Hóa học - Chapter 22: Condensations and alpha substitutions of carbonyl compounds

Chapter 22Copyright © 2010 Pearson Education, Inc.Organic Chemistry, 7th EditionL. G. Wade, Jr.Condensations and Alpha Substitutions of Carbonyl CompoundsChapter 22*Alpha SubstitutionAlpha 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*Condensation with an Aldehyde or KetoneThe enolate ion attacks the carbonyl group to form an alkoxide. Protonation of the alkoxide gives the addition product: a b-hydroxy carbonyl compound.Chapter 22*Condensation with EstersThe enolate adds to the ester to form a tetrahedral intermediate. Elimination of the leaving group (alkoxide) gives the substitution product (a b-carbonyl compound).Chapter 22*Keto–Enol TautomersTautomerization 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*Base–Catalyzed TautomerismIn the presence of strong bases, ketones and aldehydes act as weak proton acids. A proton on the a 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*Acid-Catalyzed TautomerismIn acid, a proton is moved from the a-carbon to oxygen by first protonating oxygen and then removing a proton from the carbon.Chapter 22*RacemizationFor 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*Acidity of  HydrogenspKa 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*Formation and Stability of Enolate IonsThe equilibrium mixture contains only a small fraction of the deprotonated, enolate form.Chapter 22*Energy Diagram of Enolate ReactionEven 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*Synthesis of Lithium Diisopropylamine (LDA)LDA is made by using an alkyllithium reagent to deprotonate diisopropylamine.Chapter 22*Enolate of CyclohexanoneWhen LDA reacts with a ketone, it abstracts the a-proton to form the lithium salt of the enolate.Chapter 22*The a Halogenation of KetonesWhen a ketone is treated with a halogen and a base, an a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*Base-Promoted Halogenation MechanismThe 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*Multiple HalogenationsThe -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.OHClCl2OH , H2O_OClClOClClClOClClClClChapter 22*Bromoform ReactionA 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*Mechanism of Haloform FormationThe 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*Positive Iodoform Test for AlcoholsThe 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*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 1SolutionChapter 22*Acid-Catalyzed α HalogenationKetones also undergo acid-catalyzed a 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*Mechanism of Acid-Catalyzed α HalogenationThe 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*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 2SolutionChapter 22*Hell–Volhard–Zelinsky (HVZ) ReactionThe HVZ reaction replaces a hydrogen atom with a bromine atom on the alpha-carbon of a carboxylic acid (a-bromoacid).The acid is treated with bromine and phosphorus tribromide, followed by hydrolysis.Chapter 22*Hell–Volhard–Zelinski Reaction: Step 1The 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*Hell–Volhard–Zelinski Reaction: Step 2The 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*Alkylation of Enolate IonsBecause the enolate has two nucleophilic sites (the oxygen and the a carbon), it can react at either of these sites. The reaction usually takes place primarily at the acarbon, forming a new C—C bond.Chapter 22*a Alkylation of Enolate IonsLDA 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*Enamine FormationKetones or aldehydes react with a secondary amine to form enamines.The enamine has a nucleophilic a-carbon, which can be used to attack electrophiles.Chapter 22*Mechanism of Enolate FormationAn enamine results from the reaction of a ketone or aldehyde with a secondary amine.Chapter 22*Electrostatic Potential Map of an EnamineThe electrostatic potential map (EPM) of a simple enamine shows a high negative electrostatic potential (red) near the a-carbon atom of the double bond. This is the nucleophilic carbon atom of the enamine.Chapter 22*Alkylation of an EnamineEnamines 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*Acylation of EnaminesThe enamine attacks the acyl halide, forming an acyl iminium salt.Hydrolysis of the iminium salt produces the b-diketone as the final product.Chapter 22*Aldol CondensationUnder 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 b-hydroxy carbonyl compound can be isolated. Heating will dehydrate the aldol product to the a-b unsaturated compound.Chapter 22*Base-Catalyzed Aldol Condensation: Step 1During Step 1, the base removes the a-proton, forming the enolate ion.The enolate ion has a nucleophilic a-carbon.Chapter 22*Base-Catalyzed Aldol Condensation: Step 2The enolate attacks the carbonyl carbon of a second molecule of carbonyl compound.Chapter 22*Base-Catalyzed Aldol Condensation: Step 3Protonation of the alkoxide gives the aldol product.Chapter 22*Dehydration of Aldol ProductsHeating a basic or acidic aldol dehydration of the alcohol functional group. The product is a a,b-unsaturated conjugated aldehyde or ketone. An Aldol condensation, followed by dehydration, forms a new carbon–carbon double bond.Chapter 22*Crossed Aldol CondensationsChapter 22*Successful Crossed Aldol CondensationsChapter 22*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 3SolutionChapter 22*Aldol CyclizationIntramolecular 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*Retrosynthesis of Aldol CondensationChapter 22*Claisen CondensationThe Claisen condensation results when an ester molecule undergoes nucleophilic acyl substitution by an enolate.Chapter 22*Dieckman CondensationChapter 22*Crossed ClaisenTwo 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*Crossed Claisen CondensationIn a crossed Claisen condensation, an ester without a hydrogens serves as the electrophilic component.Chapter 22*Crossed Claisen Condensation with Ketones and EstersCrossed 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*Crossed Claisen MechanismThe ketone enolate attacks the ester, which undergoes nucleophilic acyl substitution, and thereby, acylates the ketone.Chapter 22*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 4SolutionChapter 22*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*Show what ester would undergo Claisen condensation to give the following b-keto ester.First, break the structure apart at the a, b bond (a, b to the ester carbonyl). This is the bond formed in the Claisen condensation.Solved Problem 5SolutionChapter 22*Next, replace the a 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*Chapter 22*Malonic Ester SynthesisThe 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*Decarboxylation of the Alkylmalonic AcidDecarboxylation takes place through a cyclic transition state, initially giving an enol form that quickly tautomerizes to the product.Chapter 22*Example of the Malonic SynthesisChapter 22*Dialkylation of Malonic EsterChapter 22*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 6SolutionChapter 22*Acetoacetic Ester SynthesisThe acetoacetic ester synthesis is similar to the malonic ester synthesis, but the final products are ketones.Chapter 22*Alkylation of Acetoacetic EsterEthoxide ion completely deprotonates acetoacetic ester. The resulting enolate is alkylated by an unhindered alkyl halide or tosylate to give an alkylacetoacetic ester.Chapter 22*Hydrolysis of Alkylacetoacetic EsterAcidic hydrolysis of the alkylacetoacetic ester initially gives an alkylacetoacetic acid, which is a b-keto acid. The keto group in the b-position promotes decarboxylation to form a substituted version of acetone.Chapter 22*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 7SolutionChapter 22*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*Conjugate Additions: The Michael Reactiona,b-unsaturated carbonyl compounds have unusually electrophilic double bonds. The b-carbon is electrophilic because it shares the partial positive charge of the carbonyl carbon through resonance.Chapter 22*1,2-Addition and 1,4-AdditionWhen 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 AcceptorsChapter 22*Chapter 22*1,4-Addition of an Enolate to Methyl Vinyl Ketone (MVK)An enolate will do a 1,4-attack on the a,b-unsaturated ketone (MVK).Chapter 22*Show how the following diketone might be synthesized using a Michael addition.A Michael addition would have formed a new bond at the b carbon of the acceptor. Therefore,we break this molecule apart at the b,g bond.Solved Problem 8SolutionChapter 22*Robinson AnnulationWith 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*Robinson Mechanism