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.)

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.