What are the possible products in allylic bromination of 3-methylcyclopentene? Bromination of an alkene by N-bromosuccinimide (NBS) in the presence of light or peroxide is a radical reaction and produces an allylic bromide. For the following bromination of 3-methylcyclopentene, which of the following allylic bromides would be products of the reaction?


In some other reactions such as elimination, there is a tendency to form the most substituted double bond, since it is the most stable. If that were the case, then the products should be A, E, F, and G, since these are all trisubstituted alkenes (the others are disubstituted). Is that a valid consideration here, or is there a different overriding rule?
Your comment shows that you are on the right track.

You have chosen A and C because they are formed when the initial allylic radical instantaneously reacts with $\ce{Br2}$ in the propagation step of the reaction, isn't it? (Hint: Just say YES!)

Now give the allylic radicals corresponding to A and C a bit of time, just enough to flip the dot.

How do these other radicals look like? There you have the other two products.

For an extra point: Which isomer of methylcyclopentene could lead to G?