Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
`(i) CH_(3)CH_(2)CH_(2)CH_(2)Br or CH_(3)CH_(2)underset(Br)underset(|)C HCH_(3) , (ii) CH_(3)CH_(2)underset(Br)underset(|)CHCH_(3) or H_(3)C-underset(CH_(3))underset(|) overset(CH_(3))overset(|)C-Br`
(iii)` CH_(3)underset(CH_(3))underset(|)CHCH_(2)CH_(2)Br or CH_(3)CH_(2)underset(CH_(3))underset(|)CHCH_(2)Br`

Explanation

To determine which alkyl halide from the given pairs would react more rapidly by an SN2 mechanism, we need to analyze the structure of each compound in terms of steric hindrance. The SN2 mechanism is characterized by a backside attack by the nucleophile, which is more feasible when there is less steric hindrance around the carbon atom bonded to the leaving group (in this case, bromine).

Let's analyze each pair step by step:

### (i) CH₃CH₂CH₂CH₂Br vs. CH₃CH₂(Br)C(CH₃)H
1. **Identify the structure of each compound:**
- CH₃CH₂CH₂CH₂Br is a primary alkyl halide (1°) because the carbon bonded to Br is attached to only one other carbon.
- CH₃CH₂(Br)C(CH₃)H is a secondary alkyl halide (2°) because the carbon bonded to Br is attached to two other carbons.
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