Chapter 15, Problem 15.24P

When $\text{HBr}$ adds to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}={\text{CH}}_{\text{2}}$ under radical conditions, two radicals are possible products in the first step of chain propagation. Draw the structure of both radicals and indicate which one is formed. Then draw the preferred product from $\text{HBr}$ addition under radical conditions.