Ethers are class of organic compounds that contain ether group – an oxygen atom connected to two alkyl groups or aryl groups. They have the general formula R – O – R′, where R and R′ represents the alkyl or aryl groups. Ether, like water have a tetrahedral geometry i.e., oxygen is sp3 hybridised. The C – O – C bond angle in ethers is slightly greater than the tetrahedral angle due to repulsive interactions between the two bulky groups when they are attached to oxygen.

Which of the following cannot be made by using Williamson Synthesis?

Di-tert-Butyl ether

The correct answer is option 4. Di-tert-Butyl ether.

3°-alkylhalide do not form ether actually they undergo elimination reaction. Here, out of the given options only Di-tert-Butyl ether is a 3°-alkylhalide and will not undergo Williamson synthesis.

In the Williamson Ether Synthesis, an alkyl halide (or sulfonate, such as a tosylate or mesylate) undergoes nucleophilic substitution $(S_N2)$ by an alkoxide to give an ether. Being an $S_N2$ reaction, best results are obtained with primary alkyl halides or methyl halides. Tertiary alkyl halides give elimination instead of ethers. Secondary alkyl halides will give a mixture of elimination and substitution. The alkoxide $RO^–$ can be those of methyl, primary, secondary, or tertiary alcohols. The reaction is often run with a mixture of the alkoxide and its parent alcohol (e.g. $NaOEt/EtOH$or $CH_3ONa/CH_3OH$). Alternatively, a strong base may be added to the alcohol to give the alkoxide. Sodium hydride (NaH) or potassium hydride $(KH)$ are popular choices. When an alkoxide and alkyl halide are present on the same molecule, an intramolecular reaction may result to give a new ring. This works best for 5- and 6-membered rings. When planning the synthesis of ethers using the Williamson, take care to select the best starting materials for an $S_N2$ reaction.

The Williamson ether synthesis is a substitution reaction, where a bond is formed and broken on the same carbon atom. In this substitution reaction, a new $C-O$ bond is formed, and a bond is broken between the carbon and the leaving group (LG) which is typically a halide or sulfonate.

It proceeds through an $S_N2$ mechanism (nucleophilic substitution, bimolecular) where the nucleophile approaches the carbon atom from the backside of the carbon-leaving group bond.

A pair of electrons from the nucleophile are donated into the sigma* (antibonding) orbital of the C-leaving group bond.

This requires that the nucleophile actually makes its way to the orbital on the backside of the carbon! For this reason the $S_N2$ is fastest for methyl and primary alkyl halides, and does not occur on tertiary alkyl halides due to the fact that nucleophiles can’t make their way through the tangled thicket of alkyl groups on the backside.

Substitution reactions of alkoxides with secondary alkyl halides can occur, but often occur with significant elimination through the $E_2$ pathway.