Question

The following ethers can, in principle, be synthesized by two different combinations of haloalkane or halocycloalkane and metal alkoxide. Show one combination that forms ether bond (1) and another that forms ether bond (2). Which combination gives the higher yield of ether?

Short answer

Answer: The combination of a primary haloalkane or halocycloalkane (R1-X) and metal alkoxide (R2'-O-M) forming ether bond (1) (-R1-O-R2'-) tends to give a higher yield. However, actual yields may vary due to practical factors and competing reactions.

Step-by-step solution

Identifying the ether structure and bond (1) and bond (2)

Looking carefully at the given ethers, we can identify the ether bond (1) (-R1-O-R2-) and ether bond (2) (-R2'-O-R1'-).

Determining possible combinations of haloalkane or halocycloalkane and metal alkoxide

In the Williamsons Ether Synthesis, the general reaction can be represented as follows: Primary Haloalkane or Halocycloalkane (R1-X) + Metal Alkoxide (R2'-O-M) -> Ether (R1-O-R2'-) Primary Haloalkane or Halocycloalkane (R2-X') + Metal Alkoxide (R1'-O-M') -> Ether (R2'-O-R1'-) Where X and X' are halogens, and M and M' are metal ions. Note that for higher yield, we want to choose primary haloalkanes because secondary and tertiary ones have higher tendencies for an elimination reaction.

Combination for Ether Bond (1)

Let R1 be a primary alkyl halide and M be a metal ion. Combining R1-X with R2'-O-M will give us the ether having bond (1): R1-X + R2'-O-M -> R1-O-R2'- For example, let R1-X be ethyl bromide (CH3CH2-Br) and R2'-O-M be potassium methoxide (CH3-O-K), the ether formed with bond (1) will be ethyl methyl ether (CH3CH2-O-CH3).

Combination for Ether Bond (2)

Similarly, let R2' be a primary alkyl halide and M' be a metal ion. Combining R2-X' with R1'-O-M' will give us the ether having bond (2): R2-X' + R1'-O-M' -> R2'-O-R1'- For example, let R2-X' be methyl bromide (CH3-Br) and R1'-O-M' be potassium ethoxide (CH3CH2-O-K), the ether formed with bond (2) will be ethyl methyl ether (CH3CH2-O-CH3).

Determining the Higher Yield Combination

Both combinations are similar in their theoretical yield; however, it is important to start with the primary alkyl halides when possible. The combination with R1-X being primary and starting with R2'-O-M results in a higher yield. So, the first combination that forms an ether bond (1) tends to give a higher yield of ether than the second combination. Remember that practical yields may differ due to various factors like reaction conditions, purity of reactants, and competing reactions like the elimination reaction.

Key concepts

Haloalkane

Haloalkanes, also known as alkyl halides, are organic compounds containing one or more halogen atoms (like chlorine, bromine, iodine, or fluorine) attached to a carbon atom. These compounds play a crucial role in organic chemistry, particularly in reactions like Williamson's Ether Synthesis.

Haloalkanes are classified based on the carbon they are attached to:

• Primary haloalkane: The carbon attached to the halogen is connected to only one other carbon atom.
• Secondary haloalkane: The carbon attached to the halogen is connected to two other carbon atoms.
• Tertiary haloalkane: The carbon attached to the halogen is connected to three other carbon atoms.

In Williamson's Ether Synthesis, primary haloalkanes are preferred as they are less prone to elimination reactions. Secondary and tertiary haloalkanes are more likely to participate in side reactions, reducing the yield of ether.

Metal Alkoxide

Metal alkoxides are compounds containing a metal ion bonded to an alkoxide ion (RO⁻). This ionic character is crucial in Williamson's Ether Synthesis, as it facilitates the nucleophilic substitution reaction required to form ethers.

Metal alkoxides are generally strong bases and can be prepared by reacting an alcohol with a corresponding metal. For example, potassium methoxide can be formed by combining methanol and potassium.

In the synthesis process:

• The alkoxide ion acts as a nucleophile, attacking the haloalkane's carbon atom attached to the halogen.
• This leads to the formation of an ether by replacing the halogen with the alkoxy group.
• Metal alkoxides need careful handling, as they are often reactive and sensitive to moisture.

Primary Alkyl Halide

In Williamson's Ether Synthesis, the selection of a primary alkyl halide is a strategic choice to optimize the reaction's yield and efficiency. Because primary alkyl halides have less steric hindrance, they are more susceptible to nucleophilic attack from the alkoxide ion.

The mechanism unfolds as follows:

• The alkoxide ion approaches the primary alkyl halide's carbon atom attached to the halogen.
• This carbon-halogen bond is weaker and more accessible than in secondary or tertiary alkyl halides.
• As a result, the nucleophilic substitution process proceeds swiftly, forming the ether and expelling the halogen as a leaving group.

Choosing a primary alkyl halide minimizes competing side reactions such as elimination, ensuring the primary product is the desired ether.

Elimination Reaction

Elimination reactions represent a common competing pathway in organic reactions involving haloalkanes, especially with secondary and tertiary alkyl halides. These reactions often result in the formation of alkenes instead of the desired ether.

They occur when:

• A base abstracts a proton adjacent to the carbon attached to the halogen.
• The former halogen leaves simultaneously, resulting in the formation of a double bond.
• Consequently, an alkene is formed rather than progressing towards ether synthesis.

In Williamson's Ether Synthesis, using primary haloalkanes is advantageous because they are less likely to undergo elimination under typical reaction conditions. This preference helps in maintaining a higher yield of the desired ether product by avoiding pathways that lead to alkene formation.