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

(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.

the answer is true.

thats how its supposed to be