✓ Correct answer: E. 1-bromopentane and sodium iodideS<sub>N</sub>2 reactions involve a backside nucleophilic attack on an electrophilic carbon. As a result, less steric congestion for this backside attack results in a faster reaction, meaning that S<sub>N</sub>2 reactions proceed fastest for primary carbons. In addition, beta-branching next to a primary carbon results in a slower reaction, as does a poorer leaving group (i.e. chloride instead of bromide).<br/><br/>1-bromopentane has a good leaving group (bromine) attached to a primary carbon with no beta-branching, meaning it will proceed the fastest.
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