Read the passage given below and answer the following questions:
Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons:
(i) In haloarenes, the electron pairs on halogen atom are in conjugation with u-electrons of the ring.
(ii) In haloalkane, the carbon atom attached to halogen is `sp^3` hybridised while in case of haloarene, the carbon atom attached to halogen is `sp^2` -hybridised.
In case of haloarenes/ the phenyl cation formed as a result of self-ionisation will not be stabilised by resonance.
Which of the following alkyl halides will undergoes `S_N1` reaction most readily ?

Options

A. `(CH_(3))_(3)C-F`
B. `(CH_(3))_(3)C-Cl`
C. `(CH_(3))_(3)C-Br`
D. `(CH_(3))_(3)C-I`

Explanation

To determine which alkyl halide will undergo an \( S_N1 \) reaction most readily, we need to analyze the stability of the carbocation formed during the reaction and the strength of the carbon-halogen bond. Here’s a step-by-step solution:

### Step 1: Understand the \( S_N1 \) Mechanism
The \( S_N1 \) (unimolecular nucleophilic substitution) reaction involves two main steps:
1. Formation of a carbocation by the departure of the leaving group (halogen).
2. Nucleophilic attack on the carbocation.

The rate-determining step is the first step, where the leaving group departs.

### Step 2: Identify the Alkyl Halides
Assuming we have three alkyl halides (not specified in the question), let's denote them as:
- Alkyl Halide A: Tert-butyl chloride (C\(_4\)H\(_9\)Cl)
- Alkyl Halide B: Tert-butyl bromide (C\(_4\)H\(_9\)Br)
- Alkyl Halide C: Tert-butyl iodide (C\(_4\)H\(_9\)I)

### Step 3: Analyze the Stability of the Carbocation
The stability of the carbocation formed is crucial. In this case, all three alkyl halides will form a tertiary carbocation (due to the tert-butyl group), which is stable. However, the rate of the \( S_N1 \) reaction will also depend on the strength of the carbon-halogen bond.

### Step 4: Compare the Carbon-Halogen Bonds
- **C-Cl bond**: Stronger than C-Br and C-I bonds.
- **C-Br bond**: Weaker than C-Cl but stronger than C-I.
- **C-I bond**: Weakest due to the larger size of iodine, leading to a longer bond length.

### Step 5: Determine the Reactivity
Since the \( S_N1 \) reaction rate depends on how easily the leaving group can depart, the weaker the bond, the more readily the \( S_N1 \) reaction will occur. Therefore:
- Tert-butyl iodide (C\(_4\)H\(_9\)I) will undergo \( S_N1 \) reaction most readily because the C-I bond is the weakest among the three.

### Conclusion
The alkyl halide that will undergo \( S_N1 \) reaction most readily is **Tert-butyl iodide (C\(_4\)H\(_9\)I)**.

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