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 79

Chapter 18, Problem 79

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.)

Verified step by step guidance

Step 1: The hydroxide ion (HO⁻) acts as a base and removes a proton from the α-carbon that is not bonded to the bromine atom. This deprotonation generates a carbanion intermediate at the α-carbon.

Step 2: The carbanion intermediate undergoes intramolecular nucleophilic attack on the carbonyl carbon, forming a three-membered cyclic intermediate known as a cyclopropanone.

Step 3: The bromine atom is expelled from the cyclopropanone intermediate, resulting in the formation of a new carbanion at the α-carbon adjacent to the carbonyl group.

Step 4: The hydroxide ion (HO⁻) acts as a nucleophile and attacks the carbonyl carbon of the intermediate, leading to the opening of the three-membered ring and formation of a tetrahedral intermediate.

Step 5: The tetrahedral intermediate undergoes proton transfer, resulting in the formation of the carboxylic acid product with the hydroxide ion regenerating as the leaving group.

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Key Concepts

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

Favorskii Reaction

The Favorskii reaction is a rearrangement reaction involving α-haloketones that leads to the formation of carboxylic acids. In this reaction, a hydroxide ion (HO-) acts as a base and nucleophile, facilitating the conversion of the α-haloketone into a carboxylic acid through a series of steps, including deprotonation and ring formation.

Nucleophilic Attack

Nucleophilic attack is a fundamental concept in organic chemistry where a nucleophile, which is an electron-rich species, attacks an electron-deficient center, typically a carbon atom in a carbonyl group. In the Favorskii reaction, the hydroxide ion acts as a nucleophile, attacking the electrophilic carbon in the three-membered ring formed during the reaction, leading to the final product.

Three-Membered Ring Formation

The formation of a three-membered ring, often referred to as an epoxide or cyclic intermediate, is a key step in many organic reactions, including the Favorskii reaction. This intermediate is formed when a nucleophile attacks an electrophilic carbon, resulting in a cyclic structure that can stabilize the transition state and facilitate further reactions, ultimately leading to the desired product.

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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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Show how the following compounds can be synthesized. The only carbon-containing compounds available to you for each synthesis are shown.

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Propose a mechanism for each of the following reactions:

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

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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

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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