Question

Consider the nucleophilic attacks given below. Select in each pair that shows the greater $S_{N^2}$ reaction rate.

• A. A$\to$(I) ; B$\to$(IV) ; C$\to$(VI) ; D$\to$(VIII)
• B. A$\to$(II) ; B$\to$(III) ; C$\to$(V) ; D$\to$(VIII)
• C. A$\to$(I) ; B$\to$(III) ; C$\to$(V) ; D$\to$(VIII)
• D. A$\to$(I) ; B$\to$(III) ; C$\to$(V) ; D$\to$(VII)

Answer 1

The correct option is C A$\to$(I) ; B$\to$(III) ; C$\to$(V) ; D$\to$(VIII)

(A) $S_{N}2$ reaction depends on steric hindrance, more the steric hindrance less will be chances of Nucleophilic substitution $\left(S_{N}2\right)$ .

$\left(A\right)\to I$

(B) A species with a negative charge is a stronger base and a better nucleophile than a species that has the same attacking atom but it is neutral. Thus ${}^{-}SH$ is stronger base and a better nucleophile than $C{H}_{3}SH$ .

Better the nucleophile, more will be the rate of reaction of $S_{N}2$

as Rate= $K$[Reactant] [Nucleophile]

thus, $\left(B\right)\to \left(III\right)$

(C) An aprotic polar solvent does not have any hydrogen with partial positive changes to form ion-dipole interactions.

The molecule of an aprotic polar solvent (shown as DMSO) have a partial negative charge on their surface that can solvate cations, but the partial positive charge is on the inside of the molecule and therefore less accessible to solvate anion.

as $C{l}^{-}$ is a good nucleophile than $I^{-}$ thus will have high rate of reaction in presence of DMSO.

$\left(C\right)\to \left(V\right)$

When a negatively charged species is placed in a protic solvent (such as $H_{2}O$ or methanol, the ion becomes solvated. The interaction between the ion and the dipole of the protic solvent is called an ion-dipole interaction.

Because the solvent shields the nucleophile

Weak base interact weakly with protic solvents, whereas strong bases interact strongly.

Thus, it is easier to break the ion-dipole interaction between a $I^{-}$ ion (weak base) and the solvent than between a $C{l}^{-}$ ion (strong base) and the solvent. In protic solvent, $I^{-}$ ion is a better nucleophile. Even though it is a better base.

Thus, Rate of $S_{N}2$ reaction will be higher in $\left(VIII\right)$

$\left(D\right)\to \left(VIII\right)$