How man different alkene-products, including stereoisomers, are possible when this molecule reacts in an E2 elimination reaction?
What will happen here is that I leaves the molecule, while a new double bond is created. The new bond can be created on two places, so that we have two different alkenes. We also have two stereoisomers (E and Z) for each of these solutions. I expect the answer to be 4.
However, the correct answer is 5. What's the last product?
