Predict the major products when isopropylbenzene is irradiated by light and treated with excess br2.

The reaction of isopropylbenzene with bromine under light is a radical reaction. The light initially reactions with the bromine in the initiation step to form the bromine radicals:

Br₂ + hv → 2Br·

The next step is the first propagation step where a bromine radical reacts with the isopropylbenzene to abstract the hydrogen from the tertiary carbon in the isopropyl group:

                CH₃                          CH₃
                 |                               |
Br· + Ph--C--H  → H-Br + Ph--C·
                 |                               |
                CH₃                          CH₃

We have now formed a radical in the benzylic position of the isopropylbenzene structure. Now this radical will react with another molecule of bromine in a second propogation step to form the final brominated product:

       CH₃                        CH₃
        |                             |
Ph--C·   + Br-Br → Ph--C--Br  + Br·
        |                             |
       CH₃                        CH₃

The bromine radical can terminate in a reaction with another bromine radical or the benzylic radical to give more of the final product. The final product is shown and is called (2-bromopropan-2-yl)benzene.

I’m not sure what your trying to say here.. can you clarify again?