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

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Answer 1 · 2017-03-19T22:37:07

Here's what I get.

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

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