Question

Starting with any alcohols, outline all steps in the synthesis of \(n\) -hexyl isopropyl ether, using the Williamson method.

Short answer

To synthesize n-hexyl isopropyl ether using the Williamson method, start with 1-hexanol and 2-propanol as the starting alcohols. Convert 1-hexanol into hexyl alkoxide by reacting it with sodium hydride (NaH): \[ 1\text{-}C_6H_{13}OH + NaH \rightarrow C_6H_{13}O^{-}Na^{+} \] Next, convert 2-propanol into isopropyl chloride by reacting it with hydrochloric acid (HCl): \[ 2\text{-}C_3H_{7}OH + HCl \rightarrow C_3H_{7}Cl + H_2O \] Finally, perform the Williamson ether synthesis by reacting hexyl alkoxide and isopropyl chloride via an SN2 reaction: \[ C_6H_{13}O^{-}Na^{+} + C_3H_{7}Cl \rightarrow C_6H_{13}OC_3H_7+ NaCl \] The final product is n-hexyl isopropyl ether: \[ C_6H_{13}OC_3H_7 = n\text{-}hexyl\;isopropyl\;ether \]

Step-by-step solution

Choose the alcohols

To synthesize n-hexyl isopropyl ether, we need to choose two alcohols: one with a hexyl group and one with an isopropyl group. In this case, we will use 1-hexanol and 2-propanol as our starting alcohols.

Convert one alcohol to an alkoxide

We need to convert one of the alcohols into its corresponding alkoxide. We will convert 1-hexanol into hexyl alkoxide. To do this, we will react 1-hexanol with a strong base such as sodium hydride (NaH): \[ 1\text{-}C_6H_{13}OH + NaH \rightarrow C_6H_{13}O^{-}Na^{+} \]

Convert the other alcohol to an alkyl halide

We need to convert the other alcohol, 2-propanol, into an alkyl halide. We can do this by reacting 2-propanol with hydrochloric acid (HCl): \[ 2\text{-}C_3H_{7}OH + HCl \rightarrow C_3H_{7}Cl + H_2O \]

Perform the Williamson synthesis

Now that we have prepared the hexyl alkoxide and isopropyl chloride, we can perform the Williamson ether synthesis by reacting them together via an SN2 reaction: \[ C_6H_{13}O^{-}Na^{+} + C_3H_{7}Cl \rightarrow C_6H_{13}OC_3H_7+ NaCl \] The final product, n-hexyl isopropyl ether, has now been synthesized using the Williamson method: \[ C_6H_{13}OC_3H_7 = n\text{-}hexyl\;isopropyl\;ether \]

Key concepts

Alkoxide Formation

The first step in the synthesis of ethers using the Williamson method is the formation of an alkoxide ion. This involves converting an alcohol into its corresponding alkoxide salt by deprotonation, which means removing a hydrogen atom from the alcohol's hydroxyl (–OH) group.

The choice of a strong base is crucial for successful alkoxide creation. Sodium hydride (NaH) is often used because it is a potent base that can effectively deprotonate alcohols without introducing water, which could lead to side reactions. The general reaction for alkoxide formation from an alcohol (ROH) looks like this:
\[ \text{ROH} + \text{NaH} \rightarrow \text{RO}^- \text{Na}^+ + \text{H}_2 \]
For instance, when 1-hexanol (C6H13OH) is treated with NaH, hexyl alkoxide (C6H13ONa) is formed along with the evolution of hydrogen gas. This reaction sets the stage for the next step in the Williamson ether synthesis, where an alkoxide reacts with an alkyl halide.

Alkyl Halide Preparation

The second step in the Williamson ether synthesis is the preparation of an alkyl halide, which will later react with an alkoxide ion to form an ether. An alkyl halide, also known as a haloalkane, contains a halogen atom (such as chlorine or bromine) attached to an alkyl group.

Converting an alcohol to an alkyl halide can be accomplished using hydrochloric acid (HCl), among other halide sources. The reaction involves substitution of the hydroxyl group with a halide:
\[ \text{ROH} + \text{HCl} \rightarrow \text{RCl} + \text{H}_2\text{O} \]
In practice, for producing n-hexyl isopropyl ether, 2-propanol reacts with hydrochloric acid to yield isopropyl chloride. This conversion is critical as the halogen atom will participate in the nucleophilic substitution reaction that forms the ether bond in the final step.

SN2 Reaction

The heart of the Williamson ether synthesis is the \bSN2 reaction (Substitution Nucleophilic Bimolecular), where the alkoxide ion acts as a nucleophile and the alkyl halide as an electrophile. The '2' in SN2 indicates that the rate of reaction depends on the concentration of both the nucleophile and electrophile.

In this process, the nucleophile (alkoxide ion) attacks the electrophilic carbon atom that is bonded to the halide. The reaction proceeds in one concerted step, with the nucleophile pushing out the leaving group (halide). Importantly, SN2 reactions occur with backside attack, which means the nucleophile approaches from the opposite side of the leaving group, often leading to an inversion of stereochemistry at the reaction center.
\[ \text{R}-\text{O}^- + \text{R}^{\backprime}-\text{X} \rightarrow \text{R}-\text{O}-\text{R}^{\backprime} + \text{X}^- \]
For our specific case, the hexyl alkoxide nucleophile combines with the isopropyl chloride to form n-hexyl isopropyl ether, illustrating the importance of the SN2 mechanism in producing ethers.

n-Hexyl Isopropyl Ether Synthesis

Following the previous steps, we move on to the final stage, which is the synthesis of n-hexyl isopropyl ether. This is a practical example of how the Williamson ether synthesis is applied to create a specific ether compound.

We've already prepared our reactants: hexyl alkoxide from 1-hexanol and isopropyl chloride from 2-propanol. Now, in the presence of a suitable solvent—often an aprotic solvent that doesn't interfere with the reaction—the alkoxide ion performs a nucleophilic attack on the alkyl halide. This results in the departure of the chloride ion (Cl-) and the formation of our desired ether linkage. Below is the simplified representation of this synthesis:
\[ \text{C}_6\text{H}_{13}\text{O}^- \text{Na}^+ + \text{C}_3\text{H}_7\text{Cl} \rightarrow \text{C}_6\text{H}_{13}\text{O}\text{C}_3\text{H}_7 + \text{NaCl} \]
Finally, n-hexyl isopropyl ether (\text{C}_6\text{H}_{13}\text{OC}_3\text{H}_7) is obtained as the product. It's worth noting that the reagents chosen in the beginning—1-hexanol and 2-propanol—and the sequence of steps are crucial for the success of the synthesis without encountering significant side reactions.