Ch. 17 - Reactions at the Alpha-Carbon

Bruice8th EditionOrganic ChemistryISBN: 9780135213711Not the one you use?Change textbook

Bruice 8th Edition

Ch. 17 - Reactions at the Alpha-Carbon

Problem 85

Chapter 18, Problem 85

The following reaction is known as the benzoin condensation. The reaction does not take place if sodium hydroxide is used instead of sodium cyanide. Propose a mechanism for the reaction and explain why the reaction does not occur if hydroxide ion is the base.

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The benzoin condensation is a reaction between two aldehydes (typically aromatic aldehydes like benzaldehyde) in the presence of a cyanide ion (CN⁻) as a catalyst. The reaction forms a benzoin product, which is a hydroxy ketone. The mechanism involves nucleophilic addition and subsequent rearrangement steps.

Step 1: The cyanide ion (CN⁻) acts as a nucleophile and attacks the carbonyl carbon of one benzaldehyde molecule. This forms a tetrahedral intermediate, which is stabilized by the resonance of the cyanohydrin structure.

Step 2: The intermediate formed in Step 1 generates a carbanion at the alpha-carbon (adjacent to the carbonyl group) due to the electron-withdrawing effect of the cyanide group. This carbanion is stabilized by resonance, making it a strong nucleophile.

Step 3: The carbanion attacks the carbonyl carbon of a second benzaldehyde molecule, forming a new carbon-carbon bond. This step results in the formation of another tetrahedral intermediate.

Step 4: The intermediate undergoes proton transfer and elimination of the cyanide ion, regenerating the catalyst and forming the final benzoin product. The reaction does not occur with hydroxide ion (OH⁻) because hydroxide lacks the ability to stabilize the intermediate carbanion through resonance, unlike cyanide, which has a strong electron-withdrawing effect.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Benzoin Condensation

Benzoin condensation is a reaction between two molecules of benzaldehyde in the presence of a base, leading to the formation of benzoin. This reaction involves nucleophilic addition where one benzaldehyde acts as a nucleophile, attacking the carbonyl carbon of another benzaldehyde. The reaction is typically catalyzed by cyanide ions, which facilitate the formation of the intermediate that ultimately leads to the product.

Role of Base in Organic Reactions

In organic chemistry, the choice of base can significantly influence the outcome of a reaction. Bases can deprotonate compounds, generating nucleophiles or stabilizing intermediates. In the case of benzoin condensation, sodium cyanide acts as a strong nucleophile, while sodium hydroxide, being a stronger base, does not provide the necessary nucleophilic character required for the reaction to proceed effectively.

Nucleophilicity vs. Basicity

Nucleophilicity refers to the ability of a species to donate an electron pair to form a chemical bond, while basicity is a measure of a species' ability to accept protons. In the context of the benzoin condensation, cyanide ion is a good nucleophile, allowing it to effectively attack the carbonyl carbon of benzaldehyde. In contrast, hydroxide ion, while a strong base, is less nucleophilic in this context, leading to the failure of the reaction when it is used.

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Related Practice

Textbook Question

A compound known as Hagemann's ester can be prepared by treating a mixture of formaldehyde and ethyl acetoacetate first with base and then with acid and heat. Write the structure for the product of each of the steps.

b. The second step is a Michael addition.

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Textbook Question

Propose a mechanism for each of the following reactions:

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Amobarbital is a sedative marketed under the trade name Amytal. Propose a synthesis of amobarbital, using diethyl malonate and urea as two of the starting materials.

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Textbook Question

A Cannizzaro reaction is the reaction of an aldehyde that has no a-hydrogens with concentrated aqueous sodium hydroxide. In this reaction, half the aldehyde is converted to a carboxylic acid and the other half is converted to an alcohol. Propose a mechanism for the following Cannizzaro reaction:

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Textbook Question

Alkylation of the following compound with methyl iodide under two different conditions forms two different ketoesters (A and B). Each ketoester forms a cyclic diketone (C and D) when treated with methoxide ion in methanol. a. Draw the structures of A and B, and indicate the conditions used in the alkylation reaction that cause that ketoester to be formed.

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Textbook Question

A carboxylic acid is formed when an a-haloketone reacts with hydroxide ion. This reaction is called a Favorskii reaction. Propose a mechanism for the following Favorskii reaction. (Hint: In the first step, HO^- removes a proton from the a-carbon that is not bonded to Br; a three-membered ring is formed in the second step; and HO^- is a nucleophile in the third step.)

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