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

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Question

(D) The reaction proceeds in two steps:$1$. Reaction of $(R)-2$-butanol with $p$-toluenesulphonyl chloride $(TsCl)$ in pyridine forms $(R)-2$-butyl to

Vedclass · Accepted Answer

$(S)-2$-butyl bromide

Vedclass · Answer

(D) The reaction proceeds in two steps:$1$. Reaction of $(R)-2$-butanol with $p$-toluenesulphonyl chloride $(TsCl)$ in pyridine forms $(R)-2$-butyl tosylate. This step involves the breaking of the $O-H$ bond,not the $C-O$ bond,so the configuration at the chiral center is retained.$2$. The subsequent reaction of $(R)-2$-butyl tosylate with $LiBr$ is an $S_N2$ reaction. The $Br^-$ ion attacks the chiral carbon from the side opposite to the tosylate group,leading to an inversion of configuration.Therefore,the final product is $(S)-2$-butyl bromide.

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

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