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Aldol additions also occur in the presence of acidic catalysts. For example, 2-propanone with dry hydrogen chloride slowly yields \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOCH}_{3}\) (mesityl oxide) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOCH}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (phorone). Write mechanisms for the formation of these products, giving particular attention to the way in which the new carbon-carbon bonds are formed.

Short Answer

Expert verified
The mechanism involves protonation, enol formation, nucleophilic attack, and dehydration leading to mesityl oxide and phorone.

Step by step solution

01

Protonation of the Carbonyl Group

The mechanism begins with the protonation of the carbonyl oxygen of 2-propanone by the acidic hydrogen chloride. This makes the carbonyl carbon more electrophilic, facilitating nucleophilic attack.
02

Formation of the Enol

The alpha hydrogen of the protonated 2-propanone is removed by a chloride ion, leading to the formation of an enol. This structure allows for isomerization between keto and enol forms.
03

Nucleophilic Attack on Protonated Carbonyl

Another molecule of 2-propanone undergoes protonation and the enol form acts as a nucleophile, attacking the electrophilic carbon of the protonated carbonyl group. This forms a new C-C bond.
04

Formation of beta-hydroxy ketone intermediate

The intermediate formed is a beta-hydroxy ketone. In acidic conditions, dehydration occurs, leading to the formation of a double bond and resulting in mesityl oxide.
05

Formation of Phorone

For phorone formation, mesityl oxide undergoes further aldol condensation with another molecule of 2-propanone. Similar steps are followed—protonation, enol formation, nucleophilic attack, and additional dehydration—producing phorone.

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Key Concepts

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

Acidic Catalysis
In the Aldol Addition Reaction, acidic catalysis plays an essential role. When an acid, like hydrogen chloride, is present, it protonates the carbonyl oxygen in 2-propanone. This step is vital as it increases the electrophilicity of the carbonyl carbon. By doing so, it makes the molecule more reactive and susceptible to further chemical reactions. Protonation essentially adds a positive charge, which makes the carbonyl group a better target for nucleophiles. This enhanced reactivity sets the stage for new bond formations and is key to the sequential steps that follow in the aldol reaction process.
Carbon-Carbon Bond Formation
One of the most interesting aspects of aldol reactions is the formation of new carbon-carbon bonds. This is a hallmark of the aldol addition. After the enol form of 2-propanone is generated, which we will discuss later, it acts as a nucleophile. This nucleophile then attacks the carbonyl carbon of another protonated 2-propanone molecule. The result is the creation of a new C-C bond. Such new bonds are the backbone for synthesizing complex molecules from simpler ones. The capability of forming these bonds is precisely why aldol reactions are a highly prized mechanism in organic chemistry, allowing scientists to build larger, more complex molecular structures.
Protonation Mechanism
Protonation is a critical mechanism in the aldol addition process, particularly under acidic conditions. The procedure begins with the addition of a proton (H⁺) to the carbonyl oxygen of 2-propanone. This protonation is facilitated by the acidic environment provided by hydrogen chloride. By making the carbon more electrophilic, it becomes a more attractive site for nucleophilic attack. This alteration in reactivity helps to guide the reaction forward, ensuring that subsequent steps like enol formation and nucleophile attack can smoothly proceed. The protonation steps in acidic catalysis underscore the fundamental nature of how acids enhance reaction pathways by altering molecular electronics.
Enol Formation
Enol formation in the aldol addition begins after hydrogen chloride facilitates the protonation of the carbonyl group. Enols are a special tautomeric form of ketones, often resulting from the removal of an alpha hydrogen, helped along by chloride ions. In the mechanism, this formation entails shifting to a configuration that has the double-bond character moved to a carbon-carbon link adjoining the carbonyl group. Enolization plays a vital role as it places the molecule in an arranged state for nucleophilic attack. It acts as a nucleophile, participating actively in the subsequent step of bond formation by attacking the electrophilic carbonyl carbon. This duality, where the molecule can be present in keto or enol form, adds flexibility and dynamic reactivity to the aldol addition process.

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Most popular questions from this chapter

Other groups in addition to carbonyl groups enhance the acidities of adjacent \(\mathrm{C}-\mathrm{H}\) bonds. For instance, nitromethane, \(\mathrm{CH}_{3} \mathrm{NO}_{2}\), has \(\mathrm{p} K_{a}=10\); ethanenitrile, \(\mathrm{CH}_{3} \mathrm{CN}\), has a \(\mathrm{p} K_{a} \cong 25 .\) Explain why these compounds behave as weak acids. Why is \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) a stronger acid than \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3} ?\)

On what basis can you account for the fact that \(\mathrm{HCN}\) adds to the carbonyl group of 3 -butenal and to the double bond of 3 -buten-2-one? Would you expect the carbonyl or the double-bond addition product of \(\mathrm{HCN}\) to 3 -buten- 2 -one to be more thermodynamically favorable? Give your reasoning.

If you wished to prepare the methyl ether of 4 -hydroxy-3-penten- 2 -one by \(\mathrm{O}\) -alkylation, what base and which fo the methylating agents listed would you choose? \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{I}, \mathrm{CH}_{3} \mathrm{OCO}_{2} \mathrm{OCH}_{3},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{OBF}_{4}\), or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O} .\) Give your reasoning.

It is just as important to be able to recognize reactions which do not work as it is to recognize reactions that do work. The following equations represent "possible" synthetic reactions. Consider each carefully and decide whether it will proceed as written. Show your reasoning. If you think a different reaction will take place, write an equation for it. a. \(\mathrm{CH}_{3} \mathrm{COCH}_{3}+6 \mathrm{Br}_{2}+8 \mathrm{NaOH} \rightarrow 2 \mathrm{CHBr}_{3}+\mathrm{Na}_{2} \mathrm{CO}_{3}+6 \mathrm{NaBr}+6 \mathrm{H}_{2} \mathrm{O}\) b. \(\mathrm{CH}_{3} \mathrm{CHO}+\mathrm{NaNH}_{2}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl} \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CHO}+\mathrm{NH}_{3}+\mathrm{NaCl}\) c. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOCH}_{3}+\mathrm{CH}_{2}=\mathrm{O} \stackrel{\mathrm{Ca}(\mathrm{OH})_{2}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}\left(\mathrm{CH}_{2} \mathrm{OH}\right) \mathrm{COCH}_{3}\) d. \(\mathrm{CH}_{3} \mathrm{CHO}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5} \stackrel{\text { ? }^{\circ}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\) e. \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COCH}_{3}+\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{COOC}\left(\mathrm{CH}_{3}\right)=\mathrm{CHCOCH}_{3}\)

If the keto form of 2,4 -pentanedione is more stable than the enol form in water solution, why does it also have to be a weaker acid than the enol form in water solution?

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