$C H_{3} B r C H_{3} F C H_{3} O H C H_{3} O S O_{2} C F_{3}$
$(I) (I I) (I I I) (I V)$
The correct order of decreasing reactivity of the above compounds towards $C H_{3} O^{-}$ in an $S_{N} 2$ reaction is :

I > I V > I I > I I I

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I V > I > I I > I I I

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I V > I > I I I > I I

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I V > I I > I > I I I

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Solution

The correct option is B $I V > I > I I > I I I$
The correct order of decreasing reactivity of the above compounds towards $C H_{3} O^{-}$ in an $S_{N} 2$ reaction is $I V > I > I I > I I I$

or $C H_{3} O S O_{2} C F_{3} > C H_{3} B r > C H_{3} F > C H_{3} O H$
$^{-} O S O_{2} C F_{3}$ is the best leaving group as the negative charge on $O$ is stabilized by resonance with other $O$ atoms attached to $S$ and due to electron-withdrawing $(- I)$ effect of trifluoromethyl group. Hence, the reactivity of $I V$ is maximum.
The reactivity of $I I I$ is minimum because hydroxide ion is a poor leaving group and there is no stabilization for the negative charge on $O$ atom.
Also, the stability of $B r^{-}$ is more compared to $F^{-}$ which indicates more reactivity of $C H_{3} B r$ over $C H_{3} F$ .

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CH3BrCH3FCH3OHCH3OSO2CF3(I)(II)(III)(IV)The correct order of decreasing reactivity of the above compounds towards CH3O− in an SN2 reaction is :

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