When methyl bromide is treated with sodium tert-butoxide, the compound formed is:

tert-Butyl methyl ether

The correct answer is option 2. tert-Butyl methyl ether.

When methyl bromide (\(\text{CH}_3\text{Br}\)) is treated with sodium tert-butoxide (\(\text{Na}^+\text{(t-BuO)}^-\)), the reaction is likely to produce tert-butyl methyl ether. This is because sodium tert-butoxide is a strong base and a good nucleophile, which can facilitate the formation of an ether through an \(S_N2\) reaction mechanism.

Sodium tert-butoxide (\(\text{t-BuO}^-\)) acts as a nucleophile and attacks the methyl bromide (\(\text{CH}_3\text{Br}\)). In this reaction, the tert-butoxide ion displaces the bromide ion (\(\text{Br}^-\)) from methyl bromide. The result of this nucleophilic substitution is the formation of tert-butyl methyl ether (\(\text{t-BuOCH}_3\)).

The reaction can be summarized as:

Therefore, the correct answer is: tert-Butyl methyl ether.