Question

How does concentration affect ${{\text{S}}_N}{\text{1}}$ reactions?

Answer 1

Hint : These reactions are a nucleophilic substitution reaction. That is, the functional group already present in the compound is replaced by another functional group which is a nucleophile. Nucleophiles are electron-rich species. And these reactions take place in the presence of a polar solvent.

Complete Step By Step Answer:
The rate-determining step is the slowest step in a reaction mechanism. The rate-determining step determines the rate law of the overall reaction as it limits the overall rate. Observe that in case of ${{\text{S}}_N}{\text{1}}$ reactions, the bond between the substrate and the leaving group is broken when the leaving group departs with the pair of electrons that previously composed the bond. As a result, the carbon atom in which the previous bond was formed now has a positive charge. The positive charge on a carbon atom is known as a carbocation, which comes from the words carbon and cation, which means positively charged atom. The subsequent step which is the formation of bond between nucleophile and carbocation, occurs rapidly. Since, the slow step of the reaction involves only the substrate, the reaction is unimolecular. And since only the substrate is present in the transition state, the rate of the reaction depends only on its concentration, and not on the concentration of the nucleophile.
Hence, the rate of reaction is unchanged by increasing the concentration of the nucleophile. The rate increases as the substrate concentration grows. However, if we have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affect the distribution of products that you will get. For example, ${\left( {{\text{C}}{{\text{H}}_3}} \right)_3}{\text{CCl}}$ reacting in water and formic acid where the water and formic acid are competing nucleophiles, you will get two different products: ${\left( {{\text{C}}{{\text{H}}_3}} \right)_3}{\text{COH}}$ and ${\left( {{\text{C}}{{\text{H}}_3}} \right)_3}{\text{COCOH}}$ . The relative yields of these products depend on the concentrations and relative reactivities of the nucleophiles.

Note :
Note that ${{\text{S}}_N}2$ reactions are bimolecular and so concentrations of both the reactants determine the rate of the reaction. Other factors affecting the rate of the ${{\text{S}}_N}{\text{1}}$ reactions are the strength of the leaving group and the presence of the polar solvent.