# The major product which results when 2-chloro-2-methylpentane is heated in ethanol is an ether. What is the mechanism by which this ether forms?

May 20, 2015

The ether forms by an $\text{S"_"N} 1$ mechanism.

The overall reaction is

$\text{CH"_3"CH"_2"CH"_2"C(CH"_3")"_2"Cl" + "CH"_3"CH"_2"OH" ⇌ "CH"_3"CH"_2"CH"_2"C(CH"_3")"_2"(OCH"_2"CH"_3")" + "HCl}$

The substrate is a 3° halide, and the ethanol is both a weak nucleophile and a polar protic solvent, so the reaction will go by an $\text{S"_"N} 1$ mechanism.

There will be no rearrangements, because the carbocation is already 3°.

THE MECHANISM

Step 1. Loss of ${\text{Cl}}^{-}$ to form a 3° carbocation.

${\text{CH"_3"CH"_2"CH"_2"C(CH"_3")"_2"Cl" → "CH"_3"CH"_2"CH"_2"C(CH"_3")"_2^+ + "Cl}}^{-}$

Step 2. Nucleophilic attack by ethanol

${\text{CH"_3"CH"_2"CH"_2"C(CH"_3")"_2^+ + "CH"_3"CH"_2"OH" → "CH"_3"CH"_2"CH"_2"C(CH"_3)_2"-O"^+"(H)CH"_2"CH}}_{3}$

Step 3. Deprotonation of the oxonium ion

${\text{CH"_3"CH"_2"CH"_2"C(CH"_3)_2"-"^+"O(H)CH"_2"CH"_3 + "CH"_3"CH"_2"-OH" → "CH"_3"CH"_2"CH"_2"C(CH"_3)_2"-OCH"_2"CH"_3 + "CH"_3"CH"_2"-OH}}_{2}^{+}$