Stereochemistry of a Free Radical Reaction

A substitution reaction at a stereogenic center can lead to a racemic mixture of products. For example, in the free radical reaction of bromine with (5) l-chloro-2-methylbutane, a bromine atom replaces a hydrogen atom at the tertiary stereogenic center to give a racemic mixture of R) and (5)-2-bromo-l-chloro-2-methylbutane. 

A free radical intermediate is achiral because it has a plane of symmetry. A bromine molecule can therefore attack with equal probability from above or below the plane to give a 50 50 mixture of enantiomers. The 2p orbital is half-occupied, and there is a 50% probability of finding an electron above or below the nodal plane of the orbital. 

Free radical chlorination of (5)-2-bromobutane yields a mixture of compounds with chlorine substituted at any of the four carbon atoms. Write the structure of the 2-bromo-l-chlorobutane formed. Determine the configuration (s) of the stereogenic center(s). Is the product optically active  

Based on the data for the conversion of (i )-2-bromooctane into (i)-2-octanol using NaOH, predict the product of the reaction of (i)-2-bromooctane with NaOH. 

Nucleophilic attack at the side opposite the bond of the displaced leaving group from (5)-2-bromooctane gives a product with inversion of configuration. Thus, the enantiomeric R compound should react likewise and gives an inverted product, (i)-2-octanol. 

---