Explain $S N^{1}$ and $S N^{2}$ reactions.

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Solution

(1) SN $^{1}$ reaction:
Consider the alkaline hydrolysis of tert butyl bromide to give tert butanol.
$C H_{3} C (C H_{3})_{2} - B r + N a O H Δ - \to C H_{3} C (C H_{3})_{2} - O H + N a B r$
The rate of the reaction is proportional to the concentration of tert butyl bromide only and is independent of the concentration of hydroxide ion.
Rate $= K [C H_{3} C (C H_{3})_{2} - B r]$
The order of the reaction is 1. In the rate determining step, carbon to Br bond breaking is followed by the carbon to hydroxide ion bond formation. Hence, the reaction is unimolecular as only tert butyl bromide is involved in the rate determining step.
Thus SN $^{1}$ reaction is unimolecular nucleophilic substitution reaction.
(2) SN $^{2}$ reaction
Consider the alkaline hydrolysis of methyl bromide to give methanol.
$C H_{3} - B r + N a O H Δ - \to C H_{3} - O H + N a B r$
The rate of the reaction is proportional to the concentration of methyl bromide as well as the concentration of hydroxide ion.
Rate $= K [C H_{3} - B r] [O H^{-}]$
The order of the reaction is 2. In the rate determining step, the carbon to hydroxide ion bond formation and carbon to Br bond breaking are simultaneous processes. Hence, the reaction is bimolecular as both methyl bromide and NaOH are involved in the rate determining step.
Thus SN $^{2}$ reaction is bimolecular nucleophilic substitution reaction.

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S_{N} 1

S_{N} 2

S_{N} 1

S_{N} 2

Consider the reactions

(i) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{5} O H - ---- \to (C H_{3})_{2} C H - C H_{2} O C_{2} H_{5} + H B r$

(ii) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{S} O^{-} - ---- \to (C H_{3})_{2} C H C H_{2} O C_{2} H_{5} + B r^{-}$

The mechanism of reactions (i) & (ii) are respectively

{S N}^{1}

{S N}^{2}

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