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(D) The Williamson ether synthesis involves the reaction of an alkoxide ion with an alkyl halide. For the reaction to proceed via $S_N2$ mechanism to

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$iso-BuO^{\ominus} Na^{\oplus} + sec-BuBr \rightarrow$ Elimination product

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(D) The Williamson ether synthesis involves the reaction of an alkoxide ion with an alkyl halide. For the reaction to proceed via $S_N2$ mechanism to form an ether,the alkyl halide should ideally be primary. If the alkyl halide is secondary or tertiary,the strong base (alkoxide) will prefer to undergo an $E2$ elimination reaction rather than $S_N2$ substitution. In option $A$,${}^{t}BuO^{\ominus}$ is a bulky base and $EtBr$ is a primary halide,so $S_N2$ is possible,but often elimination competes. However,in option $D$,$iso-BuO^{\ominus}$ is a base reacting with $sec-BuBr$ (a secondary alkyl halide). Due to steric hindrance and the nature of the secondary halide,the $E2$ elimination reaction is highly favored over $S_N2$ substitution,leading to alkene formation as the major product instead of the desired ether.

Which one of the following reactions will not lead to the desired ether formation in major proportion? (iso-$Bu$ $\Rightarrow$ isobutyl,sec-$Bu$ $\Rightarrow$ sec-butyl,$nPr$ $\Rightarrow$ $n$-propyl,${}^{t}Bu$ $\Rightarrow$ tert-butyl,$Et$ $\Rightarrow$ ethyl)

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