Answer:
D) sodium t-butoxide + bromomethane
Explanation:
The alkoxide ion is a strong nucleophile, that unlike alcohols, will react with primary alkyl halides to form ether. This general reaction is known as the Williamson synthesis, and is a SN₂ displacement. The alkyl halide must be primary so the back side attack is not hindered, and the alkoxide ion must be formed with the most hindered group.
The mechanism can be seen in the attachment.
The reaction of sodium t-butoxide and bromomethane gives the higher yield. Thus, option D is correct.
For the production of t-butyl methyl ether in higher yield, there has been the Williamson ether synthesis. The synthesis has been carried out with the reaction of alkylhalide with the alkoxide, that results in the formation of an ether.
The Williamson reaction has been a SN2 reaction that has been carried out in two steps with the addition of nucleophile and removal of the halide for the formation of ether.
From the given reactants, the higher yield has been obtained with the reaction of alkoxide and alkyl halide. Thus, reaction of sodium t-butoxide and bromomethane gives the higher yield. Thus, option D is correct.
For more information about the Williamson synthesis, refer to the link:
https://brainly.com/question/14491871
