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 i

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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).

	$S_{N}1$	$S_{N}2$
$(a)$	Retention of configuration	Inversion of configuration
$(b)$	Racemisation	Inversion of configuration
$(c)$	Inversion of configuration	Retention of configuration
$(d)$	Racemisation	Retention of configuration

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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