This reaction involves the addition of HBr to an alkene, but with a key twist: it proceeds via free radical addition due to the presence of peroxides (ROOR). This is known as the Anti-Markovnikov addition or Kharasch effect.

Normally, HBr adds to alkenes following Markovnikov's rule (hydrogen adds to the less substituted carbon). However, in the presence of peroxides, the reaction mechanism changes to free radical addition, resulting in anti-Markovnikov addition.

Step-by-Step Mechanism:

Step 1: Peroxide (ROOR) undergoes homolytic cleavage to form alkoxy radicals (RO•).

R O - O R \to 2 R O \cdot

Step 2: The alkoxy radical abstracts a hydrogen from HBr, forming a bromine radical (Br•).

R O \cdot + H - B r \to R O H + B r \cdot

Step 3: The bromine radical adds to the alkene. In anti-Markovnikov addition, Br• adds to the less substituted carbon of the double bond, forming a more stable (more substituted) carbon radical.

For the given alkene (1-methylcyclohexene), the double bond is between C1 and C2, where C1 is tertiary (more substituted) and C2 is secondary (less substituted). The bromine radical adds to C2 (less substituted), creating a tertiary radical on C1.

Step 4: The carbon radical abstracts a hydrogen from another HBr molecule, forming the product and regenerating Br• to continue the chain reaction.

Carbon \cdot + H - B r \to Carbon - H + B r \cdot

The final product is 1-bromo-1-methylcyclohexane (bromine on the less substituted carbon, which was originally C2). This corresponds to option P.

Why not other options?

Options like Q or R would result from Markovnikov addition (without peroxides).
Option S might imply a different regiochemistry or side product, which is not favored here.

Related Topics:

Free Radical Addition
Markovnikov vs. Anti-Markovnikov Addition
Reactivity of Alkenes

Key Formulae/Concepts:

Anti-Markovnikov Addition: HBr + Peroxide → Br adds to less substituted carbon.
Stability of Radicals: Tertiary > Secondary > Primary.