Oganic II - Klein - chapter 22

22.1 Introduction to 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 carbon.

Copyright 2012 John Wiley & Sons, Inc.
22-1 Klein, Organic Chemistry 1e

• 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 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, we see there is a minor (but
significant) resonance contributor with a nucleophilic
carbon atom.

• Practice with CONCEPTUAL CHECKPOINTs 22.1
through 22.3.

• In the presence of a strong base, an ENOLATE forms.

• The enolate is much more nucleophilic than in 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.

• Removing the aldehyde proton, or the beta or gamma
proton, will NOT yield a resonance stabilized
intermediate.
• Practice with SKILLBUILDER 22.1.

• 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 the
alpha protons.
• The appropriate base depends on how acidic the alpha
protons are .
• What method do we have to 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.


• If you want the carbonyl to react irreversibly, a stronger
base, such as H–, is necessary.

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


• Lithium diisopropylamide (LDA) is an even stronger base
that is frequently used to promote irreversible enolate
formation.

• Why is the reaction affectively irreversible?

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

• When a proton is alpha to two different carbonyl
groups, its acidity is increased.

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


• 2,4-pentanedione is acidic enough 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.

22.2 Alpha Halogenation of Enols
and Enolates
• H3O+ catalyzes the ketoenol tautomerism. HOW?
• The enol tautomer can attack a halogen molecule.

• The process is AUTOCATALYTIC:
– The regenerated acid can catalyze another tautomerization
and halogenation.

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

• The major product results from the more stable (more
substituted) enol.
• A mixture of products is generally unavoidable.


and Enolates
• 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, can also
be used in the second step.
• Practice with CONCEPTUAL CHECKPOINTs 22.9 and
22.10.

and Enolates
• The Hell-Volhard Zelinsky 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 CONCEPTUAL CHECKPOINTs 22.11
and 22.12.

and Enolates
• 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
α proton more or less acidic?


and Enolates
• 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.


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

• The last step is practically irreversible. WHY?


and Enolates
• 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.
22.14.


and Enolates
• Give the major product for the reaction below. Be
careful of stereochemistry.


22.3 Aldol Reactions
• Recall that when an aldehyde is treated with hydroxide
(or alkoxide), an equilibrium forms where 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.


• 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 for a ketone generally does NOT favor
the β-hydroxy ketone 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, often it is
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.


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


• Substrates can react in a CROSSED aldol or MIXED aldol
reaction. Predict the four possible products in the
reaction below.

• Such a complicated mixture of products is not
very synthetically practical. 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 a synthesis necessary 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.

22.4 Claisen Condensations
• Esters also undergo reversible condensations reactions.

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

• Esters also undergo reversible condensations reactions.

• The resulting doubly-stabilized enolate must be treated
with an acid in the last step. WHY?
• A beta-ketoester 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 Claisen
condensations 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 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.
• Practice with CONCEPTUAL CHECKPOINTs
22.31 and 22.32.

• Give reagents necessary to synthesize the following
molecules.


22.5 Alkylation of the Alpha Position
• 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.


• 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.



• What is meant by kinetic and thermodynamic enolate?


• 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 in an irreversible manner.
• NaH is not quite as strong, and if heat is available, the
system will be reversible.
22.24.


• Give necessary reagents to synthesize the compound
below starting with carbon fragments with five carbons
or less.


• 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, and using heat.


• 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:

• The acetoacetic ester synthesis is a very similar process.


• Give a complete mechanism for the process below.



22.6 Conjugate Addition Reactions
• 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?

• 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 does the
nucleophile generally
favor attacking 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 a Michael addition
followed by an aldol condensation.

• Practice CONCEPTUAL CHECKPOINTs 22.49
and 22.50.

22.7 Synthetic Strategies
• 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.


• 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.


Oganic II - Klein - chapter 22

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