Answer: (CH 3 ) 3 CONa + CH 3 Br.
- A (CH<sub>3</sub>)<sub>3</sub>CBr + CH<sub>3</sub>ONa
- B CH<sub>3</sub>Br + CH<sub>3</sub>ONa
- C (CH<sub>3</sub>)<sub>3</sub>CBr + CH<sub>3</sub>OH
- D (CH<sub>3</sub>)<sub>3</sub>CONa + CH<sub>3</sub>Br
Correct answer: D. (CH<sub>3</sub>)<sub>3</sub>CONa + CH<sub>3</sub>Br
Explanation: Williamson works best with a methyl or primary halide; the tertiary halide would only eliminate, so the tertiary alkoxide plus methyl bromide is chosen.
The phenoxide ion's negative charge spreads into the aromatic ring through resonance, lowering its energy and making phenol give up its proton more easily; an alkoxide ion (from a plain alcohol) has nowhere to delocalise the charge, making alcohols far less acidic.
Concept context
Oxygen-containing organic compounds. Understand the acidic nature of phenols, reactions of alcohols (dehydration, oxidation, esterification), and properties of ethers. Includes industrial applications and named reactions.