[UpbeatWhiz]

Hello Everyone!
 I am trying to do this E2 Reaction problem where it asks me to deduce the best starting material for this. When I submitted my answer to the assignment it said I was partly right but missing one thing, can someone please explain to me what might be wrong! I feel like I did everything right. I was thinking that maybe the double bonded rings in the cyclohexane were removed or maybe I'm suppose to use the dash and wedge to show cis and trans but I'm not sure what would get the wedge/dash if that was even the case!

Here is the picture of the question, my answer is on the left hand side!

http://img707.imageshack.us/img707/2576/helpmej.png

Thanks everyone!

[CaverKat]

You may need to show the absolute configuration of the stereogenic centres, so that you get the hydrogen anti-periplanar to the Cl leaving group when the two large groups arrange themselves to minimise steric interactions.

[orgopete]

You may need to show the absolute configuration of the stereogenic centres, so that you get the hydrogen anti-periplanar to the Cl leaving group when the two large groups arrange themselves to minimise steric interactions.

Almost correct. You need to show the relative configurations (you will get the same product from the racemic mixture as the optically active isomer indicated by an absolute configuration). In this case, you should have added hash and wedges to the bonds.

While CaverKat is correct about the conformations, they may not control the products. In this case, since elimination is possible from a single hydrogen, that hydrogen needs to be anti-periplanar as CaverKat noted. If the opposite relative stereochemistry were present, the conformations would be altered and the preferred product would be the Z-isomer.

While anti-eliminations can be much faster than syn (~500x (or more?)), the conformations and rates of the proposed reactions will probably not be equal. Thus, the elimination to give the E-isomer will probably be faster because of conformational preferences. You may find small amounts of the syn-elimination product in each reaction, but the ratio will be much higher (less syn-elimination) for the E-product.