Sure, let's analyze the situation step by step.
We have a reaction involving tert-butyl chloride ([tex]\((CH_3)_3CCl\)[/tex]) and iodide ion ([tex]\(I^-\)[/tex]) to form tert-butyl iodide ([tex]\((CH_3)_3CI\)[/tex]) and chloride ion ([tex]\(Cl^-\)[/tex]):
[tex]\[
\left( CH _3\right)_3 C - Cl + I ^{\ominus} \longrightarrow \left( CH _3\right)_3 C - I + Cl ^{\ominus}
\][/tex]
The rate of the reaction depends on the mechanism by which it occurs. The two possible mechanisms for this nucleophilic substitution reaction are SN1 and SN2:
1. SN1 Mechanism:
- In an SN1 reaction, the rate-determining step is the loss of the leaving group (in this case, [tex]\( Cl^- \)[/tex]), forming a carbocation intermediate.
- The rate of the reaction depends only on the concentration of the substrate (tert-butyl chloride).
- Therefore, doubling the concentration of iodide ion would have no effect on the rate of an SN1 reaction.
2. SN2 Mechanism:
- In an SN2 reaction, the rate-determining step involves a single concerted attack where the nucleophile ( [tex]\( I^- \)[/tex] ) attacks the substrate (tert-butyl chloride) at the same time that the leaving group ( [tex]\( Cl^- \)[/tex] ) departs.
- The rate of the reaction depends on both the concentration of the substrate (tert-butyl chloride) and the concentration of the nucleophile (iodide ion).
- Therefore, if the concentration of the iodide ion is doubled, the rate of the reaction will also double.
Considering these mechanisms, if we assume the reaction follows the SN2 pathway, we can predict the outcome given that the rate is dependent on both the substrate and the nucleophile's concentrations.
Thus, when the concentration of the iodide ion ([tex]\( I^- \)[/tex]) is doubled, the rate of forming tert-butyl iodide ([tex]\((CH_3)_3CI\)[/tex]) will increase by a factor of 2.
So, the rate of forming tert-butyl iodide will:
[tex]\[
\text{Double}
\][/tex]
The final answer is:
[tex]\[
2
\][/tex]
