Drawing Resonance Products

Question

We are presented with the compound (I think it's called) 2-ethylcyclohex-1-ene. It's a hexagon shape, there's a double bond starting on C1 then there's an ethyl group branching off of C2. The problem asks:

Draw the product formed when the compound shown below undergoes a reaction with Br₂ in CH₂Cl₂

While also drawing the correct stereochemistry. I understand a Br ion will go on each side of the double bond to break the pi bond, but how do I know which bond goes in which direction? Am I even approaching this correctly?

Answer 1

This is a picture of how the reaction proceeds:

A Br is added to each side of the double bond. The stereochemistry of the reaction is that the new bonds are on opposite faces of the reacting π bond. In a flat drawing of the cyclohexane ring there will be one dashed and one wedged C–Br bond.