Which of the following molecule would have a carbon halogen bond most susceptible to nucleophilic substitution ?

Options

A. 2-Fluorobutane
B. 2-Chlorobutane
C. 2-Bromo butane
D. 2-Iodobutane

Answer

To determine which molecule has a carbon-halogen bond most susceptible to nucleophilic substitution, we need to analyze the leaving group ability of the halogens involved. The ability of a leaving group to depart from the molecule is crucial in nucleophilic substitution reactions.

### Step-by-Step Solution:

1. **Identify the Halogens**: The halogens we are considering are iodine (I), bromine (Br), chlorine (Cl), and fluorine (F).

2. **Understand Leaving Group Ability**: The tendency of a leaving group to leave is a key factor in nucleophilic substitution. The better the leaving group, the more susceptible the carbon-halogen bond is to nucleophilic attack.

3. **Rank the Halogens**: The leaving group ability of halogens decreases in the following order:
- Iodine (I) > Bromine (Br) > Chlorine (Cl) > Fluorine (F)
This means that iodine is the best leaving group, followed by bromine, chlorine, and finally fluorine.

4. **Select the Molecule**: Based on the information above, the molecule that contains iodine as the halogen will have the carbon-halogen bond that is most susceptible to nucleophilic substitution.

5. **Conclusion**: If we are given a list of molecules, the one that contains iodine (e.g., 2-iodobutane) will be the correct answer.

### Final Answer:
The molecule with the carbon-halogen bond most susceptible to nucleophilic substitution is **2-iodobutane**.