Question 1

Why does bromoethane form an ether in Williamson synthesis?

Accepted Answer

Bromoethane is a primary alkyl halide. Ethoxide can approach the electrophilic carbon from the backside, displace bromide in an SN2 reaction, and form ethoxyethane.

Question 2

Why does 2-bromo-2-methylpropane favor elimination?

Accepted Answer

Its tertiary carbon is too crowded for a practical backside SN2 attack. Ethoxide removes a beta hydrogen while bromide leaves in E2, producing 2-methylpropene instead of an ether.

Question 3

What is the main product when methoxyethane reacts with HI?

Accepted Answer

The ether oxygen is protonated first. Iodide attacks the less hindered methyl carbon, producing CH3I and C2H5OH.

Question 4

Why does anisole cleave at O–CH3 instead of its aromatic C–O bond?

Accepted Answer

The aromatic C–O bond has partial double-bond character because the methoxy oxygen interacts with the ring. Cleavage occurs at O–CH3, giving phenol and CH3I; iodobenzene is not formed.

Question 5

Why is 4-bromoanisole the major ring bromination product?

Accepted Answer

The methoxy group activates ortho and para positions by resonance. The para product is generally major because it has less steric crowding than the ortho product.

Ethers & Anisole: Williamson, HI Cleavage and Ring Attack

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Practice