Which of the following alkyl halide will undergo \(S_N1\) reaction most readily:

\((CH_3)_3C-I\)

The correct answer is option 4. \((CH_3)_3C-I\).

An \(S_N1\) reaction is a substitution nucleophilic unimolecular reaction. It follows a two-step mechanism:

Formation of a carbocation: The leaving group departs from the alkyl halide, forming a carbocation intermediate.

\(\text{R-X} \rightarrow \text{R}^+ + X^- \) 

A nucleophile then attacks the carbocation to form the product.

\(\text{R}^+ + \text{Nu}^- \rightarrow \text{R-Nu} \)

Key Factors that Influence the \(S_N1\) Reaction:

Carbocation Stability:

The \(S_N1\) mechanism depends heavily on the stability of the carbocation formed after the leaving group departs.  Tertiary carbocations \((CH_3)_3C^+\) are the most stable because they are stabilized by the inductive effect and hyperconjugation from the three surrounding methyl groups.

Leaving Group Ability:

The rate-determining step of the \(S_N1\) reaction is the formation of the carbocation, which requires the leaving group (halide) to depart. A good leaving group is one that can stabilize itself well after leaving. The better the leaving group, the faster the \(S_N1\) reaction. The quality of the leaving group among halogens follows this trend:

\(\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-\)

This means iodide (I⁻) is the best leaving group, followed by bromide (Br⁻), chloride (Cl⁻), and fluoride (F⁻) is the worst.

Comparing the Alkyl Halides:

Each of the given alkyl halides has the same tertiary carbocation backbone, \((CH_3)_3C^+\), but different halogens (F, Cl, Br, I) as leaving groups. Since the carbocation formed is the same in each case, the main deciding factor for the reaction rate will be the quality of the leaving group:

1. \((CH_3)_3C-F\):

Fluoride \((F⁻)\) is a very poor leaving group because it is highly electronegative and forms a very strong bond with carbon. This makes it difficult for fluoride to leave, so the \(S_N1\) reaction will be very slow.

2. \((CH_3)_3C-Cl\):

Chloride \((Cl⁻)\) is a better leaving group than fluoride, but it is still not ideal. The bond between carbon and chlorine is relatively strong, so the reaction will be moderately slow.

3. \((CH_3)_3C-Br\):

Bromide \((Br⁻)\) is a good leaving group. The bond between carbon and bromine is weaker than in the case of chlorine or fluorine, so bromide can leave more easily. The reaction will proceed faster than with chloride.

4. \((CH_3)_3C-I\):

Iodide \((I⁻)\) is the best leaving group because it forms the weakest bond with carbon among the halogens, and it can stabilize itself very well after leaving. This makes the \(S_N1\) reaction fastest in this case.

Conclusion:

The alkyl halide \((CH_3)_3C-I\), with iodide as the leaving group, will undergo the \(S_N1\) reaction most readily because iodide is the best leaving group. The trend in leaving group ability (and thus the rate of the \(S_N1\) reaction) is:

\(I^- > Br^- > Cl^- > F^-\)

Thus, the correct answer is 4. \((CH_3)_3C-I\).