Final answer:
The compound that yields (S)-2-bromopentane when treated with tosyl chloride (TsCl) and sodium bromide (NaBr) is (S)-2-pentanol.
Step-by-step explanation:
The compound that provides (S)-2-bromopentane upon exposure to TsCl (tosyl chloride) and NaBr is (S)-2-pentanol.
Here's the explanation:
TsCl converts alcohols to tosylates: When (S)-2-pentanol reacts with TsCl, it forms (S)-2-pentyl tosylate, a more reactive leaving group compared to the hydroxyl group of the alcohol.
NaBr performs nucleophilic substitution: The nucleophile in this reaction is the bromide ion (Br⁻) from NaBr. It attacks the carbon atom of the tosylate, displacing the tosylate group and forming (S)-2-bromopentane.
Here's the overall reaction:
(S)-2-Pentanol + TsCl → (S)-2-Pentyl tosylate + HCl
(S)-2-Pentyl tosylate + NaBr → (S)-2-Bromopentane + TsONa
Since the reaction proceeds through an SN2 (bimolecular nucleophilic substitution) mechanism, the stereochemistry of the chiral center (carbon atom with the hydroxyl group) is retained. Therefore, the final product is (S)-2-bromopentane.