Chapter 17, Problem 1Q

What would be the effect of carrying out the sodium iodide in acetone reaction with the alkyl halides using an iodide solution half as concentrated?

Interpretation Introduction

Interpretation: The effect of carrying out the sodium iodide in acetone reaction with the alkyl halide using iodide solution half as concentrated is to be interpreted.

Concept introduction: The reaction of alkyl halide with sodium iodide results the formation of alkyl iodide. The reaction occurs in the presence of acetone as the solvent. It is named as Finkelstein reaction and is followed by ${\text{S}}_{\text{N}}{}^{\text{2}}$ mechanism.

Answer to Problem 1Q

In the presence of half concentrated iodide solution, the ${\text{S}}_{\text{N}}{}^{\text{2}}$ mechanism would be slow and will move with half rate that will decrease the amount of product and make only half precipitate.

Explanation of Solution

The mechanism of the given reaction can be shown as given below:

  

Here the iodide ions act as nucleophile and involve in the rate determining step as reaction follows the ${\text{S}}_{\text{N}}{}^{\text{2}}$ mechanism.

Therefore, with decrease in the concentration of iodide ion, the rate of reaction will also halves that yield only half of the product.

Conclusion

In ${\text{S}}_{\text{N}}{}^{\text{2}}$ mechanism, the rate of reaction depends on the concentration of iodide ions.