
How does concentration affect $ {S_N}2 $ reactions?
Answer
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Hint: This type of reaction mechanism involves nucleophilic substitution reaction of groups like halide or other withdrawing groups with the nucleophile in a given organic compound. Here $ {S_N}2 $ stands for Substitution Nucleophilic Bimolecular.
Complete answer:
The Nucleophile and the organic compound are the two species on which the rate-determining step of this reaction depends upon. In the $ {S_N}2 $ reaction mechanism, the nucleophile attacks from the back side of the carbon atom. This reaction is a nucleophilic substitution reaction within which one bond is broken and another is formed coincidently. The rate determining step involved both the reacting species of the reaction. Thus this reaction is bimolecular. Since both substrate and nucleophile are present in the transition state, the rate of the reaction depends upon both of their concentrations.
Hence, by increasing the concentration of any of the both will change the rate of the reaction. Let us consider a reaction of bromomethane with hydroxide ion. As both substrate and nucleophile are involved in the transition state, so this reaction is bimolecular.
The rate law expression is:
$ r = k\left[ {C{H_3}Br} \right]\left[ {O{H^ - }} \right] $
This shows that the rate of reaction is directly proportional to $ {\left[ {OH} \right]^ - } $ and $ \left[ {C{H_3}Br} \right] $ .
If we increase the concentration of any of the reactants, the rate of reaction will also increase. If we increase the concentration of nucleophile, i.e. $ O{H^ - } $ then the rate of reaction will also increase as there will be more $ O{H^ - } $ ions for attacking the substrate.
By increasing the concentration of substrate, i.e. $ C{H_3}Br $ then the rate of reaction will also increase as there will be more $ C{H_3}Br $ molecules present to be attacked.
Note:
Remember that $ {S_N}1 $ reactions are unimolecular which means only concentration of substrate will determine the rate of the reaction. This is because only substrate is available in the transition state, so the rate of reaction depends upon the substrate, not on the concentration of the nucleophile.
Complete answer:
The Nucleophile and the organic compound are the two species on which the rate-determining step of this reaction depends upon. In the $ {S_N}2 $ reaction mechanism, the nucleophile attacks from the back side of the carbon atom. This reaction is a nucleophilic substitution reaction within which one bond is broken and another is formed coincidently. The rate determining step involved both the reacting species of the reaction. Thus this reaction is bimolecular. Since both substrate and nucleophile are present in the transition state, the rate of the reaction depends upon both of their concentrations.
Hence, by increasing the concentration of any of the both will change the rate of the reaction. Let us consider a reaction of bromomethane with hydroxide ion. As both substrate and nucleophile are involved in the transition state, so this reaction is bimolecular.
The rate law expression is:
$ r = k\left[ {C{H_3}Br} \right]\left[ {O{H^ - }} \right] $
This shows that the rate of reaction is directly proportional to $ {\left[ {OH} \right]^ - } $ and $ \left[ {C{H_3}Br} \right] $ .
If we increase the concentration of any of the reactants, the rate of reaction will also increase. If we increase the concentration of nucleophile, i.e. $ O{H^ - } $ then the rate of reaction will also increase as there will be more $ O{H^ - } $ ions for attacking the substrate.
By increasing the concentration of substrate, i.e. $ C{H_3}Br $ then the rate of reaction will also increase as there will be more $ C{H_3}Br $ molecules present to be attacked.
Note:
Remember that $ {S_N}1 $ reactions are unimolecular which means only concentration of substrate will determine the rate of the reaction. This is because only substrate is available in the transition state, so the rate of reaction depends upon the substrate, not on the concentration of the nucleophile.
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