# 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 and Wikipedia 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.

• In the mechanism of Beckmann rearrangement, as $H_2O$ leaves the nitrogen atom and adjacent carbon forms another bond with Nitrogen. At that point the migrating group forms a new bond with N leaving C. So, to form a strong covalent bond, the migrating groups orbital has to form a strong overlap with Nitrogen's orbital containing orbital. At the same time the orbital of Nitrogen which was overlaping with O before, has to now overlap with C. If the migrating group is at cis to $OH$ both the overlaps can't be done effectively. I think, this may be the reason. Feb 14, 2018 at 8:03
• @SoumikDas Thanks! Are you saying that if both the migrating alkyl group and the leaving $\ce{-OH}$ group are on the same side of the pi bond, then the overlap will be difficult to achieve, as compared to when the alkyl group migrates from one side and the $\ce{-OH}$ leaves from the other side? Feb 14, 2018 at 8:23
• Yes. I actually tried to say that. Feb 14, 2018 at 9:08

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.
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.