See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The reaction involves two sequential steps. First, treatment of an alcohol with p-toluenesulfonyl chloride (TsCl) in pyridine converts the -OH group into a tosylate (-OTs) leaving group. Second, treatment with LiBr (a source of Br⁻, a good nucleophile) displaces the tosylate via an SN2 mechanism. Step 2 - Step 1 (Tosylation): R-2-butanol reacts with TsCl in pyridine. This reaction proceeds with retention of configuration at the stereocenter because the C-O bond is NOT broken; only the O-H bond is broken and the oxygen is derivatized. Therefore, R-2-butanol gives R-2-butyl tosylate. Step 3 - Step 2 (SN2 with LiBr): Br⁻ attacks the carbon bearing the tosylate in an SN2 reaction. SN2 proceeds with inversion of configuration at the stereocenter (Walden inversion). Starting from R-2-butyl tosylate, inversion gives S-2-butyl bromide. Step 4 - Overall result: R-2-butanol → (TsCl/pyridine, retention) → R-2-butyl tosylate → (LiBr, SN2 inversion) → S-2-butyl bromide. Step 5 - Why other options fail: (a) R-2-butyl bromide would require retention in the SN2 step, which does not occur. (b) S-2-butyl tosylate would require inversion during tosylation, which does not occur since the C-O bond is not broken. (c) R-2-butyl tosylate is an intermediate but not the final product after LiBr treatment. Therefore, the correct answer is D.