Question
Explain bimolecular nucleophilic substitutio $(SN^2)$ reaction with example.
✓
Answer
$\rightarrow$ In the year $1937$, Edward Davies Hughes at Sir Christopher Ingold proposed a mechanis for an $SN^2$ reaction.
$\rightarrow$ The reaction between $CHCl_3$ and hydroxide in to yield methanol and chloride ion follows second order kinetics.
$\rightarrow$ The rate depends upon the concentration of both the reactants.
$\rightarrow$ Rate $= \ce{K [CH_3CI]^1 [OH]^1}$
$\rightarrow$ The incoming nucleophile interacts with alkyl halide causing the carbon$-$halide bond to break and a new bond is formed between carbon and attacking nucleophile.
$\rightarrow$ Here it is $C-O$ bond formed between $C$ and $-OH.$ These two processes take place simultaneously in a single step and no intermediate is formed.
$\rightarrow$ As the reaction progresses and the bond between the incoming nucleophile and the carbon atom starts forming, the bond between carbon atom and leaving group weakens.
$\rightarrow$ As this happens, the three carbon$-$hydrogen bonds of the substrate start moving away from the attacking nucleophile.
$\rightarrow$ In transition state all the three $C-H$ bonds are in the same plane and the attacking and leaving nucleophiles are partially attached to the carbon.
$\rightarrow$ As the attacking nucleophile approaches closer to the carbon, $C-H$ bonds still keep on moving in the same direction till the attacking nucleophile attaches to carbon and leaving group leaves the carbon.
$\rightarrow$ As a result configuration is inverted, this process is called inversion of configuration.
$\rightarrow$ In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group.
$\rightarrow$ Such structures are unstable and cannot be isolated. Thus, in the transition state, carbon is simultaneously bonded to five atoms.
$\rightarrow$ The reaction between $CHCl_3$ and hydroxide in to yield methanol and chloride ion follows second order kinetics.
$\rightarrow$ The rate depends upon the concentration of both the reactants.
$\rightarrow$ Rate $= \ce{K [CH_3CI]^1 [OH]^1}$
$\rightarrow$ The incoming nucleophile interacts with alkyl halide causing the carbon$-$halide bond to break and a new bond is formed between carbon and attacking nucleophile.
$\rightarrow$ Here it is $C-O$ bond formed between $C$ and $-OH.$ These two processes take place simultaneously in a single step and no intermediate is formed.
$\rightarrow$ As the reaction progresses and the bond between the incoming nucleophile and the carbon atom starts forming, the bond between carbon atom and leaving group weakens.
$\rightarrow$ As this happens, the three carbon$-$hydrogen bonds of the substrate start moving away from the attacking nucleophile.
$\rightarrow$ In transition state all the three $C-H$ bonds are in the same plane and the attacking and leaving nucleophiles are partially attached to the carbon.
$\rightarrow$ As the attacking nucleophile approaches closer to the carbon, $C-H$ bonds still keep on moving in the same direction till the attacking nucleophile attaches to carbon and leaving group leaves the carbon.
$\rightarrow$ As a result configuration is inverted, this process is called inversion of configuration.
$\rightarrow$ In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group.
$\rightarrow$ Such structures are unstable and cannot be isolated. Thus, in the transition state, carbon is simultaneously bonded to five atoms.
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