List the following compounds in order of decreasing reactivity toward ${C H}_{3} O H$ in an $S_{N} 2$ reaction carried out in ${C H}_{3} O H$ ; ${C H}_{3} F$ , ${C H}_{3} C l$ , ${C H}_{3} B r$ , ${C H}_{3} I$ , ${C H}_{3} O {S O}_{2} {C F}_{3}$ , $^{14} {C H}_{3} O H$

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Solution

(1) $S N_{2}$ Reaction depends on nature of leaving group

(2) The best leaving groups are those that become either a relatively stable anion on a neutral molecule when they depart

Here the order is

$C{ H }_{ 3 }OS{ O }_{ 2 }C{ F }_{ 3 }>C{ H }_{ 3 }I>C{ H }_{ 3 }Br>C{ H }_{ 3 }Cl>C{ H }_{ 3 }F>C{ H }_{ 3 }OH$

$C H_{3} O S O_{2} C F_{3}$ -Most reactive

$C H_{3} O H$ -Least reactive

$C F_{3} S O_{3}^{-}$ is best leaving groups because it form a stable anion

It is a conjugate base of $C F_{3} S O_{3} H$ (strong acid)

whereas -OH (hydroxide ion) is poor leaving group because it leaves as $O H^{-}$ which is highly reactive and unstable

Among the halogens, iodide is the best leaving group and fluoride is poorest
$I^{-} > {B r}^{-} > {C l}^{-} > F^{-}$ (leaving group)

It is because $C - I$ is the weakest bond due to longer bond whereas $C - F$ is the shortest bond. $∴ I^{-}$ leaves easily whereas $F^{-}$ requires more energy

Therefore the order is
$C H_{3} O S O_{2} C F_{3} > C H_{3} I > C H_{3} B r > C H_{3} C l > C H_{3} F > C H_{3} O H$

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Similar questions

C H_{3} B r + O H^{-} C_{2} H_{5} O H - ---- \to C H_{3} O H + B r^{-}

C H_{3} B r + O H^{-} (C H_{3})_{2} S O - ----- \to C H_{3} O H + B r^{-}

C H_{3} C l, C H_{3} B r, C H_{3} I

C H_{3} F

S_{N}^{2}

C H_{3} (C H_{2})_{3} C H_{2} B r

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Compare the dipole moment of the following.

CH₃F,CH₃Cl,CH₃Br, CH3 I.

Arrange in decreasing order of basicity,

$H_{2} O, O H^{-}, C H_{3} O H, C H_{3} O^{-}$

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