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In the reaction: CH3-CH2-S-CH2-CH2-Br + HOH (water) -> [X], the product [X] is:

1. CH3-CH2-S^(+)-CH2-CH2-OH with Br^(-) counter ion (sulfonium salt stays, then hydrolysis gives sulfide alcohol)
2. CH3-CH2-SH + HOCH2CH2 species
3. An episulfonium ion intermediate that opens to give CH3-CH2-S-CH2-CH2-OH
4. CH3-CH2-OH and CH2=CH2 via elimination

Correct answer: An episulfonium ion intermediate that opens to give CH3-CH2-S-CH2-CH2-OH

Solution

The thioether sulfur (nucleophilic) attacks the carbon bearing Br intramolecularly to form a cyclic episulfonium ion (three-membered S-containing ring). Water then acts as a nucleophile, opening the ring to give CH3-CH2-S-CH2-CH2-OH (2-(ethylthio)ethanol) as product X. The original options A and B both show CH3-CH2-S-CH2-CH2-OH, which is the correct connectivity of the final product.

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