Given the reaction of ${\mathrm{CH}}_3\mathrm{CH}=\mathrm{CH}2$ (propene) with $\mathrm{HBr}$ in the presence of peroxides, what would be the major product?

• A. ${\mathrm{CH}}_3\mathrm{CH}2$CH2Br (1-bromopropane)
• B. ${\mathrm{CH}}_3\mathrm{CH}=\mathrm{CH}2$ (no reaction)
• C. ${\mathrm{CH}}_3\mathrm{CH}\mathrm{Br}{\mathrm{CH}}_3$ (2-bromopropane)
• D. None of the above

Answer: (A)

Verified step by step guidance

1. Identify the type of reaction: The reaction of an alkene (propene) with HBr in the presence of peroxides is a classic example of a radical addition reaction, specifically an anti-Markovnikov addition due to the peroxide effect (also known as the Kharasch effect).
2. Recall the mechanism: In the presence of peroxides, the reaction proceeds via a radical chain mechanism rather than the usual ionic mechanism. The peroxide initiates the formation of bromine radicals (Br•), which add to the alkene.
3. Determine regioselectivity: The bromine radical adds to the less substituted carbon of the double bond to form the more stable carbon radical intermediate. This is opposite to the Markovnikov rule, which applies in the absence of peroxides.
4. Complete the radical chain: The carbon radical formed then reacts with HBr, abstracting a hydrogen atom to form the bromoalkane and regenerate the bromine radical, propagating the chain reaction.
5. Predict the major product: Since the bromine adds to the less substituted carbon, the major product is 1-bromopropane, where the bromine is attached to the terminal carbon of propene.