[Ian]

http://img413.imageshack.us/img413/1608/rxn1hm3.jpg

For this one, I guessed this product because when the base takes the Hydrogen, it creates the most stable alkene product (unhindered base).  The CH3 should be on the same side as the long branch, so the long branches are trans from each other. 

http://img413.imageshack.us/img413/1927/rxn2pi3.jpg

For this one, I guessed this product because when the Br first gets trapped by the double bond, the nucleophile comes and attaches to the most substituted carbon.

Am I close?  =)

[Sam (NG)]

Second reaction is fine, Explanation of Electrophilic Addition: http://www.chemicalforums.com/index.php?topic=13736.msg63477#msg63477, your product is racemic.

[Ian]

Does that mean the first is wrong? 

[Sam (NG)]

No, sorry, it means i'm re-reading elimination reactions because i haven't studied them in 2 years, i'll get there in a second.

[Ian]

lol thanks bro 

[Sam (NG)]

Looks like the first one is correct to me. For an explanation of why it goes to that product, i figure that in the polar solvent, you are going to get E1.  Once you have the carbocation intermediate, then the double bond (pi) must form by the Hydrogen being removed being in line with the p orbital of the carbocation,  two transition states are possible here.  The first has an ethyl and a methyl on the same face, while the other has greater steric interaction of 2 ethyl groups together, meaning that the first transition state is lower energy.

If i am wrong and it's E2, then antiperiplanar elimination will result in the same product so the result is the same either way.

(attached picture of E1 possible transition states, the first being lower energy) whoops, the methyl group in the plane of the monitor should actually be a hydrogen that is being eliminated.

[Ian]

Thanks alot bro.  I really appreciate it.