# 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?

• I think it should be A,C,F,D ,though I am not sure. I'll do some review and get back. Feb 8 '14 at 5:28
• Thank you, I agree with A, C, and F wondering why it cant be G or E :/ confusing. Feb 8 '14 at 6:13

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?

Alkene substitution is only really important in a couple of cases: firstly when the reaction is entirely thermodynamically controlled (e.g. aldol reactions), and secondly if the transition state in the rate-determining step contains strong double bond character (i.e. more substituted double bond = lower-energy transition state).

Neither of that is the case here. The double bond is constrained by its initial position, and cannot jump around all that much. In theory, there are $$2 \times 2 = 4$$ possible products (A, C, D, and F) because:

• the initial abstraction of a hydrogen atom can occur at two different positions, and

• the allylic radical thus formed can pick up a bromine atom at two different positions.

The mechanistic possibilities are drawn out fully below:

In practice, there would likely be one product that is preferred over the others. Note the use of resonance arrows, $$\ce{<->}$$: these are not flipping about between the two different resonance structures. Rather, the reality is that the allylic radical is somewhere in-between the two extremes. We only draw out the two extremes here because it's a convenient way of looking at its reactivity (which is sort of a mixture of both extremes).