Concept used: Williamson ether synthesis

Ethers are commonly prepared by Williamson synthesis: $\mathrm{R'O^-Na^+ + R-X \rightarrow R-O-R' + NaX}$

This proceeds mainly by an $\mathrm{S_N2}$ mechanism, so it works best when the alkyl halide is primary.

Key limitation

If the alkyl halide is secondary or tertiary, then with a strong base like alkoxide ion:

• $\mathrm{S_N2}$ becomes difficult due to steric hindrance.
• Elimination ($\mathrm{E2}$) becomes major.

Therefore, for major ether formation, the halide should preferably be primary.

Examine the given visible options

Option B

$\mathrm{iso\text{-}BuO^-Na^+ + sec\text{-}BuBr \rightarrow sec\text{-}Bu-O-iso\text{-}Bu}$

Here:

• Nucleophile: isobutoxide, a strong alkoxide base
• Alkyl halide: sec-butyl bromide, which is a secondary halide

Since the substrate is secondary, $\mathrm{S_N2}$ is hindered and $\mathrm{E2}$ elimination will compete strongly, in fact generally dominate.

So this reaction will not give the desired ether in major proportion.

Option D

$\mathrm{^tBuO^-Na^+ + EtBr \rightarrow ^tBu-O-Et}$

Here the alkyl halide is ethyl bromide, a primary halide. Even though tert-butoxide is bulky, a primary halide can still undergo Williamson substitution to form ether. This is much more favorable than using a secondary halide.

So this can lead to ether formation.

Conclusion

The reaction that will not lead to the desired ether formation in major proportion is the one using a secondary alkyl halide in Williamson synthesis, i.e. Option B.

Comparison with stored answer

Stored correct answer: B

Derived answer: B

Hence, the derived answer agrees with the stored answer.