When 1-iodo-1-methylcyclohexane is treated with NaOCH2CH3 as the base, the more highly substituted alkene product predominates. When KOC(CH3)3 is used as the base, the less highly substituted alkene predominates. Why? What are the structures of the two products?
1 Answer
Mar 3, 2015
The less highly substituted alkene predominates with potassium t-butoxide because the t-butoxide ion is sterically hindered.
Explanation:
1-iodo-1-methylcyclohexane is a tertiary alkyl halide.
It tends to undergo E1 eliminations to give the most stable product — the most highly substituted alkene — 1-methylcyclohexene.
The t-butoxide ion is a strong base, but it is bulky. It is highly sensitive to steric interactions.
In an E2 mechanism, it attacks the least hindered, most accessible α-hydrogen it can find — an
The product is methylenecyclohexane.
