Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

a. Propose a mechanism to show why free-radical halogenation occurs almost exclusively at the benzylic position.
b. Draw the two stereoisomers that result from monobromination at the benzylic position.
c. Assign R and S configurations to the asymmetric carbon atoms in the products.

Verified step by step guidance

1. Step 1: Understand the reaction conditions. Free-radical bromination occurs under the presence of bromine (Br₂) and light (hv). The benzylic position is highly reactive due to the stability of the benzylic radical formed during the reaction.
2. Step 2: Propose the mechanism for free-radical bromination. The reaction begins with the homolytic cleavage of Br₂ under light to form two bromine radicals. The benzylic hydrogen is abstracted by a bromine radical, forming a benzylic radical. This radical is stabilized by resonance with the aromatic ring, making the benzylic position the preferred site for bromination.
3. Step 3: The benzylic radical reacts with another bromine molecule (Br₂), leading to the formation of the monobrominated product at the benzylic position. This product has a new chiral center due to the introduction of bromine.
4. Step 4: Draw the two stereoisomers resulting from monobromination. Since the benzylic position is now a chiral center, two stereoisomers (R and S configurations) are formed. These stereoisomers differ in the spatial arrangement of the substituents around the chiral center.
5. Step 5: Assign R and S configurations to the products. Use the Cahn-Ingold-Prelog priority rules to determine the configuration of the chiral center in each stereoisomer. Assign priorities based on atomic number and analyze the arrangement of substituents to label each stereoisomer as R or S.