See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Concept - Tosylation of an alcohol with p-toluenesulphonyl chloride (TsCl) in pyridine converts the OH group into a tosylate (OTs) leaving group. 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 replaced by O-Ts. Therefore, R-2-butanol gives R-2-butyl tosylate as an intermediate. Step 2: Reaction with LiBr - Bromide (Br-) is a good nucleophile and acts via an SN2 mechanism on the tosylate. SN2 reactions proceed with INVERSION of configuration (Walden inversion) at the stereocenter. Step 3: Stereochemical outcome - Starting with R-2-butanol, tosylation gives R-2-butyl tosylate (retention). SN2 attack by Br- on the R-tosylate gives inversion, yielding S-2-butyl bromide. Step 4: Why other options fail: - (a) R-2-butyl bromide: This would require retention in the SN2 step, which does not occur. - (b) S-2-butyl tosylate: Tosylation retains configuration; the tosylate intermediate is R, not S. - (c) R-2-butyl tosylate: This is the intermediate formed after tosylation (correct stereochemistry for intermediate), but it is not the final product after LiBr treatment. Therefore, the correct answer is D.