JEE Main 2025 Apr 8 Shift 2, Chemistry Q12: Preparation of Ether

Step 1: Principle of Williamson Ether Synthesis

Williamson ether synthesis proceeds via an ${\mathrm{S}}_{\mathrm{N}}2$ reaction between an alkoxide (or phenoxide) and an unhindered alkyl halide.

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Step 2: Analyse Option (4)

In option (4), sec-butyl bromide is a secondary alkyl halide and the nucleophile is the bulky iso-butoxide ion.

This sterically hindered combination makes ${\mathrm{S}}_{\mathrm{N}}2$ attack difficult. Instead, the bulky base preferentially removes a $\beta$-hydrogen, leading to E2 elimination.

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Step 3: Other Options

The remaining combinations involve either a methyl/primary alkyl halide or a less hindered nucleophile, making ${\mathrm{S}}_{\mathrm{N}}2$ substitution the major reaction and forming the desired ether.

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Conclusion

Only option (4) predominantly undergoes E2 elimination instead of ether formation.

$\boxed{\text{Correct Answer: Option (4)}}$