[kelaklub]

The following reaction is a halogenation reaction and the answer is stated as choice C. Is there a possible typo with the lone hydrogen still being attached to the most substituted isopropyl carbon? Thanks.

[Albert]

Answer C is clearly the right one, except for that hydrogen, which is obviously misprinted.

[kelaklub]

That's what I thought. Thanks.

[victor]

is it because that there's no FeBr₃ catalyst present?(so Br group will get attached into isopropyl group?)

[Albert]

is it because that there's no FeBr₃ catalyst present?(so Br group will get attached into isopropyl group?)

I would just say it's a radical halogenation of an alkane.

[Donaldson Tan]

When there's UV, you gonna have Bromine radicals everywhere. The alkyl group is most susceptible to radical attacks compared to the benzene ring which is reasonance stabilised. Your main product will be (c)

[movies]

The reason that it goes to that particular position is because the benzyllic radical can be stabilized by the adjacent aromatic ring.  It's the difference between stabilizing an intermediate and stabilizing the transition state that gets you there.

If it was just the difference between the availability of hydrogens on the aromatic ring versus those not on the aromatic ring, you would expect to see bromination on the methyl groups as well.