[areiken]

Hi all!

I've been doing some organic chemistry practice exams and happened across http://puu.sh/gpeeA/827c9c6d3c.png this question. 

Now, I know that the answer is http://puu.sh/gpezu/2bff95c91a.png this.  Up to now, in every elimination reaction I've dealt with it's a Hydrogen trans to the leaving group that supplies the electron for the double bond.  Because one Br is trans to the other, can it act similarly to an H and give its electron to the double bond after being attacked by the Iodide?

Also, based on I^-, which is a weak base and good nucleophile, I reason it to be an Sn2 reaction.  However, the answer is the result of E2.  Is it because of how large the iodide is?

Maybe I haven't gotten this far yet, but I would appreciate some insight.

[OrgXemProf]

Yes, this is an E2-type of elimination. The reactive conformation is that in which the two C-Br bonds are mutually trans, which sets them up for concerted elimination of Br₂ via an anti, coplanar transition state. Iodide ion functions as a nucleophile to attack one of the bromine atoms, which at the time of reaction is polarized toward Br(delta+) [positive halogen, electrophilic; the leaving group is polarized toward Br(delta -)]. The dehalogenation reaction proceeds through a transition state analogous to that associated with concerted anti, coplanar E2 elimination. The products are the corresponding alkene, IBr [polarization I(delta +) Br(delta -)]and bromide ion.

For additional information see Goering, H. L.; Espy, H. H. J. Amer. Chem. Soc. 1955, 77. 5023-5026.

Hope this helps.

[areiken]

Thanks for the *delete me*  Now, the question asked for the major product.  Iodide is a good nuleophile and weak base, and from what I've read this should favour Sn2 over E2.  Why is the E2 product the major one?

[orgopete]

You need to look carefully at the electron movements in the mechanism and less attention to the term "E2 elimination". We often use E2-elimination to refer to a dehydrohalogenation. This is a different reaction. In this case iodide is an electron donor. The product will also contain an I₃^- ion. You would get the similar reaction if you did a reductive elimination with a dissolving metal.