Free-radical bromination of the following compound introduces bromine primarily at the b...

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.

(d) What is the relationship between the two isomeric products?

(e) Will these two products be produced in identical amounts? That is, will the product mixture be exactly 50:50?

(f) Will these two stereoisomers have identical physical properties like boiling point, melting point, solubility, etc.?

Could they be separated (theoretically, at least) by distillation or recrystallization?