8.1 Introduction to Elimination - WordPress.com...The Zaitsev product is generally favored over the Hofmann product, unless a sterically hindered base is used, in which case the Hofmann

• 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

Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 8 -1

• Consider –OH, which can act as a base or a nucleophile:

8.1 Introduction to Elimination


Attack at the α-carbon

Attack at the β-hydrogen

ALKENE β or 1,2

• Important alkenes:

8.2 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


8.2 Alkenes


• Alkenes are named using the same procedure we used in Chapter 4 to name alkanes with minor modifications.

1. Identify the parent chain, which should include the C=C double bond.

2. Identify and name the substituents. 3. Assign a locant (and prefix if necessary) to each substituent. Give

the C=C double bond the lowest number possible. 4. 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 Nomenclature


1. 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.



2. Identify and name the substituents. 3. 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.



4. 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.



• Name the following molecule.



• 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.



• 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 the resulting geometry.



• 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.



• Assigning E or Z to a stereoisomers: 1. Prioritize the groups attached to the C=C double bond based

on atomic number.



• 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 zusammen, 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).




• 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?





• Consider the following stability trend:

• What pattern do you see? • Practice with SKILLBUILDER 8.3.



• 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



• In general, an H atom and a leaving group are eliminated.

• To understand the mechanism of elimination, first recall the four mechanistic steps we learned in Chapter 7.

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

8.6 Elimination Reactions in Detail


• Recall the four mechanistic steps we learned in Chapter 7.

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



• 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?



• 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?




• In the mechanism below (E2), how will a change in [base] or [substrate] affect the reaction rate?

– Write a reasonable rate law for the mechanism. – What do the E and the 2 of the E2 notation represent?

• Practice with CONCEPTUAL CHECKPOINT 8.13.

8.7 Elimination by 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 Elimination by E2 – The Effect of Substrate


• 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.



• How 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.




• 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.

• What is the relationship between the alkene products?

• REGIOSELECTIVITY occurs when one product is formed predominantly over the other.

8.7 Elimination by E2 – Regioselectivity


• The identity of the base can affect the REGIOSELECTIVITY.



• 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 nonnucleophilic) are useful in many reactions.




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

8.7 Elimination by E2 – Stereoselectivity


• 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. • 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.

8.7 Elimination by E2 – Stereospecificity


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

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



• To rationalize the stereospecificity 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 elimination reaction? WHY?

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



• If each of the molecules below were to undergo dehydrohalogenation, what would the products be? Draw all of the possible products.




• Consider both regioselectivity 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 two-step process.

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

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

8.9 Elimination by E1 – The 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? • Practice with CONCEPTUAL CHECKPOINT 8.26.

8.9 Elimination by E1 – The 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 to the right, once the water leaving group leaves the carbocation, what base should be used to complete the elimination?




• 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 an E2 reaction?




• In the last step of the mechanism, a proton is removed from a β carbon adjacent to the sp2 hybridized C+.

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




• Considering stereochemistry and regiochemistry, predict the products for the following reaction.



OH

• Recall the similarities between E1 and SN1.

• 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.



• Consider the energy diagram for the mechanism on the previous slide.

• Assess each free energy change.



• The mechanism shows the formation of the major products.

• Predict the minor elimination products as well.




• In E2, the base removes the β-proton as the leaving group (LG) leaves.

• Will such a reaction require a relatively strong base?

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




• Substitution and elimination are always in competition.

• Sometimes, products are only observed from either substitution or elimination.

• Sometimes a mixture of products is observed.

8.12 Substitution vs. Elimination


• To predict whether substitution or elimination will predominate, consider the factors below:

1. 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 Substitution vs. Elimination


1. Determine the function of the reagent: assess 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 Substitution vs. Elimination – Reagent Function: Nucleophilicity


1. Determine the function of the reagent: assess 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

-?

– Are Cl- and HSO4- relatively strong or weak?

8.12 Substitution vs. Elimination – Reagent Function: Basicity


1. Determine the function of the reagent: assess 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 Substitution vs. Elimination – Reagent Function: Basicity


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

polarizable and/or they have very strong conjugate acids.

8.12 Substitution vs. Elimination – Basicity vs. Nucleophilicity


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



• 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?




1. Analyze the function of the reagent (nucleophile and/ or base).

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

8.13 Predicting Substitution vs. Elimination
















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

8.14 Predicting Products


REGIOCHEMICAL OUTCOME STEREOCHEMICAL OUTCOME SN2 The nucleophile attacks the α

position, where the leaving group is connected.

The nucleophile replaces the leaving group with inversion of configuration.

SN1 The nucleophile attacks the carbocation, which is where the leaving group was originally connected, unless a carbocation rearrangement took place.

The nucleophile replaces the leaving group with racemization.


8.14 Predicting Products


REGIOCHEMICAL OUTCOME STEREOCHEMICAL OUTCOME E2 The Zaitsev product is generally

favored over the Hofmann product, unless a sterically hindered base is used, in which case the Hofmann product will be favored

This process is both stereoselective and stereospecific. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene. When the β position of the substrate has only one proton, the stereoisomeric alkene resulting from anti-periplanar elimination will be obtained (exclusively, in most cases).

E1 The Zaitsev product is always favored over the Hofmann product.

The process is stereoselective. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene.

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8.1 Introduction to Elimination - WordPress.com...The Zaitsev product is generally favored over the Hofmann product, unless a sterically hindered base is used, in which case the Hofmann