By a proper choice of reagents, both symmetrical and unsymmetrical ethers can be prepared by Williamson synthesis which involves the reaction between an alkyl halide and an alkoxide ion. The reverse process involving the cleavage of ethers to give back the original alkyl halide and the alcohol can be carried out by heating the ether with HI at 373K.

Which of the following ethers can be prepared by Williamson synthesis?  

Benzyl methyl ether

The correct answer is option 1. Benzyl methyl ether.

The Williamson ether synthesis involves the reaction of an alkoxide ion (derived from an alcohol) with a primary alkyl halide or a sulfonate ester. In the case of benzyl methyl ether:

Alkoxide ion: Methoxide ion (\(\text{CH}_3\text{O}^{-}\)) is derived from methanol (\(\text{CH}_3\text{OH}\)).

 Alkyl halide: Benzyl chloride (\(\text{PhCH}_2\text{Cl}\)).

The reaction proceeds as follows:

This reaction forms benzyl methyl ether (\(\text{PhCH}_2\text{OCH}_3\)).

Therefore, the correct answer is indeed: Benzyl methyl ether.