Ah here we go. I remember this problem. It's a tricky one. First you have to know a little something about this reaction. It is Bromination / addition across a double bond that adds anti. So when Br2 adds and no chiral centers are created you have 1 product. When Br2 adds across a double bond generating 1 chiral center you have 2 products every time. When 2 chiral centers are created you can have at most 2 products; 2 enantiomers. ("At most" 2 products means you could have created a meso compound and in that case you would have 1 product). Anyway adding Br2 to choices A, B, D, will generate 0, 1, and 2 chiral centers respectively. However, a Br2 to compound C will result in a product with 3 stereocenters. Check out what it looks like in the solutions and convince yourself that choice C has 3 stereocenters: 2 that contain Br and a 3rd containing the ETHYL substituent. Now becasuse you have 3 chiral centers, you can have diastereomers. Rem enantiomer = opp config at EVERY chiral center. Look at the eg in the solution. You'll see those are diastereomers.
This problem doesn't give you the configuration of the ethyl group in choice C, which makes it a little awkward but it does let you practice a bunch of different concepts at once. Which is why destroyer is so badass. keep it up.
