Why do E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane?

E2 reactions ([dehydrohalogenation](https://en.wikipedia.org/wiki/Dehydrohalogenation)) only occur when the hydrogen and leaving group are in the same plane.


I can't think of a reason that they would need to be in the same plane, so what is the reason for this strange phenomenon?




When the $\ce{C-X}$ and $\ce{C-H}$ $\ce{sp^3}$bonds break, the orbitals left behind on the two carbons will rehybidize into p orbitals and form the pi bond in the forming double bond. For these two p orbitals to form a pi bond they must be in the same plane (or very close). 


This is illustrated in the following figure. This is called the anti-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~180° with respect to each other) and is the preferred geometry for E2 elimination.


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


[image source](http://www.mhhe.com/physsci/chemistry/carey/student/olc/graphics/carey04oc/ref/ch05eliminationreactions.html)


Elimination can also occur from the syn-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~0° with respect to each other) as shown below, but this conformation is slightly less favored.


![enter image description here](https://i.stack.imgur.com/3dNND.gif)


This is why in molecules where the $\ce{C-X}$ and $\ce{C-H}$ bonds are locked in a conformation (perhaps by an inflexible ring system) where they are oriented 90° apart, E2 elimination becomes very difficult, if not impossible.


 9