What is Williamson Ether Synthesis?
An organic reaction in which an organohalide and a deprotonated alcohol forms ether is known as Williamson ether synthesis. Alexander Williamson developed the Williamson ether synthesis in 1850. The formation of ether in this synthesis is an SN2 reaction.
What is Ether?
Ether is an organic compound in which two alkyl or aryl groups are attached by an oxygen atom. The general formula of ether can be given as follows:
Where, R maybe alkyl group or aryl group.
Williamson Ether Synthesis
An organic reaction in which an organohalide and a deprotonated alcohol forms ether is known as Williamson Ether Synthesis. Alexander Williamson developed the Williamson ether synthesis in 1850. The formation of ether in this synthesis is an SN2 reaction.
The general reaction of Williamson ether synthesis can be given as
Mechanism of Williamson Ether Synthesis
The reaction mechanism of the Williamson ether synthesis follows the SN2 pathway. In the SN2 reaction, an attack of the nucleophile occurs from the backside of the carbon atom, giving the product with inversion of configuration.
An example of the Williamson ether synthesis is the reaction of sodium ethoxide with chloroethane. The products formed are diethyl ether and sodium chloride.
It is a concerted reaction which occurs in single step. The detailed mechanism is given below:
Alkoxide Used as a Nucleophile
Alkoxide ion is formed by the treatment of alcohols with base or alkali metal. Alkoxide ion is a good nucleophile in replacing the halide ion in alkyl halide. This results in the formation of a new bond between carbon and oxygen atoms.
Alkyl halide used as substrate
Williamson Ether synthesis follows SN2 reaction mechanism. The SN2 reaction does not occurs when the substrate is sterically hindered. The order of rate of reaction of SN2 is methyl halides, primary alkyls, secondary alkyls. Whereas, SN2 reaction does not occur in tertiary alkyls due to steric hindrance.
Methyl halide and primary alkyl halides have less steric hindrance and are good substrates for Williamson ether synthesis.
Formation of ether from methyl halide
Formation of ether from primary alkyl halide
Williamson synthesis is a bit difficult to perform with secondary alkyl halides than primary alkyl halides due to steric hindrance.
Alkoxide ions are strong bases. When a sterically hindered substrate is used, elimination reaction may occur. To avoid the elimination reaction and to encourage the reaction, polar aprotic solvent is used. Solvent increases the nucleophilicity of alkoxide favoring the reaction. The solvents used in the synthesis are acetonitrile N,N-dimethyl formamide, DMSO (dimethyl sulfoxide), HMPA (hexamethylphosphoric triamide).
Formation of ether from secondary alkyl halide in presence of solvent
Tertiary alkyl halides do not undergo Williamson synthesis. This is due to heavy steric hindrance in tertiary alkyl halides. Hence, tertiary alkyl halides do not give substitution products, whereas forms elimination product. The reactions can be given as follows:
Base Used to Form Alkoxide Ion
The base used for the formation of alkoxide from alcohol should be strong enough to deprotonate the alcohol. Some bases deprotonated alcohol to form alkoxide but the conjugate acid formed will react with the substrate forming byproducts. To avoid such byproducts, the base used to form the alkoxide should be ideal for the reaction.
Example for a good base to form alkoxide is sodium or potassium hydride. When sodium hydride is added to alcohol, it deprotonates alcohol forming alkoxide ion and hydrogen gas. Hydrogen gas evolves as bubbles and do not interfere in reaction with substrate. Thus, no byproducts are formed. The reaction can be given as follows:
Example for a bad base to form alkoxide is sodium amide. When sodium amide is used as a base, it forms alkoxide ion and ammonia. Ammonia reacts with alkyl halide and forms amine byproducts. Thus, the desired reaction may yield less product. The reaction with sodium amide base is given as below:
The conjugate acid of alkoxide is used as a solvent in Williamson synthesis. To avoid elimination reaction, polar aprotic solvent is used.
When propanol is added as a solvent for sodium ethoxide, sodium ethoxide deprotonates propanol to form sodium prop oxide and ethanol. When alkyl halide is added, two different ethers are formed in the reaction. Hence, the solvent used most commonly in Williamson synthesis is conjugate acid of alkoxide.
Conditions for Williamson Ether Synthesis
- Alkoxide ions are prepared in In-situ conditions as it is very reactive.
- In industrial synthesis, phase transfer catalyst is used.
- In laboratory preparation, carbonate base or potassium hydroxide are used.
- The temperature required for the reaction to occur is 50 - 100°C .
- The reaction completes in 1 to 8 hours.
- The yield of product will be 50 to 95% .
- Limitations of Williamson synthesis
- Williamson synthesis does not occur in tertiary alkyl halides and sterically hindered alkyl halides. In steric hindered alkyl halides, elimination reaction occurs.
- In some cases, such as sodium phenoxide, an alkali phenoxide undergoes C-C alkylation in addition to O-C alkylation.
- Williamson ether synthesis is limited to prepare ethers for synthetic organic chemists.
Uses of Williamson Ether Synthesis
- Williamson synthesis is used mostly to prepare ethers in laboratories and industries.
- Both symmetrical and asymmetrical ethers can be formed through this process.
- In Williamson synthesis, ethers can be formed by using two alcohols. One of the alcohols in two transforms into a leaving group (tosylate group) and reacts with alcohol forming ether.
- The preferred alkylating agent is primary whereas alkoxide can be primary or secondary.
- Sulfonate ester can be created for the purpose of reaction instead of halide as the leaving group.
Intramolecular Williamson Ether Preparation
Cyclic ethers also can be formed by Williamson synthesis. In intramolecular synthesis, two functional groups present in a single molecule reacts to form the product. To form cyclic ethers, intramolecular reaction takes place in a single molecule which have hydroxyl group attached to one carbon atom and halide group to another carbon atom. On reaction, alkyl halide gets eliminated forming cyclic ether. The reaction can be given as follows:
- The functional group of ether R-O-R is and not -C=O.
- Only alkyl oxides are suitable to form ether and aryl oxides forms C-C bond.
Write the product of the following reaction:
In presence of solvent DMSO, pentan-3-olate reacts with 2-bromopropane to form 3-isopropoxypentane.
Context and Applications
This topic is significant in the professional exams for both undergraduate and graduate courses, especially for
B.Sc. in Chemistry
M.Sc in Chemistry
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