The answer is (B) benzyl chloride, C₆H₅–CH₂Cl.

The reagent is NaI in DMSO, which strongly favors an SN2 reaction.

Step 1: Compare the substrates

• A: tertiary alkyl chloride → highly hindered ❌
• B: benzyl chloride, C₆H₅–CH₂Cl → primary benzylic chloride ✅
• C: chlorobenzene, C₆H₅–Cl → aryl chloride, SN2 doesn’t occur ❌
• D: cyclohexylmethyl chloride → primary, but not benzylic

Step 2: Why is B exceptionally fast?

In benzyl chloride:

C₆H₅–CH₂–Cl

The carbon attacked by I⁻ is the CH₂ next to the benzene ring.

During SN2, the transition state has some electron deficiency at this carbon. The benzene ring can stabilize this transition state by resonance.

So:

benzyl chloride → very fast SN2

Also, DMSO is a polar aprotic solvent, which makes I⁻ a strong nucleophile and favors SN2.

Important order to remember

For SN2 reactivity:

benzyl/allyl halide > primary > secondary >> tertiary

And aryl/vinyl halides do not normally undergo SN2.