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Question
Write the mechanism of the following reaction:
\[\ce{{n}BuBr + KCN ->[EtOH-H2O] {n}BuCN}\]
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Solution 1
This reaction is a first-order nucleophilic substitution (SN1). The mechanism can be stated as:
Step 1: Generation of nucleophile:
\[\ce{KCN ->[EtOH-H2O]K+ + \overset{-}{C} ≡ N}\]
Step 2: Nucleophilic attack and formation of the transition state:
Step 3: Generation of product:
Therefore, we have
\[\ce{CH3 - CH2 - CH2 - CH2 - Br + KCN ->[EtOH/H2O]CH3CH2CH2CH2CN + KBr}\]
Solution 2
KCN is the resonating hybrid of the following structures:
\[\ce{K^+ [^- ^{\bullet}_{\bullet}C ≡ N^{\bullet}_{\bullet} <-> ^{\bullet}_{\bullet}C = \overset{\bullet\bullet}{N}^{\bullet}_{\bullet}^-]}\]
Therefore, CN− acts as an ambident nucleophile. It can attack the carbon atom of the C–Br bond in n-BuBr through either the carbon (C) or nitrogen (N) atom. Since the C–N bond is weaker than the C–C bond, the attack occurs at the carbon atom, leading to the formation of n-butyl cyanide.
\[\ce{K^+CN^- + \underset{n-butyl bromide}{CH3CH2CH2\overset{\delta+}{C}H2 - \overset{\delta-}{B}r} -> \underset{n-butyl cyanide}{CH3CH2CH2CH2CN} + KBr}\]
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