Which one of following compounds undergoes `EI` reaction most readily?

Options

A. `CH_3-undersetunderset(Br)(|)oversetoverset(CH_3)(|)C-CH_2-CH_3`
B. `CH_3-CH_2-CH_2-Br`
C. `CH_3-CH_2-CH_2-I`
D. `CH_3-undersetunderset(I)(|)oversetoverset(CH_3)(|)C-CH_2-CH_3`

Answer

To determine which compound undergoes an E1 reaction most readily, we need to analyze the stability of the carbocations formed during the reaction. The E1 mechanism involves two steps: the formation of a carbocation and then the elimination of a proton to form an alkene.

### Step-by-Step Solution:

1. **Understanding E1 Mechanism**:
- E1 stands for unimolecular elimination. In this mechanism, the first step involves the departure of a leaving group (like a halide) to form a carbocation. The second step involves the removal of a proton by a base, leading to the formation of an alkene.

2. **Carbocation Stability**:
- The stability of the carbocation is crucial for the E1 reaction. The order of stability for carbocations is:
- Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (0°).
- Therefore, the more stable the carbocation, the more readily the E1 reaction will occur.

3. **Analyzing the Options**:
- We need to evaluate the given compounds (not provided in the question) to determine which one forms the most stable carbocation.
- If options include primary, secondary, and tertiary alkyl halides, we can eliminate primary and secondary options since they form less stable carbocations.

4. **Identifying the Best Leaving Group**:
- The leaving group also plays a significant role in the rate of carbocation formation. The better the leaving group, the faster the carbocation forms.
- Among halides, the order of leaving group ability is: Iodide (I⁻) > Bromide (Br⁻) > Chloride (Cl⁻) > Fluoride (F⁻).

5. **Comparing Compounds**:
- If two compounds both form tertiary carbocations, we then compare their leaving groups. The compound with the better leaving group (iodine being the best) will undergo the E1 reaction more readily.

6. **Conclusion**:
- After analyzing the options, we conclude that the compound that forms the most stable carbocation (preferably a tertiary carbocation) and has the best leaving group (iodine) will undergo the E1 reaction most readily.

### Final Answer:
- The compound that undergoes the E1 reaction most readily is the one that forms a tertiary carbocation with iodine as the leaving group.