Step 1: Requirement for an \(S_N2\) reaction.
The \(S_N2\) reaction proceeds through a single-step mechanism involving backside attack of the nucleophile. Therefore, the carbon atom attached to the leaving group should be easily accessible and should not be sterically hindered. Primary alkyl halides generally undergo \(S_N2\) reactions most readily, whereas tertiary alkyl halides do not.
Step 2: Analysis of Option (1).
tert-Butyl bromide is a tertiary alkyl halide. The carbon bearing bromine is surrounded by three alkyl groups, causing severe steric hindrance. As a result, backside attack is not possible. Hence, it does not readily undergo an \(S_N2\) reaction.
Step 3: Analysis of Option (2).
1-Chloro-1-methylcyclopentane is also a tertiary halide. Due to steric crowding around the carbon attached to chlorine, \(S_N2\) attack is highly unfavorable. Therefore, this compound does not readily participate in an \(S_N2\) reaction.
Step 4: Analysis of Option (3).
2-Phenylethyl iodide has the structure:
\[
\mathrm{C_6H_5-CH_2-CH_2-I}
\]
The carbon attached to iodine is a primary carbon. Primary alkyl halides are ideal substrates for \(S_N2\) reactions. In addition, iodide is an excellent leaving group because the C–I bond is relatively weak. Hence, this compound readily undergoes an \(S_N2\) reaction.
Step 5: Analysis of Option (4).
Vinyl halides generally do not undergo \(S_N2\) reactions because the carbon attached to the halogen is \(sp^2\)-hybridized. Backside attack on an \(sp^2\)-hybridized carbon is geometrically difficult, making \(S_N2\) substitution unfavorable.
Step 6: Final conclusion.
Among the given compounds, only 2-phenylethyl iodide is a primary alkyl halide with a good leaving group and minimal steric hindrance. Therefore, it readily undergoes an \(S_N2\) reaction.
\[
\boxed{\text{2-Phenylethyl iodide}}
\]