Which among the following is most reactive via \(S_N2\) mechanism?
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For \(S_N2\) reactions, steric hindrance is the most important factor.
\[
CH_3X \gt 1^\circ RX \gt 2^\circ RX \gg 3^\circ RX
\]
Less crowded carbon atoms undergo faster backside attack by the nucleophile.
Step 1: Recall the characteristics of the \(S_N2\) mechanism.
The \(S_N2\) reaction is a bimolecular nucleophilic substitution reaction that occurs through a single-step mechanism.
\[
\text{Rate}=k[\text{Alkyl Halide}][\text{Nucleophile}]
\]
The nucleophile attacks from the backside of the carbon atom bonded to the leaving group.
Step 2: Understand the effect of steric hindrance.
Since backside attack is required, steric hindrance around the reaction center greatly affects the reaction rate.
As steric crowding increases, the nucleophile finds it more difficult to approach the carbon atom.
Therefore, the reactivity order is
\[
\text{Methyl} \gt 1^\circ \gt 2^\circ \gt 3^\circ
\]
Step 3: Analyze the given structures.
The second structure is a primary alkyl bromide. The carbon attached to bromine experiences the least steric hindrance among the given compounds.
The first and fourth structures are secondary alkyl bromides, which are less reactive than primary alkyl halides in \(S_N2\) reactions.
The third structure is highly hindered and therefore reacts very slowly through the \(S_N2\) pathway.
Step 4: Compare all options.
Because \(S_N2\) reactions favor less substituted carbon atoms, the primary alkyl bromide reacts the fastest.
Hence,
\[
\text{Primary alkyl bromide}
\gt
\text{Secondary alkyl bromide}
\gt
\text{Tertiary alkyl bromide}
\]
Step 5: Final conclusion.
Therefore, the compound represented in option (2) is the most reactive towards the \(S_N2\) mechanism.
\[
\boxed{\text{Option (2)}}
\]
Hence, option (2) is correct.
