When 2-chloro-2,3-dimethylbutane is treated with different strong bases then elimination causes the formation of two isomers (A and B) with formula C₆H₁₂. When sodium hydroxide is used as a base then the major product is isomer A and when tert-butoxide is used as a base then the isomer B is formed in a higher quantity.
a. Determine the structures of the two isomers using the NMR spectra given below.
b. Explain why isomer A predominates when sodium hydroxide is used while isomer B predominates when tert-butoxide is used.

• A. With NaOH as base, the less substituted alkene, isomer B would predominate while with tert-butoxide as a hindered, bulky base, the highly substituted alkene, isomer A, would predominate.
• B. With NaOH as base, the more highly substituted alkene, isomer A would predominate while with tert-butoxide as a hindered, bulky base, the less substituted alkene, isomer B, would predominate.
• C. With NaOH as base, the less substituted alkene, isomer B would predominate while with tert-butoxide as a hindered, bulky base, the highly substituted alkene, isomer A, would predominate.
• D. With NaOH as base, the more highly substituted alkene, isomer A would predominate while with tert-butoxide as a hindered, bulky base, the less substituted alkene, isomer B, would predominate.

Answer: (D)