Why are oxime geometrical isomers stable?

231st page of Clayden OC 2nd edition: 



> 
> Unlike the geometrical isomers of alkenes, however, those of an imine
>  usually interconvert quite rapidly at room temperature. The
>  geometrical isomers of oximes on the other hand are stable and can
>  even be separated.
> 
> 
> 


Why are oxime geometrical isomers much more stable than the geometrical isomers of immines?




It is the same reasoning that is used to predict relative barrier heights in **a**mines. 


In the example pictured below, the imine starts with the nitrogen lone pair in "more or less" an $\ce{sp^2}$ orbital. In the transition state for interconversion, the lone pair is in a $\ce{p}$ orbital. 


![enter image description here](https://i.stack.imgur.com/i0wRA.jpg)


The term "more or less" is important. If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit more p character than $\ce{sp^2}$ (say $\ce{sp^{2.2}}$ for example), then it is already closer to the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be reduced. On the other hand, If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit less p character (more $\ce{s}$ character) than $\ce{sp^2}$ (say $\ce{sp^{1.8}}$ for example), then it is further from the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be greater.


In the case of oximes ($\ce{X=OH}$) or other molecules where the substituent on nitrogen is electronegative (electron withdrawing), the electron density in the $\ce{N-X}$ bond is **shifted towards** the electronegative atom. Consequently there is less electron density in the nitrogen portion of this bond. As a result the nitrogen rehybridizes. Since $\ce{s}$ orbitals are lower in energy than $\ce{p}$ orbitals, orbitals with increased $\ce{s}$ character are better able to stabilize electrons. Since there is less electron density in the orbital nitrogen is using for this bond, $\ce{s}$ character is removed from the orbital. This "extra" $\ce{s}$ character is used to stabilize the orbital containing the lone pair of electrons. The lone pair orbital now has more $\ce{s}$ character and is further away from the $\ce{p}$ hybridized transition state - the barrier to reach this transition state increases. If our substituent is an alkyl group (X=R, an imine), the carbon is less electronegative then nitrogen, there will be increased electron density in the nitrogen orbital used to form the $\ce{N-R}$ bond. The nitrogen will hybridize such that there is more $\ce{s}$ character in this orbital so as to better stabilize these electrons. As a consequence, there will less $\ce{s}$ character (more $\ce{p}$ character) in the lone pair orbital. The lone pair orbital is therefore closer to the $\ce{p}$ hybridized transition state and less energy will be required to reach the transition state.


As a result of this reasoning $\ce{NF3}$ has a higher inversion barrier then $\ce{NH3}$ and oximes have a higher barrier to inversion than imines.


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It's a matter of energetic barrier between the $E$ and $Z$ stereoismers. It's sufficiently low for imines. So, it's relatively difficult to isolate $E$ or $Z$ stereoismer.


In the case of oximes, the energetic barrier of interconversion between the two stereoisomers is much higher and allows the isolation of the two stereoisomers.


For more details, please see <http://chemweb.bham.ac.uk/~coxlr/Teaching/1st_Year/Carbonyl_Group/Carbonyl_pdfs/lecture%207%20%28student%20HO%29.pdf>


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