Question

What is the product of the reaction between 2-bromo-2-methylpentane and sodium ethoxide in ethanol?

Answer 1

Here's what I get.

Explanation:

The tertiary substrate favours an `"S"_text(N)1//"E1"` mechanism.

The strong base favours elimination, so we draw an `"E1"` mechanism.

The strongest base in the reaction mixture is the ethoxide ion, formed by the reaction

`underbrace("CH"_3"CH"_2"O-H")_color(red)("ethanol") + "OH"^"-" ⇌ underbrace("CH"_3"CH"_2"O"^"-")_color(red)("ethoxide ion") + "H-OH"`

The mechanism is

Step 1. The `"Br"` atom leaves as in a typical `"S"_text(N)1` ionization.

Step 2. The strong base removes a β-hydrogen atom to form the most stable (most highly-substituted) alkene.