Favorskii Rearrangement

Here, the Favorskii Rearrangement is described with examples, detailed mechanisms, evidence in support of the mechanism, applications, problems, and solutions.

Favorskii Rearrangement

1. What is Favorskii Rearrangement?
2. Mechanism
3. Evidence in Support of the Mechanism  
4. Stereochemistry Involved in the Favorskii Rearrangement
5. Quasi-Favorskii Rearrangement
6. Applications
7. Problems
8. References

What is Favorskii Rearrangement?

The Favorskii Rearrangement is the rearrangement of the α-haloketone to the corresponding carboxylic acids or carboxylic acid derivatives in the presence of a strong base (e.g, alkoxide base, hydroxide base, amine base). 

In the presence of an alkoxide base, the rearrangement leads to the formation of an ester. However, in the presence of hydroxide and amine base, the rearrangement leads to the formation of carboxylic acid and amide, respectively. 

Favorskii Rearrangement occurs both in cyclic and acyclic α-haloketones. In cyclic α-halo ketones, the rearrangement leads to ring contraction. 

Examples:

Mechanism of the Favorskii Rearrangement

The reaction mechanism of Favorskii Rearrangement involves the following steps:

1. In the first step, the base abstracts an 𝞪’-H atom and form an enolate ion. 
2. In the second step, 𝞪’-carbanion attacks on the 𝞪-carbon atom and the chloride anion leave to produce a cyclopropanone intermediate. This step is the intramolecular S_N2 displacement of halogen by 𝞪’-carbanion. 
3. In the third step, the cyclopropanone intermediate undergoes the ring-opening to form a stable carbanion. In the case of an unsymmetrical cyclopropanone, the direction of ring-opening is determined by which is the more stable carbanion of the two possible carbanions that can be formed.  
4. In the final step, the carbanion abstracts a proton from the solvent to form corresponding carboxylic acids or carboxylic acid derivatives. 

Evidence in Support of the Mechanism  

1. The direct evidence supporting the Favorskii rearrangement mechanism has been obtained from ¹⁴C-labelling studies. When 2-chlorocyclohexanone (I) labelled with ¹⁴C at the chlorine bearing carbon is subjected to the Favorskii rearrangement, the product (II) was found to contain half the ¹⁴C at the 𝞪-carbon and half at the 𝛃-carbon atom. 

This result demonstrates that the rearrangement occurs through the symmetrical cyclopropanone intermediate (III) where two carbon atoms (C-2 and C-6) are alpha to the carbonyl group become equivalent and the cyclopropanone intermediate (III) can open up in either side of the carbonyl group with equal probability. 

 2. When two isomeric 𝞪-heloketones (A and B) are treated with the same base, both the 𝞪-heloketones give the same product. This indicates that the rearrangement occurs through the same cyclopropanone intermediate (C) for both the isomeric 𝞪-heloketones (A and B). 

3. In the Favorskii rearrangement, when the reaction is carried out in presence of a strong base and furan, a cyclopropanone intermediate can also be trapped. For example, when 𝞪-heloketone (A) is treated with a base (EtONa) in the presence of furan, a Diels-Alder product (B) is obtained. This indicates that the Favorskii rearrangement reaction proceeds through the cyclopropanone intermediate. 

Stereochemistry Involved in the Favorskii Rearrangement 

Favorskii rearrangement involves the S_N² displacement of the halogen atom. Hence, an inversion of configuration occurs at the center containing a halogen atom, and the reaction is considered to be a stereospecific reaction. An example is shown below.

Semi-Benzilic Mechanism 

There are cases in which α-haloketones with no 𝞪’-H atom undergo the Favorskii rearrangement and give the same type of rearranged products. The reaction proceeds through the semi-benzilic-like mechanism. This type of reaction is usually called the quasi-Favorskii rearrangement.

Applications

1. One of the most common uses of the Favorskii rearrangement is the ring contractions for cyclic compounds. For example,

2. From Favorskii rearrangement, heavily branched carboxylic acids and esters can be prepared easily which are hard to prepare by any other methods. For example,

3. By quasi-Favorskii rearrangement, pethidine (a pain killer) can be synthesized easily.

Questions on Favorskii Rearrangement

Problems: Predict the product of the following reactions and explain mechanistically.

References

1. A. Favorskii and V. Boshowski, J. Russ. Phys. Chem. Soc., 46, 1098, 1914.
2. A. Favorskii, J. Prakt. Chem. 88 (1), 641–698, 1913.