Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.

Answer

To explain why ethyl iodide undergoes an SN2 reaction faster than ethyl bromide, we can break down the reasoning into clear steps:

### Step-by-Step Solution:

1. **Identify the Compounds**: We are comparing two compounds: ethyl iodide (C2H5I) and ethyl bromide (C2H5Br).

2. **Understand SN2 Mechanism**: The SN2 reaction mechanism involves a nucleophile attacking the carbon atom that is bonded to the leaving group (iodine or bromine in this case) and simultaneously displacing the leaving group.

3. **Determine the Leaving Groups**: In ethyl iodide, the leaving group is iodide ion (I⁻), while in ethyl bromide, the leaving group is bromide ion (Br⁻).

4. **Evaluate Leaving Group Ability**: The ability of a leaving group to depart from the molecule is crucial in determining the rate of the SN2 reaction. A better leaving group will lead to a faster reaction.

5. **Compare Iodide and Bromide**: Iodide (I⁻) is a better leaving group than bromide (Br⁻). This is due to the larger size and lower electronegativity of iodine compared to bromine, which makes it more stable when it leaves as an ion.

6. **Conclusion**: Since I⁻ is a better leaving group than Br⁻, ethyl iodide (C2H5I) will undergo the SN2 reaction faster than ethyl bromide (C2H5Br).

### Final Answer:
Ethyl iodide undergoes SN2 reaction faster than ethyl bromide because iodide (I⁻) is a better leaving group than bromide (Br⁻).

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