Final answer:
When 2-methylhex-2-ene is treated with HBr in the presence of peroxides, the reaction follows a free radical mechanism. The reaction predominantly forms a trans product with a new chiral center. This result is due to the restriction of rotation around the double bond caused by the methyl group at the 2-position of the hexene ring.
Step-by-step explanation:
To draw the products formed when 2-methylhex-2-ene is treated with HBr in the presence of peroxides, we need to consider the reaction mechanism. When an alkene reacts with HBr in the presence of peroxides, a free radical mechanism occurs.
The first step is the initiation step, where the peroxide breaks down into two radicals:
R-O-O-R → 2R•
Next is the propagation step, where the alkene reacts with the HBr radical to form a bromoalkane:
R• + HBr → RBr + H•
The produced radical H• can then react with another HBr molecule to regenerate HBr, and the reaction can continue.
This process can result in the formation of different stereoisomers. However, in the case of 2-methylhex-2-ene and (S)-2,4-dimethylhex-2-ene, the presence of a methyl group at the 2-position of the hexene ring restricts the rotation around the double bond.
As a result, when treated with HBr in the presence of peroxides, these compounds will predominantly form a single stereoisomer, a trans product where the bromine atom adds to the carbon of the double bond, creating a new chiral center.