[rleung]

Hey,

I am slightly confused about this one thing.  It asks me to predict the elimination product by debromination of the following molecule.  I understand that since the two Br's are coplanar, they need to be anti-coplanar in order to debrominate.  However, the solutions manual contorts the molecule so that one Br is facing up and the other Br is facing down (both in equatorial positions), and they when they return it to its original shape after debromination and formation of the double bond, they get the trans-cyclodecene structure on the right.  I am unsure as to how it is ONLY trans-cyclodecene.  

Depending on how you decide to return it to its original shape, couldn't it be cis-cyclohexene as well (like in the molecule below this post)?  It is just a matter of where the double bond lies.

Thanks.

Ryan

[rleung]

Here is an example of cis that I think can just as well be a product of debromination.

[movies]

What are the conditions for the debromination reaction?  Must be radical, huh?

[AWK]

http://www.mr.mnscu.edu/maresch/Chapter%207/322a_Ch_7_11.pdf

[rleung]

There were no conditions for the reaction.  This was a problem in the introductory section about debromination.  The book only starts covering debromination reactions when it gives us this problem.  I understand how the molecule has to be contorted so that both Br's are anti-coplanar with each other (in axial positions), but I just don't understand how the product of debromination HAS to be trans and nothing else.  Is it only a matter of how you return the molecule to its original shape (as in where you choose to place the newly formed double bond)?  In my molecular model kit, I can just as easily make the product cis as I can make it trans.

Ryan

[movies]

Think about the transition state for the reaction.  Remember that once you form a C-C double bond you can't freely rotate about it anymore.  What consequences does that have?

[rleung]

Ahh....thank you so much.  I get it now.