Propene, also known as propylene, is an unsaturated hydrocarbon with the chemical formula C3H6. When propene reacts with HBr (hydrogen bromide) in the presence of peroxide, it undergoes a radical addition reaction to form n-propyl bromide (n-C3H7Br).

Here is a detailed explanation of the reaction:

1. Radical Initiation:
The presence of peroxide (often represented as ROOR, where R is an alkyl group) initiates the reaction by undergoing homolytic cleavage to produce two alkyl radicals, represented as R•. In this case, the peroxide could be tert-butyl peroxide (C(CH3)3OO•).

ROOR → 2R•

2. Radical Propagation:
The alkyl radical (R•) reacts with propene (C3H6) to form a more stable secondary radical intermediate.

R• + C3H6 → RCH2CH2•

The secondary radical intermediate then reacts with HBr to produce the alkyl bromide.

RCH2CH2• + HBr → RCH2CH2Br

This reaction step can occur at any position on the propene molecule, resulting in different possible alkyl bromide products.

3. Radical Termination:
The reaction can also undergo radical termination steps where two alkyl radicals combine to form a non-radical species. These termination steps play a role in consuming any excess alkyl radicals and stopping the chain reaction.

R• + R• → R-R

In the case of propene reacting with HBr in the presence of peroxide, the major product formed is n-propyl bromide (n-C3H7Br). This is because the secondary radical intermediate reacts preferentially with HBr at the terminal carbon of propene, resulting in the formation of n-propyl bromide. The other possible products, such as allyl bromide (C3H5Br), isopropyl bromide (C3H7Br), and 3-bromopropane (C3H7Br), are less favored due to the stability of the radical intermediate and the reactivity of HBr at the terminal carbon.

In summary, when propene reacts with HBr in the presence of peroxide, the major product obtained is n-propyl bromide (n-C3H7Br). This reaction follows a radical mechanism, initiated by the homolytic cleavage of peroxide and proceeds through radical propagation steps to form the alkyl bromide product.