Write the product of addition of HBr to propene in the presence and absence of peroxide (anti-Markovnikov) — JEE Chemistry

$$CH_3-CH=CH_2 + HBr$$

Without peroxide (ionic, Markovnikov):

The H adds to C1 (more H), forming 2° carbocation on C2:

$$\rightarrow CH_3-CHBr-CH_3 \quad 2-bromopropane$$

With peroxide (free radical, anti-Markovnikov):

Br• adds to C1 (less substituted, forms more stable 2° radical on C2... wait — radical adds to terminal carbon):

$$\rightarrow CH_3-CH_2-CH_2Br \quad 1-bromopropane$$