Question

Reaction of $HBr$ with propene in the presence of peroxide gives

• A. iso-propyl bromide
• B. 3-bromo propane
• C. allyl bromide
• D. n-propyl bromide

Answer 1

The Correct Option is D

The reaction of HBr with propene in the presence of peroxide is a classic example of an anti-Markovnikov addition, also known as the Kharasch effect. In this reaction, the presence of peroxides leads to a free-radical mechanism which results in the anti-Markovnikov product.

• Normally, HBr adds to an alkene in a Markovnikov fashion, where the hydrogen atom attaches to the less substituted carbon, and the bromine atom attaches to the more substituted carbon. This mechanism follows an ionic pathway.
• However, the presence of peroxides changes the mechanism to a free-radical pathway. The peroxide radical initiates the formation of bromine free radicals, which then react with the alkene.
• In the free-radical mechanism, the bromine radical attaches to the less substituted carbon atom, leading to the formation of a more stable free radical intermediate.
• This intermediate then reacts with a molecule of HBr to give the final product, where the hydrogen atom attaches to the free radical site.

Applying this concept to propene, we get the following step-by-step process:

1. Propene is treated with bromine radical (formed from peroxide-initiated decomposition of HBr).
2. The bromine radical adds to the less substituted terminal carbon of propene (anti-Markovnikov addition).
3. This generates a free radical at the 1-position.
4. This free radical then reacts with another HBr molecule, where the hydrogen attaches to the radical site, forming n-propyl bromide.

Therefore, the correct product of the reaction of HBr with propene in the presence of peroxide is n-propyl bromide.

Correct Answer: n-propyl bromide