What is the major product formed when ${\text{(CH}}_3{)}_3\text{CCH}={\text{CH}}_2$ reacts with $\text{HBr}$ in the presence of a peroxide?

Solution

1. The reaction of an alkene with $\text{HBr}$ in the presence of peroxide follows a free radical mechanism.
2. The bromine radical adds to the terminal carbon to form the more stable secondary radical.
3. This mechanism avoids the formation of a carbocation, thus preventing rearrangement of the neopentyl group.
4. The final product is the anti-Markovnikov addition product, ${\text{(CH}}_3{)}_3{\text{CCH}}_2{\text{CH}}_2\text{Br}$.
5. Hence the answer is (A).