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Question 62

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)

In Williamson ether synthesis, an alkoxide ion reacts with an alkyl halide by the $$\mathrm{S_N2}$$ mechanism to give the ether.
For the $$\mathrm{S_N2}$$ path to dominate, the alkyl halide should be primary and the attacking alkoxide should not be highly hindered.

If the halide is secondary or tertiary, or if the nucleophile is very bulky, the $$\mathrm{E2}$$ elimination pathway competes strongly and often becomes the major reaction, giving an alkene instead of the ether.

Case A:

$$^tBuO^-Na^+ + EtBr \; \longrightarrow \; ^tBu-O-Et$$
The halide carbon in $$EtBr$$ is primary. Even though $$^tBuO^-$$ is bulky, $$\mathrm{S_N2}$$ on a primary centre proceeds fast. Desired ether forms in good yield.

Case B:

$$PhO^-Na^+ + CH_3Br \; \longrightarrow \; Ph-O-CH_3$$
$$CH_3Br$$ is a primary (methyl) halide; phenoxide is not excessively bulky. Efficient $$\mathrm{S_N2}$$ gives the required anisole in high yield.

Case C:

$$Na^+O^-nPr + nPrBr \; \longrightarrow \; nPr-O-nPr$$
Both nucleophile and electrophile are unhindered primary species. $$\mathrm{S_N2}$$ dominates and the symmetrical ether is obtained in major proportion.

Case D:

$$iso\text{-}BuO^-Na^+ + sec\text{-}BuBr \; \longrightarrow \; sec\text{-}Bu-O-iso\text{-}Bu$$
Here the alkyl halide $$sec\text{-}BuBr$$ is secondary. The attacking alkoxide $$iso\text{-}BuO^-$$ is branched and sterically hindered. Under these conditions the nucleophile abstracts a $$\beta$$-hydrogen much faster than it can perform backside attack on the secondary carbon. Consequently, the $$\mathrm{E2}$$ elimination that produces 2-butene becomes the chief reaction, and very little ether is formed.

Therefore, the reaction that will not give the desired ether as the main product is found in Option D (Option 4).

Answer - Option D (4)

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