free radical halogenation

a) There are five different carbons in 2-methylbutane, so hypothetically, you should get five different products, 20% A, 20% B, 20% C, 20% D, and 20% E.

But notice that A and B are the same product! So only four products are possible.

b) Based on statistics alone, we would expect product A (or B) to be the major product, encompassing 40% of all the products (20% A + 20% B).

But this is not what happens in real life because A is primary, and for free-radical halogenation reactions, more substituted products are favored. So the major product will be C, which is tertiary.

This happens because the intermediate is a radical, and radical stability goes 3° > 2°  > 1°.

c) Chlorine reacts faster than bromine, and so is less selective than bromine.

For example, we know that the 3° halide will be the major product. But with chlorine, the proportion might turn out to be ~70% 3° , 20% 2°, and 10% 1°.

But since bromine reacts more slowly, you will get better selectivity. The proportion of products will might be something like ~95% 3° , 4% 2°, and 1% 1°.

While on this topic of reactivity, it's interesting to note that iodine reacts too slowly to be useful, and fluorine reacts so violently it's dangerous!

This is a good synthesis to know, because vinyl benzene is a good starting point for many synthesis problems you will encounter down the road.

An ethyl group can be added to benzene via Friedel-Crafts alkylation. 

To add that alkene, we you will have to do some sort of elimination reaction. The easiest way to do this is via bromination. Free radical bromination (with NBS or Br₂/hν) will add a bromine to the position of the most stable radical, which is the benzylic position (due to resonance).

Then adding a strong base like potassium tert-butoxide will do an E2 reaction to form the alkene.

This is a free radical halogenation. The hint is that you see light (hν) in the reaction arrow, indicating radicals are involved.

The more substituted the carbon atom, the more stable the corresponding radical, so the major product of this reaction is the 3º alkyl halide.

Note that the radical will form on the 3° carbon only, as this is the most stable radical.