Which of the following compounds undergo `E2 reaction with maximum rate?

Options

A. `CH_(3)C(CH_(3))(Br)CH_(2)CH_(3)`
B. `CH_(3)(CH_(2))_(2)CH_(2)Cl`
C. `CH_(3)(CH_(2))_(2)CH_(2)I`
D. `CH_(3)C(CH_(3))(I)CH_(2)CH_(3)`

Answer

To determine which of the given compounds undergoes an E2 reaction with the maximum rate, we need to consider several factors that influence the rate of E2 reactions, including the structure of the alkyl halide and the nature of the leaving group.

### Step-by-Step Solution:

1. **Understand E2 Reaction Mechanism**:
- E2 (bimolecular elimination) reactions are characterized by a single-step mechanism where a strong base abstracts a proton while the leaving group departs simultaneously. The rate of the reaction depends on the concentration of both the substrate and the base.

2. **Identify the Leaving Group**:
- The ability of the leaving group is crucial in E2 reactions. The order of leaving group ability is typically: Iodine (I) > Bromine (Br) > Chlorine (Cl). This means that compounds with iodine as a leaving group will generally react faster than those with bromine or chlorine.

3. **Evaluate the Structure of the Compounds**:
- The structure of the alkyl halide also plays a significant role. Tertiary alkyl halides react faster in E2 reactions compared to secondary and primary alkyl halides due to steric hindrance and stability of the transition state.

4. **Analyze the Given Options**:
- Without the specific compounds listed in the question, we can still apply the principles:
- If one compound is a tertiary alkyl halide with iodine as the leaving group, it will likely undergo the E2 reaction at the maximum rate.
- If another compound is a primary or secondary alkyl halide with a weaker leaving group (like Cl or Br), it will react slower.

5. **Conclusion**:
- Based on the analysis, the compound that is a tertiary alkyl halide with iodine as the leaving group will undergo the E2 reaction with the maximum rate.

### Final Answer:
The compound that undergoes the E2 reaction with maximum rate is the tertiary alkyl halide with iodine as the leaving group.