Here, the Favorskii Rearrangement is described with examples, detailed mechanism, evidence in support of the mechanism, applications, problems, and solutions.
- What is Favorskii Rearrangement?
- Evidence in Support of the Mechanism
- Stereochemistry Involved in the Favorskii Rearrangement
- Quasi-Favorskii Rearrangement
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.
Mechanism of the Favorskii Rearrangement
The reaction mechanism of Favorskii Rearrangement involves the following steps:
- In the first step, the base abstracts an 𝞪’-H atom and form an enolate ion.
- 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 SN2 displacement of halogen by 𝞪’-carbanion.
- 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.
- 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 SN2 displacement of the halogen atom. Hence, an inversion of configuration occurs at the centre containing a halogen atom and the reaction is considered to be a stereospecific reaction. An example is shown below.
There are cases in which α-haloketones with no 𝞪’-H atom undergo the Favorskii rearrangement and gives the same type of rearranged products. The reaction proceeds through the semi-benzilic like mechanism. This type of reactions is usually called the quasi-Favorskii rearrangement.
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 synthesised easily.
Problems: Predict the product of the following reactions and explain mechanistically.
- A. Favorskii and V. Boshowski, J. Russ. Phys. Chem. Soc., 46, 1098, 1914.
- A. Favorskii, J. Prakt. Chem. 88 (1), 641–698, 1913.