ch22_hg15(3)

Chapter 22Alpha Carbon Chemistry: Enols and Enolates

Organic ChemistrySecond Edition

David Klein

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

22.1 Introduction Alpha Carbon Chemistry: Enols and Enolates

• For carbonyl compounds, Greek letters are often used to describe the proximity of atoms to the carbonyl center.*

• This chapter will primarily explore reactions that take place at the alpha (α) carbon.

*This method is also used elsewhere, β-phenylethanol (PhCH2CH2OH), omega-3 (ω-3) fats have a double bond 3 atoms from the chain’s end, …

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 22-2 Klein, Organic Chemistry 2e

The reactions we will explore proceed though either an enol or an enolate intermediate:



• Trace amounts of acid or base catalyst provide equilibriums in which both the enol and keto forms are present

• How is equilibrium different from resonance?

• At equilibrium, >99% of the molecules exist in the keto form. WHY?



• In rare cases such as the example below, the enol form is favored in equilibrium

• Give two reasons to explain WHY the enol is favored

• The solvent can affect the exact percentages



Phenol is an example where the enol is vastly favored over the keto at equilibrium. WHY?



The tautomerization mechanism depends on whether it is catalyzed by acid or base (very slow without). Acid catalysis:



*

*Note: the most plentiful base in aqueous acid is water; don’t invoke hydroxide under acid catalysis.

For example, in an aqueous solution at pH 2, water’s concentration is 55 M, while [OHˉ] is 10-12 M.

The mechanism for the tautomerization depends on whether it is acid catalyzed or base catalyzed



• As the tautomerization is practically unavoidable, some fraction of the molecules will exist in the enol form.

• Analyzing the enol form, there is a minor (but significant) resonance contributor with a nucleophilic carbon atom

• Practice with conceptual checkpoints 22.1 through 22.3



double bond

single bond

• In the presence of a strong base, an enolate forms

• The enolate is much more nucleophilic than the enol. WHY?



• The enolate can undergo C-attack or O-attack.

• Enolates generally undergo C-attack. WHY?



• Alpha protons are the only protons on an aldehyde or ketone that can be removed to form an enolate

http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_disp.cgi?sdbsno=1925

• Removing the aldehyde proton or the beta or gamma proton will NOT yield a resonance stabilized intermediate

• Practice with SkillBuilder 22.1



HC(=O)-CH2-CH2-CH3Assign. Shift(ppm)

aldehyde 9.764

α 2.37

β 1.64

γ 0.97


sp2–hybridized carbon, so no overlap between

aldehyde hydrogen and p orbital on the carbonyl C.

Electrostatic charges

of enolate atoms

HOC

CH

H

HOC

CH

H

Natural charges

of enolate atoms

Better predictor

Draw all possible enolates that could form from the following molecule:



• Why would a chemist want to form an enolate?

• To form an enolate, a base must be used to remove an alpha proton.

• The best choice of base depends on the acidity of the alpha protons.

• How do we quantify how acidic something is?



Let’s compare some pKa values for some alpha protons



• When pKa values are similar, both products and reactants are present in significant amounts:

• Which side will this equilibrium favor?



• In this case, it is an advantage to have both enolate and aldehyde in solution so they can react with one another

• Show how the electrons might move in the reaction between the enolate and the aldehyde.



• To have the carbonyl react irreversibly, a stronger base (needing to have low nucleophilicity) such as H- is needed.

• When is it synthetically desirable to convert all of the carbonyl into an enolate?



*

* Note: the author’s slides often omit the enolate resonance form

having the negative charge on oxygen (which is the major contributor),

perhaps to emphasize that most reactions take place on the carbon.

• LDA is an even stronger base that is frequently used to promote irreversible enolate formation:

• Why is the reaction effectively irreversible?

• LDA has two bulky isopropyl groups. Why would such a bulky base be desirable?



(see box on

Previous slide)

• When a proton is next to two carbonyl groups, its acidity is increased:

• Draw the resonance contributors that allow 2,4-pentanedione to be so acidic



• 2,4-Pentanedione is sufficiently acidic that hydroxide or alkoxides can deprotonate it irreversibly.

• Figure 22.2 summarizes the relevant factors you should consider when choosing a base.

• Practice with conceptual checkpoints 22.6 through 22.8.



• H3O+ catalyzes ketoenol tautomerism. HOW?

• The enol tautomer can attack a halogen molecule:

• The process is autocatalytic –the regenerated acid can catalyze another tautomerization and halogenation.

22.2 Alpha Halogenation of Enols and Enolates


• When an unsymmetrical ketone is used, bromination occurs primarily at the more substituted carbon…

• Because the major product results from the more stable (more substituted) enol.

• A mixture of products is generally unavoidable.



• This provides a two step synthesis for the synthesis of an α,β-unsaturated ketone:

• Give a mechanism that shows the role of pyridine.

• Other bases such as potassium tert-butoxide (why this one?) may also be used in the second step.

• Practice with conceptual checkpoints 22.9 and 22.10.



• The Hell-Volhard-Zelinski(ii,y) reaction brominates the alpha carbon of a carboxylic acid:

• PBr3 forms the acyl bromide, which more readily forms the enol and attacks the bromine.

• Hydrolysis of the acyl bromide is the last step.

• Draw a complete mechanism.

• Practice checkpoints 22.11 and 22.12Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 22-26 Klein, Organic Chemistry 2e


• Alpha halogenation can also be achieved under basic conditions:

• The formation of the enolate is not favored, but the equilibrium is pushed forward by the second step.

• Will the presence of the α bromine make the remaining αproton more or less acidic?



• Controlled monosubstitution is not possible. WHY?

• Methyl ketones can be converted to carboxylic acids using excess halogen and hydroxide.

• Once all three α protons are substituted, the CBr3 group becomes a decent leaving group.



Once all three α protons are substituted, the CBr3 group becomes a decent leaving group

The last step below is practically irreversible. WHY?



• The carboxylate produced on the last slide can be protonated with H3O+

• The reaction works well with Cl2, Br2, and I2, and it is known as the haloform reaction.

• The iodoform reaction may be used to test for methyl ketones, because iodoform can be observed as a yellow solid when it forms.




Give the major product for the reaction below. Be careful with regards to stereochemistry (?!):



• Recall that treating an aldehyde with hydroxide or alkoxideyields an equilibrium in which significant amounts of both enolate and aldehyde are present.

• If the enolate attacks the aldehyde, an aldol reaction occurs:

• The product features both aldehyde and alcohol groups.

• Note the location of the –OH group on the beta carbon.

22.3 Aldol Reactions



• The aldol mechanism


• The aldol reaction is an equilibrium process that generally favors the products:

• How might the temperature affect the equilibrium?



• A similar reaction with ketones generally does NOT favor the β-hydroxyketone product:

• Give a reasonable mechanism for the retro-aldol reaction.




• Predict the products for the follow reaction, and give a reasonable mechanism. Be careful of stereochemistry



• When an aldol product is heated under acidic or basic conditions, an α,β-unsaturated carbonyl forms:

• Such a process is called an aldol condensation, because water is given off.

• The elimination reaction above is an equilibrium, which generally favors the products.

• WHY? Consider enthalpy and entropy.



• The elimination of water can be promoted under acidic or under basic conditions.

• Give a reasonable mechanism for each



• When a water is eliminated, two products are possible.

• Which will likely be the major product? Use the mechanism to explain.



• Because the aldol condensation is favored, it is often impossible to isolate the aldol product without elimination:


• Condensation is especially favored when extended conjugation results.


• At low temperatures, condensation is less favored, but the aldol product is still often difficult to isolate in good yield


• Practice with SkillBuilder 22.3.


Predict the major product of the following reaction. Be careful of stereochemistry



• Substrates can react in a crossed aldol or mixed aldol reaction. Predict 4 possible products in the reaction below

• Such a complicated mixture of products is not very practical synthetically. WHY?



Practical crossed aldol reactions can be achieved through one of two methods:

1. One of the substrates is relatively unhindered and without alpha protons:



1. One of the substrates is relatively unhindered and without alpha protons



Practical crossed aldol reactions can be achieved through one of two methods:

2. One substrate is added dropwise to LDA forming the enolate first. Subsequent addition of the second substrate produces the desired product:




Describe an aldol route to yield the following compound:



• Cyclic compounds can be formed through intramolecular aldol reactions.


• One group forms an enolate that attacks the other group.

• Recall that 5 and 6-membered rings are most likely to form. WHY?

• Practice conceptual checkpoints 22.25 through 22.27.


• Esters also undergo reversible condensations reactions

• Unlike a ketone or aldehyde, an ester has a leaving group...

22.4 Claisen Condensations


• Esters also undergo reversible condensations reactions

• The resulting doubly-stabilized enolate must be treated with an acid in a last step. WHY?

• A beta-keto ester is produced



• There are some limitations to the Claisen condensation:

1. The starting ester must have two alpha protons, because removal of the second proton by the alkoxide ion is what drives the equilibrium forward.

2. Hydroxide cannot be used as the base to promote Claisencondensations, because a hydrolysis reaction occurs between hydroxide and the ester.

3. An alkoxide equivalent to the –OR group of the ester is a good base, because nontrivial transesterification is avoided.

• Practice conceptual checkpoints 22.28 and 22.29.



• Crossed Claisen reactions can also be achieved using the same strategies employed in crossed aldol reactions

• Practice with conceptual checkpoint 22.30



• Intramolecular Claisen condensations can also be achieved

• This Diekmann cyclization proceeds through the expected 5-membered ring transition state. DRAW it.




Draw reactants needed to synthesize the following molecules:



• The alpha position can be alkylated when an enolate is treated with an alkyl halide:

• The enolate attacks the alkyl halide via an SN2 reaction:

22.5 Alkylation of the Alpha Proton


• When 2° or 3° alkyl halides are used, the enolate can act as a base in an E2 reaction. SHOW a mechanism.

• The aldol reaction also competes with the desired alkylation, so a strong base, such as LDA, must be used.

• Regioselectivity is often an issue when forming enolates.

• If the compound below is treated with a strong base, two enolates can form – see next few slides:



• What is meant by kinetic and thermodynamic enolate? See next few slides for details



For clarity, the kinetic and thermodynamic pathways are exaggerated below


Explain the energy differences below using steric and stability arguments


• LDA is a strong base, and at low temperatures, it will react effectively irreversibly.

• NaH is not quite as strong, and if heat is available, the system will be reversible:


Practice with conceptual checkpoints 22.33 and 22.24.


Give reagents necessary to synthesize the compound below starting with carbon fragments having five or fewer carbons:



• The malonic ester synthesis allows a halide to be converted into a carboxylic acid with two additional carbons:

• Diethyl malonate is first treated with a base to form a doubly-stabilized enolate:



• The enolate is treated with the alkyl halide

• The resulting diester can be hydrolyzed with acid or base using heat

• And then…



• One of the resulting carboxylic acid groups can be decarboxylated with heat through a pericyclic reaction:

• Why isn’t the second carboxylic acid group removed?



Here is an example of the overall synthesis



• Double alkylation can also be achieved:

• Practice with SkillBuilder 22.5.

• The acetoacetic ester synthesis is a very similar process:



• Give a complete mechanism for the process below




• Recall that α,β-unsaturated carbonyls can be made easily through aldol condensations

• α,β-unsaturated carbonyls have three resonance contributors:

• Which contributors are electrophilic?

22.6 Conjugate Addition Reactions


• Grignard reagents generally attack the carbonyl position of α,β-unsaturated carbonyls yielding a 1,2 addition

• In contrast, Gilman reagents generally attacks the beta position giving 1,4 addition or conjugate addition



• Conjugate addition of α,β-unsaturated carbonyls starts with attack at the beta position:


• WHY is the nucleophile generally favored to attack the beta position?


• More reactive nucleophiles (e.g., Grignard) are more likely to attack the carbonyl directly. WHY?

• Enolates are generally less reactive than Grignards, but more reactive than Gilman reagents, so enolates often give a mixture of 1,2- and 1,4- addition products

• Doubly-stabilized enolates are stable enough to react primarily at the beta position



When an enolate attacks a beta carbon, the process is called a Michael addition:



• Give a mechanism showing the reaction between the two compounds shown below:

• Practice with conceptual checkpoints 22.44 through 22.46.



• Because singly-stabilized enolates do not give high yielding Michael additions, Gilbert Stork developed a synthesis using an enamine intermediate.

• Recall the enamine synthesis from chapter 20:



• Enolates and enamines have reactivity in common:

• The enamine is less nucleophilic and more likely to act as a Michael donor



• Water hydrolyzes the imine and tautomerizes and protonates the enol



• Give reagents necessary to synthesize the molecule below using the Stork enamine synthesis




• The Robinson Annulation utilizes the a Michael addition followed by an aldol condensation

• Practice checkpoints 22.49 and 22.50



• Most of the reactions in this chapter are C-C bond forming

• Three of the reactions yield a product with two functional groups

• The positions of the functional groups in the product can be used to design necessary reagents in the synthesis – see next few slides


22.7 Synthetic Strategies


• Stork enamine synthesis 1,5-dicarbonyl compounds

• Aldol and Claisen 1,3-difunctional compounds



• We have learned two methods of alkylation

1. The alpha position of an enolate attacks an alkyl halide

2. A Michael donor attacks the beta position of a Michael acceptor

• These two reactions can also be combined

• Give a reasonable mechanism




Give reagents necessary for the following synthesis



Explain why an enolate is more likely to produce products resulting from attack by the alpha carbon than by direct attack by the oxygen. Is the argument kinetic, thermodynamic or both?

Additional Practice Problems


Give the major products for the reaction below



Using both kinetic and thermodynamic arguments, explain why aldol reactions involving a ketone are less product favored than those involving aldehydes



Give reagents necessary to produce the product below using aldol chemistry:



Give reagents necessary for the synthesis below



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