Why does the alkyl group anti to the hydroxyl migrate in Beckmann rearrangement?

**Background:** Many organic reactions involve the migration of an alkyl group from one position to the adjacent one. The migration of the alkyl group is decided by its *migratory aptitute* *i.e.* electron-richness. It generally follows the priority order of hydride > phenyl > higher alkyl > methyl.


**Main question:** The Beckmann rearrangement also involves an alkyl migration. However, this migration is *not* governed by migratory aptitude. In fact, the $\ce{-R}$ which is in *anti*-position to the hydroxyl group in the oxime migrates, irrespective of its migratory aptitude! My question is why is this so? And are there other organic reactions which have such a rule for alkyl migration?


[MasterOrganicChemistry](https://www.masterorganicchemistry.com/reaction-guide/beckmann-rearrangement/) and [Wikipedia](https://en.wikipedia.org/wiki/Beckmann_rearrangement) don't even mention the word "trans" or "anti" anywhere. My textbook obviously doesn't mention anything either, hence I am driven to ask this question. 




Oximes can undergo the Beckmann rearrangement. Oximes also exist as stable *syn* and *anti* isomers. In the figure below, $\ce{R\_1}$ is *anti* to the hydroxyl group; another isomer exists where it is *syn* the the hydroxyl group.


In the first step of the Beckmann rearrangement the protonated hydroxyl group makes for a good leaving group. As the hydroxyl group begins to depart, the $\ce{R}$ group which is *anti* to the leaving group begins to migrate and form a bond with nitrogen.


[![Wikipedia](https://i.stack.imgur.com/mgLW1.png)](https://i.stack.imgur.com/mgLW1.png)


([Wikipedia](https://en.wikipedia.org/wiki/Beckmann_rearrangement))


This step is said to be concerted (rather than stepwise) with the $\ce{N-O^{+}H2}$ bond being broken and the new $\ce{R-N}$ bond being formed at (more or less) the same time. Whether you think of the $\ce{R}$ group as participating in something similar to an SN2 reaction or whether you think of the $\ce{R}$ group as starting to bond to the available $\sigma^{\*}$ orbital of the elongating $\ce{N-O}$ bond (equivalent descriptions), you see that having the $\ce{R}$ group approach the nitrogen from a 180° angle from the breaking $\ce{N-O}$ bond would be preferred. **Hence, the $\ce{R}$ group *anti* to the $\ce{N-O}$ bond migrates preferentially**.


There are many reactions showing this geometric preference, the [Baeyer-Villiger oxidation](https://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation) being an example.


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