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Condensation Reactions I

Building Bridges to Knowledge

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Structures analogous to the example below have hydrogen atoms attached to carbon atoms, adjacent to carbonyl carbon atoms. These hydrogen atoms are acidic due to the resonance stabilization of the resulting carbanions when the hydrogen atoms are abstracted. Compounds containing an α-hydrogen atom adjacent to carbonyl groups are referred to as carbon acids.

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The hydrogen atom on the α carbon atom is acidic because the resulting anion, the conjugate base, is resonance stabilized. Therefore, aldehydes and ketones containing hydrogen atoms on the α carbon are weakly acidic. The acid-base equation for the abstraction of an acidic hydrogen atom is illustrated by the following equation.

As indicated earlier, the resulting carbanion is resonance stabilized. Resonance stabilization in carbon acids is illustrated by the following equation.

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enolate

Resonance through the formation of an enolate anion stabilizes the anion produced by the abstraction of a hydrogen atom attached to the α carbon atom of a carbonyl compound. The pKa of the carbonyl compound containing an alpha carbon atom with a hydrogen atom attached, reflects the weak acidity of carbonyl compound.

Table 15.1 is a list of pKa values of selected carbonyl compounds.

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Table 15.1 pKa values of carbonyl compounds

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As indicated in the above table, the pKa of a compound with hydrogen atoms attached to the α carbon atom is affected by the electronegativity of the substituent attached to the carbonyl group. The greater the electronegativity of the group or groups attached to the alpha carbon atom of a carbonyl compound, the more acidic the carbon acid.

As demonstrated by table 15.1, the pKas of compounds with hydrogen atoms attached to carbon atoms, flanked by two carbonyl groups, decrease. A decrease in pKa of a carbon acid leads to an increase in acidity of the carbon acid. Consequently, according to table 15.1, the acidity of carbon acids increase with added carbonyl groups. Resonance clearly rationalizes the increase in acidity for carbon acids sandwiched by two carbonyl groups.

The resulting carbanion is resonance stabilized:

Which of the following hydrogen atoms would be acidic, and give a rationale for your answer?

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The Aldol Condensation

When a hydrogen atom on an α carbon atom is abstracted, the resulting carbanion can react with a second carbonyl molecule. Such a reaction is called the Aldol Condensation. The following set of equations is an illustration of the Aldol Condensation reaction.

The series of elementary steps that rationalize the formation of the β-hydroxyaldehyde produced in the Aldol Condensation are described below.

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The first step in the mechanism is the abstraction of an acidic α hydrogen atom by a strong base to form a resonance stabilized carbanion via the bimolecular reaction between the designated aldehyde and a base.

The second step is the self condensation of the aldehyde via bimolecular nucleophilic attack of the enolate anion on the carbonyl of another aldehyde molecule. This results in the production of the salt of the β-hydroxyaldehyde molecule (with its carbon skeleton doubled)

This two-step mechanism creates the salt of the β-hydroxyaldehyde. This salt then undergoes hydrolysis to produce the β-

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hydroxyaldehyde.

The Aldol Condensation reaction has multiple applications in organic syntheses. For example, the Aldol Condensation reaction can be used in the synthesis for 3-hydroxy-2,4-diphenylbutanal.

3-hydroxy-2,4-diphenylbutanal

The precursor aldehyde for this reaction can be determined by mentally cleaving the molecule at the α-carbon atom of the β-hydroxyaldehyde.

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This would result in the decision to use phenylethanal for the aldehyde in the Aldol Condensation for the synthesis of 3-hydroxy-2,4-diphenylbutanal.

(1)

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(2)

Hydrolysis of the salt produces 3-hydroxy-2,4-diphenylbutanal.

In the above reaction schema, a compound with two chiral centers is generated. Consequently, four stereochemical products are theoretically possible in this reaction process.

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The four possible compounds are:

(1) The (2S,3S) stereoisomer

(2) The (2R,3S) stereoisomer

(3) The (2S,3R) stereoisomer

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(4) The (2R,3R) stereoisomer

The β-hydroxyaldehyde can undergo dehydration on heating to form an α, β-unsaturated aldehyde. The chiral centers disappear with the formation of 2, 4-diphenyl-2-butenal, an α,β-unsaturated aldehyde. At about 100oC, 3-hydroxy-2, 4-diphenylbutanal is converted into 2, 4-diphenyl-2-butenal.

3-hydroxy-2, 4-diphenylbutanal α, β-unsaturated aldehydes

β-hydroxyaldehyde 2, 4-diphenyl-2-butenal

β-Hydroxyaldehydes, formed in aldol condensations, can be

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reduced to a 1,3-diol. For example, 3-phenylpropanal can undergo an Aldol Condensation to form 2-benzyl-3-hydroxy-5-phenylpentanal. 2-Benzyl-3-hydroxy-5-phenylpentanal can be reduced to 2-benzyl-5-phenyl-1,3-pentandiol using sodium borohydride.

2-benzyl-3-hydroxy-5-phenylpentanal

2-benzyl-5-phenyl-1,3-pentandiol

Aldol type condensations are easily accomplished, and similar

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reactions in relatively weak bases can occur intramolecularly. For example, 1,6-Cyclodecanedione, a diketone, can undergo an internal condensation (i.e., an intramolecular condensation analogous to the Aldol condensation) in the presence of a base to form bicylo[5.3.0]dec-1 (7)-en--2-one.

1, 6-Cyclodecanedione

bicylo[5.3.0]decan-2-on-7-ol

bicyclo[5.3.0]dec-1(7)-en-2-one

Mixed Aldol Condensations are useful when one of the reactants can form an enolate or one of the reactants is more reactive toward nucleophilic addition to the carbonyl group than the other.

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Benzaldehyde or formaldehyde cannot form an enolate. A mixture of benzaldehyde and phenylethanal, an aldehyde with a hydrogen atom attached to an α carbon atom, would produce 2, 3-diphenyl-3-hydroxypropanal.

The nucleophilic carbanion and the carbonyl substrate can be used to determine the precursors for the synthesis of the β-hydroxyaldehyde.

The choice of reagents for the mixed aldol condensation in this case would be phenylethanal and benzaldehyde. Phenylethanal serves as the carbon acid for the abstraction of a hydrogen atom on the α carbon atom. Benzaldehyde is not a carbon acid; therefore, it is the substrate in the reaction.

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In the reaction between phenylethanal and benzaldehyde in base, benzaldehyde is more susceptible toward nucleophilic attack than the self - condensation of phenylethanal; therefore, the carbonyl of benzaldehyde is attacked preferentially over the carbonyl of phenylethanal.

(1)

(2)

(3)

Since the OH group is adjacent to two carbon atoms with phenyl

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groups attached, the β-hydroxyaldehyde spontaneously dehydrates to the α,β-unsaturated aldehyde, (Z)-2, 3-diphenyl-2-propenal, the major product.

(Z)-2, 3-Diphenyl-2-propenal

(Z)-2,3-Diphenyl-2-propenal is resonance stabilized.

The charge separation gives 2,3-diphenyl-2-propenal a higher dipole moment than comparable molecular mass aldehydes or ketones. This observation holds true for α,β-unsaturated aldehydes and ketones in general.

Nucleophilic Attack on α, β-Unsaturated Aldehydes and Ketones

Since the double bonds of α, β-unsaturated aldehydes and ketones

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are in resonance with the carbonyl groups, then addition reactions to the double bond of α, β-unsaturated aldehydes or ketones are less reactive than addition reactions to alkenes. However, the β carbon atoms of α,β-unsaturated aldehydes or ketones are susceptible to nucleophilic attacks by nucleophilic reagents.

Grignard reagents, organolithium compounds, and lithium aluminum hydride attack the carbonyl of the α, β-unsaturated aldehyde or ketone.

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The resulting α, β –alcohol (the 1,2-addition product) will dehydrate to form a conjugated diene when heated:

an α,β –alcohol

This is an example of a strong base attacking the carbonyl carbon of an α, β-unsaturated aldehydes to give the 1,2-addition product.

Weaker bases like -CN will react with α, β-unsaturated aldehydes and ketones via a 1,4 addition process. The following reaction is an example of a weak nucleophile (weak base) attacking the β carbon atom of the α, β-unsaturated aldehydes and ketones.

(1)

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(2)

(3)

Following is a general reaction of the 1,2-addition product and the 1, 4-addition product. M+ Y- represents a strong base for the formation of the 1,2-addition product. M+ Y- represents a weak base for the formation of the 1,4-addition product.

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There are several characteristics of 1,2-addition products to the α,β-unsaturated aldehydes and ketones and the 1,4-addition product to α, β-unsaturated aldehydes and ketones. These characteristics are

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listed in Table 15.2.

Table 15.2 makes a comparison between the 1,2 addition product to the α, β-unsaturated aldehyde or ketone and the 1,4-addition product to the α, β-unsaturated aldehyde or ketone.

1,2-Addition Product 1,4-Addition Product

kinetically controlled reaction thermodynamically controlled reaction faster reaction slower reaction less stable product more stable product formed with strong bases, e.g., NaNH2 formed with weak bases, eg., NaCN products contains a double bond product doesn’t contain a double bond Table 15.2

The formation of the 1,4-addition product undergoes hydrolysis to form the enol compound. The enol compound undergoes tautomerization to form the more stable keto compound.

Following is a general reaction for enol-keto tautomerism.

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Carbon acids can be converted to the enol structure by a base catalyzed reaction or an acid catalyzed reaction.

The mechanism for base catalyzed enolization of a carbon acid includes the following steps.

Slow Step

Fast Step

The mechanism for acid catalyzed enolization of a carbon acid includes the following steps.

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Fast Step

Slow Step

Once the enol structure is formed, it is converted to the more stable keto structure. For example, the equilibrium between the keto structure of an acetaldehyde and the enol structure is about 3 x 10-7. In the event that the enol structure of the molecule is produced in a reaction process, it will usually undergo tautomerism to form the more stable keto structure of the molecule unless the arrangement of the molecule stabilizes the enol structure. For example, the concentration of the enol structure of β-diketones is about equal to the concentration of the keto structure, because intramolecular association (hydrogen bonding) stabilizes the enol structure.

.

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The enol structure of 1-phenyl-1,3-butanedione would be more stable than the keto structure.

The hydrogen atoms on the alpha carbon can be replaced on aldehydes and ketone to produce alpha halogenated products. The reaction is regiospecific, i.e., it replaces only a hydrogen atom on the α-carbon atom; no other hydrogen atoms are replaced. Such reactions can be represented by the following transformation.

The mechanism is a four-step process beginning with a fast equilibrium bimolecular step.

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(1) Fast equilibrium

intermediate1

(2) The next step is a slow equilibrium step involving the abstraction of a hydrogen atom on the α carbon atom.

Slow equilibrium

Intermediate1 enol

(3) Followed by a fast step leading to the protonated α-bromoketone.

Fast Step

Enol intermediate2

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Finally, the proton on the protonated α bromoketone is transferred to water to give the desired product.

(4) Fast Step

intermediate2 product

Recall that mechanisms are based on the kinetics of reactions. According to the above mechanism, the rate of the reaction depends on the slow step of the proposed mechanism. Step 2 is the slow step of the mechanism; therefore, it is the rate-controlling step. In this step, the rate of the reaction depends on intermediate1 and water. The rate expression for step 2 is represented by

(1)

The concentration of intermediate 1 cannot be measured, but the following equation from step 1 can be used to determine the concentration of the ketone:

(2)

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Substituting equation (2) into equation (1) gives equation (3)

(3)

Where intermediate1 represents

And the ketone is

The rate of the halogenation reaction is second order, first order in the ketone and first order in the acid. The overall order of the

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reaction is second order, and the rate of the reaction is independent of the concentration of the halogen.

Reaction of Aldehydes with Ketones in Base

Ketones don’t readily undergo self-condensation. For example, a mixture of acetone in formaldehyde would form 4-hydroxy-2-butanone instead of undergoing self condensation to produce 4-hydroxy-4-methyl-2-pentanone. The following is the chemical equation indicating that a mixture of acetone and formaldehyde would prefer to form 4-hydroxy-2-butanone rather than 4-hydroxy-4-methyl-2-pentanone.

4-hydroxy-2-butanone

Enolate Anions as Nucleophiles

Enolate anions are nucleophiles that readily undergo SN2 type reactions. For example, an enolate anion can attack a primary alkyl halide leading to the displacement of the halogen by way of backside attack.

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The reaction is faster with β-diketones.

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The Haloform Reactions/The Iodoform Reaction

Ketones with methyl groups attached will react with halogens in the presence of sodium hydroxide to form iodoform, a yellow precipitate. Also, this reaction works for bromine to produce bromoform and with chlorine to produce chloroform. Iodine is use, because the formation of a yellow precipitate is a visual indication of the presence

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of methyl ketones. The iodoform reaction was discussed in the paper titled “Aldehydes and Ketones, Building Bridges to Knowlege.” In review, the following general chemical equation represents the iodoform reaction.

The series of elementary steps that explain the formation of haloform are described below.

(1) Fast Step

(2) Fast Step

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(3) Fast Step

(4) Fast Step

(5) Fast Step

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(6) Fast Step

(7) Slow Step, the rate-determining step

(8) Fast Step

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(9) Fast Step

Reaction of p-Methoxyphenyl Methyl Ketone with Deuterium Oxide

Following is the 11HNMR spectrum for p-methoxyphenyl methyl

ketone.

p-methoxyphenyl methyl ketone

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When p-methoxyphenyl methyl ketone is treated with D2O in dilute base under 100oC, the following 1HNMR signal is obtained.

These results can be explained by suggesting that the following structure would be formed when p-methoxyphenyl methyl ketone is treated with deuterium oxide.

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The hydrogen atoms on the methyl group attached to the carbonyl group are replaced with deuterium atoms via the following series of elementary steps.

(1)

(2)

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(3)

(4)

(5)

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(6)

The sum of the equations would be:

The Michael Condensation and the Robinson Annulation Reaction

Certain nucleophilic reagents can add to the β carbon of α,β-unsaturated ketones. Such reactions are characteristic of β-diketoenolates, and the Michael Condensation is an example of the carbanion of β-diketoenolates reacting with α,β-unsaturated ketones.

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(1)

(2)

(3)

The product of the Michael Condensation can undergo an intramolecular Aldol condensation to form a fused ring cyclohexenone. This reaction is the Robinson Annulation (building a ring onto another molecule) reaction.

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(1)

(2)

(3)

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(4)

The reaction is carried out in one synthetic operation without isolating the intermediate.

A stepwise analysis of the Michael condensation followed by an intramolecular Aldol Condensation (the Robinson Annulation Reaction) is provided in the following illustration of the synthesis of 3-methyl-2, 6-diphenyl-2-cyclohexenone from dibenzyl ketone and methyl vinyl ketone in base. Again, the reaction is carried out in one synthetic operation without isolating any intermediates.

3-methyl-2,6-diphenyl-2-cyclohexenone

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The first phase of the synthesis is the Michael Condensation reaction.

(1)

(2)

(3)

The second phase of the reaction is the Robinson Annulation

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reaction.

(1)

(2)

(3)

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(4)

The Addition of the Gilman Reagent to α,β-Unsaturated Carbonyl Compounds

Lithium dialkylcuprate reagents (the Gilman reagent) can be added to α,β-unsaturated carbonyl compounds. Following is an example of adding the Gilman reagent to 4-ethyl-3-hexen-2-one, an α,β-unsaturated carbonyl compound.

The mechanism of the reaction has not been definitively established, but similar to organometallic reactions, the mechanism could involve the transfer of electrons. If so, the following could be the series of

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elementary steps that may rationalize the formation of 4,4-diethyl-2-heptanone from reacting the Gilman reagent, lithium dipropylcuprate, with 4-ethyl-3-hexen-2-one, the α,β-unsaturated ketone.

(1)

(2)

(3)

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(4)

The second process in the reaction, i.e., hydrolysis, would involve reacting the base-catalyzed tautomerization of the enolate with water to give the final product, 4,4-diethyl-2-heptanone.

The sum of the elementary steps 1 through 4 is

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.

Following is the overall equation for this reaction.

The addition of the Gilman reagent to α,β-Unsaturated Carbonyl Compounds could be used for the alkylation of α,β-unsaturated ketones.

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Problems

Condensation Reactions I

1. Suggest products for (a) - (f), and suggest IUPAC names for the products of (a) - (e).

(a)

(b)

(c)

(d)

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(e)

(f)

2. Suggest a synthesis for the following compound from the indicated starting material and any necessary organic or inorganic materials.

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3. Suggest a synthesis for the following compound from the indicated starting materials and any necessary inorganic materials.

4. The following sequence of reactions leads to the formation of compound I.

Compound I

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(a) Suggest structures for compounds A, C, and D.

(b) What is reagent B?

(c) Suggest a mechanism for the formation of compound A.

5. Suggest a product for the following reaction, and

Suggest a mechanism for the reaction.

6. Suggest a series of elementary steps to rationalize the formation of the following transformation.

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7. Suggest a mechanism for the following transformation.

8. Suggest a mechanism for the following reaction.

9. Using the cyclohexane-like chair conformation, suggest a series of elementary steps to rationalize the following transformation.

10. Suggest a synthesis for the following from the indicated starting material and any other necessary organic or inorganic compounds.

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11. Suggest structures for A, B, C6H12O2

12. Design a synthesis for Compound A from any necessary inorganic and organic compounds.

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compound A

13.

Yotes and Anderson reported the following transformation in the Journal of the American Chemical Society.

Suggest a series of elementary steps that could account for this transformation. In your suggested mechanism, use arrows to show how electrons flow.