Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e Chapter 8...

Chapter 8Alkenes: Structure and Preparation via

Elimination Reactions

Organic ChemistrySecond Edition

David Klein

8.1 Introduction to Elimination• Elimination reactions often compete with substitution

reactions.• What are the two main ingredients for a substitution?

– A nucleophile and an electrophile with a leaving group

• What are the two main ingredients for an elimination? – A base and an electrophile with a leaving group

• How is a base both similar and different from a nucleophile?

8.1 Introduction to Elimination• Consider –OH, which can act as a base or a nucleophile

Attack at the α Carbon

Reaction at the β Hydrogen

β or 1,2

ALKENE

8.2 Alkenes• Important alkenes

• C=C double bonds are found in a variety of compounds including pheromones and many other classes of compounds

8.2 Alkenes

• Why might it be helpful to know the chemical structure of pheromones such as those below?

8.2 Alkenes

• Alkenes are also important compounds in the chemical industry

• 70 billion pounds of propylene (propene) and 200 billion pounds of ethylene (ethene) are both made from cracking petroleum each year

8.2 Alkenes

8.3 Alkene Nomenclature• Alkenes are named using the same procedure we used

in Chapter 4 to name alkanes with minor modifications1. Identify the parent chain, which should include the C=C

double bond2. Identify and Name the substituents3. Assign a locant (and prefix if necessary) to each substituent.

Give the C=C double bond the lowest number possible4. List the numbered substituents before the parent name in

alphabetical order. Ignore prefixes (except iso) when ordering alphabetically

5. The C=C double bond locant is placed either just before the parent name or just before the -ene suffix

8.3 Alkene Nomenclature1. Identify the parent chain, which should include the C=C

double bond• The name of the parent chain should end in -ene rather

than –ane

• The parent chain should include the C=C double bond

8.3 Alkene Nomenclature2. Identify and Name the substituents3. Assign a locant (and prefix if necessary) to each

substituent. Give the C=C double bond the lowest number possible

– The locant is ONE number, NOT two. Although the double bond bridges carbons 2 and 3, the locant is the lower of those two numbers

8.3 Alkene Nomenclature4. List the numbered substituents before the parent name

in alphabetical order. Ignore prefixes (except iso) when ordering alphabetically

5. The C=C double bond locant is placed either just before the parent name or just before the -ene suffix

• Practice with SkillBuilder 8.1

8.3 Alkene Nomenclature• Name the following molecule

8.4 Alkene Isomerism• For the pi bond to remain intact, rotation around a

double bond is prohibited• As a result, cis and trans structures are not identical

• Is there a difference between cis-butane and trans-butane?

• What specific type of isomers are cis and trans butene?

8.4 Alkene Isomerism• In cyclic alkenes with less than 8 atoms in the ring, only

cis alkenes are stable. WHY?

• Draw the structure for trans-cyclooctene

• When applied to bicycloakenes, this rule is called Bredt’s rule

8.4 Alkene Isomerism• Apply Bredt’s rule to the compounds below

• The carbons of the C=C double bond and the atoms that are directly attached to them must be planar to maintain the pi bond overlap.

• A handheld model can be used to help visualize the p orbital overlap and resulting geometry

8.4 Alkene Isomerism• Cis and trans modifiers are strictly used to describe C=C

double bonds with identical groups on each carbon. Where are the identical groups in trans-2-pentene?

• For molecules with different groups attached to the C=C double bond, the E/Z notation is used instead of cis/trans notation

8.4 Alkene Isomerism• Assigning E or Z to a stereoisomers

1. prioritize the groups attached to the C=C double bond based on atomic number

8.4 Alkene Isomerism• Assigning E or Z to a stereoisomers

1. prioritize the groups attached to the C=C double bond based on atomic number

2. If the top priority groups are on the same side of the C=C double bond, it is Z (for zussamen, which means together)

3. If the top priority groups are on opposite sides of the C=C double bond, it is E (for entgegen, which means opposite)

8.5 Alkene Stability• Because of steric strain, cis isomers are generally less

stable than trans

8.5 Alkene Stability• The difference in stability can be quantified by

comparing the heats of combustion

• How does heat of combustion relate to stability?

8.5 Alkene Stability

8.5 Alkene Stability• Consider the following stability trend

• What pattern do you see?• Practice with SkillBuilder 8.3

8.5 Alkene Stability• List the following molecules in order of increasing heat

of combustion– 2,3,4-trimethyl-1,3-pentadiene– 2-isopropyl-1,4-pentadiene– 3,3-dimethyl-1,5-hexadiene– 4,5-dimethylcyclohexene

8.6 Elimination Reactions in Detail• In general, a H atom and a leaving group are eliminated

• To understand the mechanism of elimination, first recall the 4 mechanistic steps we learned in chapter 7

– Nucleophilic attack– Loss of a leaving group– Proton transfer– Rearrangement

8.6 Elimination Reactions in Detail• The 4 mechanistic steps we learned in chapter 7

• Which of the 4 steps MUST take place in every elimination mechanism?

8.6 Elimination Reactions in Detail• All elimination reactions involve both loss of a leaving

group and proton transfer• The mechanism may be a concerted (one step) process

or a step-wise process. Which process is shown below?

8.6 Elimination Reactions in Detail• All elimination reactions involve both Loss of a leaving

group and proton transfer• The mechanism of the step-wise process:

• Could the steps happen in the reverse order?

8.7 Elimination by E2• The E2 mechanism below matches the observed rate

law. Write a reasonable rate law for the mechanism

• How will a change in [base] or [substrate] affect the reaction rate?

• What do the E and the 2 of the E2 notation represent?• Practice with conceptual checkpoint 8.13

8.7 The Effect of Substrate on E2• The kinetics of E2 and SN2 are quite similar. WHY?

• However, tertiary substrates are unreactive toward SN2 while they react readily by E2. WHY?

8.7 The Effect of Substrate on E2• 3° substrates are more reactive toward E2 than are 1°

substrates even though 1° substrates are less hindered

• The 3° substrate should proceed through a more stable transition state (kinetically favored) and a more stable product (thermodynamically favored). Let’s take a look at the energy diagram – see next slide

8.7 The Effect of Substrate on E2How would both the transition state energy and the product energy be different if the substrate were 1°?

• Notice the differences in transition state and in product energies

• Practice with conceptual checkpoint 8.14

8.7 The Effect of Substrate on E2

• If there are multiple reactive sites or regions on a molecule, multiple products are possible

• In elimination reactions, there are often different β sites that could be deprotonated to yield different alkenes

8.7 Regioselectivity of E2

• What is the relationship between the alkene products?

• Regioselectivity occurs when one constitutional product is formed predominantly over the other

Hofmann product

Zaitsev product

• The identity of the base can affect the regioselesctivity

• Why does the Zaitsev product predominate when a base that is NOT sterically hindered is used?

• Is the Zaitsev product kinetically favored, thermodynamically favored, or both?

• Why does a sterically hindered base favor the Hofmann product?

• Sterically hindered bases (sometimes called non-nucleophilic) are useful in many reactions

• Consider the dehydrohalogenation (elimination of a hydrogen and a halogen) of 3-bromopentane

8.7 Stereoselectivity of E2

• Why are both the transition state and product more stable for the trans product?

• What is the difference between stereoselective and stereospecific?

• Consider dehydrohalogenation for the molecule below

8.7 Stereospecificity of E2

• There is only one available β Hydrogen to be eliminated

• You might imagine that it would be possible to form both the E and Z alkene products from this reaction. Draw both the E and the Z products

• When the reaction is actually performed, only the E product is observed

• Is the E formed exclusively because it is formed through a slightly lower transition state?

• To rationalize the stereospecificty of the reaction, consider the transition state for the reaction

• In the transition state, the C-H and C-Br bonds that are breaking must be rotated into the same plane as the pi bond that is forming

• Draw the transition state structure illustrating the coplanar geometry

• There are two coplanar options for the molecule

• Why does the reaction proceed exclusively through the Anti coplanar structure?

• To see the difference between Anti and Syn, Newman projections and hand held models can be helpful

• Evidence suggests that a strict 180° angle is not necessary for E2 mechanisms.

• Similar angles (175–179°) are sufficient• The term, anti-periplanar is generally used instead of

anti-coplanar to account for slight deviations from coplanarity

• Although the E isomer is usually more stable because it is less sterically hindered, the requirement for an anti-periplanar transition state can often lead to the less stable Z isomer

• Assuming they proceed through an anti-periplanar transition state, predict the products for the following reactions, and label them as cis or trans

• Assuming they proceed through an anti-periplanar transition state, predict all of the products for the following reaction

• Will the reaction be stereospecific or stereoselective, and what factors most affect the product distribution?

• Consider the dehydrohalogenation of a cyclohexane• Given the anti-periplanar requirement, which of the

two possible chair conformations will allow for the elimination to occur?

• Which of the two molecules below will NOT be able to undergo an E2 elimination reaction? WHY?

• It might be helpful to draw their chair structures and make a handheld model

• Draw all of the possible products if each of the molecules below were to undergo dehydrohalogenation

• Consider both regioselestivity and stereoselectivity to predict the products for the eliminations below, and draw complete mechanisms

8.8 Predicting Products for E2

ClNaOEt NaOtBu

• The E1 mechanism is a 2-step process

• Similar to SN1 (chapter 7), the reaction rate for E1 is not affected by the concentration of the base

• What does the E and the 1 stand for in the E1 notation?

8.9 The E1 Mechanism

• Given the rate law for E1, which step in the mechanism is the rate-determining slow step?

• If the second step were the slow step, how would you write the rate law?

8.9 The E1 Mechanism

• How does the substrate reactivity trend for E1 compare to the trend we discussed in chapter 7 for SN1? WHY?

• Just like we did for SN1 in chapter 7, to explain the reactivity trend above, we must compare the energy diagrams for each substrate

• To compare their energies, draw the structures for each transition state, intermediate, and product below

• Because E1 and SN1 proceed by the same first step, their competition will generally result in a mixture of products

• How might you promote one reaction over the other?

• Alcohols can also undergo elimination or dehydration by E1, but the –OH group is not a stable leaving group

• In the E1 reaction below, once the water leaving group leaves the carbocation, what base should be used to complete the elimination?

8.9 Regioselectivity for E1• The final step of the E1 mechanism determines the

regioselectivity• E1 reactions generally produce the Zaitsev product

predominantly. WHY?

• Why can’t we control the regioselectivity in this reaction like we can in and E2 reaction?

8.9 Stereoselectivity for E1• In the last step of the mechanism, a proton is removed

from a β carbon adjacent to the sp2 hybridized carbocation

• Draw the appropriate carbocation that forms in the reaction below, and rationalize the product distribution

8.9 Stereoselectivity for E1• Considering stereochemistry and regiochemistry,

predict the products if the molecule below was treated with concentrated sulfuric acid

8.10 Complete E1 Mechanisms• Recall the similarities between SN1 and E1

• After the carbocation is formed and possibly rearranged, the E1 proton transfer neutralizes the charge

8.10 Complete E1 Mechanisms

• Why is the first proton transfer necessary?• Practice with conceptual checkpoint 8.32

• Explain why the carbocation rearranges• Practice with conceptual checkpoint 8.33

• The maximum number of steps in an E1 mechanism is generally four

8.10 Complete E1 Mechanisms• Consider the

energy diagram for the mechanism on the previous slide

• Assess each free energy change

8.10 Complete E1 Mechanisms• The mechanism shows the formation of the major

products • Predict the minor elimination products as well

8.11 Complete E2 Mechanisms• In E2, the base removes the β proton as the LG leaves • Will such a reaction require a relatively strong base?

• Will E2 dehydrations likely involve a proton transfer prior to the elimination?

8.12 Substitution vs. Elimination• Substitution and Elimination are always in competition• Sometimes products are only observed from S or E

• Sometimes a mixture of products is observed

8.12 Substitution vs. Elimination• To predict whether substitution or elimination will

predominate, consider the factors below1. Determine the function of the reagent. Is it more likely

to act as a base, a nucleophile, or both?– Kinetics control nucleophilicity. WHY? HOW?– Thermodynamics control basicity. WHY? HOW?

2. Analyze the substrate and predict the expected mechanism (SN1, SN2, E1, or E2)

3. Consider relevant regiochemical and stereochemical requirements

8.12 Reagent Function: Nucleophilicy1. Assessing the strength of a nucleophile• The greater the negative charge, the more nucleophilic

it is likely to be

• The more polarizable it is, the more nucleophilic it should be

• The less sterically hindered it is, the more nucleophilic it should be. WHY?

8.12 Reagent Function: Basicity1. Assessing the strength of a base• Assess the strength of its conjugate acid quantitatively

using the pKa of its conjugate acid

• Which is a stronger base Cl- or HSO4-?

• As bases, are Cl- and HSO4- relatively strong or weak?

8.12 Reagent Function: Basicity1. Assessing the strength of a base• If the base is neutral, assess the strength of its

conjugate acid qualitatively using ARIO (atom, resonance, induction, orbital)

– Compare CH3OH and CH3NH2

• If the base carries a negative formal charge, qualitatively assess the strength of the base using ARIO

– Compare Cl- and HSO4-

8.12 Basicity vs. Nucleophilicity

• Consider each of the reagent categories• Reagents that act as nucleophiles only are either highly

polarizable and/or they have very strong conjugate acids

• Reagents that act as bases only have either very low polarizability and/or they are sterically hindered

8.12 Basicity vs. Nucleophilicity• The stronger the reagent, the more likely it is to

promote SN2 or E2. WHY? • The more sterically hindered reagents are more likely to

promote elimination than substitution. WHY?

• The weaker the reagent, the more likely it is to promote SN1 or E1. WHY?

8.13 Predicting Subst. vs. Elim.1. Analyze the function of the reagent (nucleophile and/

or base)2. Analyze the substrate (1°, 2°, or 3°)

8.14 Predicting Products1. Analyze the function of the reagent (nucleophile and/

or base)2. Analyze the substrate (1°, 2°, or 3°)3. Consider regiochemistry and stereochemistry

8.14 Predicting Products3. Consider regiochemistry and stereochemistry

Additional Practice Problems• Name the following molecules

Additional Practice Problems• Label the molecules below as either cis or trans and

either E or Z where appropriate

Additional Practice Problems• For the substrate, give both the kinetically favored E2

product and the thermodynamically favored E2 product. Explain what conditions can be used to favor each.

Additional Practice Problems• Since tertiary substrates react more readily than

secondary or primary in both E1 and E2 mechanisms, what factor(s) usually controls which mechanism will dominate and why?

• Consider both regioselestivity and stereoselectivity to predict the major product for the elimination below

Additional Practice Problems

Br NaOEt

• Predict the major product if the alcohol below were treated with concentrated sulfuric acid. Be aware of the possible rearrangements.

Additional Practice Problems

concH2SO4

• Predict the major product for the following reactions considering competing substitution and elimination pathways.

Additional Practice ProblemsCl

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e Chapter 8...

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