[griffon87]

I believe this is an E2 reaction because potassium t-butoxide favors E2. Am I suppose to have a double bond between the benzene and where the carbon which was originally bonded to Bromide? I tried removing the hydrogen on the left side and double bonded C1 and C2 but it was wrong. So, I think it's suppose to be a double bond between benzene and the carbon originally bonded to the bromide, but I'm unable to obtain the correct answer. Any helpful hints will be greatly appreciated. Thanks!

[azmanam]

E2 is right, for the reason you described.  you had the double bond right the first time.  I suspect the problem is E/Z selectivity.  Make sure you get the correct orientation of the starting material for an E2 elimination.  That specific orientation will give rise to a specific double bond geometry. 

Does that help?


also for more generalities:
http://www.chem.uky.edu/courses/che230/RBG/handouts/SNorE/SNorE_print.HTML
http://www.cem.msu.edu/~reusch/VirtTxtJml/alhalrx3.htm#hal9

[griffon87]

Thank you. It was the orientation like you said, and thanks for the 2 sites you posted. I'll be sure to check them out.

[pantsboy]

i find it helpful to draw Newman Projections to solve problems like these so I get the stereochemistry right.  First place the hydrogen antiperiplanar (opposite position) to the leaving group and you should have a transition state that will resemple the orientation of the final product.