If all the nucleophilic substitution reaction | vedclass

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(D) The reaction sequence involves two $S_N2$ steps,each resulting in an inversion of configuration at the chiral center.$1$. The starting material is

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(D) The reaction sequence involves two $S_N2$ steps,each resulting in an inversion of configuration at the chiral center.$1$. The starting material is $sec$-butyl chloride. Reaction with aqueous $NaOH$ proceeds via $S_N2$ mechanism,leading to the formation of $sec$-butyl alcohol with inversion of configuration.$2$. The alcohol then reacts with $p$-toluenesulfonyl chloride $(TsCl)$ in pyridine to form a tosylate intermediate,which is a good leaving group.$3$. Finally,the tosylate reacts with $NaI$ in acetone via another $S_N2$ reaction,resulting in a second inversion of configuration.$4$. Since there are two inversions,the final product $C_4H_9I$ will have the same configuration as the starting material $C_4H_9Cl$.$5$. Comparing the structures,option $D$ correctly represents the intermediate alcohol (inverted) and the final iodide (retained configuration relative to the starting material).

If all the nucleophilic substitution reactions at saturated carbon atoms in the above sequence of reactions follow $S_N2$ mechanism,then $\underline{E}$ and $\underline{E}$ will be respectively,

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