Question

Reaction of t-butyl bromide with sodium methoxide produces:
(A) Isobutane
(B) Isobutylene
(C) Sodium- t- butoxide
(D) t- butyl methyl ether.

Answer 1

Hint: This method is called Williamson’s synthesis. General reaction of this synthesis is:

Complete step by step answer:
When alkyl halide is heated with alc. Sodium or potassium alkoxide gives corresponding ethers. Simple ethers can be easily prepared by this method.
But for preparation of mixed ether a proper choice of reactant is necessary. Primary alkyl halide is more susceptible to $S{N}^2$ reaction. Therefore, best yield of ether obtained when alkyl halide is primary and alkoxide is tertiary.
Example: tert-butyl ether is prepared by heating ethyl bromide with sodium tert-butoxide.

But when alkyl halide is secondary or tertiary the nucleophilic attack of alkoxide ion on $\alpha -$ carbon atom becomes difficult due to crowding effect.
Since alkoxide is a stronger base and attacking $\beta -$ hydrogen is easier. Therefore $\beta -$ elimination dominates.

Therefore, in the given reaction isobutylene is formed.

So, the correct answer is “Option B”.

Note:
Williamson’s synthesis is a good method for preparation of mixed ether. But proper choice of reagent should be necessary. If we want to prepare t-butyl methyl ether then methyl bromide and sodium t-butoxide should be taken.