Question

Show how ethyl hexyl ether might be prepared by a Williamson ether synthesis.

Short answer

Question: Show how ethyl hexyl ether can be prepared using Williamson ether synthesis. Answer: Ethyl hexyl ether can be prepared by reacting sodium ethoxide (CH3CH2ONa) with hexyl bromide (CH3CH2CH2CH2CH2CH2Br) through Williamson ether synthesis. The resulting product will be ethyl hexyl ether (CH3CH2OCH2CH2CH2CH2CH2CH2) and sodium bromide (NaBr).

Step-by-step solution

Determine the structure of ethyl hexyl ether

Ethyl hexyl ether has an ethyl group (CH3CH2) and a hexyl group (CH3CH2CH2CH2CH2CH2) connected by an ether oxygen atom. As a result, the structure can be represented as CH3CH2OCH2CH2CH2CH2CH2CH2.

Choose the alkoxide ion and alkyl halide

We can choose either an ethoxide ion or a hexoxide ion as the alkoxide ion, and an ethyl halide or a hexyl halide as the alkyl halide. Let's use sodium ethoxide (CH3CH2ONa) as the alkoxide ion and a hexyl halide (hexyl bromide) as the alkyl halide. Sodium ethoxide can be prepared by reacting ethanol with sodium metal. Hexyl bromide can be prepared by reacting 1-hexanol with a halogenating agent such as PBr3 or HBr.

Write the Williamson ether synthesis reaction

Now, we can write the reaction of sodium ethoxide with hexyl bromide to form ethyl hexyl ether: CH3CH2ONa + CH3CH2CH2CH2CH2CH2Br -> CH3CH2OCH2CH2CH2CH2CH2CH2 + NaBr This completes the preparation of ethyl hexyl ether using Williamson ether synthesis.

Key concepts

Ethyl Hexyl Ether

Ethyl hexyl ether is an organic compound that belongs to the class of ethers. Ethers consist of two hydrocarbon groups connected by an oxygen atom. In ethyl hexyl ether, one of these groups is an ethyl group \(CH_3CH_2\), and the other is a hexyl group \(CH_3CH_2CH_2CH_2CH_2CH_2\).
This structure gives it the chemical formula \(CH_3CH_2OCH_2CH_2CH_2CH_2CH_2CH_3\). The ether is created by a substitution reaction where two alcohol groups join together by losing a molecule of water.
In practical applications, such ether compounds are often used as solvents due to their relatively low reactivity and their ability to dissolve a variety of chemical substances.

Alkoxide Ions

Alkoxide ions are negatively charged species derived from alcohols. They form when the hydrogen atom of the hydroxyl group \(OH\) in an alcohol is removed, resulting in a structure represented as \(RO^-\), where \(R\) is an alkyl group.
Sodium ethoxide (\(CH_3CH_2ONa\)) is a common alkoxide ion used in organic synthesis, including the Williamson ether synthesis. Sodium ethoxide is prepared from ethanol by reacting it with sodium metal. This reaction generates hydrogen gas and the alkoxide ion:

• Ethanol \(CH_3CH_2OH\) reacts with sodium (Na) to produce sodium ethoxide \(CH_3CH_2ONa\) and hydrogen gas \(H_2\).

While alkoxides are strong bases and good nucleophiles, it's their nucleophilicity that plays a key role in reactions like the Williamson ether synthesis. They attack the positively charged carbon atom in alkyl halides during this synthesis process.

Alkyl Halides

Alkyl halides, or haloalkanes, are compounds where one or more hydrogen atoms in an alkane are replaced by halogen atoms, such as bromine (Br), chlorine (Cl), or iodine (I). Their general formula is \(RX\), where \(R\) is an alkyl group and \(X\) is a halogen.
Hexyl bromide is an example of an alkyl halide used in the Williamson ether synthesis to produce ethyl hexyl ether. It can be synthesized by converting 1-hexanol into hexyl bromide using a halogenating agent such as phosphorus tribromide \(PBr_3\) or hydrobromic acid \(HBr\).

• Hexyl bromide serves as the electrophile in the Williamson ether synthesis.
• The electrophilic carbon in hexyl bromide is susceptible to attack by nucleophiles, such as alkoxide ions.

The versatility of alkyl halides in substitution reactions makes them invaluable in synthetic organic chemistry, allowing for the formation of various compounds.