Question

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. Give the structures of the two products and offer an explanation.

Answer 1

Answer:

See explanation

Explanation:

In this case, we have 2 types of reactions. $CH_3CH_2ONa$ is a strong base but only has 2 carbons therefore we will have less steric hindrance in this base. So,  the base can remove hydrogens that are bonded on carbons 1 or 6, therefore, we will have a more substituted alkene (1-methylcyclohex-1-ene).

For the  $KOC(CH_3)_3$ we have more steric hindrance. So, we can remove only the hydrogens from carbon 7 and we will produce a less substituted alkene (methylenecyclohexane).

See figure 1

I hope it helps!