Read the passage given below and answer the following questions:
Nucleophilic substitution reaction of haloalkane can be conducted according to both `S_(N)^(1)` and `S_(N)^(2)` mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent.
Influences of halogen : No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom, that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is `I^(-) lt Br^(-) lt Cl^(-) lt F^(-)` and the order of their leaving tendency should be `I^(-) gt Br^(-) gt Cl^(-) gt F^(-)` . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is `RI gt RBr gt RCl gt RF` . In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on `S_(N)^(1)` mechanism. If the leaving group is not easy to leave, the reaction is based on `S_(N)^(2)` mechanism.
Influences of solvent polarity: In `S_(N)^(1)` reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In `S_(N)^(2)` reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate `(S_(N)^(1))` of tertiary chlorobutane in `25^(@)C` water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate `(S_(N)^(2))` of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both `S_(N)^(1)` and `S_(N)^(2)` reactions, but with different results. Generally speaking, weak polar solvent is favorable for `S_(N)^(2)` reaction, while strong polar solvent is favorable for `S_N^(1)` reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on `S_(N)^(1)` mechanism in solvents with a strong polarity (for example, ethanol containing water).
(Ding, Y. (2013). A Brief Discussion on Nucleophilic Substitution Reaction on Saturated Carbon Atom. In Applied Mechanics and Materials (Vol. 312, pp. 433-437). Trans Tech Publications Ltd.)
`S_(N)^(1)` reaction will be fastest in case of:

Options

A. 1-Chloro-2-methyl propane
B. 1-Iodo-2-methyl propane
C. 1-Chlorobutane
D. 1-Iodobutane

Answer

To determine which nucleophilic substitution reaction will be fastest in the case of the given haloalkanes, we need to analyze the factors influencing the reaction rates based on the SN1 and SN2 mechanisms.

### Step-by-Step Solution:

1. **Identify the Haloalkanes**:
The haloalkanes in question are:
- 1-chloro-2,2-dimethylpropane
- 1-iodo-2,2-dimethylpropane
- 1-bromo-2,2-dimethylpropane
- 1-fluoro-2,2-dimethylpropane

2. **Understand the Mechanism**:
- For SN1 reactions, the rate is influenced by the stability of the carbocation formed after the leaving group departs. A more stable carbocation leads to a faster reaction.
- For SN2 reactions, the reaction rate depends on the strength of the leaving group, as the nucleophile attacks the substrate simultaneously as the leaving group departs.

3. **Evaluate Leaving Groups**:
The leaving group ability follows the trend:
- Iodine (I⁻) > Bromine (Br⁻) > Chlorine (Cl⁻) > Fluorine (F⁻)
This means that I⁻ is the best leaving group, followed by Br⁻, Cl⁻, and F⁻.

4. **Determine the Fastest SN1 Reaction**:
- In an SN1 reaction, the formation of a stable carbocation is crucial.
- The 1-iodo-2,2-dimethylpropane will form a more stable carbocation than the others because iodine is the best leaving group, allowing the reaction to proceed faster.

5. **Conclusion**:
Based on the above analysis, the fastest nucleophilic substitution reaction will occur with **1-iodo-2,2-dimethylpropane** due to the stability of the carbocation formed and the excellent leaving ability of the iodide ion.

### Final Answer:
**1-iodo-2,2-dimethylpropane** will undergo the fastest nucleophilic substitution reaction.

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