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प्रश्न
Write the mechanism of the following reaction:
\[\ce{{n}BuBr + KCN ->[EtOH-H2O] {n}BuCN}\]
उत्तर १
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 transition state:
Step 3: Generation of product:
Therefore, we have
\[\ce{CH3 - CH2 - CH2 - CH2 - Br + KCN ->[EtOH/H2O]CH3CH2CH2CH2CN + KBr}\]
उत्तर २
KCN is the resonating hybrid of the following structures:
\[\ce{K^+ [^- :C ≡ N: ↔ :C = \overset{\bullet\bullet}{N} :^-]}\]
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.
\[\ce{K^+CN^- + \underset{n-butyl bromide}{CH3CH2CH2\overset{δ+}{C}H2 - \overset{δ-}{B}r} -> \underset{n-butyl cyanide}{CH3CH2CH2CH2CN} + KBr}\]
Notes
Students can refer to the provided solutions based on their preferred marks.
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