What is the correct decreasing order of reactivity of the following alkyl halides towards sodium methoxide in a Williamson synthesis? I. CH₃-CH₂-CH₂-CH(Br)-CH₃ II. CH₃-CH₂-CH(CH₃)-CH₂-Br III. C₆H₅-CH₂-Br IV. CH₃-CH₂-CH₂-CH₂-Br

Step-by-step solution

Williamson ether synthesis proceeds via an S_N2 mechanism. The rate of an S_N2 reaction is inversely proportional to the steric hindrance around the electrophilic carbon and is enhanced by the stabilization of the transition state. Compound III ( C₆H₅-CH₂-Br ) is a benzylic halide. Its S_N2 transition state is highly stabilized by the adjacent phenyl ring through conjugation, making it the most reactive. Compound IV ( CH₃-CH₂-CH₂-CH₂-Br ) is an unbranched primary alkyl halide with minimal steric hindrance. Compound