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 83

Chapter 18, Problem 83

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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Step 1: Identify the key features of the reaction. The Cannizzaro reaction involves an aldehyde that lacks α-hydrogens reacting with concentrated sodium hydroxide. In this case, the aldehyde is 4-nitrobenzaldehyde, which lacks α-hydrogens due to the absence of hydrogens on the carbon adjacent to the carbonyl group.

Step 2: Initiate the reaction mechanism. The hydroxide ion (OH⁻) from the concentrated NaOH attacks the carbonyl carbon of the aldehyde, forming a tetrahedral intermediate. This step is nucleophilic addition.

Step 3: The tetrahedral intermediate undergoes rearrangement. One molecule of the aldehyde donates a hydride ion (H⁻) to another molecule of the aldehyde. This hydride transfer is the key step in the Cannizzaro reaction, leading to the formation of two distinct products.

Step 4: Formation of products. The molecule that receives the hydride ion is reduced to an alcohol (4-nitrobenzyl alcohol), while the molecule that donates the hydride ion is oxidized to a carboxylic acid (4-nitrobenzoate ion).

Step 5: Final step involves stabilization. The carboxylic acid product is deprotonated by the hydroxide ion in the basic medium, forming the carboxylate ion (4-nitrobenzoate ion). The alcohol product remains neutral.

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Key Concepts

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

Cannizzaro Reaction

The Cannizzaro reaction is a redox reaction that occurs with aldehydes lacking alpha-hydrogens when treated with a strong base, such as sodium hydroxide. In this reaction, one molecule of the aldehyde is oxidized to a carboxylic acid, while another molecule is reduced to an alcohol. This reaction is significant in organic chemistry as it provides a method for synthesizing both alcohols and carboxylic acids from simple aldehydes.

Mechanism of the Cannizzaro Reaction

The mechanism of the Cannizzaro reaction involves the formation of a hydride ion transfer between two aldehyde molecules. Initially, one aldehyde molecule is deprotonated by the base, forming an alkoxide ion. This alkoxide then attacks another aldehyde molecule, leading to the transfer of a hydride ion, resulting in the formation of a carboxylate ion and an alcohol. The carboxylate can then be protonated to yield the carboxylic acid.

Role of Sodium Hydroxide

Sodium hydroxide (NaOH) acts as a strong base in the Cannizzaro reaction, facilitating the deprotonation of the aldehyde. This step is crucial as it generates the alkoxide ion, which is a key intermediate in the reaction mechanism. The presence of concentrated NaOH not only drives the reaction forward but also helps stabilize the carboxylate ion formed during the process, allowing for the successful conversion of the aldehyde into both the carboxylic acid and the alcohol.

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Related Practice

Textbook Question

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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Textbook Question

Propose a mechanism for each of the following reactions:

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Textbook Question

A Mannich reaction puts a -group on the α-carbon of a ketone. Propose a mechanism for the reaction.

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Textbook Question

A student tried to prepare the following compounds using aldol condensations. Which of these compounds was she successful in synthesizing? Explain why the other syntheses were not successful.

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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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