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Aren't alpha halo carbonyl compounds extremely fast and reactive via $S_N2$ (the fastest i guess)? If this is so, why isn't the major product in the Favorskii rearrangement an alpha hydroxy carbonyl compound?

And if the alpha carbon(to which the halide is attached) is tertiary, there should be a high chance of an elimination product via $E_2$ (leading to the formation of alpha beta unsaturated carbonyl), right?

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This is an old question, but in case anyone comes down this road...

Aren't alpha halo carbonyl compounds extremely fast and reactive via $\ce{S_{N}2}$ (the fastest i guess)?

I don't think so. In an $\ce{S_{N}2}$ reaction a nucleophile is attacking a positively polarized carbon atom. The more positive charge on the carbon, the more the nucleophile is attracted. Look at structure I below, it shows one of the resonance structures for a carbonyl that leads to a polarized carbonyl with positive charge on the carbonyl carbon. Now to whatever degree the adjacent C-Cl bond is polarized with positive charge on carbon, we wind up with an unfavorable electrostatic situation of adjacent positive charges. Said differently, the presence of a carbonyl group suppresses polarization of the adjacent C-Cl bond, which will make the C-Cl carbon less prone to nucleophilic attack.

enter image description here

And if the alpha carbon(to which the halide is attached) is tertiary, there should be a high chance of an elimination product via E 2 (leading to the formation of alpha beta unsaturated carbonyl), right?

No, you want to remove hydrogen D in the diagram below. Hydrogen D is much less acidic than hyrdrogen A because removal of the latter results in a resonance stabilized carbanion. The difference in acidity between hydrogens A and D is so large that effectively only hydrogen A will be removed by a nucleophile.

These two arguments effectively eliminate pathways C and D from consideration. Pathway B is simply an equilibrium that goes nowhere, so it has no real effect on the outcome of the reaction. This leaves pathway A, the Favorskii reaction, as the only viable alternative. Pathway A, besides being the only game in town, is actually favored by two factors, 1) as noted above, hydrogen A is acidic and it's removal produces a resonance stabilized carbanion, and 2) completion of the Favorskii reaction is strongly favored by entropic factors since the carbanion and the C-Cl carbon are contained in the same molecule and held relatively close together.

enter image description here

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  • $\begingroup$ I mostly agree with your analysis, but I think there may be some caveats worth noting: (a) while I agree with you about the inductive effect of the carbonyl group on $\ce{C-Cl}$ bond polarity, in the case of a halogen on a primary carbon, $\mathrm{S_N2}$ actually seems plausible to me; (b) I'm inclined to think $\mathrm{E_2}$ is quite possible, though perhaps only favorable under reflux conditions (and obviously dependent on reactivity of the hydrogen $\beta$ to the halogen vs. the $\alpha '$ hydrogen). Of course, I'm referring to different substrates than the one pictured. $\endgroup$ – Greg E. Jul 30 '14 at 1:59
  • $\begingroup$ I was also thinking about the possibility of aldol condensations (given a less bulky molecule), especially since the carbon bearing the halogen has the more acidic hydrogens, but it's hard to imagine a set of conditions favorable for aldol condensation that wouldn't just give rise to elimination products. If you have any thoughts on that, I'd be very glad to have your input. $\endgroup$ – Greg E. Jul 30 '14 at 2:12
  • $\begingroup$ @Greg E a) if you agree with the carbonyl inductive effect argument, then you should agree that the $\ce{S_{N}2}$ would be at least slowed down, if not eliminated - right? b) If you agree that H(a) is much more acidic then H(d), then wouldn't a nucleophile strongly prefer to interact with H(a)? What do you mean by "I'm referring to different substrates than the one pictured" $\endgroup$ – ron Jul 30 '14 at 3:42
  • $\begingroup$ @Greg E One thought your comment about removing the alpha hydrogen attached to the C-Cl carbon generated is why not eliminate that alpha hydrogen and chlorine to form a carbene(oid). It could then insert into H(beta) to produce the "elimination" product and\or insert into the other C(alpha)-CO bond to produce cyclopentane with a ketene attached (methylenecyclopentane with a carbonyl stuck on the exo-methylene - sorry its late for me), which then reacts with the nucleophile to give the Favorskii product. $\endgroup$ – ron Jul 30 '14 at 3:51
  • $\begingroup$ What I mean by "I'm referring to different substrates than the one pictured" is that my comments were about alternate reaction pathways for $\alpha$-halo ketones in general in the presence of appropriate nucleophiles/bases, not cyclohexanone specifically. As far as $S_N2$ reactions, while I think the carbonyl could plausibly reduce the polarization of the $\ce{C-Cl}$ bond, I would be surprised if that effect were large. Furthermore, the $\delta^+$ of the carbonyl carbon might attract nucleophiles, so even if pathway B in your rxn. scheme leads nowhere, the proximity effect could be relevant. $\endgroup$ – Greg E. Jul 30 '14 at 4:54
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In extension to ron's answer, there are two reasons, why I think a $\ce{S_{N}2}$ reaction is unlikely to occur.

  1. The LUMO, where the nucleophile will attack, of 2-Chlorocyclohexanone is the antibonding orbital of the carbonyl bond. Also the computed natural charges show a much more polarised carbonyl carbon (DF-BP86/dfe2-SVP).
    lumocharge
    On the german wikipedia page I found a proposed mechanism, that would make use of the LUMO instead of the acidity of the $\alpha$ hydrogen for the rearrangement. I would suspect, that it heavily depends on the substrate and the base, which one is favoured.
    reaction mechanism
  2. The most stable conformation of this compound is where the chlorine is in equatorial position. The antibonding $\ce{C-Cl}$ bond orbital is sterically shielded by the ring framework. The molecule has to flip the ring to make it accessible. The energy difference is about $1.5~\mathrm{kcal\cdot{}mol^{-1}}$ (DF-BP86/def2-SVP).
    I do not have the time to calculate the activation barrier, but I think it is fair to assume an unfavoured equilibrium for the $\ce{S_{N}2}$ state, at room temperature about $1:9$. This would make this reaction not very likely.
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