Reaction of (R)-2-butanol with p-toluenesulph | vedclass

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Question

(D) The reaction proceeds in two steps:$1$. The reaction of $(R)-2$-butanol with $p$-toluenesulphonyl chloride $(TsCl)$ in pyridine converts the hydro

Vedclass · Accepted Answer

$(S)-2$-butyl bromide

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(D) The reaction proceeds in two steps:$1$. The reaction of $(R)-2$-butanol with $p$-toluenesulphonyl chloride $(TsCl)$ in pyridine converts the hydroxyl group into a good leaving group,the tosylate group $(-OTs)$. This reaction occurs without breaking the $C-O$ bond,so the configuration at the chiral center remains unchanged,resulting in $(R)-2$-butyl tosylate.$2$. The subsequent reaction with $LiBr$ involves a nucleophilic substitution $(S_N2)$ reaction. The bromide ion $(Br^-)$ attacks the chiral carbon from the side opposite to the tosylate group,leading to an inversion of configuration (Walden inversion). Thus,$(R)-2$-butyl tosylate is converted into $(S)-2$-butyl bromide.

Reaction of $(R)-2$-butanol with $p$-toluenesulphonyl chloride in pyridine followed by reaction with $LiBr$ gives

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