When optically active 2-chloro-3-methylhexane is treated with aqueous NaOH , what is the structure and stereochemical nature of the major product formed?

Step-by-step solution

The reaction of a secondary alkyl halide in a polar protic solvent (aqueous NaOH ) proceeds via an S _ N 1 mechanism. First, ionization of optically active 2-chloro-3-methylhexane forms a secondary carbocation at C2: CH ₃ -CH(Cl)-CH(CH ₃ )-CH ₂ -CH ₂ -CH ₃ CH ₃ -CH ^+ -CH(CH ₃ )-CH ₂ -CH ₂ -CH ₃ Next, a 1,2-hydride shift from the adjacent tertiary C3 carbon occurs to generate a more stable tertiary carbocation at C3: CH ₃ -CH ^+ -CH(CH ₃ )-CH ₂ -CH ₂ -CH ₃ CH ₃ -CH ₂ -C ^+ (CH ₃ )-CH ₂ -CH ₂ -CH ₃ The tertiary carb