In the Favorskii rearrangement, why doesn't the base grab the alpha hydrogen of the carbon bearing the halogen?
Since it is much easier to discuss based on actual schemes, see scheme 1 for a general 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.


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 &#x3b1;-chloroketone is not included in this data.

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

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.