Question

For the reaction –
\[\text{R-X}+\text{O}{{\text{H}}^{-}}\to \text{ROH}+{{\text{X}}^{-}}\]
The rate is given by –
\[\text{Rate}=5.0\times {{10}^{-5}}\left[ \text{R-X} \right]\left[ \text{O}{{\text{H}}^{-}} \right]+0.20\times {{10}^{-5}}\left[ \text{R-X} \right]\]
What percentage of R-X reaction is by ${{\text{S}}_{\text{N}}}2$ mechanism when $\left[ \text{O}{{\text{H}}^{-}} \right]=1.0\times {{10}^{-2}}\text{M}$ ?
(A) $96.1%\text{ }$
(B) $3.9%\text{ }$
(C) $80%\text{ }$
(D) $20%\text{ }$

Answer 1

Hint: The ${{\text{S}}_{\text{N}}}2$ mechanism of the reaction follows second order kinetics in which the rate of the reaction depends upon the concentration of the two reactants. The percentage can be calculated by dividing the concentration of reactant involved by total concentration.

Complete step by step solution: The given reaction proceeds through a ${{\text{S}}_{\text{N}}}2$ mechanism. This is a nucleophilic substitution bimolecular reaction in which the rate of the reaction depends upon the concentration of the two reactant species.
But the reaction may proceed to ${{\text{S}}_{\text{N}}}1$ mechanism sometimes based upon the concentration of reactants involved. So, the rate of reaction part that will follow ${{\text{S}}_{\text{N}}}1$ mechanism is given as:
\[
   {{\text{r}}_{{{\text{S}}_{\text{N}}}1}}=0.20\times {{10}^{-5}}\left[ \text{R-X} \right] \\
  \text{And, }{{\text{r}}_{{{\text{S}}_{\text{N}}}2}}=5.0\times {{10}^{-5}}\left[ \text{R-X} \right]\left[ \text{O}{{\text{H}}^{-}} \right] \\
\]
The reactant species in the given reaction are $\text{R-X}$ and $\text{O}{{\text{H}}^{-}}$. The overall rate equation involves the rate of a part of the reaction that follows ${{\text{S}}_{\text{N}}}2$ mechanism as well as the rate of a small part of the equation that follows ${{\text{S}}_{\text{N}}}1$ mechanism.
\[\text{Overall Rate}=5.0\times {{10}^{-5}}\left[ \text{R-X} \right]\left[ \text{O}{{\text{H}}^{-}} \right]+0.20\times {{10}^{-5}}\left[ \text{R-X} \right]\]
So, to find the percentage of reaction that is following ${{\text{S}}_{\text{N}}}2$ mechanism, we have to divide the corresponding rate by the overall rate of the reaction as shown below:
\[
   %\text{ of reaction by }{{\text{S}}_{\text{N}}}2=\left[ \dfrac{{{\text{r}}_{{{\text{S}}_{\text{N}}}2}}}{{{\text{r}}_{{{\text{S}}_{\text{N}}}1}}+{{\text{r}}_{{{\text{S}}_{\text{N}}}2}}} \right]\times 100 \\
  \Rightarrow %\text{ of reaction by }{{\text{S}}_{\text{N}}}2=\left[ \dfrac{5.0\times {{10}^{-5}}\left[ \text{R-X} \right]\left[ \text{O}{{\text{H}}^{-}} \right]}{5.0\times {{10}^{-5}}\left[ \text{R-X} \right]\left[ \text{O}{{\text{H}}^{-}} \right]+0.20\times {{10}^{-5}}\left[ \text{R-X} \right]} \right]\times 100 \\
  \Rightarrow %\text{ of reaction by }{{\text{S}}_{\text{N}}}2=\left[ \dfrac{5.0\times {{10}^{-5}}\left[ \text{O}{{\text{H}}^{-}} \right]}{5.0\times {{10}^{-5}}\left[ \text{O}{{\text{H}}^{-}} \right]+0.20\times {{10}^{-5}}} \right]\times 100 \\
  \because \left[ \text{O}{{\text{H}}^{-}} \right]=1.0\times {{10}^{-2}}\text{M} \\
  \therefore %\text{ of reaction by }{{\text{S}}_{\text{N}}}2=\left[ \dfrac{\left( 5.0\times {{10}^{-5}} \right)\left( 1.0\times {{10}^{-2}}\text{M} \right)}{5.0\times {{10}^{-5}}\left( 1.0\times {{10}^{-2}}\text{M} \right)+0.20\times {{10}^{-5}}} \right]\times 100 \\
  \therefore %\text{ of reaction by }{{\text{S}}_{\text{N}}}2=96.1%\text{ } \\
\]

Hence, the correct answer is (A), $96.1%\text{ }$

Note: The ${{\text{S}}_{\text{N}}}2$ mechanism follows second order kinetics because the rate-determining step involves a species where a nucleophile attaches to a reactant and simultaneously leaving group detaches from it. Thus, it is a bimolecular reaction.