In the Favorskii rearrangement, the base (like hydroxide) always abstracts the alpha hydrogen of the carbon not bearing the halogen. But to me the hydrogens on the carbon directly connected to the halogen should be more acidic due to the electron pulling inductive effect of the halogen (which indeed does happen in the haloform reaction). This should result in an aldol condensation type reaction. Why is this not so?

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    $\begingroup$ The crux of the matter is that deprotonation opposite the halogen leads to a faster (intramolecular) reaction, so that's what you're going to observe. Never take the existence of a mechanism to mean that those steps are the only things that can happen in solution. Undoubtedly some deprotonation at the halogen position does occur. Whether it leads to a sufficiently fast reaction is another matter altogether. $\endgroup$ Nov 3, 2016 at 16:09
  • $\begingroup$ I agree with orthocresol. Both alpha positions will be roughly similar in acidity but the intramolecular reaction that occurs when the non-halogenated alpha position gets deprotonated will be just much faster than the intermolecular aldol condensation that could in principle happen on the other side. $\endgroup$ Nov 3, 2016 at 16:25

1 Answer 1


Since it is much easier to discuss based on actual schemes, see scheme 1 for a general Favorskii rearrangement.

Scheme of the Favorskii rearrangement
Scheme 1: General Favorskii rearrangement reaction conditions. Image taken from Wikipedia where a full list of authors is available.

And because it is also helpful, scheme 2 shows one of the most accepted mechanisms of the Favorskii rearrangement — the one used by both Wikipedia and Kürti-Czakó.[1] One should note, however, that a second mechanistic proposal[2] explains the reaction outcome well and is favoured by some calculations.[3] Because it does not require any deprotonation, it is outside of the scope of this answer.

Mechanism of the Favorskii rearrangement
Scheme 2: ‘Cyclopropanone’-mechanism of the Favorskii rearrangement. Image taken from Wikipedia where a full list of authors is available.

Your question asks why carbon 6 (from the usual numbering of 2-chlorocyclohexanone) is deprotonated to form the corresponding enolate and not carbon 2.

The most important notice here is the base used $\ce{NaOH}$ and the conditions employed (not mentioned in the image, but an example on organic-chemistry.org mentions room temperature). For comparison, table 1 shows a few $\mathrm{p}K_\mathrm{a}$ values from the Bordwell data — unfortunately, an α-chloroketone is not included in this data.

$$\textbf{Table 1: }\mathrm{p}K_\mathrm{a}\text{ values of different relevant acids in }\ce{H2O}\\ \begin{array}{ccc}\hline \text{acid} & \mathrm{p}K_\mathrm{a}\left(\ce{H2O}\right) & \mathrm{p}K_\mathrm{a} \left(\ce{DMSO}\right)\\ \hline \ce{Me2CO} & 19.3 & 26.5\\ \ce{PhCOMe} & 18.3 & 24.7\\ \ce{PhCO-CH2F} & & 21.7\\ \ce{Et2CO} & & 27.1\\ \text{3,3-dimethylcyclohexanone} & & 24.8\\ \ce{H2O} & 15.7^* & 31.4\\ \hline \end{array}\\ ^*\text{ Other sources say 14.}$$

Obviously from this data, the choice of solvent is import when considering the acidity of a given substance — note the marked decrease in acidity of water in $\ce{DMSO}$. However generally, water is a weaker acid than carbonyl compounds; Professor Mayr taught us to estimate $\mathrm{p}K_\mathrm{a}$ values: alcohols $\approx 15$, ketones $\approx 20$. That means, the deprotonation is in no way complete; it is in equilibrium. This equilibrium will lead to a thermodynamic product distribution: carbon 2 will be deprotonated slightly more than carbon 6, but the most significant species will be the unenolised ketone.

Now as soon as carbon 6 has enolised, it is able to attack carbon 2 to form the bicyclo[3.1.0]hexan-6-one. This is an intramolecular attack and thus very fast. Once this has happened, the formation of the hydrate (6-hydroxybicyclo[3.1.0]hexanolate) is favoured due to the strained three-membered ring. And the subsequent intramolecular migration is also very fast. This leads to a stable product with little or no means for any back reaction; as soon as the initial attack has taken place, things are pretty much settled.

On the other hand, an aldol addition would not only require the enolate (doesn’t matter which one) to be around for some time; it is also inherently prone to retro-aldol dissociation. Usually in organic chemistry, aldol reactions are performed selectively by using kinetic deprotonation (e.g. $\ce{LDA}$; $\mathrm{p}K_\mathrm{a}(\ce{HN(iPr)2}) \approx 35$) ensuring that the forward reaction is strongly favoured. Hydroxide bases may be used in aldol reactions, but then generally the reaction is heated to reflux and the final product is the condensation rather than the addition.

Thus, to sum it up:

  • Both enolisations occur
  • The conditions do not favour aldol addition; they will revert readily
  • Once the Favorskii path has been entered, the rearrangement is basically irreversible.

Notes and References:

[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 164.

[2]: The alternative, semi-benzilic mechanism is initiated by the nucleophilic attack of a hydroxide ion on the carbonyl group. The thus formed 2-chloro-1-hydroxycyclohexanolate anion undergoes 1,2-migration with chloride displacement to form the product cyclopentyl carboxylic acid in a single step.

[3]: V. Moliner, R. Castillo, V. S. Safont, M. Oliva, S. Bohn, I. Tuñón, J. Andrés, J. Am. Chem. Soc. 1997, 119, 1941. DOI: 10.1021/ja962571q.

  • $\begingroup$ But why does this not happen in the haloform reaction? There the base deprotonates the alpha carbon having the halogen, and not the other alpha carbon(once the first halogen has been added) $\endgroup$
    – Newton
    Nov 4, 2016 at 2:33
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    $\begingroup$ Btw. both of your used images are in the public domain, there is no need for citing the source ;) $\endgroup$ Nov 4, 2016 at 4:16
  • $\begingroup$ I have doubts about the proposed reaction mechanism and find them supported by J. Am. Chem. Soc., 1997, 119 (8), pp 1941–1947. Tl;dr: I would have assumed nucleophilic attack at the carbonyl carbon first, then rearrangement with dissociation of chloride. $\endgroup$ Nov 4, 2016 at 13:31
  • $\begingroup$ @Martin-マーチン Hm, interesting. Kürti-Czakó show the mechanism that I wrote down here. But to be perfectly honest, the semi-benzilic rearrangement looks very plausible, too; especially in the light of other, similar rearrangements. I just skimmed across the paper right to the conclusions, but I’ll take a deeper look at it later™. $\endgroup$
    – Jan
    Nov 5, 2016 at 15:43
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    $\begingroup$ Before I left the office I found a paper with a title among the lines of 'a new mechanism for the Fav. Re.' I have not had the time to read it. I think it's safe to assume that no ultimate truth and it probably depends heavily on many different factors, like base, solvent, temperature, ring size, etc. When I did a few calculations (chex), the PA of the 2 or 6 position were not very much different, certainly within the accuracy of the method. $\endgroup$ Nov 5, 2016 at 16:08

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