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

1 Answer
Mar 19, 2017

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

Mechanism

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.