Write the mechanism of the following reaction:
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$
(N/A) The given reaction is:
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$
This reaction proceeds via an $S_{N}2$ mechanism.
$1$. The nucleophile $CN^{-}$ attacks the electrophilic carbon atom of $n-butyl$ bromide $(n-BuBr)$ from the side opposite to the leaving group $(Br^{-})$.
$2$. This results in the formation of a transition state where the $C-CN$ bond is partially formed and the $C-Br$ bond is partially broken.
$3$. Finally,the $Br^{-}$ ion leaves,resulting in the formation of $n-butyl$ cyanide $(n-BuCN)$.
$4$. The $CN^{-}$ ion is an ambident nucleophile,but in the presence of $KCN$,the attack occurs primarily through the carbon atom due to the higher nucleophilicity of the carbon center.
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$
This reaction proceeds via an $S_{N}2$ mechanism.
$1$. The nucleophile $CN^{-}$ attacks the electrophilic carbon atom of $n-butyl$ bromide $(n-BuBr)$ from the side opposite to the leaving group $(Br^{-})$.
$2$. This results in the formation of a transition state where the $C-CN$ bond is partially formed and the $C-Br$ bond is partially broken.
$3$. Finally,the $Br^{-}$ ion leaves,resulting in the formation of $n-butyl$ cyanide $(n-BuCN)$.
$4$. The $CN^{-}$ ion is an ambident nucleophile,but in the presence of $KCN$,the attack occurs primarily through the carbon atom due to the higher nucleophilicity of the carbon center.
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Similar Questions
Write a note on $S_{N}1$ reaction.
Medium
View SolutionFor the following reactions:
$CH_{3}CH_{2}CH_{2}Br + Z^{-} \xrightarrow[substitution]{k_s} CH_{3}CH_{2}CH_{2}Z + Br^{-}$
$CH_{3}CH_{2}CH_{2}Br + Z^{-} \xrightarrow[elimination]{k_e} CH_{3}CH=CH_{2} + HZ + Br^{-}$
where
$Z^{-} = CH_{3}CH_{2}O^{-} (A)$ or $(CH_{3})_{3}CO^{-} (B)$
$k_s$ and $k_e$ are,respectively,the rate constants for the substitution and elimination,and $\mu = \frac{k_s}{k_e}$,the correct option is:
$CH_{3}CH_{2}CH_{2}Br + Z^{-} \xrightarrow[substitution]{k_s} CH_{3}CH_{2}CH_{2}Z + Br^{-}$
$CH_{3}CH_{2}CH_{2}Br + Z^{-} \xrightarrow[elimination]{k_e} CH_{3}CH=CH_{2} + HZ + Br^{-}$
where
$Z^{-} = CH_{3}CH_{2}O^{-} (A)$ or $(CH_{3})_{3}CO^{-} (B)$
$k_s$ and $k_e$ are,respectively,the rate constants for the substitution and elimination,and $\mu = \frac{k_s}{k_e}$,the correct option is:
Identify '$X$' and '$Y$' in the following sequence of transformations: $CH_3-CH_2-CH_2Br$ $\xrightarrow{X} \text{Product}$ $\xrightarrow{Y} CH_3-CH(Br)-CH_3$.
Difficult
View SolutionThe rate of $S_N2$ reaction will be negligible in:
Medium
View SolutionSodium tertiary butoxide on reaction with methyl bromide produces the product $P$.
Sodium methoxide upon reaction with tertiary butyl bromide generates the product $Q$.
The products $P$ and $Q$ are
Sodium methoxide upon reaction with tertiary butyl bromide generates the product $Q$.
The products $P$ and $Q$ are
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