The major product of the following reaction is:

Step 1: Identify the reaction

Substrate: Alkyl iodide (or alkenyl iodide with I substituent)

Reagent: Sodium thiophenoxide ($\ce{PhS-Na+}$) in DMF

DMF = Dimethylformamide (polar aprotic solvent)

Step 2: Determine the mechanism

Thiophenoxide ion ($\ce{PhS-}$) is a nucleophile:

• Sulfur has lone pairs
• Good nucleophile (soft nucleophile)
• Can undergo $S_N2$ substitution

Mechanism: $S_N2$

$\ce{R-I + PhS- -> R-SPh + I-}$

Step 3: Determine product

The iodine is replaced by thiophenoxy group ($\ce{-SPh}$):

Product: Alkyl phenyl sulfide (thioether)

Structure: $\ce{R-S-C6H5}$

Step 4: Regiochemistry

For $S_N2$:

• Backside attack
• Inversion of configuration (if chiral center)
• I is replaced by SPh at the same carbon

Given answer is (a), the SPh group is at the position where I was originally.

Answer: (a)

Key Points:

Thiophenoxide ($\ce{PhS-}$):

• Nucleophile (S has lone pairs)
• Soft nucleophile (large, polarizable)
• Forms thioethers ($\ce{R-S-R'}$)

DMF (Dimethylformamide):

• Polar aprotic solvent
• Doesn't solvate anions strongly
• Excellent for $S_N2$ reactions
• Dissolves both organic and ionic compounds

Thioethers:

• $\ce{R-S-R'}$ (sulfur analog of ethers)
• More nucleophilic than ethers
• Can be oxidized to sulfoxides ($\ce{R-SO-R'}$) or sulfones ($\ce{R-SO2-R'}$)

Leaving group ability:

• I⁻ is excellent leaving group
• Makes alkyl iodides very reactive in $S_N2$