Question

Draw the product formed by the reaction of potassium t-butoxide with (1s,2s)-1-bromo-2-methyl-1-phenylbutane (shown below). (draw the correct stereoisomer of the product.)

Answer 1

 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.

Answer 2

Answer:

(Z)-(2-methylbut-1-en-1-yl)benzene

Explanation:

In the E2 elimination reaction, we will have a single-step mechanism. In this single step, we need an anti-peri-planar geometry. In other words, we need a specific configuration for the leaving group and the H that would be removed. These 2 atoms must have an opposite configuration when the base attacks the hydrogen.

With this in mind, we have to rotate the molecule in order to obtain the desired geometry (see figure). When we do this, the methyl group would be placed on top and the ethyl group at the bottom.  

Finally, when the elimination takes place (blue arrows) the configuration of the final alkene would be "Z" due to the previous rotation that we did in order to obtain the correct configuration for the elimination.