<span>You have to use a Newman projection to make sure that the H on C#2 is anti-coplanar with the Br on C#1. (Those are the two things that are going to be eliminated to make the alkene.)
My Newman projection looks like this when it's in the right configuration:
Front carbon (C#2) has ethyl group straight up, H down/left, and CH3 down/right
Back carbon (C#1) has H straight down, Ph up/left, and Br up/right.
Then when you eliminate the H from C#2 and the Br from C#1, you will have Ph and the ethyl group on the same side of the molecule, and you'll have the remaining H and CH3 on the same side of the molecule.
This is going to give you (Z)-2-methyl-1-phenyl-1-butene.</span>
Explanation:
o2- Due to lowest Zeff.
I had that question and my answer was this. because you did not put options.
