4-(tert-butyl)-2-methylcyclohexane. b) Conformational analysis for the reaction.​

The previous question is an example of a stereospecific reaction. The stereochemistry of the reactant determines the stereochemistry of the product. Only one stereospecific product could be formed because only one specific β hydrogen was available for the elimination reaction to occur.

Can you do the next question? Draw the chair configurations to determine whether the β hydrogens are diaxial to the leaving group.


Q10.12 - Level 3

What is the major product formed in the following reaction?

question description
A

1

B

2

Following is the explanation to Question 10.12.


Q10.13 - Level 3

What is the major product produced in the following reaction?

question description
A

1)

B

2)

C

3)

D

4)


Q10.14 - Level 3

Click on the hydrogen(s) that could be removed in an E2 reaction to produce cyclohexene. Click in the center of the circle.


10.3.4 Zaitsev's Rule: Regiochemistry and Stereochemistry

The following reaction demonstrates the regioselectivity and stereoselectivity of the E2 reaction.

The reaction of 2-bromobutane with a strong base generates three alkenes and, as expected, the more substituted internal double bond dominates, 82.2% (Zaitsev’s rule). However, two disubstituted stereoisomers are formed, which are cis and trans diastereomers. The trans isomer is produced in the greatest yield; therefore, this reaction is not only regioselective but also stereoselective in that a single reactant can produce two or more stereoisomers and one of these products is preferred over the other.

Figure 10.27. Regiochemical and stereochemical outcome of E2 elimination of 2-bromobutane​

Question: Why are two stereoisomers formed?

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


Q10.15 - Level 2

Match the mechanism to the alkene that is produced, trans-butene or cis-butene.

question description
Premise
Response
1

1

A

ciscis-butene

2

2

B

transtrans-butene


The type and number of stereoisomers of  alkenes that are produced in elimination reaction depends on the number of hydrogens removed. In the case of 2-bromobutane, two β hydrogens can be removed, therefore, two stereoisomers are formed. Elimination of the β hydrogen atom Ha produces the trans product and abstraction of Hb produces the cis product.

Figure 10.28. Stereoisomeric outcome of E2 reaction of 2-bromobutane​

Notice that the less stable monosubstituted alkene is produced in greater yield (17.8%) than the cis product (15.3%). Could you think of a reason for that outcome? Why is the trans alkene produced in larger yield?

Shown below is the Free Energy–Reaction Coordinate diagram. As you can see, the formation of the cis product is energetically less favored from start to finish. The initial conformation, transition state, and final product are all less stable than the trans reaction.

Figure 10.29​a. Reaction mechanism for the formation of cis and trans-but-2-ene from 2-bromobutane. ​


Figure 10.29b. Free energy diagram for the formation of cis-but-2-ene (A) and trans-but-2-ene (B) from 2-bromobutane​


Q10.16 - Level 3

The major alkene obtained when 2-bromo-4-methylhexane is heated with potassium ethoxide is:

question description
A

1)

B

2)

C

3)

D

4)


Q10.17 - Level 2

Three conformations of 2-bromopentane are depicted. Click on the hydrogen that would be removed in an E2 reaction to produce (Z)-pent-2-ene. Click in the center of the circle.


10.3.5 Hofmann Elimination: Bulky Bases and Regioselectivity

This is a special β-elimination reaction in which the less stable alkene becomes the dominant product.

Examine the table below. As the bulkiness of the attacking base increases, the less stable or Hofmann product will be predominant.

Figure 10.30. Product distribution of elimination of 2-bromo-2-methylbutane with increasingly bulky bases​


This is confusing. Was it not said earlier that the transition state in any E2 reaction has a double bond character? Thus, in the above example, the trisubstituted alkene-like transition state will be more stable and have a lower activation energy (Ea) than the disubstituted alkene-like transition state. Since the Ea for the formation of the trisubstituted alkene (2-methylbut-2-ene) has a lower activation energy, does this mean that the trisubstituted alkene (2-methylbut-2-ene) will be the majority product and will form faster?

Experiments have shown a different trend when elimination reactions are carried out with bulky bases: the more stable product has a smaller yield. This means that the transition state for the more stable product is higher in energy than the less stable product. What would cause this?


Q10.18 - Level 2

Hofmann elimination produces the less stable alkene as the major product. What would cause the transition state for the less stable product to lower its energy when compared with the energy levels of the more stable alkene?

question description
A

Hyperconjugation between the forming double bond and the base in the transition state.

B

Steric interaction is greater between base and substrate for the abstraction of a primary proton than a secondary proton.

C

Steric interaction is greater between base and substrate for the abstraction of a secondary proton than a primary proton.

Click on the video for a visual explanation to question 10.18 and Hofmann elimination.

The Potential Energy–Reaction Coordinate diagram below illustrates the energies involved in the Hofmann elimination reaction. As explained in the video; there is greater steric interaction between the base and the substrate in the transition state when the bulky base is abstracting a secondary proton compared with abstracting a primary proton. The energy of the transition state that produces the more thermodynamically stable alkene is greater than the transition state that produces the less stable alkene. Increasing the potential energy of the transition state increases the energy barrier to the reaction. A larger Ea decreases the rate of reaction and causes the more stable alkene to become a minor product.

Figure 10.31. Hofmann elimination. a) More stable transition state leading to major product. b) Less stable transition state leading to minor product. c) Potential energy diagram of the Hofmann elimination reaction of 2-bromo-2-methylbutane leading to the less stable alkene as the major product

To get more insight into this reaction, see Sec. 4.0 Kinetics vs. Thermodynamics.


Q10.19 - Level 2

Which reagent would you use to produce the desired alkene?

question description
A

1)

B

2)

C

3)

D

4)


Q10.20 - Level 3

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

question description
Premise
Response
1

1)

A

Major

2

2)

B

Minor


10.3.5.1 Hofmann Elimination: Bulky and Poor Leaving Groups and Regioselectivity

Hofmann products are also observed in elimination reactions with quaternary ammonium salts, tertiary sulfonium salts, and with poor leaving groups, such as fluoride.

Bulky and poor leaving groups: Previously, we have seen that steric interaction between the bulky base and the substrate increases the activation energy of the pathway toward the more stable alkene to such an extent that the less stable alkene becomes the major product. Similarly, a substrate that has a bulky leaving group may impede the approach of a base to the secondary β protons. This interaction would cause the transition state energy of the Zaitsev’s product to become larger than the transition state energy of the Hofmann product.

Figure 10.32. Elimination product distribution for substrates with bulky leaving groups

Poor leaving groups: Electronic factors play an important role in elimination reactions. An example of this is represented on the table below, where an increase in the Hofmann product is observed with an increase of the basicity of the leaving group, such as in the case of fluorine.

Figure 10.33. Elimination product distribution as a function of leaving group basicity

Again, consider the transition state. When an E2 reaction occurs with a good leaving group, the leaving group departs as the proton is being removed by the base. The transition state resembles the forming alkene. However, with a poor leaving group, like fluorine, the proton C–H bond and the C–F bond may not be breaking simultaneously. Since the C–F bond is quite strong, it is conceivable that in the transition state the C–H sigma bond breaking is farther advanced than the breaking of C–F sigma bond. In this transition state, a partial negative charge is forming on the carbon bonded to the proton that is being removed. The energy of this transition state is thus determined by the stability of the carbon gaining the partial negative charge.

Figure 10.34. Comparison of the transition states of 2-bromobutane and 2-fluorobutane as examples of good and poor leaving groups​

In Sec. 2.0 it was shown that carbocation stability increases with the number of attached alkyl groups. This was explained by the electron donating ability of the adjacent C–C or C–H sigma bonds into the empty carbocation’s p orbital. Alkyl groups donate electron density inductively through σ bond conjugation or hyperconjugation.

Figure 10.35a. Carbocation substitution and relative stability.​


Figure 10.35b. Hyperconjugation stabilization of carbocation. Dashed lines represent donation of electron density through hyperconjugation

Conversely, carbanions are negatively charged and are destabilized by electron donating alkyl groups. In solution, smaller negatively charged ions are more readily solvated (stabilized by the solvent). Therefore, methyl carbanions are more stable than tertiary carbanions.


Figure 10.36. Carbanion stability as a function of substitution

The reaction of 2-fluoropentane with sodium methoxide produces the less stable alkene in a 70% yield.

Figure 10.37. Elimination reaction of 2-fluoropentane with sodium methoxide

Similar to the Hofmann product with a bulky base, the transition state leading to the less stable product must be more stable than the transition state leading to the more stable product. In this reaction, the transition state that is developing a partial negative charge on the primary carbon is more stable than the transition state that is developing a partial negative charge on the secondary carbon, which leads to an increased yield of the less stable alkene or Hofmann product.

Figure 10.38​. Comparison of the transition states of Hofmann and Zaitsev elimination products of 2-fluoropentane with sodium methoxide​


Fluorouracil

Fluorouracil (Adrucil™) is an anti-cancer drug used to treat breast, skin, stomach, pancreatic and colon cancers. It is on the World Health Organization's List of Essential Medicines, which is a list of the most important medications known to mankind. 

Did you know this wonder drug’s mechanism of action relates to elimination reactions?

Thymidylate synthase

To understand how fluorouracil works, let’s first take a look at how a cancer cell divides.

An enzyme called thymidylate synthase must convert deoxyuridine monophosphate (dUMP) to deoxythymidine monophosphate (dTMP). One of the steps in this multistep process is an elimination reaction, which is shown below.

Keeping it Real Q10.1

Keeping it Real Q10.1 - Level 1

Identify the role of the nitrogen substituent shown in blue.

question description
A

Nucleophile

B

Base

C

Leaving group

D

Enzyme