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Chapter 22
Copyright © 2010 Pearson Education, Inc.
Organic Chemistry, 7th EditionL. G. Wade, Jr.
Condensations and Alpha Substitutions of Carbonyl Compounds
Chapter 22 2
Alpha Substitution
Alpha substitution is the substitution of one of the hydrogens attached to the alpha-carbon for an electrophile.
The reaction occurs through an enolate ion intermediate.
Chapter 22 3
Condensation with an Aldehyde or Ketone
The enolate ion attacks the carbonyl group to form an alkoxide.
Protonation of the alkoxide gives the addition product: a -hydroxy carbonyl compound.
Chapter 22 4
Condensation with Esters
The enolate adds to the ester to form a tetrahedral intermediate.
Elimination of the leaving group (alkoxide) gives the substitution product (a -carbonyl compound).
Chapter 22 5
Keto–Enol TautomersO
H
H
OH
H
keto form(99.99%)
enol form(0.01%)
Tautomerization is an interconversion of isomers that occur through the migration of a proton and the movement of a double bond.
Tautomers are not resonance form.
Chapter 22 6
Base–Catalyzed Tautomerism
In the presence of strong bases, ketones and aldehydes act as weak proton acids.
A proton on the carbon is abstracted to form a resonance-stabilized enolate ion with the negative charge spread over a carbon atom and an oxygen atom.
The equilibrium favors the keto form over the enolate ion.
Chapter 22 7
Acid-Catalyzed Tautomerism
In acid, a proton is moved from the -carbon to oxygen by first protonating oxygen and then removing a proton from the carbon.
Chapter 22 8
Racemization
For aldehydes and ketones, the keto form is greatly favored at equilibrium.
If a chiral carbon has an enolizable hydrogen atom, a trace of acid or base allows that carbon to invert its configuration, with the enol serving as the intermediate. This is called racemization.
Chapter 22 9
Acidity of Hydrogens
pKa for H of aldehyde or ketone ~20.
Much more acidic than alkane or alkene (pKa > 40) or alkyne (pKa = 25).
Less acidic than water (pKa = 15.7) or alcohol (pKa = 16–19).
Only a small amount of enolate ion is present at equilibrium.
Chapter 22 10
Formation and Stability of Enolate Ions
The equilibrium mixture contains only a small fraction of the deprotonated, enolate form.
Chapter 22 11
Energy Diagram of Enolate Reaction
Even though the keto–enol tautomerism equilibrium favors the keto form, addition of an electrophile shifts the equilibrium toward the formation of more enol.
Chapter 22 12
Synthesis of Lithium Diisopropylamine (LDA)
LDA is made by using an alkyllithium reagent to deprotonate diisopropylamine.
Chapter 22 13
Enolate of Cyclohexanone
When LDA reacts with a ketone, it abstracts the -proton to form the lithium salt of the enolate.
Chapter 22 14
The Halogenation of Ketones
When a ketone is treated with a halogen and a base, an halogenation reaction occurs.
The reaction is called base-promoted, rather than base-catalyzed, because a full equivalent of the base is consumed in the reaction.
Chapter 22 15
Base-Promoted Halogenation Mechanism
The base-promoted halogenation takes place by a nucleophilic attack of an enolate ion on the electrophilic halogen molecule.
The products are the halogenated ketone and a halide ion.
Chapter 22 16
Multiple Halogenations
The -haloketone produced is more reactive than ketone because the enolate ion is stabilized by the electron-withdrawing halogen.
The second halogenation occurs faster than the first. Because of the tendency for multiple halogenations
this base-promoted halogenation is not widely used to prepare monohalogenated ketones.
O
H
ClCl2
OH , H2O_
O
Cl
Cl
O
Cl
ClCl
O
Cl
ClClCl
Chapter 22 17
Bromoform Reaction
A methyl ketone reacts with a halogen under strongly basic conditions to give a carboxylate ion and a molecule of haloform.
The trihalomethyl intermediate is not isolated.
Chapter 22 18
Mechanism of Haloform Formation
The trihalomethyl ketone reacts with hydroxide ion to give a carboxylic acid.
A fast proton exchange gives a carboxylate ion and a haloform.
When Cl2 is used, chloroform is formed; Br2 forms bromoform ; and I2 forms iodoform.
Chapter 22 19
Positive Iodoform Test for Alcohols
The iodine oxidizes the alcohol to a methyl ketone and it will give a positive iodoform test.
Iodoform (CHI3) is a yellow solid that will precipitate out of solution.
Chapter 22 20
Propose a mechanism for the reaction of 3-pentanone with sodium hydroxide and bromine to give 2-bromo-3-pentanone.
In the presence of sodium hydroxide, a small amount of 3-pentanone is present as its enolate.
The enolate reacts with bromine to give the observed product.
Solved Problem 1
Solution
Chapter 22 21
Acid-Catalyzed α Halogenation
Ketones also undergo acid-catalyzed halogenation. Acidic halogenation may replace one or more alpha
hydrogens depending on how much halogen is used. Acetic acid serves as both the solvent and the acid
catalyst.
Chapter 22 22
Mechanism of Acid-Catalyzed α Halogenation
The mechanism of acid-catalyzed halogenation involves attack of the enol form of the ketone on the electrophile halogen molecule.
Loss of a proton gives the haloketone and the hydrogen halide.
Chapter 22 23
Propose a mechanism for the acid-catalyzed conversion of cyclohexanone to 2-chlorocyclohexanone.
Under acid catalysis, the ketone is in equilibrium with its enol form.
The enol acts as a weak nucleophile, attacking chlorine to give a resonance-stabilized intermediate. Loss of a proton gives the product.
Solved Problem 2
Solution
Chapter 22 24
Hell–Volhard–Zelinsky (HVZ) Reaction
The HVZ reaction replaces a hydrogen atom with a bromine atom on the alpha-carbon of a carboxylic acid (-bromoacid).
The acid is treated with bromine and phosphorus tribromide, followed by hydrolysis.
Chapter 22 25
Hell–Volhard–Zelinski Reaction: Step 1
The enol form of the acyl bromide serves as a nucleophilic intermediate.
The first step is the formation of acyl bromide, which enolizes more easily than does the acid.
Chapter 22 26
Hell–Volhard–Zelinski Reaction: Step 2
The enol is nucleophilic, so it attacks bromine to give the alpha-brominated acyl bromide.
In the last step of the reaction, the acyl bromide is hydrolyzed by water to the carboxylic acid.
Chapter 22 27
Alkylation of Enolate Ions
Because the enolate has two nucleophilic sites (the oxygen and the carbon), it can react at either of these sites.
The reaction usually takes place primarily at the carbon, forming a new C—C bond.
Chapter 22 28
Alkylation of Enolate Ions
LDA forms the enolate. The enolate acts as the nucleophile and attacks the
partially positive carbon of the alkyl halide, displacing the halide and forming a C—C bond.
Chapter 22 29
Enamine Formation
Ketones or aldehydes react with a secondary amine to form enamines.
The enamine has a nucleophilic -carbon, which can be used to attack electrophiles.
Chapter 22 30
Mechanism of Enolate Formation
An enamine results from the reaction of a ketone or aldehyde with a secondary amine.
Chapter 22 31
Electrostatic Potential Map of an Enamine
The electrostatic potential map (EPM) of a simple enamine shows a high negative electrostatic potential (red) near the -carbon atom of the double bond.
This is the nucleophilic carbon atom of the enamine.
Chapter 22 32
Alkylation of an Enamine
Enamines displace halides from reactive alkyl halides, giving alkylated iminium salts.
The alkylated iminium salt can be hydrolyzed to the ketone under acidic conditions.
Chapter 22 33
Acylation of Enamines
The enamine attacks the acyl halide, forming an acyl iminium salt.
Hydrolysis of the iminium salt produces the -diketone as the final product.
Chapter 22 34
Aldol Condensation
Under basic conditions, the aldol condensation involves the nucleophilic addition of an enolate ion to another carbonyl group.
When the reaction is carried out at low temperatures, the -hydroxy carbonyl compound can be isolated.
Heating will dehydrate the aldol product to the unsaturated compound.
Chapter 22 35
Base-Catalyzed Aldol Condensation: Step 1
During Step 1, the base removes the -proton, forming the enolate ion.
The enolate ion has a nucleophilic -carbon.
Chapter 22 36
Base-Catalyzed Aldol Condensation: Step 2
The enolate attacks the carbonyl carbon of a second molecule of carbonyl compound.
Chapter 22 37
Base-Catalyzed Aldol Condensation: Step 3
Protonation of the alkoxide gives the aldol product.
Chapter 22 38
Dehydration of Aldol Products
Heating a basic or acidic aldol dehydration of the alcohol functional group.
The product is a ,-unsaturated conjugated aldehyde or ketone.
An Aldol condensation, followed by dehydration, forms a new carbon–carbon double bond.
Chapter 22 39
Crossed Aldol Condensations
Chapter 22 40
Successful Crossed Aldol Condensations
Chapter 22 41
Propose a mechanism for the base-catalyzed aldol condensation of acetone (Figure 22-2).
The first step is formation of the enolate to serve as a nucleophile.
The second step is a nucleophilic attack by the enolate on another molecule of acetone. Protonation gives the aldol product.
Solved Problem 3
Solution
Chapter 22 42
Aldol Cyclization
Intramolecular aldol reactions of diketones are often used for making five- and six-membered rings.
Rings smaller or larger than five or six members are not favored due to ring strain or entropy.
Chapter 22 43
Retrosynthesis of Aldol Condensation
Chapter 22 44
Claisen Condensation
The Claisen condensation results when an ester molecule undergoes nucleophilic acyl substitution by an enolate.
Chapter 22 45
Dieckman Condensation
Chapter 22 46
Crossed Claisen
Two different esters can be used, but one ester should have no hydrogens.
Useful esters are benzoates, formates, carbonates, and oxalates.
Ketones (pKa = 20) may also react with an ester to form a -diketone.
Chapter 22 47
Crossed Claisen Condensation
In a crossed Claisen condensation, an ester without hydrogens serves as the electrophilic component.
Chapter 22 48
Crossed Claisen Condensation with Ketones and Esters
Crossed Claisen condensation between ketones and esters are also possible.
Ketones are more acidic than esters, and the ketone component is more likely to deprotonate and serve as the enolate component in the condensation.
Chapter 22 49
Crossed Claisen Mechanism
The ketone enolate attacks the ester, which undergoes nucleophilic acyl substitution, and thereby, acylates the ketone.
Chapter 22 50
Propose a mechanism for the self-condensation of ethyl acetate to give ethyl acetoacetate.
The first step is formation of the ester enolate. The equilibrium for this step lies far to theleft; ethoxide deprotonates only a small fraction of the ester.
The enolate ion attacks another molecule of the ester; expulsion of ethoxide ion gives ethyl acetoacetate.
Solved Problem 4
Solution
Chapter 22 51
In the presence of ethoxide ion, ethyl acetoacetate is deprotonated to give its enolate. This exothermic deprotonation helps to drive the reaction to completion.
When the reaction is complete, the enolate ion is reprotonated to give ethyl acetoacetate.
Solved Problem 4 (Continued)Solution (Continued)
Chapter 22 52
Show what ester would undergo Claisen condensation to give the following -keto ester.
First, break the structure apart at the bond ( to the ester carbonyl). This is the bond formed in the Claisen condensation.
Solved Problem 5
Solution
Chapter 22 53
Next, replace the proton that was lost, and replace the alkoxy group that was lost from the carbonyl. Two molecules of methyl 3-phenylpropionate result.
Now draw out the reaction. Sodium methoxide is used as the base because the reactants are methyl esters.
Solved Problem 5 (Continued)Solution (Continued)
Chapter 22 54
Chapter 22 55
Malonic Ester Synthesis
The malonic ester synthesis makes substituted derivatives of acetic acids.
Malonic ester is alkylated or acylated on the carbon that is alpha to both carbonyl groups, and the resulting derivative is hydrolyzed and allowed to decarboxylate.
Chapter 22 56
Decarboxylation of the Alkylmalonic Acid
Decarboxylation takes place through a cyclic transition state, initially giving an enol form that quickly tautomerizes to the product.
Chapter 22 57
Example of the Malonic Synthesis
Chapter 22 58
Dialkylation of Malonic Ester
Chapter 22 59
Show how the malonic ester synthesis is used to prepare 2-benzylbutanoic acid.
2-Benzylbutanoic acid is a substituted acetic acid having the substituents Ph–CH2– and CH3CH2–.
Adding these substituents to the enolate of malonic ester eventually gives the correct product.
Solved Problem 6Solution
Chapter 22 60
Acetoacetic Ester Synthesis
The acetoacetic ester synthesis is similar to the malonic ester synthesis, but the final products are ketones.
Chapter 22 61
Alkylation of Acetoacetic Ester
Ethoxide ion completely deprotonates acetoacetic ester.
The resulting enolate is alkylated by an unhindered alkyl halide or tosylate to give an alkylacetoacetic ester.
Chapter 22 62
Hydrolysis of Alkylacetoacetic Ester
Acidic hydrolysis of the alkylacetoacetic ester initially gives an alkylacetoacetic acid, which is a -keto acid.
The keto group in the -position promotes decarboxylation to form a substituted version of acetone.
Chapter 22 63
Show how the acetoacetic ester synthesis is used to make 3-propylhex-5-en-2-one.
The target compound is acetone with an n-propyl group and an allyl group as substituents:
Solved Problem 7
Solution
Chapter 22 64
Hydrolysis proceeds with decarboxylation to give the disubstituted acetone product.
With an n-propyl halide and an allyl halide as the alkylating agents, the acetoacetic ester synthesis should produce 3-propyl-5-hexen-2-one. Two alkylation steps give the required substitution:
Solved Problem 7 (Continued)Solution (Continued)
Chapter 22 65
Conjugate Additions: The Michael Reaction
,-unsaturated carbonyl compounds have unusually electrophilic double bonds.
The -carbon is electrophilic because it shares the partial positive charge of the carbonyl carbon through resonance.
Chapter 22 66
1,2-Addition and 1,4-Addition
When attack occurs at the carbonyl group, protonation of the oxygen leads to a 1,2-addition.
When attack occurs at the β-position, the oxygen atom is the fourth atom counting from the nucleophile, and the addition is called a 1,4-addition.
Donors and Acceptors
Chapter 22 67
Chapter 22 68
1,4-Addition of an Enolate to Methyl Vinyl Ketone (MVK)
An enolate will do a 1,4-attack on the -unsaturated ketone (MVK).
Chapter 22 69
Show how the following diketone might be synthesized using a Michael addition.
A Michael addition would have formed a new bond at the carbon of the acceptor. Therefore,we break this molecule apart at the bond.
Solved Problem 8
Solution
Chapter 22 70
Robinson Annulation
With enough base, the product of the Michael reaction undergoes a spontaneous intramolecular aldol condensation, usually with dehydration, to give a six-membered ring—a conjugated cyclohexenone.
Chapter 22 71
Robinson Mechanism