enols and enolates (and enamines) carbon nucleophiles · 2019. 12. 30. · enols and enolates :...

Enols and Enolates :

Enols and Enolates (and Enamines) Carbon Nucleophiles • The hydrogen atoms on the carbon that is alpha- to (a-, i.e. adjacent to) the carbon of a C=O bond are the ones involved in keto/enol tautomerization and they are unusually acidic for an sp3 hybridized carbon. • These hydrogens are enolizable (although we didn't call them that when we first discussed enols).

• Alpha (a-), beta (b-), gamma (g-) terminology refers to relative position, alpha means next to, beta means one position further away etc., in this context, alpha (a-) means the carbon next to the C=O bond. 1 Bronsted Acidity of Enolizable Hydrogens Which bases can be used to make an enolate anion?

• In the equilibrium above, water is stronger acid, therefore the equilibrium lies mostly on ketone side, which means that the hydroxide anion can not be used to irreversibly make an enolate. • However, hydroxide could be used to produce a small amount of enolate that is in equilibrium with the neutral aldehyde/ketone. We need a Stronger Base: Recall some strong bases that we know:

• Lithium diisopropylamide is a very strong bulky base:

• In the acid/base reaction above, the ketone on the left is a much stronger acid than the amine that is formed on the right, therefore equilibrium lies essentially completely on enolate side. • LDA can be used to irreversibly deprotonate an aldehyde/ketone and form an enolate anion.

enol

R

O

H

α− β− γ-positions

H

enolizable

acid or base

catalyzedR

O

H HR

O

H

H

enolizable

CH3C

H3C

O

CH2C

H3C

O+ OH + H2O

weaker acidpKa ~19

stronger acidpka ~15weaker base stronger base

N

new stronger bulky base

O H

strong base

NH H

stronger base

O

strong bulky base

hydroxide

t-butoxide

amide

lithiumdiisopropylamide

LDA

Li

CH3C

H3C

O

stronger acidpKa ~19

CH2C

H3C

ON

HN

weaker acidpKa ~40

weaker base100% in enolate form

stronger baseLDA

+

essentiallyirreversible

+Li Li

XCH3C

H3C

ON+

Li

does NOT add to the C=O, strong base but weak nucleophile because it is BULKYLDA


• LDA will deprotonate an aldehyde or ketone at a carbon alpha-to the C=O, but it will not add to the C=O the same way that other strong nucleophiles do, e.g. the acetylide anion, Grignard reagent etc., because it is bulky and therefore sterically hindered. Summary: • Use –OH if you don't care about forming enolate reversibly. • Use LDA if you want to form the enolate irreversibly, using LDA, all of the carbonyl is consumed. 2 Relative Acidities of Carbonyl Compounds • Compare an aldehyde, a ketone, an ester, and a new structure, a b-dicarbonyl (beta-dicarbonyl):

• The acidity order is thus esters (least acidic) < ketones < aldehydes. • The b-dicarbonyl is by far the strongest acid because the enolate anion is doubly resonance stabilized. And so:

CH

O

CH3C

H

O

CH2

CC

C

O

O

O

ORR

CC

C

O

O

O

ORR

CC

C

O

O

O

ORR

CC

C

O

O

O

ORR

aldehyde, pKa ~17

ester, pKa ~25

β-dicarbonyl, pKa ~13

+ H3O

+ H3O

H2O

H2O

CH

O

CH2

CR

O

CH3C

R

O

CH2

ketone, pKa ~20

+ H3OH2O

CR

O

CH2

CRO

O

CH3C

RO

O

CH2

+ H3OH2O

CRO

O

CH2

weakly donating R group destabilizes enolate

strongly donating RO- group further destabilizes enolate

most resonance contributors

wins!!

S.D.

W.D.

H

H

H

H H

strongest Bronsted acid

decreasingacidity

CH3C

H3C

O

CH2C

H3C

O+ OEt + EtOH

weaker acidpKa ~19

stronger acidpka ~15

equilibrium on THIS side


• Adding a base such as an alkoxide (ethoxide is shown above) or hydroxide results in formation of only a very small amount of the enolate of an aldehyde or a ketone, most of the carbonyl is not deprotonated. But……

• Adding a base such as an alkoxide to a b-dicarbonyl results in essentially complete formation of the enolate anion. 3 Enolizable Hydrogens: What is this all about? • We understand that the majority of the reactions we encounter can be understood terms of Lewis acid/base and electrophile/nucleophile theory, the Lewis base/nucleophile provides the electrons to make a new bond • Carbon nucleophiles/Lewis bases are difficult to make, however, because carbon is not electronegative, it is not possible to have a simple carbon anion, the non-bonding electrons need to be stabilized somehow. Some stabilized Carbon Nucleophiles/Lewis Bases we have already seen:

• The electrons in the acetylide anion are stabilized by sp hybridization, those in the Grignard reagent are stabilized in a weak bond to magnesium. New carbon nucleophiles/Lewis bases on carbons based on enolizable Hydrogen atoms: • Enols and enamines (see previously) and the enolate anion

• The electrons on the nucleophilic carbon in the enolate anion are stabilized by resonance. • The nucleophilic carbon in the enamine and enol has a partial negative charge by resonance. • Enolate anions, enamines and enols are Lewis bases with nucleophile carbon atoms in the position next to (alpha to) the carbonyl carbon of an aldehyde or ketone. • Enolates are formed by deprotonation of an enolizable hydrogen using a Bronsted base. • Enamines are formed in reaction between an aldehyde or ketone with a secondary amine (with acid catalysis) and deprotonation of an enolizable hydrogen (we just didn't call it that previously). • Enols are structural isomers of aldehydes/ketones where the position of one enolizable hydrogen is changed, we just didn't call it that previously. 4 Alkylation Reactions of Enols/Enolates 4.1 Alkylation of Aldehydes/Ketones via the Enolate Anion • Making C-C bonds is a very important reaction for enols, enolates and enamines. A Basic Principle Underlying Enol/Enolate and related reactions:

+ OEt

stronger acidpKa ~13

weaker acidpka ~15

CC

CO

OEt

O

EtOH H

+ EtOHCC

CO

OEt

O

EtO

H

equilibrium on THIS side

malonic ester

CH2R CR C CH2R MgBrδ

acetylide Grignardδ

nucleophilic nucleophilictoo reactive

need tostablize a

negative chargeon a carbon atom

CO

R CH2CO

R CH2

enolate anion

RCOH

CH2

enol

nucleophilicnucleophilic

RCOH

CH2 RCNR2

CH2

enamine

nucleophilic

RCNR2

CH2


• The enamine and enolate are reactive (nucleophilic) enough to do SN2 with an appropriate alkyl halide, and, these reactions make new C–C bonds! Alkylation via the Enolate Anion

• In principle this should work, however, can't use -OH to make the enolate because it also acts as a nucleophile, this problem can be fixed, see below. • Addition occurs mainly at the carbon of the enolate anion rather than the oxygen. • One way that we can understand this is that addition at oxygen forms a less stable enol ether (similar to keto/enol equilibrium, although it is actually more complicated than this). Recall:

How to Fix the Problem of the base competing with the Enolate in the SN2 reaction? • Use LDA!

X

X

COH

R CH2

CH3-Br

No reactionalkene not strong enough Lewis BaseC

R

R CH2

CH3-Br

No reactionenol not strong enough Lewis Base

LB

CO

R CH2

CH3-Br

CNR2

R CH2

CH3-Br CNR2

R CH2CH3

CO

R CH2CH3

iminium salt

LA

LBLA

LALB

LALB

increasingLewis base

Nucleophilicitystrength stronger D

strongest D

alkene

enol

enamine

enolateSN2

SN2

CH3-Br+ OH

CH3-Br

CH3OHUnwanted side reaction

CO

R CH3 CO

R CH2

CO

R CH2CH3

hardly formedCO

R CH2

CH3

SN2 SN2

SN2

CO

R CH2

CH3-Br

SN2

SN2major product

enol ether

new C-C bond!

CO

R CH2CH3CO

R CH2

CH3CO

R CH2CH3CO

R CHCH3

H

keto-isomer favored enol-isomer not favored ketone favored enol ether not favored

similarly

1 Equiv. LDACH3-BrO O

CH3

O O

SN2H

H

Li

ALL ketone consumed

irreversible


• One Equivalent of LDA completely deprotonates the carbonyl to form a lithium enolate (actually, the lithium ion may play a small role in controlling the reactions of the enolate). • Both the carbonyl and the LDA are completely consumed when one equivalent is used. • Alkyl halide should be 1° or allylic to ensure SN2. Examples

• Alpha-alkylation (a-alkylation) and formation of a new C-C bond is accomplished! • It is best if there is only one kind of enolizable hydrogen atom or mixtures of products may result:

4.2 Alkylation of Aldehydes/Ketones via Enamines • LDA works well if the rest of the molecule can withstand the presence of the extremely strong base. But in many cases we need a milder, more subtle approach, enamines allow such an approach. Alkylation via an Enamine: The Stork Reaction. • An enamine is a stronger nucleophile than an enol, but less nucleophilic than an enolate. • The Stork enamine alkylation reaction avoids LDA, which is a very strong base and also avoids enolate anion intermediates which are also strong Bronsted bases. As we will see, enolates can also react with the carbonyl structures they are formed from. The Stork enamine reaction sequence: It seems a bit complicated, but it is actually relatively straightforward and is very much usually preferred over the LDA method. 1. Convert carbonyl to enamine. 2. React the halide with the enamine (SN2). 3. Hydrolyze enamine to carbonyl (acid workup). Examples: H+ could be HCl, TFA, TsOH etc., any organic acid, but not H3O+.

O O

1. LDA

2. CH3CH2I

H

only 1 enolizable H (±)*

O

1. LDA

2. CH3CH2I

H3 enolizable H's (total) HH

but 2 KINDS of enolizable H, thus two possible kinds of product

O O

+(±)*

H+ cat.

CH3Br

PhCH2Br

H3O

H3O

O

N

pyrrolidine

N

NCH3

OCH3

NCH2Ph

OCH2Ph

H

iminium+

iminium

hydrolysis

(±)

(±)SN2

SN2


• Alkylation is accomplished after hydrolysis of the iminium salt that is formed in the SN2 reaction. • The iminium salt has a carbon with two bonds to heteroatoms (nitrogen in this case), and we have learned previously that molecules with this structural feature can be hydrolyzed with water under acidic conditions. • The hydrolysis mechanism is very similar to those that we have seen previously, in this case the iminium salt starts with a formal positive charge, and so protonation is not needed in the first step to get the reaction started, it is a strong enough LA/Electrophile to react directly with water as the weak LB/Nucleophile. The Iminium Hydrolysis mechanism

Example

5 Aldol and Claisen Reactions of Enols/Enolates 5.1 Aldol Condensation of Aldehydes/Ketones Definition: Condensation reaction liberates a small neutral molecule product, often water, that can be "condensed". Definition: Aldol reaction is another nucleophilic addition to an aldehyde of ketone where the nucleophile is an enol or enolate. A summary of the reaction:

• The product is a conjugated "enone", formed after elimination of water via the (E1 or E2 mechanism). • The water that is formed in the reaction is removed irreversibly by heating (it can subsequently be condensed).

NCH3

O H

H

NCH3

O

H

H

OHH

NCH3

OH

O H

H

H

NCH3

OHH

OCH3

OCH3

HO H

H

H3O+

OCH3

H

(±)

(±)

TsOH(cat.)

H3OH

O N

H

NH Br

H

N

H

O

CH3C CH3

O

CH3C CH2

OC

CH3

H3C

CH3C CH

OC

CH3

heat

H+ or -OHAldol ADDITION

productAldol CONDENSATION

product

H+ or -OH

catalyst

OH CH3

+ H2O


The Aldol Condensation Reaction Mechanism: Base catalyzed via the Enolate Anion 1st part: Formation of the addition product:

2nd part: Formation of the condensation product. • Formation of addition product is reversible, but, formation of condensation product is irreversible due to elimination of water. • The elimination can be made irreversible by physically removing the water by evaporation away from the reaction by heating. This water is often condensed (hence a condensation reaction), this condensation required to make overall reaction "go" irreversibly. • First, hydroxide removes another enolizable proton to make another enolate anion, then, hydroxide leaves. This not a conventional E1 or E2 elimination, it is called an E1cb mechanism, but you don't need to know that.

• Every time we see an oxygen anion as a leaving group the structure it leaves from is itself an anion (an enolate anion in this case), we need to get high energy electrons into the structure that "kicks out" the hydroxide (it must be an anion), and that means starting the reaction with high energy electrons, usually in the form of strong base. For example, when we reduced an ester with LiAlH4, the intermediate anion was able to have an oxygen anion as the leaving group, again we needed to start with a strong base.

• A conventional one-step E2 elimination does not occur because the hydroxide oxygen anion is too poor a leaving group.

CH3C CH3

O

H3C CH2

OC

H3C CH3

O

CH3C C

H2

OC

CH3

OC

H3C CH2

OC

CH3

OHOH

Addition Product

CH3CH3

base makes enolate

H OH

another molecule of

starting ketone

CH3C

HC

O

CCH2

OH

Addition Product

CH3C C

H

OC

CH3

CH3CH3

H OH

CH3C C

H

OC

CH3

OHCH3

+ H2O (g)

enolate ANION

waterformedin thisstep

heat

+ OH catalyst reformed

ORCO

HAl HHH

ORCO

H

HCO

addition elimination OR+

ANIONstrong baseoxygen anions ONLY eliminate when

we START with an anion reagent

CH3C

HC

OC

CH2

OH

CH3C C

H

O

CCH3

CH3CH3

H OH

E2

+ OHXto poor a leaving group


Aldol Condensation Mechanism: Acid catalyzed via the enol:

• The acid catalyzed reaction proceeds via the enol. • The water elimination reaction could proceed via either the E1 or the E2 mechanism (E1 shown here). Examples

• In base catalyzed reactions, the reaction proceeds via the enolate anion. • In acid catalyzed reactions, the reaction proceeds via the enol. 5.2 Rationalizing the Various Possible Products, How do Carbonyls Know What to do? • So far we have seen the following acid and base catalyzed reactions for carbonyl compounds"

CH3C C

H2

OC

CH3

OH

Addition Product

CH3C

HC

OC

CH3

OH2 CH3CH3

H

CH3C C

H

OC

CH3

Condensation Product

CH3C

H3CHC

OC

CH3

CH3

H

CH3C CH3

O

acid catalyzed - enol!

H–OTs

CH3C C

H2

OH

H

OTs

CH3C CH2

OH

CH3C CH3

O

CH3C C

H2

OHC

CH3

OCH3

H–OTs

CH3C C

H2

OC

CH3

OHCH3

H

OTs

H–OTs

TsO

Na OHHeat

HCl HCl /Heat

O O

OH

O

O

H

OH

OH

Na OH

addition product via enolate

condensation product

addition product via enol

condensation products

OH

OH+

OOH

O

OHHO

O OHH+ or –OH H+ or –OH

H+ or –OHH2O

hydrate

enolate

enol Aldol addition

O

Aldol condensation

–H2O

All accumulates here if the water is removed

heat

–OH

not a "product"

not a "product" not a

"product"

not USUALLY a "product"


How does the carbonyl know which one to do? Answer: It doesn't, so it does them all! Depending upon the amount of water, acid and bases catalyst, we would expect some unreacted carbonyl, some enolate, some enol, some hydrate and some Aldol addition product, all in equilibrium (all are formed reversibly). However, if the water is removed (for example by heating), then everything can eventually be "driven" towards the Aldol condensation product, which is formed irreversibly. • Note that the enolate can only be made in the presence of base, but both enol and enolate (and also hydrate) can form in the presence of acid and base, however, in the presence of base, the enol has to be formed VIA the enolate, in other words both enol and enolate are present in base, and the enolate is considerably more reactive than the enol, which is why the enolate is the active Lewis base/nucleophile under basic conditions, under acid conditions the reactive Lewis base/nucleophile is the enol. 5.3 Crossed Aldol Condensations • Aldol reactions between different carbonyl compounds have a potential major problem.....

• More than one possible product, both crossed Aldol and self-Aldol reactions can occur. • How to control these reactions and direct towards a controlled crossed Aldol reaction? Select Appropriate Conditions:

• We can't make an enolate of benzaldehyde, it can't do a self-Aldol and it can't do an Aldol with acetaldehyde • Benzaldehyde is in excess to ensure the enolate from acetaldehyde reacts with it, and to minimize the acetaldehyde self-Aldol reaction. Examples

• An abbreviated (incomplete) mechanism is shown above.

H+ or -OH

Self AldolCrossed Aldol

CH

OC

H3C H

O+

TWO carbonyls

CH

O

H

HCH

O

H

H3CH

+

heat

benzaldehyde

Excess

acetaldehyde

CH

O

CH3C H

O+

heat

Na+ -OH CH2C H

OC

H

O

OHC

H

O

heat

Na+ -OH CH

ONO enolizable

H atoms

Cinnamaldehyde

O

H

O

H

OH O

HOH OH, Heat

+

no enolizable H'sExcess

O

H

O O O O

HOO O

OH OH, Heat


• There are two kinds of enolizable hydrogens in the molecule above, but only one gives the favored six-membered ring. • Intramolecular reaction wins out over any intermolecular reaction.

• An abbreviated (incomplete) mechanism is shown above. 5.4 Claisen Condensations : Enolates of Esters • Revisit acidity of carbonyl compounds:

The Claisen Reaction: • Every time we have had a strong nucleophile reacting with an ester, for example a Grignard, hydroxide, hydride in the form of LiAlH4, etc. the reaction has always been addition/elimination. • In the Claisen reaction an enolate anion is the nucleophile, which is a strong nucleophile. • The Claisen reaction is of a strong nucleophile reaction with an ester, the mechanism, therefore, is addition/elimination.

• Again, the addition/elimination mechanism. • The pKa of R'OH is ~ 15, thus the R'O– must irreversibly deprotonate the b-dicarbonyl product that is formed initially, which has a pKa of ca. 11.

O

O

O

O

OH

O OOH OH, Heat

CC

C

O

O

O

ORR C

CC

O

O

O

ORR

ester, pKa ~25

β-dicarbonyl, pKa ~13most acidic

+ H3O

CR

O

CH3C

R

O

CH2ketone, pKa ~20

least acidic

+ H3OCR

O

CH2

CRO

O

CH3C

RO

O

CH2+ H3OC

RO

O

CH2

three resonance contributors

S.D.

W.D.

HH H

CC

C

O

O

O

ORR

H

CC

C

O

O

O

ORR

H

CH2RC

O

OR'

CHRCO

OR'

CH2RCO

OR'

CHC

OO

R'

CH2RC OO

R'

R

HC

CO

OR'

CH2RCO

R

CCO

OR'

CH2RCO

R

H3OC

CO

OR

CH2RCO

RH

R'OH/R'O

R'O

final product

base makes enolate β-dicarbonyl

this deprotonation can not be prevented

addition elimination

stable anion

pKa ~ 25pKa ~ 11

same

(acid workup)


• Therefore, H3O+ must be added as a last step to reprotonate the final product. This acid workup step only has to protonate the b-dicarbonyl enolate, and so simple addition of aqueous acid works. • However, this last deprotonation of the beta-dicarbonyl makes reaction overall irreversible. • The base used is the same alkoxide of the ester (R'O–) to ensure that trans-esterification does not occur. Example 1: A Self-Claisen with only 1 kind of enolizable proton:

Example 2: An intramolecular Claisen (called a Dieckmann condensation) with only 1 kind of enolizable proton:

Example 3: A crossed Claisen reaction with only one kind of enolizable proton:

Example 4: A crossed Claisen with two esters:

CH3CO

OEt

CH2CO

OEt

CH3CO

OEt

H2CCO

OEt

CH3

C OO

EtCH2

CO

OEt

CH3CO

CH

CO

OEt

CH3CO

1. EtOH/EtO

2. H3O (acid workup)

H3O

CH2

CO

OEt

CH3CO

EtO

EtO

same

unavoidabledeprotonation

requiredreptotonation

additionelimination

O

MeO

O OMe

O

MeO

O OMe O

MeOH3O

O12

3 4

56

7 7-membered ring(±)

12

3 45

67

NaOMe

MeOH

H3CH2CCO

OCH3 C

CO

OCH3

no enolizable hydrogens

Ph C

O

CH3

O

OCH3H

H3C

Ph OCO

CH3

+

ExcessPh O

CO

CH3

(H3O+)NaOCH3

CH3OH H(±)

OCH3C

O

H3CO+

OCH3C

O

OCH3

OCH3CO

+H3CO

C COCH3

OO

H3O

no enolizable H's

OCH3CO

H3CO

1. NaOCH3/CH3OH

2. H3O+ H3COC C

OCH3

OOExcess


Example 5: A crossed Claisen with a ketone:

• Both structures have enolizable protons, but the ketone is more acidic, thus the enolate of the ketone forms and reacts preferentially. 5.5 Aldol and Claisen Reactions Using Heuristics • Heuristics are skills students develop that allow them to solve the problems more quickly than by working them out from first principles, by skipping steps that become obvious. • Heuristics can't be taught, each student develops their own heuristics as a result of repeatedly solving a specific type of problem. • Nevertheless, we will now show you how some students will solve Aldol and Claisen problems heuristically. Please appreciate the context in which we tell you this,. We are not trying to give you "short-cuts" or "tricks", and we definitely don't want you to start learning Aldol or Claisen reactions in this way in the hope that you can save some time by skipping the details. You still need to understand exactly how and why these reactions work, you need to know the mechanisms also! However, it is also important to understand that there is nothing "wrong" with solving problems this way, all skilled students develop them naturally, here we are just "pointing the way". Example Aldol Problem: Aldol reaction of acetaldehyde is performed industrially on a scale of many thousands of tons annually to form crotonaldehyde, which is used in the synthesis of many useful chemicals, including sorbic acid, a food preservative. How do you draw the structure of the crotonaldehyde Aldol product without writing the entire mechanism? • First, no other organic structures are involved, the aldehyde must react with itself, so draw another molecule.

• Identify the two carbons involved in the bond-forming reaction, don't forget basic Lewis acid/base nucleophile/electrophile principles! The electrons come from the carbon with the enolizable hydrogens. These electrons maker the new bond to the C=O carbon. • Eventually a C=C bond forms between these two carbons so eliminate atoms as necessary. The product should be a conjugated enone.

CH3C

O

H3C+

OCH3C

O

Et

CH2CO

H3C OCH3CO

Et

CH2

CO

H3C

OCH3

C OEt

CH

CO

H3CCO

Et

CH2

CO

H3CCO

Et

OCH3 H3O

-H

ketone more acidic

1. NaOCH3/CH3OH

2. H3O+

ester less acidic

Excess

CH2

CO

H3CCO

Et

OCH3

CH3CO

HH C

OHNaOCH3

CH3OHC

CH3CO

H

crotonaldehydeacetaldehyde

nucleophilic carbon, where the electrons "come from"

this oxygenis eliminatednew C=C

bond formshere

CH3H

electrophilic carbon accepts the electrons to make the new bond

acetaldehyde 1

acetaldehyde 2


Example Claisen Problem: Claisen reaction of ethyl acetate forms ethyl acetoacetate, which is used in the synthesis of many useful chemicals, including dyes, perfumes and plastics. How do you draw the structure of the ethyl acetoacetate Claisen product without writing the entire mechanism? • First, no other organic structures are involved, the ethyl acetate must react with itself, so draw another one.

• Identify the two carbons involved in the bond-forming reaction. The electrons come from the carbon with the enolizable hydrogens. These electrons maker the new bond to the C=O carbon. • The mechanism is addition/elimination, therefore, eliminate the -OEt on the electrophilic carbon. • The product should be a b-dicarbonyl 5.6 Aldol/Claisens in Reverse Example 1:

• The bond that is made in the Aldol condensation is the C=C bond, disconnect this C=C bond (pull it apart) to determine the structures of the reacting fragments. Example 2:

• The bond that is made in the Claisen reaction is tone of the central C-C bonds between the two C=O. • Therefore, there are almost always two ways of generating a Claisen reaction product, sometimes one of these is preferred over the other. • Disconnect each of the two C-C bonds (pull them apart) to determine the two possible sets of structures of the reacting fragments. • Reaction B is slightly better because there is only one set of enolizable hydrogens in the reactants. • Reaction A is still OK, since the ketone "end" is much more acidic than the ester "end" and thus the desired reaction will probably occur to give the desired product.

CH3CO

EtOEtO CH2

O1. NaOEt/EtOH

2. H3O+

C

OEtCO

H3C

ethyl acetoacetateethyl acetate

nucleophilic carbon, where the electrons "come from"

this -OEtis eliminated

new C=Cbond forms

here

OH3C

electrophilic carbon accepts the electrons to make the new bond

ethyl acetate 1

ethyl acetate 2

β-dicarbonyl

O

Ph

H O

O Ph

Hcame from+

H+ or –OHenone - Aldolheat Excess

OO

Ph

OO

Ph

OCH3

O

H3COO

Ph

OR

+β-dicarbonyl - Claisen A

B

A

B

Excess

1. CH3O–/CH3OH2. H3O+

1. CH3O–/CH3OH2. H3O+

(±)


Example 3:

• There are almost always two ways of generating a Claisen reaction product, sometimes one of these is preferred over the other. In this case neither is preferred, either method could be used 5 Summary of Reactions of Enols/Enolates Do not start studying by trying to memorize the reactions here! Work as many problems as you can, with this list of reactions in front of you if necessary, so that you can get through as many problems as you can without getting stuck on the reagents/conditions, and so that you can learn and practice solving reaction problems.

H3CO

O O

Ph

β-dicarbonyl - Claisen

A BB

A H3CO

O

+ H3CO

O

Ph

H3CO

O

OCH3+

O

PhExcess

Excess

(±)

O1. LDA

2. Br

O

O / H+ (cat.)NH

N

2. H3O+

N OBr1.

O Oheat

TsOH or Na+–OH

PhO O

ORO

+Ph

O 1. Na+ –OR2. H3O+

enamine formation

Stork reaction

Aldol condensation, many variants

Y / N

Y / N seen previously

Y / N

Y / N

Y / N

Claisen condensation, many variants

enols and enolates (and enamines) carbon nucleophiles · 2019. 12. 30. · enols and enolates :...

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