Chapter 13.10, Problem 13.70P

Interpretation Introduction

Interpretation:

The way the given compound is prepared using Williamson ether synthesis has to be shown along with reagents and starting with 1-propanol.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups.  The R groups may be alkyl, vinyl or aryl.

Alcohols when treated with a very strong base undergoes deprotonation to form an alkoxide.  The formed alkoxide ions can acts as a strong nucleophile for the ${\text{S}}_{\text{N}}\text{2}$ reaction.  When an alkoxide and an alkyl halide is combined to form an ether, it is known as Williamson ether synthesis.  Usage of primary alkyl halide is recommended for Williamson ether synthesis because the reaction proceeds via ${\text{S}}_{\text{N}}\text{2}$ mechanism.  If secondary alkyl halides are used, the elimination reaction will predominate over substitution reaction.  Tertiary alkyl halides cannot be used at all because, it undergoes only ${\text{S}}_{\text{N}}\text{1}$ reaction.

The general scheme for Williamson ether synthesis can be given as,

Answer to Problem 13.70P

The reaction scheme can be given as,

Explanation of Solution

Given compound is,

First step is to analyse the $\text{C}-\text{O}$ bonds in the given structure.

In the problem statement it is given that the alcohol is 1-propanol.  Hence, the other $\text{C}-\text{O}$ bond has to be primary halide.

The following reaction sequence can be considered for obtaining the given compound via Williamson ether synthesis using 1-propanol.

The primary halide considered here is an iodide.  This is because alkyl iodides are more reactive than chlorides and bromides.