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

Mar 19, 2017

Here's what I get.

#### Explanation:

The tertiary substrate favours an $\text{S"_text(N)1//"E1}$ mechanism.

The strong base favours elimination, so we draw an $\text{E1}$ 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 $\text{Br}$ atom leaves as in a typical ${\text{S}}_{\textrm{N}} 1$ ionization.

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