The hydrocarbon shown is monobrominated, counting every distinct product formed by replacing any one hydrogen (ring positions with Br2/FeBr3 and side-chain positions with Br2/light; constitutional isomers only). How many products are possible?

JEE Advanced levelSingle correct optionDifficulty 3 of 3si268

The hydrocarbon shown is monobrominated, counting every distinct product formed by replacing any one hydrogen (ring positions with $Br X_{2}$ / $FeBr X_{3}$ and side-chain positions with $Br X_{2}$ /light; constitutional isomers only). How many products are possible?

The compound shown is cumene. Ring: ortho, meta and para bromides, 3. Side chain: the tertiary benzylic C-H and the six equivalent methyl H, 2. Total

3 + 2 = 5