Well, you got off to the right start. The important factor is the conformation of the ring.
Let's start at the start; here is a generic elimination reaction to form ethylene from iodoethane:
The leaving group and the proton that is pulled off by the base must be anti-periplanar to one another, meaning they are in the same plane, but pointing in opposite directions. See the Newman projection below:
I have also shown a scheme with the orbitals of the starting material and the product shown. These alignment of these orbitals is the reason that the anti-periplanar geometry is necessary for reaction by the E
2 mechanism. The sp
3 hybrid orbitals will "un-hybridize" to eventually form the pi bond in the product. As you know, for pi bonding, the two p-orbitals must be in the same plane. Forming the alkene directly is a much lower energy reaction pathway than the alternative (deprotonation of a non-anti-periplanar hydrogen followed by rotation to place the resulting anion anti-periplanar to the leaving group followed by elimination). Specifically, the electrons from the C-H bond are released from the C-H bond and then donated into the adjacent sigma anti-bonding orbital (not shown) of the C-I bond.
So, let's apply these principles to the cyclic case at hand.
In both cases there are two possible sites for deprotonation, the two carbons adjacent to the carbon bearing the bromine (labeled 1 and 2). In order to achieve the anti-periplanar geometry that is necessary, the bromine
must be axial (draw out all the possible Newman projections if you don't believe me). Let's look at carbon 2, since it is the interesting one. In this case, when we place the bromine axial we see that the proton at carbon 2 is anti-periplanar and elimination can proceed as we would expect. If you drew out the Newman projection for carbon 1, you would find that elimination can occur from that side as well. So, we could get two products from the
cis starting material.
Now for the
trans isomer:
In this case when we apply the same model we find that the bromine and the hydrogen at carbon 2 cannot achieve an anti-periplanar arrangement! So elimination cannot occur from carbon 2. Were you to draw out the Newman projection for carbon 1, you would find that a hydrogen can be anti-periplanar to the bromine so elimination can occur from carbon 1. Only one product would be expected from this starting material.
Cool, huh?
