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Chapter 21: Problem 7

Give the structure of the mixed aldol condensation product of benzaldehyde with Sample Solution (a) The enolate of acetophenone reacts with benzaldehyde to yield the product of mixed addition. Dehydration of the intermediate occurs, giving the \(\alpha, \beta\) -unsaturated ketone. As actually carried out, the mixed aldol condensation product was isolated in \(85 \%\) yield on treating benzaldehyde with acetophenone in an aqueous solution of sodium hydroxide at \(15-30^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(E)-1,3-diphenyl-2-propen-1-one is formed.

Step by step solution

01

Understand the Reaction Components

The reaction is between benzaldehyde (C鈧咹鈧匔HO) and acetophenone (C鈧咹鈧匔OCH鈧�). Benzaldehyde serves as the electrophile while acetophenone generates an enolate ion in the presence of a base.
02

Formation of the Enolate Ion

Acetophenone, in the presence of the strong base sodium hydroxide (NaOH), forms an enolate ion. This is when one of the hydrogen atoms on the alpha carbon (the carbon next to the carbonyl) is removed, creating a resonance-stabilized enolate anion.
03

Nucleophilic Addition

The enolate ion acts as a nucleophile and adds to the carbonyl carbon of benzaldehyde. This forms a beta-hydroxy ketone intermediate, which has an alcohol functional group positioned beta to a carbonyl group.
04

Dehydration

The beta-hydroxy ketone undergoes dehydration, typically under the basic conditions of the reaction. This eliminates a molecule of water, and forms a double bond, yielding an alpha, beta-unsaturated ketone. This specific process involves the removal of a hydroxyl group adjacent to the carbonyl group and forms a double bond in its place.
05

Final Structure of the Product

The product of the aldol condensation reaction is an alpha, beta-unsaturated ketone, specifically (E)-1,3-diphenyl-2-propen-1-one. This structure features a double bond between the alpha and beta carbons adjacent to a phenyl group.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enolate Ion Formation
During a mixed aldol condensation, the formation of an enolate ion is a crucial step. When acetophenone is treated with sodium hydroxide, a strong base, it results in the removal of a hydrogen atom from the alpha carbon. The alpha carbon is the one directly adjacent to the carbonyl group.
This hydrogen removal leads to the formation of an enolate ion, which is a resonance-stabilized anion. Resonance stabilization makes this ion particularly reactive and a key initial player in the aldol condensation.
Enolate ions have a negative charge, usually delocalized between the oxygen and the alpha carbon, making them very effective nucleophiles for the next steps in the reaction.
Nucleophilic Addition
Once the enolate ion is formed, it acts as a nucleophile. Nucleophiles are species that donate an electron pair to an electrophile to form a chemical bond in reaction mechanisms.
In the context of our reaction, the enolate ion attacks the carbonyl carbon of benzaldehyde. Here, benzaldehyde acts as the electrophile, a key participant that accepts electrons from the nucleophile.
This addition of the nucleophile to the carbonyl carbon is what leads to the creation of the beta-hydroxy ketone intermediate. This intermediate is a fascinating structure and is characterized by having a hydroxyl group (\(-OH\)) located on the beta carbon, relative to the carbonyl group.
Dehydration Reaction
The dehydration reaction follows the formation of the beta-hydroxy ketone intermediate. Under the same basic conditions, this intermediate undergoes a dehydration reaction, which means it loses a water molecule.
Dehydration involves the removal of the hydroxyl group and a hydrogen from the beta and alpha positions, respectively, leading to the formation of a double bond.
Through this reaction, water is removed, and the intermediate is transformed into an alpha, beta-unsaturated system. This transformation underlines how conditions can change the structure from one with only single bonds to one that includes double bonds.
Alpha, Beta-Unsaturated Ketone
The end product of the aldol condensation process is an alpha, beta-unsaturated ketone. In our specific example, this product is (E)-1,3-diphenyl-2-propen-1-one.
This compound has a distinct structure where the double bond is positioned between the carbon atoms directly adjacent to the carbonyl group. Alpha refers to the first carbon next to the carbonyl, and beta refers to the second.
The presence of this double bond is a signature of unsaturation, and it offers unique reactivity and chemical properties. This unsaturation in the presence of phenyl rings gives the compound potential in further reactions, making it an interesting compound in organic synthesis.

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Most popular questions from this chapter

(a) Give the structure of the Claisen condensation product of ethyl phenylacetate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\right)\) (b) What ketone would you isolate after saponification and decarboxylation of this Claisen condensation product? (c) What ketone would you isolate after treatment of the Claisen condensation product of ethyl phenylacetate with sodium ethoxide and allyl bromide, followed by saponification and decarboxylation? (d) Give the structure of the mixed Claisen condensation product of ethyl phenylacetate and ethyl benzoate. (e) What ketone would you isolate after saponification and decarboxylation of the product in part (d)? (f) What ketone would you isolate after treatment of the product in part (d) with sodium ethoxide and allyl bromide, followed by saponification and decarboxylation?

Show how each of the following compounds could be prepared from 3-pentanone. In most cases more than one synthetic transformation will be necessary. (a) 2-Bromo-3-pentanone (b) 1-Penten-3-one (c) 1 -Penten- \(3-01\) (d) 3-Hexanone (e) 2-Methyl-1-phenyl-1-penten-3-one

The following questions concern ethyl (2-oxocyclohexane)carboxylate. (a) Write a chemical equation showing how you could prepare ethyl (2-oxocyclohexane)-carboxylate by a Dieckmann cyclization. (b) Write a chemical equation showing how you could prepare ethyl (2-oxocyclohexane)-carboxylate by acylation of a ketone. (c) Write structural formulas for the two most stable enol forms of ethyl (2-oxocyclohexane)carboxylate. (d) Write the three most stable resonance contributors to the most stable enolate derived from ethyl (2-oxocyclohexane)carboxylate. (e) Show how you could use ethyl (2-oxocyclohexane)carboxylate to prepare 2-methylcyclohexanone. (f) Give the structure of the product formed on treatment of ethyl (2-oxocyclohexane)-carboxylate with acrolein \(\left(\mathrm{H}_{2} \mathrm{C} \square \mathrm{CHCH} \square \mathrm{O}\right)\) in ethanol in the presence of sodium ethoxide

Write structural formulas for the enol isomers of each of the following. (a) 2,2 -Dimethyl-3-pentanone (b) Acetophenone (c) 2-Methylcyclohexanone (d) Methyl vinyl ketone Sample Solution (a) Only one of the a carbons of 2,2-dimethyl-3-pentanone has an attached hydrogen, so only one constitutional isomer is possible for the enol. \(E\) and \(Z\) stereoisomers are possible.

(a) On addition of one equivalent of methylmagnesium iodide to ethyl acetoacetate, the Grignard reagent is consumed, but the only organic product obtained after working up the reaction mixture is ethyl acetoacetate. Why? What happens to the Grignard reagent? (b) On repeating the reaction but using \(\mathrm{D}_{2} \mathrm{O}\) and \(\mathrm{DCl}\) to work up the reaction mixture, it is found that the recovered ethyl acetoacetate contains deuterium. Where is this deuterium located?
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